\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 15, pp. 1--38.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/15\hfil Asymptotic behavior]
{Asymptotic behavior for a quadratic nonlinear
Schr\"{o}dinger equation}

\author[N. Hayashi, P. I. Naumkin\hfil EJDE-2008/15\hfilneg]
{Nakao Hayashi, Pavel I. Naumkin}  % in alphabetical order

\address{Nakao Hayashi \newline
Department of Mathematics\\
Graduate School of Science\\
Osaka University, Osaka\\
Toyonaka, 560-0043, Japan}
\email{nhayashi@math.wani.osaka-u.ac.jp}

\address{Pavel I. Naumkin \newline
Instituto de Matem\'{a}ticas\\
Universidad Nacional Aut\'{o}noma de M\'{e}xico, Campus Morelia, AP 61-3
(Xangari), Morelia CP 58089, Michoac\'{a}n, Mexico}
\email{pavelni@matmor.unam.mx}


\thanks{Submitted March 19, 2007. Published February 1, 2008.}
\thanks{P. I. Naumkin is partially supported by CONACYT}
\subjclass[2000]{35B40, 35Q55}
\keywords{Nonlinear Schrodinger equation; large time
 asymptotic; \hfill\break\indent self-similar solutions}

\begin{abstract}
 We study the initial-value problem for the quadratic nonlinear
 Schr\"{o}dinger equation
 \begin{gather*}
 iu_{t}+\frac{1}{2}u_{xx}=\partial _{x}\overline{u}^{2},\quad x\in
 \mathbb{R},\; t>1, \\
 u(1,x)=u_{1}(x),\quad x\in \mathbb{R}.
 \end{gather*}
 For small initial data $u_{1}\in \mathbf{H}^{2,2}$ we prove
 that there exists a unique global solution
 $u\in \mathbf{C}([1,\infty );\mathbf{H}^{2,2})$ of this Cauchy
 problem. Moreover we show that the large time asymptotic behavior
 of the solution  is defined in the region
 $|x|\leq C\sqrt{t}$ by the self-similar solution
 $\frac{1}{\sqrt{t}}MS(\frac{x}{\sqrt{t}})$ such that the total mass
 \begin{equation*}
 \frac{1}{\sqrt{t}}\int_{\mathbb{R}}MS(\frac{x}{\sqrt{t}})
 dx=\int_{\mathbb{R}}u_{1}(x)dx,
 \end{equation*}
 and in the far region $|x|>\sqrt{t}$ the asymptotic  behavior of
 solutions has rapidly oscillating structure similar to that of the cubic
 nonlinear Schr\"{o}dinger equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{\label{S1}Introduction}

We consider the quadratic nonlinear Schr\"{o}dinger equation
\begin{equation}
\begin{gathered}
iu_{t}+\frac{1}{2}u_{xx}=\partial _{x}\overline{u}^{2},
\quad x\in \mathbb{R},\; t>1, \\
u(1,x)=u_{1}(x),\quad x\in \mathbb{R}.
\end{gathered}  \label{1.1}
\end{equation}
In general, the quadratic type nonlinearities in the one dimensional case
are considered to be subcritical with respect to the large time asymptotic
behavior of solutions. Different types of the quadratic nonlinearities,
including derivatives of the unknown function were considered previously
(see  \cite{hn17,hn19,ho1,o2,to} and
references cited therein). We choose the initial time value $t=1$ for the
convenience of the forthcoming calculations (note that by the change
$t'=t-1$ it can be transformed to the usual case of the initial time
value $t'=0$).

If we replace the nonlinear term of \eqref{1.1} by $\partial _{x}u^{2}$,
then we can represent the solution $u$ by the formula (see \cite{o1})
\begin{equation*}
u(t,x)=s(t,x)-\frac{1}{2}\frac{(2s(t,x)\psi (t,x)+\partial _{x}\psi (t,x)
)\exp \big(2\int_{0}^{x}s(t,y)dy\big)}{1+\psi
(t,x)\exp \big(2\int_{0}^{x}s(t,y)dy\big)}
\end{equation*}
through the Hope-Cole transformation, where
\begin{equation*}
s(t,x)=\frac{1}{\sqrt{t}}e^{\frac{ix^{2}}{2t}}\frac{1}{C-2\sqrt{
2i}\int_{0}^{\frac{x}{\sqrt{t}}\sqrt{-\frac{i}{2}}}e^{-y^{2}}dy},
\end{equation*}
$C$ is a constant determined by
\begin{equation*}
\int u_{1}(x)dx=\int s(t,x)dx=\int e^{\frac{i\xi
^{2}}{2}}\frac{1}{C-2\sqrt{2i}\int_{0}^{\xi \sqrt{-\frac{i}{2}}}e^{-y^{2}}dy}
d\xi
\end{equation*}
and $\psi $ is the solution of the linear Schr\"{o}dinger equation $i\psi
_{t}+\frac{1}{2}\psi _{xx}=0$ with the initial data
\begin{equation*}
\psi (1,x)=
\Big(\exp \big(-2\int_{0}^{x}(u_{1}(
y)-s(1,y))dy\big)-1\Big)
\exp \big(-2\int_{0}^{x}s(1,y)dy\big).
\end{equation*}
See Appendix \ref{S8} for details. However the Hope-Cole transformation can
not be applied to our problem. Recently in \cite{hn17} we considered the
nonlinear Schr\"{o}dinger equation
\begin{equation*}
iu_{t}+\frac{1}{2}u_{xx}=\lambda (\overline{u}_{x})^{2}+\mu
u_{x}^{2},
\end{equation*}
with $\lambda \mathbf{,}\mu \in \mathbf{C.}$ We applied a method similar to
the normal forms of Shatah \cite{s}, making a transformation of the original
equation with quadratic nonlinearity to a nonlinear Schr\"{o}dinger equation
with critical cubic nonlinearity
\begin{align*}
\mathcal{L}(u-\frac{\mu }{2}u^{2}-\lambda \mathcal{G}(
\overline{u},\overline{u}))
&= -\mu u(\mathcal{L}u)+2\lambda \mathcal{G}(\overline{
\mathcal{L}u},\overline{u})\\
&= -\lambda \mu u(\overline{u}_{x})^{2}-\lambda \mu
uu_{x}^{2}+2|\lambda |^{2}\mathcal{G}(\overline{u}
,u_{x}^{2})+2\lambda \overline{\mu }\mathcal{G}(\overline{u},
\overline{u}_{x}^{2}),
\end{align*}
where $\mathcal{L}=i\partial _{t}+\frac{1}{2}\partial _{x}^{2}$, $\mathcal{G}
$ is a symmetric bilinear operator. Then for small initial data $u_{0}\in
\mathbf{H}^{3,1}$ we obtained the large time asymptotic behavior of small solutions
which has an additional logarithmic oscillation. We note that if
$\int u(t,x)dx=0$ in \eqref{1.1}, then by introducing a new variable
$v=\int_{-\infty }^{x}udx$, we have for $v$,
\begin{equation*}
iv_{t}+\frac{1}{2}v_{xx}=(\overline{v}_{x})^{2}.
\end{equation*}
Therefore in the case of $\int u_{1}(x)dx=0$, asymptotic behavior of
solutions has been shown in \cite{hn17}.

In  \cite{hn19} we studied the one dimensional quadratic nonlinear Schr
\"{o}dinger equation $iu_{t}+\frac{1}{2}u_{xx}=t^{-\alpha }|
u_{x}|^{2}$ with $\alpha \in (0,1)$. Heuristically
the solution should have a quasilinear character if
$\alpha \in (\frac{1}{2},1)$. However we showed that the asymptotic behavior
 of solutions does
not have a quasilinear character for all range $\alpha \in (0,1)
$ due to the special structure of the nonlinear term. For the case $\alpha
\in [\frac{1}{2},1)$ we proved that if the initial data
$u_{0}\in \mathbf{H}^{3,0}\cap \mathbf{H}^{2,2}$ are small then the solution
has a slow time-decay as $t^{-\frac{\alpha }{2}}$. And the derivative $u_{x}$
of the solution has a quasilinear behavior $t^{-1/2}$ as
$t\to \infty $. When $\alpha \in (0,\frac{1}{2})$, if we
assume that the initial data $u_{0}$ are analytic and small, then the same
result as for the case $\alpha \in [\frac{1}{2},1)$ holds.

The aim of the present paper is to prove existence of global solutions and
large time behavior of solutions to the Cauchy problem \eqref{1.1}.
Here we use the
method similar to the normal forms of Shatah and the transformation in
\cite{H-O}. Also we use the following factorization formulas for the free
Schr\"{o}dinger evolution group
\begin{gather*}
\mathcal{U}(t)\mathcal{F}^{-1}=M(t)\mathcal{D}_{t}
\mathcal{V}(t), \\
\mathcal{FU}(-t)=i\mathcal{V}(-t)\overline{E}
(t)\mathcal{D}_{1/t}.
\end{gather*}
Here we denote
\begin{equation*}
M(t)=e^{\frac{i}{2t}x^{2}},\quad E(t)=e^{\frac{it}{2}\xi ^{2}},
\end{equation*}
the dilation operator $({\mathcal{D}}_{a}\phi )(x)=\frac{1}{\sqrt{ia}}\phi (
\frac{x}{a})$ and $\mathcal{V}(t)=\mathcal{F}M(t)
\mathcal{F}^{-1}$. Note that ${\mathcal{D}}_{1/t}M(t)
=E(t){\mathcal{D}}_{1/t}$. So we represent the solution
\begin{equation*}
u(t)=\mathcal{U}(t)\mathcal{F}^{-1}w(
t)=\mathcal{D}_{t}E(t)v(t),
\end{equation*}
where $w=\mathcal{V}(-t)v(t)$. The direct Fourier
transform $\hat{\phi}(\xi )$ of the function $\phi (
x)$ is defined by
\begin{equation*}
\mathcal{F}\phi =\hat{\phi}=\frac{1}{\sqrt{2\pi }}\int_{\mathbb{R}}e^{-ix\xi
}\phi (x)dx,
\end{equation*}
then the inverse Fourier transformation is given by
\begin{equation*}
\mathcal{F}^{-1}\phi =\frac{1}{\sqrt{2\pi }}\int_{\mathbb{R}}e^{ix\xi }\phi
(\xi )d\xi .
\end{equation*}
Denote the usual Lebesgue space $\mathbf{L}^{p}=\{\phi \in \mathbf{S}
';\|\phi \|_{\mathbf{L}^{p}}<\infty \}$, where
the norm $\|\phi \|_{\mathbf{L}^{p}}=(\int_{\mathbb{R}
}|\phi (x)|^{p}dx)^{1/p}$ if $1\leq
p<\infty $ and $\|\phi \|_{\mathbf{L}^{\infty }}=$ ess$.
\sup_{x\in \mathbb{R}}|\phi (x)|$ if $p=\infty $.
The weighted Lebesgue norm is $\|\phi \|_{\mathbf{L}
^{p,a}}=\|\langle \cdot \rangle ^{a}\phi \|_{\mathbf{L}^{p}}$.
Weighted Sobolev space is
\begin{equation*}
\mathbf{H}^{m,a}=\{\phi \in \mathbf{S}':\|\phi \|
_{\mathbf{H}^{m,a}}\equiv \|\langle i\partial \rangle
^{m}\phi \|_{\mathbf{L}^{2,a}}<\infty \},
\end{equation*}
where $m,a\in \mathbb{R}$, $\langle x\rangle =\sqrt{1+x^{2}}$. The usual
Sobolev space is $\mathbf{H}^{m}=\mathbf{H}^{m,0}$, so the index $0$ we
usually omit if it does not cause a confusion. Different positive constants
we denote by the same letter $C$.

Denote $\mathbf{Y}=\{\phi \in \mathbf{L}^{\infty },\phi '\in
\mathbf{H}^{1,1}\}$.

\begin{theorem}\label{T1.1}
Let the initial data $M(1)u_{1}=e^{\frac{i}{2}
x^{2}}u_{1}\in \mathbf{Y}$ with a norm $\|u_{1}\|_{\mathbf{Y}
}\leq \varepsilon $, where $\varepsilon >0$ is sufficiently small. Then
there exists a unique solution $u\in \mathbf{C}([1,\infty
);\mathbf{Y})$ of the Cauchy problem \eqref{1.1}.
\end{theorem}

We denote by $\frac{1}{\sqrt{t}}M\Psi (\frac{x}{\sqrt{t}})$ a
self-similar solution of the quadratic nonlinear Schr\"{o}dinger equation (
\ref{1.1}) such that the total mass
\begin{equation*}
\frac{1}{\sqrt{t}}\int_{\mathbb{R}}M\Psi (\frac{x}{\sqrt{t}})
dx=\int_{\mathbb{R}}u_{1}(x)dx.
\end{equation*}
The following theorem states that the large time asymptotic behavior of
solution of \eqref{1.1} is defined by this self-similar solution in the
region $|x|\leq C\sqrt{t}$ and in the far region $|x|>\sqrt{t}$ it has rapidly oscillating structure similar to that of
the cubic Schr\"{o}dinger equations.

\begin{theorem}\label{T1.2}
Let the initial data $M(1)u_{1}=e^{\frac{i}{2}x^{2}}u_{1}\in \mathbf{Y}$
with a norm $\|u_{1}\|_{\mathbf{Y}}$
$\leq \varepsilon $, where $\varepsilon >0$ is sufficiently small. Then
there exist unique functions $H_{j}$ and $B_{j}\in \mathbf{{{L}^{\infty }}}$
($B_{j}$ are real-valued), $j=1,2$, such that the following asymptotic
formula is valid
\begin{align*}
u(t,x) &= \frac{1}{\sqrt{t}}M(t)\Psi (\frac{x}{\sqrt{t}}
)\\
&\quad +\frac{1}{\sqrt{t}}M(t)\sum_{j=1}^{2}H_{j}({\frac{x}{t}}
)\exp \Big(iB_{j}({\frac{x}{t}})\log \frac{1+\frac{
|x|}{\sqrt{t}}}{1+\frac{|x|}{t}}\Big)\\
&\quad +O(t^{-\frac{1}{2}-\varkappa })
\end{align*}
for $t\to \infty $ uniformly with respect to $x\in \mathbb{R}$,
where $\varkappa >0$.
\end{theorem}

We organize the rest of our paper as follows. In Sections
\ref{S2}--\ref{S6}
we prove some preliminary estimates. Section \ref{S7} is devoted to the
proof of Theorem \ref{T1.1}. We prove Theorem \ref{T1.2} in Section \ref{S8}
. Section \ref{S10} is devoted to the proof of existence of the self-similar
solution.

\section{\label{S2}Transformation of equation}

We represent the solution $u(t)=\mathcal{U}(t)
\mathcal{F}^{-1}w(t)$, where the free Schr\"{o}dinger evolution
group $\mathcal{U}(t)=\mathcal{F}^{-1}e^{\frac{it}{2}\xi ^{2}}
\mathcal{F}$. Applying the Fourier transformation to equation \eqref{1.1},
changing the dependent variable $w(t,\xi )=e^{\frac{it}{2}\xi
^{2}}\widehat{u}(t,\xi )$, we get
\begin{equation}
w_{t}(t,\xi )=\frac{\xi }{\sqrt{2\pi }}\int_{\mathbb{R}}e^{itS}
\overline{w(t,\eta -\xi )}\overline{w(t,-\eta )}
d\eta ,  \label{2.1}
\end{equation}
where $S=\frac{1}{2}(\xi ^{2}+(\xi -\eta )^{2}+\eta
^{2})$. Using the identity
\begin{equation*}
e^{itS}=\frac{d}{dt}\Big(\frac{2+itS}{(1+itS)^{2}}
te^{itS}\Big)+\frac{itS-1}{(1+itS)^{3}}e^{itS}
\end{equation*}
in view of the symmetry $\eta \leftrightarrow \xi -\eta $ we rewrite
equation (\ref{2.1}) as
\begin{equation}
\begin{aligned}
&\partial _{t}(w(t,\xi )-\int_{\mathbb{R}}e^{itS}
\overline{w(t,\eta -\xi )}\overline{w(t,-\eta )}
Ad\eta ) \\
&= t^{-1}\int_{\mathbb{R}}e^{itS}\overline{w(t,\eta -\xi )
w(t,-\eta )}\widetilde{A}d\eta   \\
&\quad +\sqrt{\frac{2}{\pi }}\int_{\mathbb{R}^{2}}e^{itQ}\overline{w(t,\eta
-\xi )}w(t,\eta -\zeta )w(t,\zeta )A\eta
d\eta d\zeta ,  \end{aligned} \label{2.2}
\end{equation}
where
\begin{gather*}
Q=\frac{1}{2}(\xi ^{2}+(\xi -\eta )^{2}-(\eta
-\zeta )^{2}-\zeta ^{2}), \\
A=\frac{\xi t(2+itS)}{\sqrt{2\pi }(1+itS)^{2}}=-
\frac{i\xi }{\sqrt{2\pi }}(1+t\partial _{t})(\frac{1}{it}
+S)^{-1}, \\
\widetilde{A}=\frac{\xi t(itS-1)}{\sqrt{2\pi }(
1+itS)^{3}}=-\frac{i\xi }{\sqrt{2\pi }}(2+t\partial _{t})
t\partial _{t}(\frac{1}{it}+S)^{-1}.
\end{gather*}
Now we return to the function $u=$ $\mathcal{U}(t)\mathcal{F}
^{-1}w(t)$ to get, from (\ref{2.2}),
\begin{equation}
(i\partial _{t}+\frac{1}{2}\partial _{x}^{2})(u-\mathcal{Q
}(\overline{u},\overline{u}))=t^{-1}\widetilde{\mathcal{Q
}}(\overline{u},\overline{u})+\mathcal{Q}(\overline{u}
,\partial _{x}u^{2}),  \label{2.3}
\end{equation}
where
\begin{align*}
&\mathcal{Q}(\overline{u},\overline{u})=\mathcal{U}(
t)\mathcal{F}^{-1}\int_{\mathbb{R}}e^{itS}\overline{w(t,\eta
-\xi )}\overline{w(t,-\eta )}Ad\eta \\
&= -\frac{1}{\sqrt{2\pi }}\partial _{x}\mathcal{F}^{-1}\int_{\mathbb{R}}
\overline{\widehat{u}(\eta -\xi )}\overline{\widehat{u}(
-\eta )}(1+t\partial _{t})(\frac{1}{it}+S)
^{-1}d\eta ,
\end{align*}
\begin{align*}
&\widetilde{\mathcal{Q}}(\overline{u},\overline{u})=\mathcal{U}
(t)\mathcal{F}^{-1}\int_{\mathbb{R}}e^{itS}\overline{w(
t,\eta -\xi )}\overline{w(t,-\eta )}\widetilde{A}d\eta \\
&= -\frac{1}{\sqrt{2\pi }}\partial _{x}\mathcal{F}^{-1}\int_{\mathbb{R}}
\overline{\widehat{u}(\eta -\xi )}\overline{\widehat{u}(
-\eta )}(3+t\partial _{t})\partial _{t}(\frac{1}{it
}+S)^{-1}d\eta .
\end{align*}
In the next lemma we give an $x$-representation of the operator
\begin{equation*}
\mathcal{I}(\phi ,\psi )=\mathcal{F}^{-1}\int_{\mathbb{R}}
\widehat{\phi }(\xi -\eta )\widehat{\psi }(\eta )
(\frac{1}{it}+S)^{-1}d\eta .
\end{equation*}
Denote the convolution with a kernel $g$
\begin{equation*}
(\psi \ast \phi )_{g}\equiv \int_{\mathbb{R}^{2}}g(
t,y,z)\psi (x-y)\phi (x-z)dydz.
\end{equation*}

\begin{lemma}\label{Lemma 2.1}
The representation $\mathcal{I}(\phi ,\psi )
=(\psi *\phi )_{g}$ is true, where
\begin{equation*}
g(t,y,z)\equiv \sqrt{\frac{2}{3\pi }}K_{0}\Big(\sqrt{\frac{4}{
3it}(y^{2}-yz+z^{2})}\Big)
\end{equation*}
and $K_{0}$ is the Macdonalds function.
\end{lemma}

\begin{proof}
We substitute the Fourier transformation
\begin{align*}
\mathcal{I}(\phi ,\psi )
&=\frac{1}{\sqrt{2\pi }}\int_{\mathbf{R
}}d\eta \widehat{\psi }(\frac{\xi }{2}-\eta )\int_{\mathbb{R}
}d\xi e^{ix\xi }\widehat{\phi }(\frac{\xi }{2}+\eta )
 \big(\frac{1}{it}+\frac{3}{4}\xi ^{2}+\eta ^{2}\big)^{-1} \\
&= \frac{8}{3(2\pi )^{\frac{3}{2}}}\int_{\mathbb{R}}d\eta \int_{
\mathbb{R}}dye^{-iy\eta }\psi (x-y)\int_{\mathbb{R}}dze^{iz\eta
}\phi (x-z)\\
&\quad \times \int_{\mathbb{R}}d\xi \cos (\frac{y+z}{2}\xi )(
\xi ^{2}+\frac{4}{3}\eta ^{2}-\frac{4i}{3t})^{-1}.
\end{align*}
By \cite{Er} we find
\begin{equation*}
\int_{0}^{\infty }\frac{\cos (z\xi )}{\xi ^{2}+a^{2}}d\xi =
\frac{\pi }{2a}e^{-a|z|}
\end{equation*}
for $\mathop{\rm Re}a>0$. Hence
\begin{align*}
\mathcal{I}(\phi ,\psi )
&=\frac{2}{\sqrt{6\pi }}\int_{\mathbf{R
}^{2}}dydz\psi (x-y)\phi (x-z)\\
&\quad \times \int_{0}^{\infty }\frac{\cos ((z-y)\eta )
}{\sqrt{\eta ^{2}-\frac{i}{t}}}e^{-\frac{1}{\sqrt{3}}|y+z|\sqrt{
\eta ^{2}-\frac{i}{t}}}d\eta .
\end{align*}
We compute by \cite{Er} for $\mathop{\rm Re}a>0$, $\mathop{\rm Re}\beta >0$
\begin{equation}
\int_{0}^{\infty }\frac{\cos ((z-y)\eta )e^{-\beta
\sqrt{\eta ^{2}+a^{2}}}d\eta }{\sqrt{\eta ^{2}+a^{2}}}=K_{0}(a\sqrt{
\beta ^{2}+(z-y)^{2}}).  \label{2.4}
\end{equation}
Then taking $a=\frac{1}{\sqrt{it}}$, $\beta =\frac{1}{\sqrt{3}}|
y+z|$ we find the representation of the lemma. The proof is complete.
\end{proof}

By Lemma \ref{Lemma 2.1} we can rewrite the operators $\mathcal{Q}$ and
$\widetilde{\mathcal{Q}}$ in equation (\ref{2.3}) as $\mathcal{Q}(\phi
,\psi )=(\psi \ast \phi )_{q}$ and $\widetilde{\mathcal{Q
}}(\phi ,\psi )=(\psi \ast \phi )_{\widetilde{q}}$
with
\begin{gather*}
q = -\frac{1}{\sqrt{2\pi }}(1+t\partial _{t})(\partial
_{y}+\partial _{z})g(t,y,z), \\
\widetilde{q} = -\frac{1}{\sqrt{2\pi }}(2+t\partial _{t})
t\partial _{t}(\partial _{y}+\partial _{z})g(t,y,z).
\end{gather*}

In the same way as in paper \cite{H-O} we change the variables $u(
t,x)=t^{-1/2}e^{\frac{it}{2}\xi ^{2}}v(t,\xi )$
and $\xi =\frac{x}{t}$ in equation (\ref{2.3}) to obtain
\begin{equation}
\mathcal{L}(v-\overline{E}(\overline{Ev}*\overline{Ev})
_{h})=\mathcal{P}_{1}-\mathcal{P}_{2},  \label{2.5}
\end{equation}
where
\begin{equation*}
\mathcal{L}=i\partial _{t}+\frac{1}{2t^{2}}\partial _{\xi }^{2},\mathcal{P}
_{1}=t^{-1}\overline{E}(\overline{Ev}*\overline{Ev})_{
\widetilde{h}},\mathcal{P}_{2}=t^{-1}\overline{E}(\overline{Ev}
*\partial _{\zeta }(E^{2}v^{2}))_{h},
\end{equation*}
and by changing $x=\xi t$, $y=\eta t$, $z=\zeta t$. Denote the operators
\begin{align*}
\mathcal{G}_{a,b}(\phi ,\psi )&= \overline{E}^{a+b}(
E^{a}\phi *E^{b}\psi )_{h} \\
&= \overline{E}^{a+b}\int_{\mathbb{R}^{2}}h(t,\eta ,\zeta )e^{
\frac{a}{2}it(\xi -\eta )^{2}}\phi (\xi -\eta )e^{
\frac{b}{2}it(\xi -\zeta )^{2}}\psi (\xi -\zeta )
d\eta d\zeta ,
\end{align*}
\begin{align*}
\widetilde{\mathcal{G}}_{a,b}(\phi ,\psi )&= \overline{E}
^{a+b}(E^{a}\phi *E^{b}\psi )_{\widetilde{h}} \\
&= \overline{E}^{a+b}\int_{\mathbb{R}^{2}}\widetilde{h}(t,\eta ,\zeta
)e^{\frac{a}{2}it(\xi -\eta )^{2}}\phi (\xi -\eta
)e^{\frac{b}{2}it(\xi -\zeta )^{2}}\psi (\xi
-\zeta )d\eta d\zeta ,
\end{align*}
\begin{align*}
\mathcal{H}_{a,b}(\phi ,\psi )&= \overline{E}^{a+b}(
E^{a}\phi *\partial _{\zeta }(E^{b}\psi ))_{h} \\
&= \overline{E}^{a+b}\int_{\mathbb{R}^{2}}h(t,\eta ,\zeta )e^{
\frac{a}{2}it(\xi -\eta )^{2}}\phi (\xi -\eta )
\partial _{\zeta }e^{\frac{b}{2}it(\xi -\zeta )^{2}}\psi (
\xi -\zeta )d\eta d\zeta
\end{align*}
with
\begin{gather*}
h(t,\eta ,\zeta )=-\frac{\sqrt{t}}{\pi \sqrt{3}}(
1+t\partial _{t})(\partial _{\eta }+\partial _{\zeta })
K_{0}\Big(\sqrt{\frac{4t}{3i}(\eta ^{2}-\eta \zeta +\zeta
^{2})}\Big), \\
\widetilde{h}(t,\eta ,\zeta )=-\frac{\sqrt{t}}{\pi \sqrt{3}}
(2+t\partial _{t})t\partial _{t}(\partial _{\eta
}+\partial _{\zeta })K_{0}\Big(\sqrt{\frac{4t}{3i}(\eta
^{2}-\eta \zeta +\zeta ^{2})}\Big).
\end{gather*}

\section{\label{S3}Preliminary estimates}

Define the weighted Lebesgue norm $\|\phi \|_{\mathbf{L}
^{p,a}}=\|\langle\cdot \rangle ^{a}\phi \|_{\mathbf{
L}^{p}}$ and define the linear operator
\begin{equation*}
\mathbb{K}\phi =\int_{\mathbb{R}}K(\xi ,\eta )(\phi
(\xi -\eta )-\phi (\xi ))d\eta .
\end{equation*}

\begin{lemma}\label{Lemma3.1}
Suppose that
\begin{equation*}
K(\xi ,\eta )=O(e^{-\langle\eta \rangle
}\langle\xi \eta \rangle ^{-\alpha })
\end{equation*}
for all $\xi \in \mathbb{R}$, $\eta \in \mathbb{R}\backslash \{
0\}$, where $\alpha >1$. Then the estimates are true
\begin{equation*}
\|\mathbb{K}\phi \|_{\mathbf{L}^{p,\theta }}\leq C\|
\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda }}
\end{equation*}
if $\alpha >\frac{3}{2}+\frac{1}{p}$, where $\theta =\frac{3}{2}+\frac{1}{p}
+\lambda $, $\lambda \geq 0$, $p=2,\infty $, and
\begin{equation*}
\|\mathbb{K}\phi \|_{\mathbf{L}^{2,1+\lambda }}\leq C\|
\phi \|_{\mathbf{L}^{2,\lambda }},
\end{equation*}
if $\alpha >1$, where $\lambda \geq 0$.
\end{lemma}

\begin{proof}
Since
\begin{equation*}
\|\phi (\cdot -\eta )-\phi (\cdot )\|
_{\mathbf{L}^{p,\lambda }}
=\big\|\int_{0}^{\eta }\partial _{z}\phi (
\cdot -z)dz\big\|_{\mathbf{L}^{p,\lambda }}
\leq C|\eta |^{\frac{1}{2}+\frac{1}{p}}\langle\eta \rangle ^{\lambda
}\|\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda }},
\end{equation*}
for $p=2,\infty $,  we find
\begin{equation*}
\|\mathbb{K}\phi \|_{\mathbf{L}^{p,\theta }}\leq C\|
\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda }}\|\int_{
\mathbb{R}}e^{-\langle\eta \rangle }\langle\eta
\rangle ^{\lambda }\langle\xi \eta \rangle ^{-\alpha
}|\eta |^{\frac{1}{2}+\frac{1}{p}}d\eta \|_{\mathbf{L}
^{\infty ,\theta -\lambda }}\leq C\|\partial _{\xi }\phi \|_{
\mathbf{L}^{2,\lambda }}
\end{equation*}
if $\alpha >\frac{3}{2}+\frac{1}{p}$. We now prove the second estimate. For
$|\xi |<1$ we have
\begin{equation*}
|\mathbb{K}\phi |\leq C\|\phi \|_{\mathbf{L}
^{2}}\Big(\int_{\mathbb{R}}e^{-\langle\eta \rangle }d\eta
\Big)^{1/2}\leq C\|\phi \|_{\mathbf{L}^{2}}.
\end{equation*}
For $|\xi |\geq 1$ we write
\begin{align*}
||\xi |^{1+\lambda }\mathbb{K}\phi |
&\leq C|\xi |^{\lambda }|\phi (\xi )||\xi
|\int_{\mathbb{R}}\langle\xi \eta \rangle ^{-\alpha
}e^{-\langle\eta \rangle }d\eta \\
&\quad +C|\xi |\int_{|\eta |>\frac{|\xi |}{
2}}\langle\xi \eta \rangle ^{-\alpha }\langle\xi -\eta
\rangle ^{\lambda }|\phi (\xi -\eta )|d\eta
\\
&\quad +C|\xi |\int_{|\eta |<\frac{|\xi |}{
2}}\langle\xi \eta \rangle ^{-\alpha }\langle\xi -\eta
\rangle ^{\lambda }|\phi (\xi -\eta )|d\eta .
\end{align*}
By the Cauchy-Schwarz inequality we find for the second summand
\begin{align*}
|\xi |\int_{|\eta |>\frac{|\xi |}{2}
}\langle\xi \eta \rangle ^{-\alpha }\langle\xi -\eta
\rangle ^{\lambda }|\phi (\xi -\eta )|d\eta
&\leq |\xi |\|\phi \|_{\mathbf{L}^{2,\lambda
}}\Big(\int_{|\eta |>\frac{|\xi |}{2}
}\langle\xi \eta \rangle ^{-2\alpha }d\eta \Big)^{1/2}\\
&\leq C|\xi |^{\frac{3}{2}-2\alpha }\|\phi \|_{
\mathbf{L}^{2,\lambda }}
\end{align*}
and for the third summand we change $\eta =\frac{x}{|\xi |}$,
\begin{align*}
&|\xi |\int_{|\eta |<\xi |/2}
\langle\xi \eta \rangle ^{-\alpha }\langle\xi -\eta
\rangle ^{\lambda }|\phi (\xi -\eta )|d\eta
\\
&= C\int_{|x|<\xi ^2/ 2} \langle x\rangle
^{-\alpha }\langle\xi -\frac{x}{|\xi |}\rangle
^{\lambda }|\phi (\xi -\frac{x}{|\xi |})
|dx.
\end{align*}
Then taking the $\mathbf{L}^{2}$ - norm we have
\begin{align*}
\|\xi \mathbb{K}\phi \|_{\mathbf{L}^{2}(|\xi|\geq 1)}
&\leq C\big\|\phi \|_{\mathbf{L}^{2,\lambda
}}+C\|\phi \|_{\mathbf{L}^{2,\lambda }}\||\xi
|^{\frac{3}{2}-2\alpha }\big\|_{\mathbf{L}^{2}} \\
&\quad +C\|\int_{|x|<\frac{\xi ^{2}}{2}}\langle
x\rangle ^{-\alpha }\langle\xi -\frac{x}{|\xi |}
\rangle ^{\lambda }|\phi (\xi -\frac{x}{|\xi |
})|dx\|_{\mathbf{L}^{2}(|\xi |\geq 1)}
\end{align*}
and
\begin{align*}
&\|\int_{|x|<\frac{\xi ^{2}}{2}}\langle
x\rangle ^{-\alpha }\langle\xi -\frac{x}{|\xi |}
\rangle ^{\lambda }|\phi (\xi -\frac{x}{|\xi |
})|dx\|_{\mathbf{L}^{2}(|\xi |\geq
1)}^{2} \\
&= \int_{|\xi |\geq 1}d\xi \int_{|x|<\frac{\xi ^{2}
}{2}}\langle x\rangle ^{-\alpha }\langle\xi -\frac{x}{
|\xi |}\rangle ^{\lambda }|\phi (\xi -\frac{x
}{|\xi |})|dx \\
&\quad \times \int_{|y|<\frac{\xi ^{2}}{2}}\langle
y\rangle ^{-\alpha }\langle\xi -\frac{y}{|\xi |}
\rangle ^{\lambda }|\phi (\xi -\frac{y}{|\xi |
})|dy \\
&= \int_{\mathbb{R}}dx\langle x\rangle ^{-\alpha }\int_{\mathbb{R}
}dy\langle y\rangle ^{-\alpha }\int_{|\xi |\geq
1,|\xi |\geq 2|x|,|\xi | \geq 2|
y|}\langle\xi -\frac{x}{|\xi |}\rangle
^{\lambda }|\phi (\xi -\frac{x}{|\xi |})
|\\
&\quad \times \langle\xi -\frac{y}{|\xi |}\rangle
^{\lambda }|\phi (\xi -\frac{y}{|\xi |})
|d\xi \\
&\leq C\|\phi \|_{\mathbf{L}^{2,\lambda
}}^{2}\Big(\int_{\mathbb{R}}\langle x\rangle ^{-\alpha
}dx\Big)^{2}
\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}^{2},
\end{align*}
since changing $z=\xi -\frac{x}{|\xi |}$ in the domain $|
\xi |\geq 1,|\xi |\geq 2|x|$ we have
\begin{equation*}
|\frac{d\xi }{dz}|=\frac{1}{1+\frac{x}{\xi ^{2}}}\leq 2.
\end{equation*}
Therefore,
\begin{equation*}
\int_{|\xi |\geq 1,|\xi |\geq 2|x|
}\langle\xi -\frac{x}{|\xi |}\rangle ^{2\lambda
}\phi ^{2}(\xi -\frac{x}{|\xi |})d\xi \leq 2\int_{
\mathbb{R}}\langle z\rangle ^{2\lambda }\phi ^{2}(z)
dz=C\|\phi \|_{\mathbf{L}^{2,\lambda }}^{2}.
\end{equation*}
The proof is complete.
\end{proof}

We now estimate the bilinear operator
\begin{equation*}
\mathcal{A}_{a,b}(\phi ,\psi )=\int_{\mathbb{R}^{2}}e^{-i\xi
(a\eta +b\zeta )}A(\eta ,\zeta )(\phi (
\xi -\eta )-\phi (\xi ))(\psi (\xi
-\zeta )-\psi (\xi ))d\eta d\zeta .
\end{equation*}

\begin{lemma}\label{Lemma3.2}
Suppose that
\begin{equation*}
|\partial _{\eta }^{k}\partial _{\zeta }^{l}A(\eta ,\zeta
)|\leq C(|\eta |+|\zeta |
)^{-s-k-l}e^{-C|\eta |-C|\zeta |}
\end{equation*}
for all $\eta ,\zeta \in \mathbb{R},$ $k,l=0,1$, where $s\geq 1$. Then the
estimates are valid
\begin{equation*}
\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{p,\theta }}\leq C\|\partial _{\xi }\phi \|_{\mathbf{L}
^{2,\lambda }}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }}
\end{equation*}
if $s<3$, $a,b\neq 0$, where $\theta =2+\lambda +\delta -\epsilon $,
$p=2,\infty $, $\lambda ,\delta \in \mathbb{R}$, and
\begin{equation*}
\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{2,\sigma }}\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}\|
\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }}
\end{equation*}
if $s<\frac{5}{2}$, $b\neq 0$, where $\sigma =1+\lambda +\delta -\epsilon $,
$\lambda ,\delta \in \mathbb{R}$, $\epsilon >0$ is small.
\end{lemma}

\begin{proof}
We integrate by parts with respect to $\eta $ and $\zeta $ via identities
$e^{-ia\xi \eta }=P\partial _{\eta }(\eta e^{-ia\xi \eta })$ and
$e^{-ib\xi \zeta }=Q\partial _{\zeta }(\zeta e^{-ib\xi \zeta })$, where
\begin{equation*}
P=(1-ia\xi \eta )^{-1},Q=(1-ib\xi \zeta )^{-1},
\end{equation*}
to get
\begin{align*}
\mathcal{A}_{a,b}(\phi ,\psi )
&=\int_{\mathbb{R}^{2}}d\eta
d\zeta e^{-i\xi (a\eta +b\zeta )}\Big(\eta \zeta PQA\partial
_{\eta }\phi (\xi -\eta )\partial _{\zeta }\psi (\xi
-\zeta ) \\
&\quad +\zeta \eta \partial _{\eta }(PQA)(\phi (\xi
-\eta )-\phi (\xi ))\partial _{\zeta }\psi (
\xi -\zeta )\\
&\quad +\eta \zeta \partial _{\zeta }(PQA)(\psi (\xi
-\zeta )-\psi (\xi ))\partial _{\eta }\phi (
\xi -\eta )\\
&\quad +(\phi (\xi -\eta )-\phi (\xi )
)(\psi (\xi -\zeta )-\psi (\xi )
)\eta \partial _{\eta }\zeta \partial _{\zeta }(PQA)\Big).
\end{align*}
Using the estimates
\begin{gather*}
|P|\leq C\langle\xi \eta \rangle ^{-1},|
Q|\leq C\langle\xi \zeta \rangle ^{-1},
\\
|\phi (\xi -\eta )-\phi (\xi )|\leq
C\langle\xi \rangle ^{-\lambda }|\eta |^{\frac{1}{2}
}\langle\eta \rangle ^{|\lambda |}\|\partial
_{\xi }\phi \|_{\mathbf{L}^{2,\lambda }},
\\
|\psi (\xi -\zeta )-\psi (\xi )|\leq
C\langle\xi \rangle ^{-\delta }|\zeta |^{\frac{1}{2}
}\langle\zeta \rangle ^{|\delta |}\|\partial
_{\xi }\psi \|_{\mathbf{L}^{2,\delta }}
\end{gather*}
and by the condition of the lemma
\begin{align*}
&|PQA|+|\eta \partial _{\eta }(PQA)|
+|\zeta \partial _{\zeta }(PQA)|+|\eta
\partial _{\eta }\zeta \partial _{\zeta }(PQA)|\\
&\leq C\langle\xi \zeta \rangle ^{-1}\langle\xi \eta
\rangle ^{-1}(|\eta |+|\zeta |)
^{-s}e^{-C|\eta |-C|\zeta |},
\end{align*}
 we obtain
\begin{align*}
&|\mathcal{A}_{a,b}(\phi ,\psi )|\\
&\leq C\int_{
\mathbb{R}^{2}}d\eta d\zeta \langle\xi \eta \rangle
^{-1}\langle\xi \zeta \rangle ^{-1}(|\eta |
+|\zeta |)^{-s}e^{-C|\eta |-C|\zeta|} \\
&\quad \times \Big(|\eta \zeta ||\partial _{\eta }\phi
(\xi -\eta )||\partial _{\zeta }\psi (\xi
-\zeta )|+\langle\xi \rangle ^{-\lambda }\|
\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda }}|\eta |
^{1/2}|\zeta ||\partial _{\zeta }\psi (\xi
-\zeta )| \\
&\quad +\langle\xi \rangle ^{-\delta }\|\partial _{\xi
}\psi \|_{\mathbf{L}^{2,\delta }}|\eta ||\zeta
|^{1/2}|\partial _{\eta }\phi (\xi -\eta )
|+\langle\xi \rangle ^{-\lambda -\delta }|\eta
\zeta |^{1/2}\|\partial _{\xi }\phi \|_{\mathbf{L
}^{2,\lambda }}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta
}}\Big).
\end{align*}
Hence by the Cauchy-Schwarz inequality
\begin{align*}
|\mathcal{A}_{a,b}(\phi ,\psi )|
&\leq C\langle\xi \rangle ^{-\lambda -\delta }\|\partial _{\xi
}\phi \|_{\mathbf{L}^{2,\lambda }}\|\partial _{\xi }\psi
\|_{\mathbf{L}^{2,\delta }} \\
&\quad \times \Big(\Big(\int_{\mathbb{R}}|\eta |
^{2-s}\langle\xi \eta \rangle ^{-2}e^{-C|\eta |
}d\eta \Big)^{1/2}\Big(\int_{\mathbb{R}}|\zeta |
^{2-s}\langle\xi \zeta \rangle ^{-2}e^{-C|\zeta |
}d\zeta \Big)^{1/2} \\
&\quad +\int_{\mathbb{R}}|\eta |^{\frac{1-s}{2}}\langle\xi
\eta \rangle ^{-1}e^{-C|\eta |}d\eta \big(\int_{\mathbf{
R}}|\zeta |^{2-s}\langle\xi \zeta \rangle
^{-2}e^{-C|\zeta |}d\zeta \big)^{1/2} \\
&\quad +\int_{\mathbb{R}}|\eta |^{\frac{1-s}{2}}\langle
\xi \eta \rangle ^{-1}e^{-C|\eta |}d\eta \int_{\mathbb{R}
}|\zeta |^{\frac{1-s}{2}}\langle\xi \zeta \rangle
^{-1}e^{-C|\zeta |}d\zeta \Big)\\
&\leq C\langle\xi \rangle ^{-\theta }\|\partial _{\xi
}\phi \|_{\mathbf{L}^{2,\lambda }}\|\partial _{\xi }\psi
\|_{\mathbf{L}^{2,\delta }}
\end{align*}
with $\theta =2+\lambda +\delta -\epsilon $, $s<3$, $\epsilon >0$. For the
case $p=2$ we use the inequality
\begin{equation*}
\|\phi (\cdot -\eta )-\phi (\cdot )\|
_{\mathbf{L}^{2,\lambda }}\leq C|\eta |\langle\eta
\rangle ^{|\lambda |}\|\partial _{\xi }\phi
\|_{\mathbf{L}^{2,\lambda }}
\end{equation*}
to get
\begin{align*}
&\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}^{2,\theta }} \\
&\leq \sup_{\xi \in \mathbb{R}}\int_{\mathbb{R}^{2}}d\eta d\zeta
\langle\xi \eta \rangle ^{-1}\langle\xi \zeta
\rangle ^{-1}(|\eta |+|\zeta |)
^{-s}e^{-C|\eta |-C|\zeta |} \\
&\quad \times \Big(\|\partial _{\xi }\phi \|_{\mathbf{L}
^{2,\lambda }}\langle\xi \rangle ^{\theta -\lambda }|\eta
\zeta ||\partial _{\zeta }\psi (\xi -\zeta )
|+\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta
}}\langle\xi \rangle ^{\theta -\delta }|\eta \zeta |
|\partial _{\eta }\phi (\xi -\eta )| \\
&\quad +\|\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda
}}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta
}}\langle\xi \rangle ^{\theta -\lambda -\delta }|\eta
||\zeta |^{1/2}\Big).
\end{align*}
Hence
\begin{align*}
\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}^{2,\theta }}
&\leq C\|\partial _{\xi }\phi \|_{\mathbf{L}
^{2,\lambda }}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta
}}\sup_{\xi \in \mathbb{R}}\langle\xi \rangle ^{\theta -\lambda
-\delta } \\
&\times \Big(\int_{\mathbb{R}}|\eta |^{1-\frac{2s}{3}
}\langle\xi \eta \rangle ^{-1}e^{-C|\eta |}d\eta
\Big(\int_{\mathbb{R}}|\zeta |^{2-\frac{2s}{3}}\langle
\xi \zeta \rangle ^{-2}e^{-C|\zeta |}d\zeta \Big)^{1/2}
 \\
&\quad +\int_{\mathbb{R}}|\zeta |^{1-\frac{2s}{3}}\langle\xi
\zeta \rangle ^{-1}e^{-C|\zeta |}d\zeta \Big(\int_{
\mathbb{R}}|\eta |^{2-\frac{2s}{3}}\langle\xi \eta
\rangle ^{-2}e^{-C|\eta |}d\eta \Big)^{1/2} \\
&\quad +\int_{\mathbb{R}}|\eta |^{1-\frac{2s}{3}}\langle
\xi \eta \rangle ^{-1}e^{-C|\eta |}d\eta \int_{\mathbb{R}
}|\zeta |^{\frac{1}{2}-\frac{s}{3}}\langle\xi \zeta
\rangle ^{-1}e^{-C|\zeta |}d\zeta \Big)\\
&\leq C\|\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda
}}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }}
\end{align*}
with $\theta =2+\lambda +\delta -\epsilon $, $s<3$, $\epsilon >0$. To prove
the second estimate of the lemma as above we integrate by parts with respect
to $\zeta $
\begin{align*}
\mathcal{A}_{a,b}(\phi ,\psi )
&= \int_{\mathbb{R}^{2}}d\eta
d\zeta e^{-bi\xi \zeta }\Big(\zeta QA(\phi (\xi -\eta )
-\phi (\xi ))\partial _{\zeta }\psi (\xi -\zeta
) \\
&\quad  +(\phi (\xi -\eta )-\phi (\xi )
)(\psi (\xi -\zeta )-\psi (\xi )
)\zeta \partial _{\zeta }(QA)\Big).
\end{align*}
Using the estimates
\begin{equation*}
|\psi (\xi -\zeta )-\psi (\xi )|\leq
C\langle\xi \rangle ^{-\lambda }|\zeta |^{\frac{1}{2
}}\langle\zeta \rangle ^{|\lambda |}\|
\partial _{\xi }\psi \|_{\mathbf{L}^{2,\lambda }}
\end{equation*}
and
\begin{equation*}
|\zeta \partial _{\zeta }(QA)|+|QA|
\leq C\langle\xi \zeta \rangle ^{-1}(|\eta |
+|\zeta |)^{-s}e^{-C|\eta |-C|\zeta
|},
\end{equation*}
 we obtain
\begin{align*}
|\mathcal{A}_{a,b}(\phi ,\psi )|
&\leq \int_{\mathbb{R}^{2}}d\eta d\zeta |(\phi (\xi -\eta
)-\phi (\xi ))|\langle\xi \zeta
\rangle ^{-1}(|\eta |+|\zeta |)
^{-s}e^{-C|\eta |-C|\zeta |} \\
&\quad \times \big(|\zeta ||\partial _{\zeta }\psi (
\xi -\zeta )|+\|\partial _{\xi }\psi \|_{\mathbf{L
}^{2,\delta }}\langle\xi \rangle ^{-\delta }|\zeta |
^{1/2}\big).
\end{align*}
Hence
\begin{align*}
\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{2,\sigma }}
&\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}\|\partial
_{\xi }\psi \|_{\mathbf{L}^{2,\delta }}\sup_{\xi \in \mathbb{R}
}\langle\xi \rangle ^{\sigma -\lambda -\delta }\int_{\mathbb{R}
}e^{-C|\eta |}d\eta \\
&\times \Big(\Big(\int_{\mathbb{R}}|\zeta |
^{2}\langle\xi \zeta \rangle ^{-2}(|\eta |
+|\zeta |)^{-2s}e^{-C|\zeta |}d\zeta
\Big)^{1/2} \\
&\quad +\int_{\mathbb{R}}|\zeta |^{1/2}\langle
\xi \zeta \rangle ^{-1}(|\eta |+|\zeta
|)^{-s}e^{-C|\zeta |}d\zeta \Big)\\
&\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}\|\partial _{\xi }\psi
\|_{\mathbf{L}^{2,\delta }}
\end{align*}
if $\sigma =1+\lambda +\delta -\epsilon $, $s<\frac{5}{2}$, $\epsilon >0$.
The proof is complete.
\end{proof}

We next estimate the operator
\begin{equation*}
\mathbb{A}(\phi ,\psi )\equiv \int_{\mathbb{R}^{2}}e^{-i\xi
(a\eta +b\zeta )}A(\eta ,\zeta )\phi (\xi
-\eta )\psi (\xi -\zeta )d\eta d\zeta ,
\end{equation*}
where $a,b\in \mathbb{R}\backslash \{0\}$. Denote
\begin{equation*}
\Lambda (\xi )=\int_{\mathbb{R}^{2}}e^{-i\xi (a\eta
+b\zeta )}A(\eta ,\zeta )d\eta d\zeta .
\end{equation*}

\begin{lemma}\label{Lemma3.3}
Suppose that
\begin{equation}
\big|\partial _{\eta }^{k}\partial _{\zeta }^{l}
\Big(A(\eta ,\zeta
)-\frac{a_{1}\eta +b_{1}\zeta }{\eta ^{2}-\zeta \eta +\zeta ^{2}}
e^{-\langle\eta \rangle -\langle\zeta \rangle
}\Big)\big|
\leq C(|\eta |+|\zeta |)^{-k-l}e^{-\langle\eta \rangle -\langle\zeta
\rangle }  \label{3.1}
\end{equation}
for all $\eta ,\zeta \in \mathbb{R},$ $k,l=0,1,2,3$, where $a_{1},b_{1}\in
\mathbf{R.}$ Then
\begin{align*}
&\|\mathbb{A}(\phi ,\psi )-\Lambda \phi \psi \|_{
\mathbf{L}^{p,\theta }} \\
&\leq C\|\phi \|_{\mathbf{L}^{p,\alpha }}\|\partial
_{\xi }\psi \|_{\mathbf{L}^{2,\delta }}+C\|\psi \|_{
\mathbf{L}^{p,\beta }}\|\partial _{\xi }\phi \|_{\mathbf{L}
^{2,\lambda }}
+C\|\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda }}\|
\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }},
\end{align*}
with
\begin{equation*}
\theta =\min (\frac{3}{2}+\frac{1}{p}+\alpha +\delta ,\frac{3}{2}+
\frac{1}{p}+\beta +\lambda ,2+\lambda +\delta -\epsilon ),\quad
p=2,\infty
\end{equation*}
and
\begin{equation*}
\|\mathbb{A}(\phi ,\psi )\|_{\mathbf{L}^{2,\sigma
}}\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}(\|\psi
\|_{\mathbf{L}^{\infty ,\beta }}+\|\partial _{\xi }\psi
\|_{\mathbf{L}^{2,\delta }})
\end{equation*}
with
$\sigma =\min (1+\lambda +\beta ,1+\lambda +\delta -\epsilon )
,\alpha ,\beta ,\lambda ,\delta \in \mathbb{R}$
$\epsilon >0$ is small.
\end{lemma}

\begin{proof}
We represent
\begin{equation*}
\mathbb{A}(\phi ,\psi )=\Lambda \phi \psi +\phi \mathbb{K}
_{1}\psi +\psi \mathbb{K}_{2}\phi +\mathcal{A}_{a,b}(\phi ,\psi
),
\end{equation*}
where the kernels of the operators $\mathbb{K}_{1}$ and $\mathbb{K}_{2}$ are
\begin{equation*}
K_{1}(\xi ,\zeta )=\int_{\mathbb{R}}e^{-i\xi (a\eta
+b\zeta )}A(\eta ,\zeta )d\eta ,\quad K_{2}(\xi
,\eta )=\int_{\mathbb{R}}e^{-i\xi (a\eta +b\zeta )
}A(\eta ,\zeta )d\zeta .
\end{equation*}
For $|\xi \zeta |<1$, $\zeta \neq 0$ changing $\eta -\frac{
\zeta }{2}=z$ we get
\begin{align*}
K_{1}(\xi ,\zeta )
&=e^{-\langle\zeta \rangle
}\int_{\mathbb{R}}e^{-ia\xi \eta -\langle\eta -\frac{\zeta }{2}
\rangle }\frac{a_{1}\eta +b_{1}\zeta }{\eta ^{2}-\zeta \eta +\zeta ^{2}
}d\eta \\
&\quad+O\Big(e^{-\langle\zeta \rangle }\int_{\mathbb{R}}(
(|\eta |+|\zeta |)^{-1}|
e^{-\langle\eta \rangle }-e^{-\langle\eta -\frac{\zeta }{2}
\rangle }|+e^{-\langle\eta \rangle })d\eta
\Big)\\
&= e^{-\frac{i}{2}a\xi \zeta -\langle\zeta \rangle }\int_{
\mathbb{R}}e^{-\langle z\rangle }(C_{1}iz\sin (a\xi
z)+C_{2}\zeta \cos (a\xi z))\frac{dz}{z^{2}+\frac{3}{4}\zeta ^{2}}
 +O(e^{-\langle\zeta \rangle })\\
&=O(
e^{-\langle\zeta \rangle }).
\end{align*}
For $|\xi \zeta |\geq 1$ integrating three times by parts we
obtain
\begin{align*}
|K_{1}(\xi ,\zeta )|
&= |(a\xi )^{-3}\int_{\mathbb{R}}e^{-ia\xi \eta }
 \partial _{\eta }^{3}A(\eta ,\zeta )d\eta |\\
&\leq C|\xi |^{-3}e^{-\langle\zeta \rangle }\int_{
\mathbb{R}}\frac{d\eta }{(|\eta |+|\zeta |)^{4}}\\
&\leq C|\xi \zeta |^{-3}e^{-\langle\zeta
\rangle }.
\end{align*}
Hence the estimates are true for all $\xi \in \mathbb{R}$,
$\zeta ,\eta \in \mathbb{R}\backslash \{0\}$
\begin{equation*}
|K_{1}(\xi ,\zeta )|\leq Ce^{-\langle\zeta
\rangle }\langle\xi \zeta \rangle ^{-3},\quad
|K_{2}(\xi ,\eta )|\leq Ce^{-\langle\eta
\rangle }\langle\xi \eta \rangle ^{-3}.
\end{equation*}
Then by Lemma \ref{Lemma3.1}, we find
\begin{gather*}
\|\psi \mathbb{K}_{2}\phi \|_{\mathbf{L}^{p,\frac{3}{2}+\frac{1
}{p}+\beta +\lambda }}\leq C\|\psi \|_{\mathbf{L}^{p,\beta
}}\|\partial _{\xi }\phi \|_{\mathbf{L}^{2,\lambda }},
\\
\|\phi \mathbb{K}_{1}\psi \|_{\mathbf{L}^{p,\frac{3}{2}+\frac{1
}{p}+\alpha +\delta }}\leq C\|\phi \|_{\mathbf{L}^{p,\alpha
}}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }}.
\end{gather*}
Applying Lemma \ref{Lemma3.2} with $s=1$ we have
\begin{equation*}
\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{p,2+\lambda +\delta -\epsilon }}\leq C\|\partial _{\xi }\phi
\|_{\mathbf{L}^{2,\lambda }}\|\partial _{\xi }\psi \|_{
\mathbf{L}^{2,\delta }}
\end{equation*}
with $\epsilon >0$, $p=2,\infty $. Hence the first estimate of the lemma
follows. To prove the second estimate of the lemma we note that by
\eqref{3.1},
\begin{align*}
\Lambda (\xi )&=\int_{\mathbb{R}^{2}}e^{-i\xi (a\eta
+b\zeta )}A(\eta ,\zeta )d\eta d\zeta \\
&= \int_{\mathbb{R}}d\eta e^{-ia\xi \eta -\langle\eta \rangle
}\int_{\mathbb{R}}e^{-ib\xi \zeta }\frac{a_{1}\eta +b_{1}\zeta }{\eta
^{2}-\eta \zeta +\zeta ^{2}}d\zeta +O(\langle\xi \rangle
^{-2})\\
&= \pi \int_{\mathbb{R}}(C_{1}+C_{2}\text{sign}\eta )
e^{-i(a+\frac{b}{2})\xi \eta -\frac{\sqrt{3}}{2}|\eta
||b\xi |-\langle\eta \rangle }d\eta +O(
\langle\xi \rangle ^{-2})\\
&= O(\langle\xi \rangle ^{-1}).
\end{align*}
Hence
\begin{equation*}
\|\Lambda \phi \psi \|_{\mathbf{L}^{2,1+\lambda +\beta }}\leq
C\|\phi \|_{\mathbf{L}^{2,\lambda }}\|\psi \|_{
\mathbf{L}^{\infty ,\beta }}.
\end{equation*}
By virtue of estimates in Lemma \ref{Lemma3.1}, we get
\begin{gather*}
\|\phi \mathbb{K}_{1}\psi \|_{\mathbf{L}^{2,\frac{3}{2}+\lambda
+\delta }}\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}\|
\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }},
\\
\|\psi \mathbb{K}_{2}\phi \|_{\mathbf{L}^{2,1+\lambda +\beta
}}\leq C\|\phi \|_{\mathbf{L}^{2,\lambda }}\|\psi
\|_{\mathbf{L}^{\infty ,\beta }}.
\end{gather*}
Then applying the second estimate of Lemma \ref{Lemma3.2} with $s=1$,
 we have
\begin{equation*}
\|\mathcal{A}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{2,1+\lambda +\delta -\epsilon }}\leq C\|\phi \|_{\mathbf{L}
^{2,\lambda }}\|\partial _{\xi }\psi \|_{\mathbf{L}^{2,\delta }}
\end{equation*}
with $\epsilon >0$. The proof is complete.
\end{proof}

Next we consider the $\mathbf{L}^{2}$-estimates of the operator
\begin{equation*}
\mathbb{H}(\phi ,\psi )=\int_{\mathbb{R}^{2}}A(\eta
,\zeta )e^{\frac{i}{2}a(\xi -\eta )^{2}}\phi (\xi
-\eta )\partial _{\zeta }\big(e^{\frac{i}{2}b(\xi -\zeta
)^{2}}\psi (\xi -\zeta )\big)d\eta \,d\zeta .
\end{equation*}

\begin{lemma}\label{Lemma3.4}
Suppose that condition \eqref{3.1} is fulfilled. Then the
estimates
\begin{gather*}
\|\mathbb{H}(\phi ,\psi )\|_{\mathbf{L}^{2}}\leq
C\|\phi \|_{\mathbf{L}^{2}}(\|\psi \|_{
\mathbf{L}^{\infty }}+\|\partial _{\xi }\psi \|_{\mathbf{L}
^{2,\lambda }}+\|\partial _{\xi }^{2}\psi \|_{\mathbf{L}
^{2}}) ,
\\
\|\mathbb{H}(\phi ,\psi )\|_{\mathbf{L}^{2}}\leq
C\|\psi \|_{\mathbf{L}^{2}}(\|\phi \|_{
\mathbf{L}^{\infty }}+\|\partial _{\xi }\phi \|_{\mathbf{L}
^{2}})
\end{gather*}
are true provided that the right-hand sides are finite, $\lambda >0$.
\end{lemma}

\begin{proof}
To prove the first estimate of the lemma we write
\begin{equation*}
\mathbb{H}(\phi ,\psi )=e^{\frac{i}{2}(a+b)\xi
^{2}}(\xi \mathbb{A}_{1}(\phi ,\psi )+\mathbb{A}
_{2}(\phi ,\psi )+\mathbb{A}_{3}(\phi ,\partial _{\xi
}\psi ))
\end{equation*}
where the kernels are
\begin{gather*}
A_{1}(\eta ,\zeta )= -ibe^{\frac{i}{2}(a\eta ^{2}+b\zeta
^{2})}A(\eta ,\zeta ),\\
A_{2}(\eta ,\zeta )=
\frac{ib}{2}e^{\frac{i}{2}(a\eta ^{2}+b\zeta ^{2})}\zeta
A(\eta ,\zeta ), \\
A_{3}(\eta ,\zeta )= -e^{\frac{i}{2}(a\eta ^{2}+b\zeta
^{2})}A(\eta ,\zeta )
\end{gather*}
respectively. Then by the second estimate of Lemma \ref{Lemma3.3} we obtain
\begin{align*}
\|\mathbb{H}(\phi ,\psi )\|_{\mathbf{L}^{2}}
&\leq \|\xi \mathbb{A}_{1}(\phi ,\psi )\|_{
\mathbf{L}^{2}}+\|\mathbb{A}_{2}(\phi ,\psi )\|_{
\mathbf{L}^{2}}+\|\mathbb{A}_{3}(\phi ,\partial _{\xi }\psi
)\|_{\mathbf{L}^{2}} \\
&\leq C\|\phi \|_{\mathbf{L}^{2}}\big(\|\psi \|
_{\mathbf{L}^{\infty }}+\|\partial _{\xi }\psi \|_{\mathbf{L}
^{2,\lambda }}+\|\partial _{\xi }\psi \|_{\mathbf{L}^{\infty
}}+\|\partial _{\xi }^{2}\psi \|_{\mathbf{L}^{2}}\big).
\end{align*}
Thus the first estimate of the lemma is true. To prove the second estimate
we denote $e^{\frac{i}{2}b\xi ^{2}}\psi (\xi )=\widetilde{\psi }
(\xi )$ and represent
\begin{equation*}
e^{\frac{i}{2}a(\xi -\eta )^{2}}\phi (\xi -\eta )
=\phi (\xi )e^{\frac{i}{2}a(\xi -\eta )^{2}}+e^{
\frac{i}{2}a(\xi -\eta )^{2}}(\phi (\xi -\eta
)-\phi (\xi )).
\end{equation*}
Then we integrate by parts with respect to $\zeta $,
\begin{align*}
\mathbb{H}(\phi ,\psi )
&= -\int_{\mathbb{R}^{2}}e^{\frac{i}{2}
a(\xi -\eta )^{2}}\partial _{\zeta }A(\eta ,\zeta )
\phi (\xi -\eta )(\widetilde{\psi }(\xi -\zeta
)-\widetilde{\psi }(\xi ))d\eta d\zeta \\
&= -\phi \mathbb{K}_{1}\widetilde{\psi }-e^{\frac{i}{2}a\xi ^{2}}\widetilde{
\mathcal{A}}_{a,0}(\phi ,\widetilde{\psi }),
\end{align*}
where the kernels are respectively
\begin{gather*}
K_{1}(\xi ,\zeta )=\int_{\mathbb{R}}e^{\frac{i}{2}a(\xi
-\eta )^{2}}\partial _{\zeta }A(\eta ,\zeta )d\eta,
\\
\widetilde{A}(\eta ,\zeta )=e^{\frac{i}{2}a\eta ^{2}}\partial
_{\zeta }A(\eta ,\zeta ).
\end{gather*}
As above we have the estimate
\begin{equation*}
|K_{1}(\xi ,\zeta )|\leq C\langle\xi
\rangle e^{-\langle\zeta \rangle }\langle\xi \zeta
\rangle ^{-3}.
\end{equation*}
Thus by Lemma \ref{Lemma3.1} we find
\begin{equation*}
\|\phi \mathbb{K}_{1}\widetilde{\psi }\|_{\mathbf{L}^{2}}\leq
C\|\phi \|_{\mathbf{L}^{\infty }}\|\psi \|_{
\mathbf{L}^{2}}.
\end{equation*}
Finally by the second estimate of Lemma \ref{Lemma3.2} with $s=2$, $\lambda
=0$ we estimate
\begin{equation*}
\|e^{\frac{i}{2}a\xi ^{2}}\widetilde{\mathcal{A}}_{a,0}(\phi ,
\widetilde{\psi })\|_{\mathbf{L}^{2}}\leq C\|\partial
_{\xi }\phi \|_{\mathbf{L}^{2}}\|\psi \|_{\mathbf{L}^{2}}
\end{equation*}
Thus the second estimate of the lemma is true. The lemma is
proved.
\end{proof}

As a consequence of Lemmas \ref{Lemma3.3}--\ref{Lemma3.4} we obtain the
estimates of the operators
\begin{equation*}
\mathcal{G}_{a,b}(\phi ,\psi )=\overline{E}^{a+b}\int_{\mathbf{R
}^{2}}h(t,\eta ,\zeta )e^{\frac{a}{2}it(\xi -\eta )
^{2}}\phi (t,\xi -\eta )e^{\frac{b}{2}it(\xi -\zeta
)^{2}}\psi (t,\xi -\zeta )d\eta d\zeta
\end{equation*}
and
\begin{align*}
\mathcal{H}_{a,b}(\phi ,\psi )
&= t^{-1/2}\overline{E}^{a+b}\int_{\mathbb{R}^{2}}h(t,\eta ,\zeta )
e^{\frac{a}{2} it(\xi -\eta )^{2}}\phi (t,\xi -\eta )\\
&\quad \times \partial _{\zeta }(e^{\frac{b}{2}it(\xi -\zeta )
^{2}}\psi (t,\xi -\zeta ))d\eta d\zeta .
\end{align*}
Denote $B_{\lambda }=\langle\xi \sqrt{t}\rangle ^{\lambda }$ and
\begin{equation*}
Q_{a,b}(t,\xi )=\int_{\mathbb{R}^{2}}e^{-it\xi (a\eta
+b\zeta )}h(t,\eta ,\zeta )d\eta d\zeta .
\end{equation*}

\begin{lemma}\label{Lemma3.5}
Suppose that
\begin{equation}
|\partial _{\eta }^{k}\partial _{\zeta }^{l}(h(t,\eta
,\zeta )-\frac{\sqrt{t}(a_{1}\eta +b_{1}\zeta )}{\eta
^{2}-\zeta \eta +\zeta ^{2}}e^{-\langle\sqrt{t}\eta \rangle
-\langle\sqrt{t}\zeta \rangle })|
\leq C(|\eta |+|\zeta |)
^{-k-l}e^{-C\langle\sqrt{t}\eta \rangle -C\langle\sqrt{t}
\zeta \rangle }  \label{3.2}
\end{equation}
for all $t\geq 1$, $\eta ,\zeta \in \mathbb{R},$ $k,l=0,1,2,3$. Then the
following estimates are true for all $t\geq 1$:
\begin{align*}
&\|B_{\theta }(\mathcal{G}_{a,b}(\phi ,\psi )
-Q_{a,b}\phi \psi )\|_{\mathbf{L}^{p}} \\
&\leq Ct^{-\frac{1}{4}}\|B_{\alpha }\phi \|_{\mathbf{L}
^{p}}\|B_{\delta }\partial _{\xi }\psi \|_{\mathbf{L}^{2}}+Ct^{-
\frac{1}{4}}\|B_{\beta }\psi \|_{\mathbf{L}^{p}}\|
B_{\lambda }\partial _{\xi }\phi \|_{\mathbf{L}^{2}} \\
&\quad +Ct^{-1/2}\|B_{\lambda }\partial _{\xi }\phi \|_{
\mathbf{L}^{2}}\|B_{\delta }\partial _{\xi }\psi \|_{\mathbf{L}
^{2}},
\end{align*}
where
\begin{equation*}
\theta =\min (\frac{3}{2}+\frac{1}{p}+\alpha +\delta ,\frac{3}{2}+
\frac{1}{p}+\beta +\lambda ,2+\lambda +\delta -\epsilon ),\quad
p=2,\infty
\end{equation*}
and
\begin{equation*}
\|B_{\sigma }\mathcal{G}_{a,b}(\phi ,\psi )\|_{
\mathbf{L}^{2}}\leq C\|B_{\lambda }\phi \|_{\mathbf{L}
^{2}}(\|B_{\beta }\psi \|_{\mathbf{L}^{\infty }}+t^{-
\frac{1}{4}}\|B_{\delta }\partial _{\xi }\psi \|_{\mathbf{L}
^{2}})
\end{equation*}
for
$\sigma =\min (1+\lambda +\beta ,1+\lambda +\delta -\epsilon )
,\alpha ,\beta ,\lambda ,\delta \in \mathbb{R}$,
$\epsilon >0$ is small. Also for all $t\geq 1$,
\begin{equation*}
\|\mathcal{H}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{2}}\leq C\|\phi \|_{\mathbf{L}^{2}}\big(\|\psi
\|_{\mathbf{L}^{\infty }}+t^{-\frac{1}{4}}\|B_{\lambda
}\partial _{\xi }\psi \|_{\mathbf{L}^{2,\lambda }}+t^{-\frac{3}{4}
}\|\partial _{\xi }^{2}\psi \|_{\mathbf{L}^{2}}\big)
\end{equation*}
and
\begin{equation*}
\|\mathcal{H}_{a,b}(\phi ,\psi )\|_{\mathbf{L}
^{2}}\leq C\|\psi \|_{\mathbf{L}^{2}}\big(\|\phi
\|_{\mathbf{L}^{\infty }}+t^{-\frac{1}{4}}\|\partial _{\xi
}\phi \|_{\mathbf{L}^{2}}\big),
\end{equation*}
where $\lambda >0$.
\end{lemma}

\begin{proof}
We make a change $\sqrt{t}\xi =\xi '$, $\sqrt{t}\eta =\eta '$ and
$\sqrt{t}\zeta =\zeta '$,
\begin{align*}
\mathcal{G}_{a,b}(\phi ,\psi )
&= \overline{E}^{a+b}\int_{
\mathbb{R}^{2}}h(t,\eta ,\zeta )e^{\frac{a}{2}it(\xi
-\eta )^{2}}\phi (t,\xi -\eta )e^{\frac{b}{2}it(
\xi -\zeta )^{2}}\psi (t,\xi -\zeta )d\eta d\zeta \\
&= \int_{\mathbb{R}^{2}}e^{-i\xi '(a\eta '+b\zeta
')}\widetilde{A}(\eta ',\zeta ')\widetilde{\psi }(t,\xi '-\eta ')
\widetilde{\phi }(t,\xi '-\zeta ')d\eta
'd\zeta ',
\end{align*}
where
\begin{gather*}
\widetilde{A}(\eta ',\zeta ')= t^{-1}e^{
\frac{a}{2}i(\eta ')^{2}+\frac{b}{2}i(\zeta
')^{2}}h(t,\frac{\eta '}{\sqrt{t}},\frac{
\zeta '}{\sqrt{t}}), \\
\widetilde{\phi }(t,\xi ')= \phi (t,\frac{\xi
'}{\sqrt{t}}),\quad \ \widetilde{\psi }(t,\xi
')=\psi (t,\frac{\xi '}{\sqrt{t}}),
\end{gather*}
Also we denote
\begin{equation*}
\widetilde{Q}(\xi )=\int_{\mathbb{R}^{2}}e^{-i\xi (a\eta
'+b\zeta ')}\widetilde{A}(\eta ',\zeta ')d\eta 'd\zeta '.
\end{equation*}
Now application of Lemma \ref{Lemma3.3} yields the first two estimates of
the lemma. As above we make a change $\sqrt{t}\xi =\xi '$, $\sqrt{t}
\eta =\eta '$ and $\sqrt{t}\zeta =\zeta '$
\begin{align*}
\mathcal{H}_{a,b}(\phi ,\psi )
&= e^{\frac{i}{2}(
a+b)(\xi ')^{2}}\int_{\mathbb{R}^{2}}\widetilde{
A}(\eta ',\zeta ')e^{\frac{i}{2}a(\xi
'-\eta ')^{2}}\widetilde{\phi }(t,\xi
'-\eta ')\\
&\quad \times \partial _{\zeta '}(e^{\frac{i}{2}b(\xi
'-\zeta ')^{2}}\widetilde{\psi }(t,\xi
'-\eta '))d\eta 'd\zeta ',
\end{align*}
then applying Lemma \ref{Lemma3.4} we find the estimates of the lemma.
The proof is complete.
\end{proof}

\section{\label{S4}Inverse transformation}

We consider the transformation
\begin{equation*}
\mathcal{I}(v)=v-\overline{E}^{3}\mathcal{G}_{-1,-1}(
\overline{v},\overline{v}),
\end{equation*}
where $E(t)=e^{\frac{it}{2}\xi ^{2}}$ and
\begin{equation*}
\mathcal{G}_{a,b}(\phi ,\psi )=\overline{E}^{a+b}\int_{\mathbf{R
}^{2}}h(t,\eta ,\zeta )e^{\frac{a}{2}it(\xi -\eta )
^{2}}\phi (t,\xi -\eta )e^{\frac{b}{2}it(\xi -\zeta
)^{2}}\psi (t,\xi -\zeta )d\eta d\zeta .
\end{equation*}
We first give estimates of the operator $\mathcal{G}_{a,b}$ in the norm
\begin{equation*}
\|\phi \|_{\mathbf{X}_{T}^{\alpha ,\lambda }}\equiv \sup_{1\leq
t\leq T}(\|B_{\alpha }\phi (t)\|_{\mathbf{L}
^{\infty }}+t^{-\frac{1}{4}}\|B_{\lambda }\partial _{\xi }\phi (
t)\|_{\mathbf{L}^{2}}+t^{-\frac{3}{4}}\|\partial _{\xi
}^{2}\phi (t)\|_{\mathbf{L}^{2}}).
\end{equation*}

\begin{lemma}\label{Lemma4.1}
Let condition \eqref{3.2} be fulfilled. Then the estimate
\begin{equation*}
\|E^{q}(\mathcal{G}_{a,b}(\phi ,\psi )-Q_{a,b}\phi
\psi )\|_{\mathbf{X}_{T}^{\theta ,\sigma }}\leq C\|\phi
\|_{\mathbf{X}_{T}^{\alpha ,\lambda }}\|\psi \|_{\mathbf{
X}_{T}^{\beta ,\delta }}
\end{equation*}
is true, where
$\theta =\min (\frac{3}{2}+\alpha +\delta ,\frac{3}{2}+\beta +\lambda
,2+\lambda +\delta -\epsilon )$,
$\alpha ,\beta ,\lambda ,\delta \in \mathbb{R}$, $\epsilon >0$ is small,
\begin{equation*}
\sigma <\min (2+\alpha ,2+\beta ,2+\lambda ,2+\delta ,1+\alpha +\delta
,1+\beta +\lambda ,\frac{3}{2}+\lambda +\delta )
\end{equation*}
with an additional condition in the case of $q\neq 0$ that $\alpha ,\beta
,\lambda ,\delta $ are such that
\begin{gather*}
\sigma <\min (\alpha +\delta ,\beta +\lambda ,\frac{1}{2}+\lambda
+\delta ),
\\
\min (\alpha +\delta ,\beta +\lambda ,\frac{1}{2}+\lambda +\delta )>1.
\end{gather*}
\end{lemma}

\begin{proof}
By the first estimate of Lemma \ref{Lemma3.5} with $p=\infty $ we have
\begin{equation*}
\|B_{\theta }(\mathcal{G}_{a,b}(\phi ,\psi )
-Q_{a,b}\phi \psi )\|_{\mathbf{L}^{\infty }}\leq C\|\phi
\|_{\mathbf{X}_{T}^{\alpha ,\lambda }}\|\psi \|_{\mathbf{
X}_{T}^{\beta ,\delta }}.
\end{equation*}
We now estimate the derivative
\begin{align*}
&\partial _{\xi }(E^{q}(\mathcal{G}_{a,b}(\phi ,\psi
)-Q_{a,b}\phi \psi ))\\
&= E^{q}(\mathcal{G}_{a,b}(\partial _{\xi }\phi ,\psi )
-Q_{a,b}\psi \partial _{\xi }\phi )+E^{q}(\mathcal{G}
_{a,b}(\phi ,\partial _{\xi }\psi )-Q_{a,b}\phi \partial _{\xi
}\psi )\\
&\quad +\sqrt{t}E^{q}(\widetilde{\mathcal{G}_{a,b}}(\phi ,\psi
)-\widetilde{Q}_{a,b}\phi \psi )+itq\xi E^{q}(\mathcal{G}
_{a,b}(\phi ,\psi )-Q_{a,b}\phi \psi ),
\end{align*}
where $\widetilde{\mathcal{G}_{a,b}}$ and $\widetilde{Q}_{a,b}$ are defined
by the kernel $\widetilde{h}(t,\eta ,\zeta )=-i\sqrt{t}(
a\eta +b\zeta )h(t,\eta ,\zeta )$. Then by the first
estimate of Lemma \ref{Lemma3.5} with $p=2$ we find the estimate
\begin{align*}
&\|B_{\sigma }\partial _{\xi }(E^{q}(\mathcal{G}
_{a,b}(\phi ,\psi )-Q_{a,b}\phi \psi ))\|_{
\mathbf{L}^{2}}\\
&\leq C\|B_{\sigma }(\mathcal{G}_{a,b}(\partial _{\xi
}\phi ,\psi )-Q_{a,b}\psi \partial _{\xi }\phi )\|_{
\mathbf{L}^{2}} \\
&\quad +C\|B_{\sigma }(\mathcal{G}_{a,b}(\phi ,\partial _{\xi
}\psi )-Q_{a,b}\phi \partial _{\xi }\psi )\|_{\mathbf{L}
^{2}} \\
&\quad +C\sqrt{t}\|B_{-\frac{1}{2}-\epsilon }\|_{\mathbf{L}
^{2}}\|B_{\sigma +\frac{1}{2}+\epsilon }(\widetilde{\mathcal{G}
_{a,b}}(\phi ,\psi )-\widetilde{Q}_{a,b}\phi \psi )
\|_{\mathbf{L}^{\infty }} \\
&\quad +Cq\sqrt{t}\|B_{-\frac{1}{2}-\epsilon }\|_{\mathbf{L}
^{2}}\|B_{\sigma +\frac{3}{2}+\epsilon }(\mathcal{G}_{a,b}(
\phi ,\psi )-Q_{a,b}\phi \psi )\|_{\mathbf{L}^{\infty }}
\\
&\leq Ct^{1/4}\|\phi \|_{\mathbf{X}_{T}^{\alpha
,\lambda }}\|\psi \|_{\mathbf{X}_{T}^{\beta ,\delta }}
\end{align*}
with the additional condition
\begin{equation*}
\sigma <\min \big(2+\alpha ,2+\beta ,2+\lambda ,2+\delta ,
\alpha +\delta ,\beta +\lambda ,\frac{1}{2}+\lambda +\delta \big)
\end{equation*}
for the case of $q\neq 0$. Finally we estimate the second derivative
\begin{align*}
&\partial _{\xi }^{2}(E^{q}(\mathcal{G}_{a,b}(\phi ,\psi
)-Q_{a,b}\phi \psi ))\\
&=E^{q}(\mathcal{G}
_{a,b}(\partial _{\xi }^{2}\phi ,\psi )-Q_{a,b}\psi \partial
_{\xi }^{2}\phi )\\
&+2E^{q}(\mathcal{G}_{a,b}(\partial _{\xi }\phi ,\partial _{\xi
}\psi )-Q_{a,b}\partial _{\xi }\phi \partial _{\xi }\psi )
+E^{q}(\mathcal{G}_{a,b}(\phi ,\partial _{\xi }^{2}\psi )
-Q_{a,b}\phi \partial _{\xi }^{2}\psi )\\
&+\sqrt{t}E^{q}(\widetilde{\mathcal{G}_{a,b}}(\partial _{\xi
}\phi ,\psi )-\widetilde{Q}_{a,b}\partial _{\xi }\phi \psi )+
\sqrt{t}E^{q}(\widetilde{\mathcal{G}_{a,b}}(\phi ,\partial _{\xi
}\psi )-\widetilde{Q}_{a,b}\phi \partial _{\xi }\psi )\\
&+tE^{q}(\widetilde{\widetilde{\mathcal{G}_{a,b}}}(\phi ,\psi
)-\widetilde{\widetilde{Q}}_{a,b}\phi \psi )+iqt(
1+iqt\xi ^{2})E^{q}(\mathcal{G}_{a,b}(\phi ,\psi )
-Q_{a,b}\phi \psi )\\
&+itq\xi E^{q}(\mathcal{G}_{a,b}(\partial _{\xi }\phi ,\psi
)-Q_{a,b}\psi \partial _{\xi }\phi )+itq\xi E^{q}(
\mathcal{G}_{a,b}(\phi ,\partial _{\xi }\psi )-Q_{a,b}\phi
\partial _{\xi }\psi )\\
&+2iqt^{\frac{3}{2}}\xi E^{q}(\widetilde{\mathcal{G}_{a,b}}(
\phi ,\psi )-\widetilde{Q}_{a,b}\phi \psi ),
\end{align*}
where $\widetilde{\widetilde{\mathcal{G}_{a,b}}}$  and $\widetilde{
\widetilde{Q}}_{a,b}$ are defined by the kernel
\begin{equation*}
\widetilde{\widetilde{h}}(t,\eta ,\zeta )=-i\sqrt{t}(
a\eta +b\zeta )\widetilde{h}(t,\eta ,\zeta ).
\end{equation*}
Then by Lemma \ref{Lemma3.5} we find the estimates
\begin{align*}
&\|\partial _{\xi }^{2}(E^{q}(\mathcal{G}_{a,b}(
\phi ,\psi )-Q_{a,b}\phi \psi ))\|_{\mathbf{L}^{2}}\\
&\leq C\|\mathcal{G}_{a,b}(\partial _{\xi }^{2}\phi ,\psi
)-Q_{a,b}\psi \partial _{\xi }^{2}\phi \|_{\mathbf{L}^{2}} \\
&\quad +C\|\mathcal{G}_{a,b}(\partial _{\xi }\phi ,\partial _{\xi
}\psi )-Q_{a,b}\partial _{\xi }\phi \partial _{\xi }\psi \|_{
\mathbf{L}^{2}}+C\|\mathcal{G}_{a,b}(\phi ,\partial _{\xi
}^{2}\psi )-Q_{a,b}\phi \partial _{\xi }^{2}\psi \|_{\mathbf{L}
^{2}} \\
&\quad +C\sqrt{t}\|\widetilde{\mathcal{G}_{a,b}}(\partial _{\xi }\phi
,\psi )-\widetilde{Q}_{a,b}\psi \partial _{\xi }\phi \|_{
\mathbf{L}^{2}}+C\sqrt{t}\|\widetilde{\mathcal{G}_{a,b}}(\phi
,\partial _{\xi }\psi )-\widetilde{Q}_{a,b}\phi \partial _{\xi }\psi
\|_{\mathbf{L}^{2}} \\
&\quad +Ct\|B_{-\frac{1}{2}-\epsilon }\|_{\mathbf{L}^{2}}\|B_{
\frac{1}{2}+\epsilon }(\widetilde{\widetilde{\mathcal{G}_{a,b}}}(
\phi ,\psi )-\widetilde{\widetilde{Q}}_{a,b}\phi \psi )
\|_{\mathbf{L}^{\infty }} \\
&\quad +Cq\sqrt{t}\|B_{1}(\mathcal{G}_{a,b}(\partial _{\xi
}\phi ,\psi )-Q_{a,b}\psi \partial _{\xi }\phi )\|_{
\mathbf{L}^{2}} \\
&\quad +Cq\sqrt{t}\|B_{1}(\mathcal{G}_{a,b}(\phi ,\partial
_{\xi }\psi )-Q_{a,b}\phi \partial _{\xi }\psi )\|_{
\mathbf{L}^{2}} \\
&\quad +Cqt\|B_{-\frac{1}{2}-\epsilon }\|_{\mathbf{L}^{2}}\|B_{
\frac{3}{2}+\epsilon }(\widetilde{\mathcal{G}_{a,b}}(\phi ,\psi
)-\widetilde{Q}_{a,b}\phi \psi )\|_{\mathbf{L}^{\infty
}} \\
&\quad +Cqt\|B_{-\frac{1}{2}-\epsilon }\|_{\mathbf{L}^{2}}\|B_{
\frac{5}{2}+\epsilon }(\mathcal{G}_{a,b}(\phi ,\psi )
-Q_{a,b}\phi \psi )\|_{\mathbf{L}^{\infty }} \\
&\leq Ct^{3/4}\|\phi \|_{\mathbf{X}_{T}^{\alpha ,\lambda }}\|\psi
\|_{\mathbf{X}_{T}^{\beta ,\delta }}
\end{align*}
with the additional condition
\begin{equation*}
\min \big(\alpha +\delta ,\beta +\lambda ,\frac{1}{2}+\lambda +\delta
\big)>1
\end{equation*}
for the case of $q\neq 0$. Lemma \ref{Lemma4.1} is proved.
\end{proof}

Now let us find the inverse transformation $\mathcal{I}^{-1}$. We consider
the equation
\begin{equation}
\phi =\mathcal{I}(v).  \label{4.1}
\end{equation}
We look for the solution of (\ref{4.1}) in the form $v=\phi +\psi _{1}+
\overline{E}^{3}\psi _{2}$ and substitute it into (\ref{4.1}), then we find
\begin{equation}
\psi _{1}+\overline{E}^{3}\psi _{2}=\overline{E}^{3}\mathcal{G}
_{-1,-1}(\overline{\phi }+\overline{\psi _{1}},\overline{\phi }+
\overline{\psi _{1}})
+2\mathcal{G}_{-1,2}(\overline{\phi }+\overline{\psi _{1}},\overline{
\psi _{2}})+E^{3}\mathcal{G}_{2,2}(\overline{\psi _{2}},
\overline{\psi _{2}}).  \label{4.2}
\end{equation}
Comparing in (\ref{4.2}) the terms with the same oscillating exponents like
$\overline{E}^{3}$ we find a system of equations
\begin{equation}
\begin{gathered}
\psi _{1}=\mathcal{G}_{-1,2}(\overline{\phi }+\overline{\psi _{1}},
\overline{\psi _{2}})+E^{3}\mathcal{G}_{2,2}(\overline{\psi _{2}
},\overline{\psi _{2}}), \\
\psi _{2}=\mathcal{G}_{-1,-1}(\overline{\phi }+\overline{\psi _{1}},
\overline{\phi }+\overline{\psi _{1}}).
\end{gathered}  \label{4.3}
\end{equation}
In the next lemma we solve this system in the space
\begin{equation*}
\mathbf{Z}_{\varepsilon }=\big\{(\psi _{1},\psi _{2})
\in (\mathbf{C}([1,T] ;\mathbf{C}^{2}(\mathbf{R
})))^{2}:\|(\psi _{1},\psi _{2})
\|_{\mathbf{Z}}\leq C\varepsilon ^{2}\big\}
\end{equation*}
with the norm
\begin{equation*}
\|(\psi _{1},\psi _{2})\|_{\mathbf{Z}}\equiv
\|\psi _{1}\|_{\mathbf{X}_{T}^{2,1+\frac{\gamma }{3}}}+\|
\psi _{2}\|_{\mathbf{X}_{T}^{1,1+\frac{\gamma }{2}}},
\end{equation*}
where $\gamma \in (0,\frac{1}{2})$.

\begin{lemma}\label{Lemma4.2}
Suppose that $\phi \in \mathbf{C}([1,T] ;
\mathbf{C}^{2}(\mathbb{R}))$ and the estimate is true
\begin{equation*}
\|\phi \|_{\mathbf{X}_{T}^{0,\gamma }}\leq \varepsilon ,
\end{equation*}
where $\varepsilon >0$ is sufficiently small. Then there exists unique
solutions $\psi _{1},\psi _{2}\in \mathbf{C}([1,T] ;
\mathbf{C}^{2}(\mathbb{R}))$ of a system (\ref{4.3})
such that $\|(\psi _{1},\psi _{2})\|_{\mathbf{Z}
}\leq C\varepsilon ^{2}$.
\end{lemma}

\begin{proof}
We solve equations (\ref{4.3}) by the contraction mapping principle in the
set $\mathbf{Z}_{\varepsilon }$. Define the transformation
$\mathcal{M}(\psi _{1},\psi _{2})=(\mathcal{M}_{1},\mathcal{M}_{2})$, where
\begin{equation}
\begin{gathered}
\mathcal{M}_{1}=\mathcal{G}_{-1,2}(\overline{\phi }+\overline{\psi _{1}
},\overline{\psi _{2}})+E^{3}\mathcal{G}_{2,2}(\overline{\psi
_{2}},\overline{\psi _{2}}), \\
\mathcal{M}_{2}=\mathcal{G}_{-1,-1}(\overline{\phi }+\overline{\psi
_{1}},\overline{\phi }+\overline{\psi _{1}})
\end{gathered}  \label{2.9}
\end{equation}
for $(\psi _{1},\psi _{2})\in \mathbf{Z}_{\varepsilon }$.
Applying Lemma \ref{Lemma4.1}, in view of the fact that $\| \phi
\|_{\mathbf{X}_{T}^{0,\gamma }}\leq \varepsilon $ and $\|
(\psi _{1},\psi _{2})\|_{\mathbf{Z}}\leq C\varepsilon
^{2}$, we obtain, by (\ref{2.9}),
\begin{align*}
\| \mathcal{M}_{1}\|_{\mathbf{X}_{T}^{2,1+\frac{\gamma }{
3}}}
&\leq \| \mathcal{G}_{-1,2}(\overline{\phi }+\overline{\psi
_{1}},\overline{\psi _{2}})-Q_{-1,2}(\overline{\phi }+\overline{
\psi _{1}})\overline{\psi _{2}}\|_{\mathbf{X}_{T}^{2,1+
\frac{\gamma }{3}}} \\
&\quad +\| E^{3}(\mathcal{G}_{2,2}(\overline{\psi _{2}},
\overline{\psi _{2}})-Q_{2,2}\overline{\psi _{2}}^{2})
\|_{\mathbf{X}_{T}^{2,1+\frac{\gamma }{3}}} \\
&\quad +\| Q_{-1,-1}(\overline{\phi }+\overline{\psi _{1}})
\overline{\psi _{2}}\|_{\mathbf{X}_{T}^{2,1+\frac{\gamma }{3}
}}+\| E^{3}Q_{2,2}\overline{\psi _{2}}^{2}\|_{\mathbf{X}
_{T}^{2,1+\frac{\gamma }{3}}} \\
&\leq C\| \psi _{2}\|_{\mathbf{X}_{T}^{1,1+\frac{\gamma }{
2}}}(\| \phi \|_{\mathbf{X}_{T}^{0,\gamma
}}+\| \psi _{1}\|_{\mathbf{X}_{T}^{0,\gamma }}+\|
\psi _{2}\|_{\mathbf{X}_{T}^{1,1+\frac{\gamma }{2}}})\leq
C\varepsilon ^{3}.
\end{align*}
In the same manner we have
\begin{align*}
\| \mathcal{M}_{2}\|_{\mathbf{X}_{T}^{1,1+\frac{\gamma }{
2}}}
&\leq \| \mathcal{G}_{-1,-1}(\overline{\phi }+\overline{
\psi _{1}},\overline{\phi }+\overline{\psi _{1}})-Q_{-1,-1}(
\overline{\phi }+\overline{\psi _{1}})^{2}\|_{\mathbf{X}
_{T}^{1,1+\frac{\gamma }{2}}}\\
&\quad +\| Q_{-1,-1}(\overline{\phi }+\overline{\psi _{1}})
^{2}\|_{\mathbf{X}_{T}^{1,1+\frac{\gamma }{2}}}\\
&\leq C(\| \phi \|_{\mathbf{X}_{T}^{0,\gamma }}+\| \psi
_{1}\|_{\mathbf{X}_{T}^{0,\gamma }})^{2}\leq C\varepsilon^{2}.
\end{align*}
Thus the mapping $\mathcal{M}(\psi _{1},\psi _{2})$ transforms
the set $\mathbf{Z}_{\varepsilon }$ into itself. In the same manner we find
\begin{equation*}
\| \mathcal{M}(\psi _{1},\psi _{2})-\mathcal{M}(
\widetilde{\psi _{1}},\widetilde{\psi _{2}})\|_{\mathbf{Z}
}\leq \frac{1}{2}\| (\psi _{1},\psi _{2})-(
\widetilde{\psi _{1}},\widetilde{\psi _{2}})\|_{\mathbf{Z}}.
\end{equation*}
Therefore $(\mathcal{M}_{1},\mathcal{M}_{2})$ is a contraction
mapping in $\mathbf{Z}_{\varepsilon }\mathbf{.}$ Hence there exist a unique
solution $(\psi _{1},\psi _{2})\in \mathbf{Z}_{\varepsilon }$
of a system of integral equations (\ref{4.3}). The proof is complete.
\end{proof}

\section{\label{S5}Estimates for derivatives}

We now make a change $\phi =\mathcal{I}(v)$ in equation (\ref{2.5})
\begin{equation}
\begin{gathered}
\mathcal{L}\phi =t^{-1}\mathcal{P}, \\
\phi (1,\xi )=\phi _{0}(\xi ),
\end{gathered}  \label{5.1}
\end{equation}
where $\mathcal{L}=i\partial _{t}+\frac{1}{2t^{2}}\partial _{\xi }^{2}$,
$E=e^{\frac{i}{2}t\xi ^{2}}$,
\begin{equation*}
\mathcal{P}=\overline{E}^{3}\widetilde{\mathcal{G}}_{-1,-1}(\overline{v
},\overline{v})-\mathcal{H}_{-1,2}(\overline{v},v^{2})
\end{equation*}
with $v=\mathcal{I}^{-1}(\phi )$ and the operator
\begin{align*}
\mathcal{H}_{a,b}(\phi ,\psi )&= t^{-1/2}\overline{E}
^{a+b}\int_{\mathbb{R}^{2}}h(t,\eta ,\zeta )e^{\frac{a}{2}
it(\xi -\eta )^{2}}\phi (t,\xi -\eta )\\
&\quad\times \partial _{\zeta }\big(e^{\frac{b}{2}it(\xi -\zeta )
^{2}}\psi (t,\xi -\zeta )\big)d\eta d\zeta .
\end{align*}
Define the norms
\begin{gather*}
\|\phi \|_{\mathbf{V}_{T}^{\alpha }}\equiv \sup_{1\leq t\leq
T}\|B_{\alpha }\phi (t)\|_{\mathbf{L}^{\infty }}, \\
\|\phi \|_{\mathbf{W}_{T}^{\gamma }}\equiv \sup_{1\leq t\leq
T}(t^{-\frac{1}{4}}\|B_{\gamma }\phi (t)\|_{
\mathbf{L}^{2}}+t^{-\frac{3}{4}}\|\partial _{\xi }\phi (t)
\|_{\mathbf{L}^{2}}).
\end{gather*}
First we state the local existence result for equation \eqref{5.1}. Denote
$\mathbf{Y}=\{\phi \in \mathbf{L}^{\infty },\phi '\in \mathbf{H}^{1,1}\}$.

\begin{theorem}\label{T5.1}
Assume that the initial data $\phi _{0}\in \mathbf{Y}$. Then for
some time $T>1$ there exists a unique solution $\phi \in \mathbf{C}(
[1,T] ;\mathbf{Y})$ of the Cauchy problem \eqref{5.1}.
\end{theorem}

In the next lemma we give a representation for the derivatives of the
operator $\mathcal{H}_{a,b}$.

\begin{lemma}\label{Lemma5.1}
Let condition \eqref{3.2} be fulfilled. Then the estimate is
true
\begin{equation*}
\|\partial _{\xi }(E^{q}\mathcal{H}_{a,b}(\phi ,\psi
))-iqbt^{\frac{3}{2}}\xi ^{2}E^{q}Q_{a,b}\phi \psi \|_{
\mathbf{W}_{T}^{\rho }}\leq C\|\phi \|_{\mathbf{X}_{T}^{\alpha
,\lambda }}\|\psi \|_{\mathbf{X}_{T}^{\beta ,\delta }},
\end{equation*}
where
\begin{equation*}
\rho =\min \big(\alpha +\delta ,\beta +\lambda ,1+\alpha ,1+\beta ,\frac{1
}{2}+\alpha +\beta -\epsilon ,\lambda +\delta -\epsilon ,1+\delta -\epsilon
,1+\lambda -\epsilon \big)
\end{equation*}
$\alpha ,\beta ,\lambda ,\delta \in \mathbb{R}$, $\epsilon >0$ is small.
Also in the case of $q\neq 0$ we assume that $\alpha ,\beta \in \mathbb{R}$,
$\lambda ,\delta >0$ are such that $\rho \geq 1$.
\end{lemma}

\begin{proof}
We note that
\begin{equation*}
\mathcal{H}_{a,b}(\phi ,\psi )=ib\sqrt{t}\xi \mathcal{G}
_{a,b}(\phi ,\psi )+t^{-1/2}\mathcal{G}_{a,b}(
\phi ,\partial _{\xi }\psi )+\mathcal{G}_{a,b}^{(1)
}(\phi ,\psi ),
\end{equation*}
where $\mathcal{G}_{a,b}^{(1)}$ \ and $Q^{(1)}$
are defined by the kernel $h^{(1)}(t,\eta ,\zeta )=
\sqrt{t}b\zeta h(t,\eta ,\zeta )$. Then we obtain by Lemma
\ref{Lemma3.5}
\begin{align*}
&\| B_{\rho }\xi tE^{q}(\mathcal{H}_{a,b}(\phi ,\psi
)-ib\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{\mathbf{L}^{2}}
\\
&\leq \sqrt{t}\| B_{\rho +1}(\mathcal{H}_{a,b}(\phi
,\psi )-ib\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{\mathbf{L
}^{2}} \\
&\leq C\sqrt{t}\| B_{2+\rho }(\mathcal{G}_{a,b}(\phi
,\psi )-Q_{a,b}\phi \psi )\|_{\mathbf{L}^{2}} \\
&\quad +C\| B_{\rho +1}\mathcal{G}_{a,b}(\phi ,\partial _{\xi }\psi
)\|_{\mathbf{L}^{2}}+C\sqrt{t}\| B_{\rho +1}
\mathcal{G}_{a,b}^{(1)}(\phi ,\psi )\|_{
\mathbf{L}^{2}} \\
&\leq Ct^{1/4}\| \phi \|_{\mathbf{X}_{T}^{\alpha
,\lambda }}\| \psi \|_{\mathbf{X}_{T}^{\beta ,\delta }}.
\end{align*}
We now estimate by Lemma \ref{Lemma3.5} the derivative,
\begin{align*}
&\| B_{\rho }\partial _{\xi }(\mathcal{H}_{a,b}(\phi
,\psi )-ib\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{\mathbf{L
}^{2}} \\
&\leq C\sqrt{t}\| B_{\rho }\partial _{\xi }\xi (\mathcal{G}
_{a,b}(\phi ,\psi )-Q_{a,b}\phi \psi )\|_{
\mathbf{L}^{2}} \\
&\quad +Ct^{-1/2}\| B_{\rho }\partial _{\xi }\mathcal{G}
_{a,b}(\phi ,\partial _{\xi }\psi )\|_{\mathbf{L}
^{2}}+C\| B_{\sigma }\partial _{\xi }\mathcal{G}_{a,b}^{(
1)}(\phi ,\psi )\|_{\mathbf{L}^{2}} \\
&\leq Ct^{1/4}\| \phi \|_{\mathbf{X}_{T}^{\alpha
,\lambda }}\| \psi \|_{\mathbf{X}_{T}^{\beta ,\delta }}.
\end{align*}
Hence
\begin{align*}
&\| \partial _{\xi }\xi tE^{q}(\mathcal{H}_{a,b}(\phi
,\psi )-ib\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{\mathbf{L}^{2}}\\
&\leq Ct\| \mathcal{H}_{a,b}(\phi ,\psi )-ib\sqrt{t}
\xi Q_{a,b}\phi \psi \|_{\mathbf{L}^{2}} \\
&\quad+Cqt\| B_{2}(\mathcal{H}_{a,b}(\phi ,\psi )-ib
\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{\mathbf{L}^{2}} \\
&\quad +C\sqrt{t}\| B_{1}\partial _{\xi }(\mathcal{H}_{a,b}(
\phi ,\psi )-ib\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{
\mathbf{L}^{2}} \\
&\leq Ct^{\frac{3}{4}}\| \phi \|_{\mathbf{X}_{T}^{\alpha
,\lambda }}\| \psi \|_{\mathbf{X}_{T}^{\beta ,\delta }}
\end{align*}
if $\rho \geq 1$ for the case $q\neq 0$. Also by Lemma \ref{Lemma3.5} we
have
\begin{align*}
&\| B_{\rho }E^{q}\partial _{\xi }\mathcal{H}_{a,b}(\phi
,\psi )\|_{\mathbf{L}^{2}}\\
&\leq C\| B_{\rho}\partial _{\xi }(\mathcal{H}_{a,b}(\phi ,\psi )
-ib\sqrt{t}\xi Q_{a,b}\phi \psi )\|_{\mathbf{L}^{2}}
 +C\sqrt{t}\| B_{\rho }\partial _{\xi }(\xi Q_{a,b}\phi \psi
)\|_{\mathbf{L}^{2}}\\
&\leq Ct^{1/4}\| \phi \|_{\mathbf{X}_{T}^{\alpha ,\lambda }}\| \psi \|
_{\mathbf{X}_{T}^{\beta ,\delta }}
\end{align*}
and
\begin{align*}
&\| \partial _{\xi }E^{q}\partial _{\xi }\mathcal{H}_{a,b}(
\phi ,\psi )\|_{\mathbf{L}^{2}}\\
&\leq \| \partial _{\xi }^{2}\mathcal{H}_{a,b}(\phi ,\psi )\|_{\mathbf{L
}^{2}}+Cq\sqrt{t}\| B_{1}E^{q}\partial _{\xi }\mathcal{H}_{a,b}(
\phi ,\psi )\|_{\mathbf{L}^{2}}\\
&\leq Ct^{3/4}\| \phi \|_{\mathbf{X}_{T}^{\alpha ,\lambda }}\|
\psi \|_{\mathbf{X}_{T}^{\beta ,\delta }}
\end{align*}
 if $\rho \geq 1$ in the case of $q\neq 0$. The lemma is proved.
\end{proof}

We now substitute the inverse transformation $v=\mathcal{I}^{-1}(\phi
)=\phi _{1}+\overline{E}^{3}\psi _{2}$ with $\phi _{1}=\phi +\psi
_{1} $ into the operator $\mathcal{P}$ to get
\begin{align*}
\mathcal{P} &= \overline{E}^{3}\mathcal{H}_{2,2}(\overline{\psi _{2}}
,\phi _{1}^{2})+2E^{3}\mathcal{H}_{-1,-1}(\overline{\phi _{1}}
,\phi _{1}\psi _{2})+E^{6}\mathcal{H}_{-1,-4}(\overline{\phi
_{1}},\psi _{2}^{2})\\
&\quad +\overline{E}^{3}\widetilde{\mathcal{G}}_{-1,-1}(\overline{\phi _{1}}
,\overline{\phi _{1}})+\mathcal{H}_{-1,2}(\overline{\phi _{1}}
,\phi _{1}^{2})+2\widetilde{\mathcal{G}}_{-1,2}(\overline{\phi
_{1}},\overline{\psi _{2}})\\
&\quad +E^{3}\widetilde{\mathcal{G}}_{2,2}(\overline{\psi _{2}},\overline{
\psi _{2}})+2\mathcal{H}_{2,-1}(\overline{\psi _{2}},\phi
_{1}\psi _{2})+E^{3}\mathcal{H}_{2,-4}(\overline{\psi _{2}}
,\psi _{2}^{2}).
\end{align*}
Denote $\varepsilon =\|\phi \|_{\mathbf{X}_{T}^{0,\gamma }}$.
Since $\psi _{2}=\mathcal{G}_{-1,-1}(\overline{\phi }+\overline{\psi
_{1}},\overline{\phi }+\overline{\psi _{1}})$ then by Lemma
\ref{Lemma3.5} we have
\begin{equation*}
\|t\xi (\psi _{2}-Q_{-1,-1}\overline{\phi }^{2})\|
_{\mathbf{W}_{T}^{1+\gamma }}\leq C\varepsilon ^{2}
\end{equation*}
Then by virtue of Lemma \ref{Lemma5.1} with $\rho =1$ we have
\begin{align*}
&\|\partial _{\xi }\overline{E}^{3}\mathcal{H}_{2,2}(\overline{
\psi _{2}},\phi _{1}^{2})+6it^{\frac{3}{2}}\xi ^{2}\overline{E}
^{3}Q_{2,2}\overline{Q_{-1,-1}}\phi ^{4}\|_{\mathbf{W}_{T}^{1}} \\
&\leq \|\partial _{\xi }(\overline{E}^{3}\mathcal{H}
_{2,2}(\overline{\psi _{2}},\phi _{1}^{2}))+6it^{\frac{3
}{2}}\xi ^{2}\overline{E}^{3}Q_{2,2}\overline{\psi _{2}}\phi
_{1}^{2}\|_{\mathbf{W}_{T}^{1}} \\
&\quad +C\|t^{\frac{3}{2}}\xi ^{2}\overline{E}^{3}Q_{2,2}\overline{\psi _{2}
}\phi _{1}^{2}-t^{\frac{3}{2}}\xi ^{2}\overline{E}^{3}Q_{2,2}\overline{
Q_{-1,-1}}\phi ^{4}\|_{\mathbf{W}_{T}^{1}} \\
&\leq C\|\psi _{2}\|_{\mathbf{X}_{T}^{1,1+\frac{\gamma }{2}
}}\|\phi _{1}^{2}\|_{\mathbf{X}_{T}^{0,\gamma }}\leq
C\varepsilon ^{2}.
\end{align*}
In the same manner
\begin{gather*}
\|\partial _{\xi }E^{3}\mathcal{H}_{-1,-1}(\overline{\phi _{1}}
,\phi _{1}\psi _{2})+3it^{\frac{3}{2}}\xi ^{2}E^{3}Q_{-1,-1}^{2}\phi
\overline{\phi }^{3}\|_{\mathbf{W}_{T}^{1}}\leq C\varepsilon ^{2},
\\
\|\partial _{\xi }E^{6}\mathcal{H}_{-1,-4}(\overline{\phi _{1}}
,\psi _{2}^{2})+24it^{\frac{3}{2}}\xi ^{2}E^{6}Q_{-1,-1}^{2}Q_{-1,-4}
\overline{\phi }^{5}\|_{\mathbf{W}_{T}^{1}}\leq C\varepsilon ^{2}.
\end{gather*}
All the other terms in the derivative $\partial _{\xi }\mathcal{P}$ can be
estimated in the norm $\mathbf{W}_{T}^{1}$, the worst term is $\mathcal{H}
_{-1,2}(\overline{\phi _{1}},\phi _{1}^{2})$ which yields the
restriction \ $\gamma <\frac{1}{2}-\epsilon $. Therefore we can represent
$\partial _{\xi }\mathcal{P}$ in the form
\begin{equation}
\partial _{\xi }\mathcal{P}=t^{1/2}\sum_{j=1}^{3}E^{\omega
_{j}}\Omega _{j}\mathcal{N}_{j}+\mathcal{R},  \label{5.2}
\end{equation}
where
$\mathcal{N}_{1}=\phi ^{4}$,
$\mathcal{N}_{2}=\phi \overline{\phi }^{3}$,
$\mathcal{N}_{3}=\overline{\phi }^{5}$,
$\omega _{1}=-3$, $\omega _{2}=3$, $\omega _{3}=6$,
\begin{equation*}
\Omega _{1}=-6it\xi ^{2}\overline{Q_{-1,-1}}Q_{2,2},\quad
\Omega _{2}=-3it\xi ^{2}Q_{-1,-1}^{2},\quad
\Omega _{3}=-24it\xi ^{2}Q_{-1,-1}^{2}Q_{-1,-4}
\end{equation*}
with the estimate of the remainder
$\|\mathcal{R}\|_{\mathbf{W}_{T}^{\gamma }}\leq C\varepsilon ^{2}$
and
\begin{equation*}
t^{-\frac{k}{2}+l}|\partial _{\xi }^{k}\partial _{t}^{l}\Omega
_{j}(t,\xi )|\leq C\langle\xi \sqrt{t}
\rangle ^{-l-k}
\end{equation*}
for all $t\geq 1$, $\xi \in \mathbb{R}$, $j=1,2,3$, $k,l=0,1,2$.

\begin{lemma}\label{Lemma5.2}
Let the initial data $\phi _{0}\in \mathbf{Y}$ and $\|
\phi _{0}\|_{\mathbf{Y}}\leq \varepsilon $, where $\varepsilon >0$ is
sufficiently small. Assume that representation (\ref{5.2}) is valid with the
estimate of the remainder
\begin{equation*}
\|\mathcal{R}\|_{\mathbf{W}_{T}^{\gamma }}\leq C\varepsilon
^{2}.
\end{equation*}
Suppose that
\begin{equation}
\|\phi \|_{\mathbf{V}_{T}^{0}}\leq \varepsilon .  \label{5.3}
\end{equation}
Then the solutions $\phi \in \mathbf{C}([1,T];\mathbf{Y}
)$ of \eqref{5.1} satisfy the estimate
\begin{equation}
\|\partial _{\xi }\phi \|_{\mathbf{W}_{T}^{\gamma
}}<10\varepsilon .  \label{5.4}
\end{equation}
\end{lemma}

\begin{proof}
We prove estimate (\ref{5.4}) by contradiction. By the continuity of $\phi $
we can find a maximal time $\widetilde{T}\in (1,T] $ such that
\begin{equation}
\|\partial _{\xi }\phi \|_{\mathbf{W}_{\widetilde{T}}^{\gamma
}}\leq 10\varepsilon .  \label{5.5}
\end{equation}
Thus we have $\|\phi \|_{\mathbf{X}_{\widetilde{T}}^{0,\gamma
}}\leq 10\varepsilon $. By a direct calculation we have
\begin{equation}
\mathcal{L}(E^{\omega _{j}}\chi _{j})=t^{-1}E^{\omega
_{j}}\big(\frac{i\omega _{j}}{2A_{j}}\chi _{j}+i\omega _{j}\xi \partial
_{\xi }\chi _{j}+t\mathcal{L}\chi _{j}\big)\label{5.6}
\end{equation}
with $A_{j}=(1+(1+\omega _{j})it\xi ^{2})^{-1}$.
This identity is useful in the case of $\omega \neq -1$.
Then we get, from (\ref{5.1}) and (\ref{5.2}),
\begin{equation}
\begin{aligned}
&\mathcal{L}\big(\phi _{\xi }+\sum_{j=1}^{3}E^{\omega _{j}}\chi_{j}\big)\\
&=t^{-1}\sum_{j=1}^{3}E^{\omega _{j}}\big(t^{1/2}\Omega
_{j}\mathcal{N}_{j}+\frac{i\omega _{j}}{2A_{j}}\chi _{j}\big)
+t^{-1}\sum_{j=1}^{3}E^{\omega _{j}}(i\omega _{j}\xi \partial _{\xi
}\chi _{j}+t\mathcal{L}\chi _{j})+t^{-1}\mathcal{R}.
\end{aligned} \label{5.7}
\end{equation}
To eliminate the first summand in the right-hand side of (\ref{5.7}) we
choose $\chi _{j}=\frac{2i}{\omega _{j}}t^{1/2}\Omega _{j}A_{j}
\mathcal{N}_{j}$ and denote $\Phi =\phi _{\xi }+\sum_{j=1}^{3}E^{\omega
_{j}}\chi _{j}$. By the identities
\begin{gather*}
\mathcal{L}(uv)=v\mathcal{L}u+\frac{1}{t^{2}}u_{\xi }v_{\xi }+u
\mathcal{L}v,
\\
\mathcal{L}\overline{\phi }=-\overline{\mathcal{L}\phi }+\frac{1}{t^{2}}
\overline{\phi }_{\xi \xi }, \\
\mathcal{L}\mathcal{N}_{j}
=\mathcal{N}_{j\phi }\mathcal{L}\phi -\mathcal{N}
_{j\overline{\phi }}\overline{\mathcal{L}\phi }
+\frac{1}{2t^{2}}(\mathcal{N}_{j\phi \phi }\phi _{\xi }^{2}+2
\mathcal{N}_{j\phi \overline{\phi }}|\phi _{\xi }|^{2}+\mathcal{
N}_{j\overline{\phi }\overline{\phi }}(\overline{\phi _{\xi }})
^{2})+\frac{1}{t^{2}}\mathcal{N}_{j\overline{\phi }}\overline{\phi
_{\xi \xi }},
\end{gather*}
in view of \eqref{5.1}, we obtain
\begin{align*}
t\mathcal{L}\chi _{j}&=\frac{2i}{\omega _{j}}\mathcal{N}_{j}t\mathcal{L}
(t^{1/2}\Omega _{j}A_{j})+t^{-1/2}\partial
_{\xi }(\Omega _{j}A_{j})\partial _{\xi }\mathcal{N}_{j} \\
&\quad +\frac{i}{\omega _{j}}t^{-1/2}\Omega _{j}A_{j}(\mathcal{N}
_{j\phi \phi }\phi _{\xi }^{2}+2\mathcal{N}_{j\phi \overline{\phi }}|
\phi _{\xi }|^{2}+\mathcal{N}_{j\overline{\phi }\overline{\phi }
}(\overline{\phi _{\xi }})^{2})\\
&\quad +\frac{2i}{\omega _{j}}t^{1/2}\Omega _{j}A_{j}(\mathcal{N}
_{j\phi }\mathcal{P}-\mathcal{N}_{j\overline{\phi }}\overline{\mathcal{P}}
)\\
&\quad +\frac{2i}{\omega _{j}}t^{-1/2}\Omega _{j}A_{j}\mathcal{N}_{j
\overline{\phi }}\Big(\overline{\Phi }_{\xi }+2it^{\frac{1}{2}
}\sum_{j=1}^{3}\frac{1}{\omega _{j}}\partial _{\xi }(\overline{E}
^{\omega _{j}}\overline{\Omega _{j}}\overline{A_{j}}\overline{\mathcal{N}_{j}
})\Big)
\end{align*}
Therefore,  from (\ref{5.7}),
\begin{equation}
\mathcal{L}\Phi =2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}
E^{\omega _{j}}\Omega _{j}A_{j}\mathcal{N}_{j\overline{\phi }}\overline{\Phi
_{\xi }}+t^{-1}\mathcal{R}_{1},  \label{5.8}
\end{equation}
where in view of \eqref{5.1} we have
\begin{align*}
\mathcal{R}_{1}&=\mathcal{R}-2\sum_{j=1}^{3}E^{\omega _{j}}
\big(\xi
\partial _{\xi }(\Omega _{j}A_{j}\mathcal{N}_{j})-\frac{i}{
\omega _{j}}\mathcal{N}_{j}t\mathcal{L}(t^{1/2}\Omega
_{j}A_{j})\big)\\
&\quad +\sum_{j=1}^{3}E^{\omega _{j}}t^{-1/2}\partial _{\xi }(
\Omega _{j}A_{j})\partial _{\xi }\mathcal{N}_{j} \\
&\quad +\sum_{j=1}^{3}E^{\omega _{j}}\frac{i}{\omega _{j}}t^{-1/2}\Omega
_{j}A_{j}(\mathcal{N}_{j\phi \phi }\phi _{\xi }^{2}+2\mathcal{N}
_{j\phi \overline{\phi }}|\phi _{\xi }|^{2}+\mathcal{N}_{j
\overline{\phi }\overline{\phi }}\overline{\phi _{\xi }}^{2})\\
&\quad +\sum_{j=1}^{3}E^{\omega _{j}}\frac{2i}{\omega _{j}}t^{1/2}\Omega
_{j}A_{j}\mathcal{N}_{j}(\mathcal{N}_{j\phi }\mathcal{P}-\mathcal{N}_{j
\overline{\phi }}\overline{\mathcal{P}})\\
&\quad -\sum_{j,l=1}^{3}E^{\omega _{j}}\frac{4}{\omega _{j}\omega _{l}}\Omega
_{j}A_{j}\mathcal{N}_{j\overline{\phi }}\partial _{\xi }(\overline{E}
^{\omega _{l}}\overline{\Omega _{l}}\overline{A_{l}}\overline{\mathcal{N}_{l}
}).
\end{align*}
Since
\begin{equation*}
|\Omega _{j}A_{j}|+|\xi \partial _{\xi }(\Omega
_{j}A_{j})|+|t\mathcal{L}(\Omega _{j}A_{j})
|\leq CB_{-2},t^{-1/2}|\partial _{\xi }(\Omega
_{j}A_{j})|\leq CB_{-3},
\end{equation*}
it follows that $\mathcal{R}_{1}$ satisfies
\begin{equation*}
\|\mathcal{R}_{1}\|_{\mathbf{W}_{\widetilde{T}}^{\gamma }}\leq
C\varepsilon ^{2}.
\end{equation*}
We multiply  (\ref{5.8}) by $(M+t\xi ^{2})^{\gamma /2}$ and
use the commutator
\[
\mathcal{L}((M+t\xi ^{2})^{\gamma /2}\Phi )
= (M+t\xi ^{2})^{\gamma /2}\mathcal{L}\Phi +\Phi
\mathcal{L}(M+t\xi ^{2})^{\gamma /2}
+2\gamma t^{-1}\xi (M+t\xi ^{2})^{\frac{\gamma }{2}
-1}\partial _{\xi }\Phi
\]
then we get $\mathcal{L}((M+t\xi ^{2})^{\gamma /2}\Phi )=\mathcal{R}_{2}$, where
\begin{align*}
\mathcal{R}_{2} &= \Phi \mathcal{L}(M+t\xi ^{2})^{\gamma /2}+2\gamma t^{-1}\xi (M+t\xi ^{2})^{\frac{\gamma }{2}-1}\Phi
_{\xi } \\
&\quad +2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}E^{\omega
_{j}}B_{\gamma }\Omega _{j}A_{j}\mathcal{N}_{j\overline{\phi }}\overline{
\Phi _{\xi }}+t^{-1}(M+t\xi ^{2})^{\gamma /2}\mathcal{R}
_{1}.
\end{align*}
Since
\begin{equation*}
|\mathcal{L}(M+t\xi ^{2})^{\gamma /2}|\leq
\frac{\gamma }{2t}\big(1+\frac{1}{M}\big)(M+t\xi ^{2})^{\gamma /2},
\end{equation*}
by (\ref{5.5}) choosing $M$ sufficiently large we have
\begin{equation*}
\|\mathcal{R}_{2}\|_{\mathbf{L}^{2}}\leq \frac{\gamma }{2t}
\|(M+t\xi ^{2})^{\gamma /2}\Phi \|_{
\mathbf{L}^{2}}+C\varepsilon ^{\frac{5}{4}}t^{-\frac{3}{4}}.
\end{equation*}
Then we apply the energy method to estimate the $\mathbf{L}^{2}$-norm of
$B_{\gamma }\Phi $,
\begin{equation*}
\frac{d}{dt}\|(M+t\xi ^{2})^{\gamma /2}\Phi
\|_{\mathbf{L}^{2}}\leq \frac{\gamma }{2t}\|(M+t\xi
^{2})^{\gamma /2}\Phi \|_{\mathbf{L}^{2}}+C\varepsilon
^{\frac{5}{4}}t^{-\frac{3}{4}}.
\end{equation*}
Hence  integration with respect to time yields
\begin{equation*}
\|(M+t\xi ^{2})^{\gamma /2}\Phi \|_{
\mathbf{L}^{2}}\leq \varepsilon +C\varepsilon ^{\frac{5}{4}}t^{1/4}
\end{equation*}
for all $t\in [1,\widetilde{T}] $ if $0<\gamma <\frac{1}{2}$.
Therefore, $\|B_{\gamma }\phi _{\xi }\|_{\mathbf{L}^{2}}\leq
\varepsilon +C\varepsilon ^{\frac{5}{4}}t^{1/4}$ for all $t\in [
1,\widetilde{T}] $.

We now differentiate (\ref{5.8}) with respect to $\xi $ to get
\begin{equation}
\mathcal{L}\Phi _{\xi }=2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}
E^{\omega _{j}}\Omega _{j}A_{j}\mathcal{N}_{j\overline{\phi }}\overline{\Phi
_{\xi \xi }}+\mathcal{R}_{3},  \label{5.9}
\end{equation}
where
\begin{equation*}
\mathcal{R}_{3}=2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}
\overline{\Phi _{\xi }}\partial _{\xi }(E^{\omega _{j}}\Omega _{j}A_{j}
\mathcal{N}_{j\overline{\phi }})+t^{-1}\partial _{\xi }\mathcal{R}
_{1}.
\end{equation*}
By (\ref{5.5}) we see that $\| \mathcal{R}_{3}\|_{\mathbf{
L}^{2}}\leq C\varepsilon ^{2}t^{-\frac{1}{4}}$. Then to estimate the
$\mathbf{L}^{2}$-norm of $\Phi _{\xi }$ we apply the energy method to
(\ref{5.9})
\begin{equation*}
\frac{d}{dt}\| \Phi _{\xi }\|_{\mathbf{L}^{2}}^{2}\leq
Ct^{-3/2}\sum_{j=1}^{3}|\int_{\mathbb{R}}\Omega
_{j}A_{j}E^{\omega _{j}}\mathcal{N}_{j\overline{\phi }}\overline{\Phi _{\xi
\xi }}\overline{\Phi _{\xi }}d\xi |+C\varepsilon ^{3}t^{\frac{1}{2
}}.
\end{equation*}
Hence integrating by parts with respect to $\xi $ we avoid the derivative
loss and obtain $\| \Phi _{\xi }\|_{\mathbf{L}^{2}}\leq
\varepsilon +C\varepsilon ^{\frac{3}{2}}t^{\frac{3}{4}}$ for all $t\in [
1,\widetilde{T}]$. Therefore
\begin{equation*}
\| \partial _{\xi }^{2}\phi \|_{\mathbf{L}
^{2}}<\varepsilon +C\varepsilon ^{\frac{3}{2}}t^{\frac{3}{4}}
\end{equation*}
for all $t\in [1,\widetilde{T}]$. Thus we have $\|
\partial _{\xi }\phi \|_{\mathbf{W}_{T}^{\gamma }}<10\varepsilon $.
The contradiction proves estimate (\ref{5.4}). Lemma \ref{Lemma5.2} is
proved.
\end{proof}

\section{\label{S6}Estimates in the uniform norm}

We now estimate $\phi $ in the norm $\| \phi \|_{\mathbf{V}
_{T}^{0}}=\sup_{1\leq t\leq T}\| \phi (t)\|_{
\mathbf{L}^{\infty }}$.

\begin{lemma}\label{Lemma6.1}
Let the initial data $\phi _{0}\in \mathbf{Y}$ and $\|
\phi _{0}\|_{\mathbf{Y}}\leq \varepsilon $, where $\varepsilon >0$ is
sufficiently small. Suppose that
\begin{equation}
\|\partial _{\xi }\phi \|_{\mathbf{W}_{T}^{\gamma
}}<10\varepsilon .  \label{6.2}
\end{equation}
Then the solutions $\phi \in \mathbf{C}([1,T];\mathbf{Y})$ of \eqref{5.1}
 satisfy the estimate
\begin{equation}
\|\phi \|_{\mathbf{V}_{T}^{0}}<10\varepsilon .  \label{6.3}
\end{equation}
\end{lemma}

\begin{proof}
We prove estimate (\ref{6.3}) by the contradiction. By the continuity of
$\phi $ we can find a maximal time $\widetilde{T}\in (1,T]$ such
that
\begin{equation}
\|\phi \|_{\mathbf{V}_{\widetilde{T}}^{0}}\leq 10\varepsilon .
\label{6.4}
\end{equation}
Now (\ref{6.4}) along with (\ref{6.2}) imply that
\begin{equation}
\|\phi \|_{\mathbf{X}_{\widetilde{T}}^{0,\gamma }}\leq
C\varepsilon .  \label{6.5}
\end{equation}
Denote $w(t)=\mathcal{V}(-t)\phi (t)$, where
\begin{equation*}
\mathcal{V}(-t)=\mathcal{F}\overline{M}\mathcal{F}^{-1}=\sqrt{t}
\int_{\mathbb{R}}d\eta e^{-\frac{it}{2}(\xi -\eta )^{2}}.
\end{equation*}
Applying operator $\mathcal{V}(-t)$ to equation \eqref{5.1}, we have
\begin{equation}
\begin{gathered}
iw_{t}=t^{-1}\mathcal{V}(-t)\mathcal{P}, \\
w(1)=\mathcal{V}(-1)\phi _{0},
\end{gathered}  \label{6.6}
\end{equation}
where
$\mathcal{P}=\overline{E}^{3}\widetilde{\mathcal{G}}_{-1,-1}(\overline{v
},\overline{v})-\mathcal{H}_{-1,2}(\overline{v},v^{2})$,
and $E=e^{\frac{i}{2}t\xi ^{2}}$. Note that $\mathcal{P}$ has the form
\begin{equation*}
\mathcal{P}=\overline{E}\partial _{\xi }(\mathbb{G}_{0}(
\overline{v},\overline{v})+\mathbb{G}_{1}(\overline{v}
,v^{2})),
\end{equation*}
where
\begin{gather*}
\mathbb{G}_{0}(\overline{v},\overline{v})
= \int_{\mathbb{R}^{2}}\widetilde{h}_{1}(t,\eta ,\zeta )
e^{-\frac{1}{2}it(\xi -\eta )^{2}}\overline{v}(\xi -\eta )
e^{-\frac{1}{2}it(\xi -\zeta )^{2}}\overline{v}(\xi -\zeta )d\eta
d\zeta \\
\mathbb{G}_{1}(\overline{v},v^{2})= \int_{\mathbb{R}
^{2}}h_{1}(t,\eta ,\zeta )e^{-\frac{1}{2}it(\xi -\eta
)^{2}}\overline{v}(\xi -\eta )e^{it(\xi -\zeta
)^{2}}v^{2}(\xi -\zeta )d\eta d\zeta
\end{gather*}
with kernels
\begin{gather*}
\widetilde{h}_{1}(t,\eta ,\zeta )= -\frac{\sqrt{t}}{\pi \sqrt{3
}}(2+t\partial _{t})t\partial _{t}K_{0}\big(\sqrt{\frac{4t}{3i
}(\eta ^{2}-\eta \zeta +\zeta ^{2})}\big), \\
h_{1}(t,\eta ,\zeta )= \frac{\sqrt{t}}{\pi \sqrt{3}}(
1+t\partial _{t})\partial _{\zeta }K_{0}\big(\sqrt{\frac{4t}{3i}
(\eta ^{2}-\eta \zeta +\zeta ^{2})}\big).
\end{gather*}
Then by the identities
\begin{align*}
\mathcal{V}(-t)\overline{E}\partial _{\xi }
&= \mathcal{F}\overline{M}\mathcal{F}^{-1}(\partial _{\xi }+it\xi )
 \overline{E}\\
&= \mathcal{F}\overline{M}(-ix+t\partial _{x})\mathcal{F}^{-1}
\overline{E}\\
&= t\mathcal{F}\partial _{x}\overline{M}\mathcal{F}^{-1}\overline{E}=it\xi
\mathcal{V}(t)\overline{E},
\end{align*}
we find
\begin{equation*}
\mathcal{V}(-t)\mathcal{P}=it\xi \mathcal{V}(-t)
\overline{E}(\mathbb{G}_{0}(\overline{v},\overline{v})+
\mathbb{G}_{1}(\overline{v},v^{2})).
\end{equation*}
Integration by parts yields
\begin{align*}
&|(B_{\gamma }\mathcal{V}(-t)-\mathcal{V}(
-t)B_{\gamma })\phi |\\
&= \sqrt{t}\big|\int_{\mathbb{R}}e^{-\frac{it}{2}(\xi -\eta )
^{2}}(\langle\xi \sqrt{t}\rangle ^{\gamma }-\langle
\eta \sqrt{t}\rangle ^{\gamma })\phi (\eta )d\eta \big|\\
&= t^{-1/2}\big|\int_{\mathbb{R}}e^{-\frac{it}{2}(\xi -\eta
)^{2}}(\xi -\eta )\partial _{\eta }(\frac{
\langle\xi \sqrt{t}\rangle ^{\gamma }-\langle\eta \sqrt{t}
\rangle ^{\gamma }}{1+it(\xi -\eta )^{2}}\phi (\eta
))d\eta \big|\\
&\leq C\|\phi \|_{\mathbf{L}^{\infty }}+Ct^{-1/4}
\|\partial _{\xi }\phi \|_{\mathbf{L}^{2}}\\
&\leq C\|\phi \|_{\mathbf{X}_{\widetilde{T}}^{0,\gamma }}.
\end{align*}
Hence
\begin{align*}
\|B_{\gamma }(w-\phi )\|_{\mathbf{L}^{\infty }}
&\leq \|(B_{\gamma }\mathcal{V}(-t)-\mathcal{V}
(-t)B_{\gamma })\phi \|_{\mathbf{L}^{\infty
}}+\|(\mathcal{V}(-t)-1)B_{\gamma }\phi
\|_{\mathbf{L}^{\infty }} \\
&\leq C\|\phi \|_{\mathbf{X}_{\widetilde{T}}^{0,\gamma }}+Ct^{-
\frac{1}{4}}\|\partial _{\xi }B_{\gamma }\phi \|_{\mathbf{L}
^{2}}\leq C\|\phi \|_{\mathbf{X}_{\widetilde{T}}^{0,\gamma }}.
\end{align*}
Thus by the estimates of Section \ref{S4} we see that
\begin{equation*}
v=\phi +O(\varepsilon B_{-1})=w+O(\varepsilon B_{-\gamma}).
\end{equation*}
Then by Lemma \ref{Lemma3.5} we get the representation for the operator
$\mathcal{P}$,
\begin{align*}
t^{-1}\mathcal{V}(-t)\mathcal{P} &= i\xi \mathcal{V}(
-t)\overline{E}(\mathbb{G}_{0}(\overline{v},\overline{v}
)+\mathbb{G}_{1}(\overline{v},v^{2}))\\
&= -it^{-1/2}\xi \widetilde{Q}|w|^{2}w+O(
\varepsilon ^{2}t^{-1/2}\xi B_{-1-\gamma }),
\end{align*}
where
\begin{align*}
\widetilde{Q} &= \int_{\mathbb{R}^{2}}e^{it\xi (\eta -2\zeta )
}h_{1}(t,\eta ,\zeta )d\eta d\zeta \\
&= \frac{\sqrt{t}}{\pi \sqrt{3}}\int_{\mathbb{R}^{2}}e^{it\xi (\eta
-2\zeta )}\frac{\zeta -\frac{\eta }{2}}{\eta ^{2}-\eta \zeta +\zeta
^{2}}e^{-\langle\eta \sqrt{t}\rangle }d\eta d\zeta +O(
\langle\xi \sqrt{t}\rangle ^{-2})\\
&= \frac{2i}{\sqrt{3}(1+\xi \sqrt{3t})}+O(\langle
\xi \sqrt{t}\rangle ^{-2}).
\end{align*}
Thus we can write  (\ref{6.6}) in the form
\begin{equation}
w_{t}=i\Theta |w|^{2}w+O(\varepsilon ^{2}t^{-\frac{1}{2}
}\xi B_{-1-\gamma }).  \label{6.7}
\end{equation}
where
\begin{equation*}
\Theta (t,\xi )=-\frac{2\xi }{\sqrt{3t}(1+\xi \sqrt{3t}
)}.
\end{equation*}
The first term in the right-hand side of (\ref{6.6}) is divergent, so we
eliminate it by the change $w(t,\xi )=\varphi (t,\xi
)\mathcal{E}_{w}$, where
\begin{equation*}
\mathcal{E}_{w}=\exp \Big(i\int_{1}^{t}\Theta (\tau ,\xi )
|w(\tau ,\xi )|^{2}d\tau \Big).
\end{equation*}
Then we get from (\ref{6.7})
\begin{equation}
\varphi _{t}=O(\varepsilon ^{2}t^{-1/2}\xi B_{-1-\gamma}).  \label{6.8}
\end{equation}
Integrating in time
\begin{align*}
|\varphi |&\leq \varepsilon +C\varepsilon ^{2}\int_{1}^{t}
\frac{|\xi |d\tau }{\sqrt{\tau }(1+|\xi |
\sqrt{\tau })^{1+\gamma }} \\
&\leq \varepsilon +C\varepsilon ^{2}\int_{|\xi |}^{|\xi
|\sqrt{t}}\frac{dz}{(1+z)^{1+\gamma }}\leq \varepsilon
+C\varepsilon ^{2}.
\end{align*}
Hence $\|\phi \|_{\mathbf{V}_{\widetilde{T}}^{0}}<10\varepsilon
$. This contradiction proves (\ref{6.3})and completes the proof.
\end{proof}

\section{\label{S7}Proof of Theorem \protect\ref{T1.1}}

By Lemma \ref{Lemma5.2} we see that the a priori estimate of $\|
\phi \|_{\mathbf{V}_{T}^{0}}$ implies the a priori estimate of
$\| \partial _{\xi }\phi \|_{\mathbf{W}_{T}^{\gamma }}$.
Vice versa by Lemma \ref{Lemma6.1} the a priori estimate of $\|
\partial _{\xi }\phi \|_{\mathbf{W}_{T}^{\gamma }}$ yields the a
priori estimate of $\| \phi \|_{\mathbf{V}_{T}^{0}}$.
Therefore the global existence of solution $v=\mathcal{I}^{-1}(\phi
)\in \mathbf{C}([1,\infty );\mathbf{Y})$ of
the Cauchy problem (\ref{2.5}) satisfying a priori estimate
\begin{equation*}
\| \phi \|_{\mathbf{X}_{\infty }^{0,\gamma }}\leq
C\varepsilon
\end{equation*}
follows by a standard continuation argument from Lemma \ref{Lemma5.2}, Lemma
\ref{Lemma6.1} and the local existence Theorem \ref{T5.1}. This yields the
solution of the Cauchy problem \eqref{1.1}. Theorem \ref{T1.1} is proved.

\section{\label{S8}Proof of Theorem \protect\ref{T1.2}}

Existence of a self-similar solution of  (\ref{2.1}) of the form
$\frac{1}{\sqrt{t}}MS(\frac{x}{\sqrt{t}})$ follows from
Appendix \ref{S10} since
\begin{equation*}
u(t)=\mathcal{U}(t)\mathcal{F}^{-1}w(
t)=\mathcal{D}_{t}E(t)v(t),w(t)
=\mathcal{V}(-t)v(t)
\end{equation*}
and $w(t)$ has the form $w(t,\xi )=MS(\xi
\sqrt{t})$. We now prove the stability of solutions in the
neighborhood of a self-similar solution of the equation (\ref{2.1}). We
consider the difference $r(t,\xi )=\phi _{1}(t,\xi
)-\phi _{2}(t,\xi )$ and may assume that $\mathcal{V}
(-t)(\phi _{1}(t)-\phi _{2}(t)
)=0$ at $\xi =0$. Define the norm

\begin{equation*}
\| r\|_{\mathbf{W}_{T}^{\mu ,\nu }}
=\sup_{1\leq t\leq T}t^{\frac{\mu }{2}}
\Big(t^{-\frac{1}{4}}\| B_{\nu }r(t)
\|_{\mathbf{L}^{2}}+t^{-\frac{3}{4}}\| \partial _{\xi
}r(t)\|_{\mathbf{L}^{2}}\Big).
\end{equation*}

\begin{lemma}\label{Lemma5.2a}
Let the initial data $\phi _{j}\in \mathbf{Y}$ and $\|
\phi _{j}\|_{\mathbf{Y}}\leq \varepsilon $, where $\varepsilon >0$ is
sufficiently small. Suppose that
$\mathcal{V}(-t)(\phi_{1}(t)-\phi _{2}(t))=0$ at $\xi =0$.
Assume that representation (\ref{5.2}) is valid. Suppose that
\begin{equation*}
\|\phi _{j}\|_{\mathbf{X}_{T}^{0,\gamma }}\leq \varepsilon .
\end{equation*}
Then the solutions $\phi _{j}\in \mathbf{C}([1,T];\mathbf{
Y})$ of \eqref{5.1} satisfy the estimate
\begin{equation*}
\|\partial _{\xi }(\phi _{1}-\phi _{2})\|_{\mathbf{W}_{T}^{\mu ,\nu }}
<C\varepsilon
\end{equation*}
for $0<\mu <\gamma $ and $\frac{1}{2}-\gamma \leq \nu <\frac{1}{2}-\mu $.
\end{lemma}

\begin{proof}
Denote $r(t)=\phi _{1}(t)-\phi _{2}(t)$, then by \eqref{5.1} we have
\begin{equation}
\mathcal{L}r=t^{-1}(\mathcal{P}(\phi _{1})-\mathcal{P}
(\phi _{2})).  \label{5.1a}
\end{equation}
We prove the estimate of the lemma by contradiction. By the continuity of
$r(t)$ we can find a maximal time $\widetilde{T}\in (1,T
]$ such that
\begin{equation}
\|\partial _{\xi }(\phi _{1}-\phi _{2})\|_{\mathbf{
W}_{\widetilde{T}}^{\mu ,\nu }}\leq C\varepsilon  \label{5.2a}
\end{equation}
Since $\mathcal{V}(-t)r(t)=w_{1}(t,\xi)-w_{2}(t,\xi )$ vanishes at
$\xi =0$ for all $t\geq 1$,
we can estimate the norm
\begin{align*}
|\mathcal{V}(-t)r(t)|
&=\big| \int_{0}^{\xi }B_{-\nu }B_{\nu }\partial _{\xi }\mathcal{V}(-t)
r(t)d\xi \big| \\
&\leq C\sqrt{|\xi |}B_{-\nu}\|B_{\nu }\partial _{\xi }\mathcal{V}(-t)r(
t)\|_{\mathbf{L}^{2}} \\
&\leq |\xi |^{\mu }\langle\sqrt{t}\xi \rangle ^{
\frac{1}{2}-\nu -\mu }t^{\frac{\mu }{2}-\frac{1}{4}}\|B_{\nu }\partial
_{\xi }r(t)\|_{\mathbf{L}^{2}}\\
&\leq C\varepsilon |
\xi |^{\mu }\langle\sqrt{t}\xi \rangle ^{\frac{1}{2}-\nu
-\mu }.
\end{align*}
Also we have
\begin{equation*}
|(1-\mathcal{V}(-t))r(t)|
\leq t^{-\frac{1}{4}}\|\partial _{\xi }r(t)\|_{
\mathbf{L}^{2}}\leq C\varepsilon t^{-\mu/2}.
\end{equation*}
Hence
\begin{align*}
|r(t)|&\leq |(1-\mathcal{V}(
-t))r(t)|+|\mathcal{V}(
-t)r(t)|\\
&\leq C\varepsilon t^{-\mu/2}+C\varepsilon |\xi |
^{\mu }\langle\sqrt{t}\xi \rangle ^{\frac{1}{2}-\nu -\mu }.
\end{align*}
Then we get from (\ref{5.1a}) in view of  (\ref{5.6}),
\begin{equation}
\begin{aligned}
\mathcal{L}\big(r_{\xi }+\sum_{j=1}^{3}E^{\omega _{j}}\chi _{j}\big)
&=t^{-1}\sum_{j=1}^{3}E^{\omega _{j}}\big(t^{1/2}\Omega _{j}
\mathcal{M}_{j}+\frac{i\omega _{j}}{2A_{j}}\chi _{j}\big)\\
&\quad +t^{-1}\sum_{j=1}^{3}E^{\omega _{j}}(i\omega _{j}\xi \partial _{\xi
}\chi _{j}+t\mathcal{L}\chi _{j})+t^{-1}(\mathcal{R}_{1}-
\mathcal{R}_{2}),
\end{aligned} \label{5.7a}
\end{equation}
where we denote $\mathcal{M}_{j}=\mathcal{N}_{j}(\phi _{1})-
\mathcal{N}_{j}(\phi _{2})$. To eliminate the first summand in
the right-hand side of (\ref{5.7a}) we choose $\chi _{j}=\frac{2i}{\omega
_{j}}t^{1/2}\Omega _{j}A_{j}\mathcal{M}_{j}$ and denote $\Phi
=r_{\xi }+\sum_{j=1}^{3}E^{\omega _{j}}\chi _{j}$. Therefore,
from (\ref{5.7a}) we obtain
\begin{equation*}
\mathcal{L}\Phi =2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}
E^{\omega _{j}}\Omega _{j}A_{j}\mathcal{M}_{j\overline{r}}\overline{\Phi
_{\xi }}+t^{-1}\mathcal{R}_{3},
\end{equation*}
where in view of (\ref{5.1a}), we have
\begin{align*}
\mathcal{R}_{3}&=\mathcal{R}_{1}-\mathcal{R}_{2}
-2\sum_{j=1}^{3}E^{\omega _{j}}(\xi \partial _{\xi }(\Omega
_{j}A_{j}\mathcal{M}_{j})-\frac{i}{\omega _{j}}\mathcal{M}_{j}t
\mathcal{L}(t^{1/2}\Omega _{j}A_{j}))\\
&\quad +\sum_{j=1}^{3}E^{\omega _{j}}t^{-1/2}\partial _{\xi }(
\Omega _{j}A_{j})\partial _{\xi }\mathcal{M}_{j} \\
&\quad +\sum_{j=1}^{3}E^{\omega _{j}}\frac{i}{\omega _{j}}t^{-1/2}\Omega
_{j}A_{j}(\mathcal{M}_{jrr}r_{\xi }^{2}+2\mathcal{M}_{jr\overline{r}
}|r_{\xi }|^{2}+\mathcal{M}_{j\overline{r}\overline{r}}
\overline{r_{\xi }}^{2})\\
&\quad +\sum_{j=1}^{3}E^{\omega _{j}}\frac{2i}{\omega _{j}}t^{1/2}\Omega
_{j}A_{j}\mathcal{M}_{j}(\mathcal{M}_{jr}(\mathcal{P}_{1}-
\mathcal{P}_{2})-\mathcal{M}_{j\overline{r}}(\overline{\mathcal{
P}_{1}}-\overline{\mathcal{P}_{2}}))\\
&\quad -\sum_{j,l=1}^{3}E^{\omega _{j}}\frac{4}{\omega _{j}\omega _{l}}\Omega
_{j}A_{j}\mathcal{M}_{j\overline{r}}\partial _{\xi }(\overline{E}
^{\omega _{l}}\overline{\Omega _{l}}\overline{A_{l}}
\overline{\mathcal{M}_{l}}).
\end{align*}
Since
\begin{equation*}
|\Omega _{j}A_{j}|+|\xi \partial _{\xi }(\Omega
_{j}A_{j})|+|t\mathcal{L}(\Omega _{j}A_{j})
|\leq CB_{-2},t^{-1/2}|\partial _{\xi }(\Omega
_{j}A_{j})|\leq CB_{-3},
\end{equation*}
then $\mathcal{R}_{3}$ can be estimated as
\begin{equation*}
\|\mathcal{R}_{3}\|_{\mathbf{W}_{\widetilde{T}}^{\mu ,\nu
}}\leq C\varepsilon ^{2}
\end{equation*}
if $\gamma \geq \frac{1}{2}-\nu $. Then as in the proof of Lemma
\ref{Lemma5.2} we multiply (\ref{5.8}) by $(M+t\xi ^{2})^{\nu/2}$
and use the commutator
\[
\mathcal{L}((M+t\xi ^{2})^{\nu /2}\Phi )
= (M+t\xi ^{2})^{\nu /2}\mathcal{L}\Phi +\Phi \mathcal{
L}(M+t\xi ^{2})^{\nu /2}
+2\nu t^{-1}\xi (M+t\xi ^{2})^{\frac{\nu }{2}-1}\partial
_{\xi }\Phi
\]
then we get
\begin{equation*}
\mathcal{L}((M+t\xi ^{2})^{\nu /2}\Phi )=
\mathcal{R}_{4},
\end{equation*}
where
\begin{align*}
\mathcal{R}_{4}
&= \Phi \mathcal{L}(M+t\xi ^{2})^{\nu /2}+2\nu t^{-1}\xi (M+t\xi ^{2})^{\frac{\nu }{2}-1}\Phi _{\xi } \\
&\quad +2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}E^{\omega
_{j}}B_{\nu }\Omega _{j}A_{j}\mathcal{N}_{j\overline{\phi }}\overline{\Phi
_{\xi }}+t^{-1}(M+t\xi ^{2})^{\nu /2}\mathcal{R}_{3}.
\end{align*}
Since
\begin{equation*}
|\mathcal{L}(M+t\xi ^{2})^{\nu /2}|\leq
\frac{\nu }{2t}(1+\frac{1}{M})(M+t\xi ^{2})^{\nu /2},
\end{equation*}
by (\ref{5.5}) choosing $M$ sufficiently large we have
\begin{equation*}
\|\mathcal{R}_{4}\|_{\mathbf{L}^{2}}\leq \frac{\nu }{2t}\|
(M+t\xi ^{2})^{\nu /2}\Phi \|_{\mathbf{L}
^{2}}+C\varepsilon ^{\frac{5}{4}}t^{\frac{\mu }{2}-\frac{3}{4}}.
\end{equation*}
Then we apply the energy method to estimate the $\mathbf{L}^{2}$ - norm of
$B_{\gamma }\Phi $
\begin{equation*}
\frac{d}{dt}\|(M+t\xi ^{2})^{\nu /2}\Phi \|
_{\mathbf{L}^{2}}\leq \frac{\nu }{2t}\|(M+t\xi ^{2})^{
\frac{\nu }{2}}\Phi \|_{\mathbf{L}^{2}}+C\varepsilon ^{\frac{5}{4}}t^{
\frac{\mu }{2}-\frac{3}{4}}.
\end{equation*}
Hence the integration with respect to time yields
\begin{equation*}
\|(M+t\xi ^{2})^{\nu /2}\Phi \|_{\mathbf{L}
^{2}}\leq \varepsilon +C\varepsilon ^{\frac{5}{4}}t^{\frac{1}{4}-\frac{\mu }{
2}}
\end{equation*}
for all $t\in [1,\widetilde{T}]$ if $0<\nu <\frac{1}{2}-\mu $.
Therefore $\|B_{\nu }r_{\xi }\|_{\mathbf{L}^{2}}\leq
\varepsilon +C\varepsilon ^{\frac{5}{4}}t^{\frac{1}{4}-\frac{\mu }{2}}$ for
all $t\in [1,\widetilde{T}]$.

We now differentiate (\ref{5.7a}) with respect to $\xi $ to get
\begin{equation*}
\mathcal{L}\Phi _{\xi }=2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}
E^{\omega _{j}}\Omega _{j}A_{j}\mathcal{M}_{j\overline{r}}\overline{\Phi
_{\xi \xi }}+\mathcal{R}_{5},
\end{equation*}
where
\begin{equation*}
\mathcal{R}_{5}=2it^{-3/2}\sum_{j=1}^{3}\frac{1}{\omega _{j}}
\overline{\Phi _{\xi }}\partial _{\xi }(E^{\omega _{j}}\Omega _{j}A_{j}
\mathcal{M}_{j\overline{r}})+t^{-1}\partial _{\xi }\mathcal{R}_{3}.
\end{equation*}
By (\ref{5.2a})\ we see that $\| \mathcal{R}_{5}\|_{
\mathbf{L}^{2}}\leq C\varepsilon ^{2}t^{-\frac{\nu }{2}-\frac{1}{4}}$. Then
to estimate the $\mathbf{L}^{2}$ - norm of $\Phi _{\xi }$ we apply the
energy method to (\ref{5.9})\
\begin{equation*}
\frac{d}{dt}\| \Phi _{\xi }\|_{\mathbf{L}^{2}}^{2}\leq
Ct^{-3/2}\sum_{j=1}^{3}|\int_{\mathbb{R}}\Omega
_{j}A_{j}E^{\omega _{j}}\mathcal{M}_{j\overline{r}}\overline{\Phi _{\xi \xi }
}\overline{\Phi _{\xi }}d\xi |+C\varepsilon ^{3}t^{\frac{1}{2}
-\nu }.
\end{equation*}
Hence integrating by parts with respect to $\xi $ we avoid the derivative
loss and obtain $\| \Phi _{\xi }\|_{\mathbf{L}^{2}}\leq
\varepsilon +C\varepsilon ^{\frac{3}{2}}t^{\frac{3}{4}-\frac{\nu }{2}}$ for
all $t\in [1,\widetilde{T}]$. Therefore
\begin{equation*}
\| \partial _{\xi }^{2}r\|_{\mathbf{L}^{2}}<\varepsilon
+C\varepsilon ^{\frac{3}{2}}t^{\frac{3}{4}-\frac{\nu }{2}}
\end{equation*}
for all $t\in [1,\widetilde{T}]$. Thus we have $\|
\partial _{\xi }r\|_{\mathbf{W}_{T}^{\mu ,\nu }}<C\varepsilon $.
This contradiction proves estimate of the lemma. Lemma \ref{Lemma5.2a} is
proved.
\end{proof}

To study the asymptotic behavior we make a change as in the proof
of Lemma \ref{Lemma6.1} $w=\overline{\mathcal{V}}\phi =\varphi \mathcal{E}_{\varphi }$
and $s=\overline{\mathcal{V}}S=g\mathcal{E}_{s}$, where
\begin{equation*}
\mathcal{E}_{s}=\exp \Big(i\int_{1}^{t}\Theta (\tau ,\xi )
|s(\tau ,\xi )|^{2}d\tau \Big).
\end{equation*}
Then from (\ref{6.6}) we get for the difference $f=\varphi -g$
\begin{align*}
if_{t} &= \overline{\mathcal{E}_{\varphi }}t^{-1}\overline{\mathcal{V}}
\mathcal{P}(\phi )-i\Theta |\varphi |^{2}\varphi -
\overline{\mathcal{E}_{g}}t^{-1}\overline{\mathcal{V}}\mathcal{P}(
S)+i\Theta |g|^{2}g \\
&= (\overline{\mathcal{E}_{\varphi }}-\overline{\mathcal{E}_{g}}
)(t^{-1}\overline{\mathcal{V}}\mathcal{P}(\mathcal{V}
s)-i\Theta |s|^{2}s)+t^{-1}\overline{\mathcal{E}
_{\varphi }}(\overline{\mathcal{V}}-1)(\mathcal{P}(
\phi )-\mathcal{P}(S))\\
&\quad -i\Theta \overline{\mathcal{E}_{\varphi }}
\Big(\overline{((
\overline{\mathcal{V}}-1)\phi )}(\overline{\mathcal{V}}
\phi )^{2}-\overline{((\overline{\mathcal{V}}-1)
S)}(\overline{\mathcal{V}}S)^{2} \\
&\quad +\overline{\phi }((\overline{\mathcal{V}}-1)\phi
)(\overline{\mathcal{V}}+1)\phi -\overline{S}(
(\overline{\mathcal{V}}-1)S)(\overline{\mathcal{V}}
+1)S\Big)\\
&\quad +\overline{\mathcal{E}_{\varphi }}(t^{-1}\mathcal{P}(\phi
)-i\Theta |\phi |^{2}\phi -t^{-1}\mathcal{P}(
S)+i\Theta |S|^{2}S)\\
&=O(\varepsilon ^{2}t^{-\frac{\mu }{2}-1}).
\end{align*}
Then we get
\begin{equation*}
|f(t)-f(t_{1})|\leq C\varepsilon
^{2}\int_{t_{1}}^{t}\tau ^{-\frac{\mu }{2}-1}d\tau \leq \varepsilon
^{2}t_{1}{}^{-\mu/2}.
\end{equation*}
Also by (\ref{6.8}),
\begin{equation*}
|\varphi (t)-\varphi (t_{1})|\leq
C\varepsilon ^{2}\int_{t_{1}}^{t}\tau ^{-1/2}\xi \langle\xi
\sqrt{\tau }\rangle ^{-1-\gamma }d\tau \leq \frac{C\varepsilon ^{2}}{
\langle\xi \sqrt{t_{1}}\rangle ^{\gamma }}
\end{equation*}
for all $t>t_{1}>1$. Therefore the limits exist
\begin{equation*}
F(\xi )=\lim_{t\to \infty }f(t,\xi )\text{, }\Phi
(\xi )=\lim_{t\to \infty }\varphi (t,\xi )\text{ and }
G(\xi )=\lim_{t\to \infty }g(t,\xi )
\end{equation*}
with the  estimates $|F(\xi )|+|
\Phi (\xi )|+|G(\xi )|\leq
C\varepsilon $. Then using the estimates
\begin{gather*}
|f(t,\xi )-F(\xi )|\leq \varepsilon
^{2}t^{-\mu/2},\quad
|\varphi (t,\xi )-\Phi (\xi )|\leq \frac{C\varepsilon ^{2}}{\langle\xi
\sqrt{t}\rangle ^{\gamma }},
\\
|g(t,\xi )-G(\xi )|\leq \frac{C\varepsilon ^{2}}{\langle\xi
\sqrt{t}\rangle ^{\gamma }},\quad
|1-\mathcal{E}_{g}\overline{\mathcal{E}_{\varphi }}|\leq
C\varepsilon ^{2}t^{-\mu/2}\langle\xi \sqrt{t}\rangle
^{\gamma },
\end{gather*}
we obtain
\begin{align*}
\phi &= S+\mathcal{E}_{\varphi }(f+(1-\mathcal{E}_{g}\overline{
\mathcal{E}_{\varphi }})g)+(1-\overline{\mathcal{V}}
)r \\
&= S+\mathcal{E}_{\varphi }(F+(1-\mathcal{E}_{g}\overline{
\mathcal{E}_{\varphi }})G)+O(\varepsilon ^{2}t^{-\frac{
\mu }{2}}).
\end{align*}
Note that
\begin{equation*}
|\varphi (t,\xi )|^{2}=|\Phi (\xi
)|^{2}+O(\varepsilon ^{2}\langle\xi \sqrt{t}
\rangle ^{-\gamma }).
\end{equation*}
We now denote
\begin{gather*}
\Psi _{1}(t)=-i\int_{1}^{t}(|\varphi (\tau ,\xi )
|^{2}-|\varphi (t,\xi )|^{2})\Theta d\tau,
\\
\Psi _{2}(t)=-i\int_{1}^{t}(|g(\tau ,\xi )|
^{2}-|\varphi (\tau ,\xi )|^{2}-|g(
t,\xi )|^{2}+|\varphi (t,\xi )|
^{2})\Theta d\tau .
\end{gather*}
We then get
\[
\Psi _{1}(t)-\Psi _{1}(t_{1}) = -i\int_{t_{1}}^{t}(|\varphi
(\tau ,\xi )|^{2}-|\varphi (t,\xi )|^{2})\Theta d\tau
+i(|\varphi (t,\xi )|^{2}-|\varphi
(t_{1},\xi )|^{2})\int_{1}^{t_{1}}\Theta d\tau
\]
and
\begin{align*}
\Psi _{2}(t)-\Psi _{2}(t_{1})
&= -i\int_{t_{1}}^{t}(|g(\tau ,\xi )|
^{2}-|\varphi (\tau ,\xi )|^{2}-|g(
t,\xi )|^{2}+|\varphi (t,\xi )|
^{2})\Theta d\tau \\
&\quad +i(|g(t,\xi )|^{2}-|\varphi (
t,\xi )|^{2}-|g(t_{1},\xi )|
^{2}+|\varphi (t_{1},\xi )|^{2})
\int_{1}^{t_{1}}\Theta d\tau
\end{align*}
for all $1<t_{1}<\tau <t$. Using these estimates we get
\begin{equation*}
|\Psi _{j}(t)-\Psi _{j}(t_{1})|\leq
C\varepsilon ^{2}\langle\xi \sqrt{t_{1}}\rangle ^{-\gamma },
\quad j=1,2.
\end{equation*}
Therefore,  there exist unique functions
$\Phi _{j+2}\in \mathbf{L}^{\infty }$, such that
\begin{equation*}
i\Phi _{j+2}=\lim_{t\to \infty }\Psi _{j}(t)
\end{equation*}
and
\begin{equation*}
|i\Phi _{j+2}-\Psi _{j}(t)|\leq C\varepsilon
^{2}\langle\xi \sqrt{t}\rangle ^{-\gamma },j=1,2.
\end{equation*}
Then we find
\begin{equation*}
-i\int_{1}^{t}|\varphi (\tau )|^{2}\Theta d\tau
=i|\Phi |^{2}\log (\xi \sqrt{t})+i\Phi
_{3}+O(\varepsilon ^{2}t^{-\mu/2})
\end{equation*}
and
\[
-i\int_{1}^{t}(|g(\tau )|^{2}-|\varphi (\tau
)|^{2})\Theta d\tau
= i(|G|^{2}-|\Phi |^{2})\log (
\xi \sqrt{t})+i\Phi _{4}+O(\varepsilon ^{2}t^{-\mu/2})
\]
with some functions $\Phi _{3}$, $\Phi _{4}\in \mathbf{{{L}^{\infty }}}$.
Hence
\begin{equation*}
\mathcal{E}_{\varphi }=\exp \Big(-i|\Phi |^{2}\log (
\frac{1+|\xi |\sqrt{t}}{1+|\xi |})-i\Phi
_{3}+O(\varepsilon ^{2}t^{-\mu/2})\Big)
\end{equation*}
and
\begin{equation*}
\mathcal{E}_{g}\overline{\mathcal{E}_{\varphi }}=\exp \Big(-i(|
G|^{2}-|\Phi |^{2})\log (\frac{1+|\xi
|\sqrt{t}}{1+|\xi |})-i\Phi _{4}+O(
\varepsilon ^{2}t^{-\mu/2})\Big).
\end{equation*}
Thus
\begin{align*}
\phi &= S+e^{-i|\Phi |^{2}\log (\frac{1+|\xi
|\sqrt{t}}{1+|\xi |})-i\Phi _{3}}\Big(F+G\big(
1-e^{-i(|G|^{2}-|\Phi |^{2})\log
(\frac{1+|\xi |\sqrt{t}}{1+|\xi |})
-i\Phi _{4}}\big)\Big)\\
&\quad +O(\varepsilon ^{3}t^{-\gamma })\\
&=S+\sum_{j=1}^{2}H_{j}e^{iB_{j}\log (\frac{1+|\xi |\sqrt{t
}}{1+|\xi |})}+O(\varepsilon ^{2}t^{-\mu/2}).
\end{align*}
The asymptotic formula follows now from the inverse transformation
\cite{H-O} of $u(t,x)=t^{-1/2}e^{\frac{it}{2}\xi ^{2}}v(
t,\xi )$ and $\xi =x/t$,
\begin{equation*}
u(t,x)=t^{-1/2}Ev(t)=t^{-\frac{1}{2}
}ES+t^{-1/2}E\sum_{j=1}^{2}H_{j}e^{iB_{j}\log (\frac{1+\frac{
|x|}{\sqrt{t}}}{1+\frac{|x|}{t}})}+O(
t^{-\frac{1}{2}-\frac{\mu }{2}}).
\end{equation*}
Theorem \ref{T1.2} is proved.

\section{\label{S9}Appendix 1, Hope-Cole transformation}

We consider the quadratic nonlinear Schr\"{o}dinger equation
\begin{equation}
\begin{gathered}
iu_{t}+\frac{1}{2}u_{xx}=\partial _{x}(u^{2}),\quad x\in
\mathbb{R},t\in \mathbb{R}, \\
u(0,x)=u_{0}(x),\quad x\in \mathbb{R}.
\end{gathered}  \label{1.1 a}
\end{equation}
There exists a self-similar solution of the form
$s(t,x)=\frac{1}{\sqrt{t}}\varphi (\xi )$, with $\xi =\frac{x}{\sqrt{t}}$.
Indeed if we substitute $\frac{1}{\sqrt{t}}\varphi (\xi )$ into
(\ref{1.1 a}) we get the ordinary differential equation for the function
$\varphi (\xi )$,
\begin{equation*}
-\frac{i}{2}(\xi \varphi )'+\frac{1}{2}\varphi
^{\prime \prime }=(\varphi ^{2})',
\end{equation*}
hence integrating with respect to $\xi $ and choosing the integration
constant as zero, we obtain the Bernoulli equation $\varphi '=i\xi
\varphi +2\varphi ^{2}$, which has a solution
\begin{equation*}
\varphi (\xi )=e^{\frac{i}{2}\xi ^{2}}\Big(C-\sqrt{2\pi i}
\mathop{\rm erf}(\xi \sqrt{-\frac{i}{2}})\Big)^{-1},
\end{equation*}
with the error function $\mathop{\rm erf}(x)=\frac{2}{\sqrt{\pi }}
\int_{0}^{x}e^{-y^{2}}dy$. We suppose that the constant $C$ have a
sufficiently large absolute value. Now we represent a solution $u(
t,x)=s(t,x)+w(t,x)$ and choose a constant
$C $ such that $\int s(t,x)dx$ $=\int \varphi (\xi )
d\xi $ $=\int u(t,x)dx$, then
\begin{equation*}
\int w(t,x)dx=0,
\end{equation*}
so we can consider $w$ as a full derivative: $w=v_{x}$. Thus we have
\begin{equation*}
iv_{t}+\frac{1}{2}v_{xx}=2sv_{x}+(v_{x})^{2}.
\end{equation*}
Now by the Hopf-Cole anzatz $v=-\frac{1}{2}\log (1+\phi )$ we
linearize the above equation
\begin{equation*}
i\phi _{t}+\frac{1}{2}\phi _{xx}=2s\phi _{x}.
\end{equation*}
We can eliminate the right-hand side of the above equation by virtue of the
change of the dependent variable $\phi (t,x)=\exp (
2\int_{0}^{x}s(t,y)dy)$ $\psi (t,x)$,
\begin{gather*}
i\psi _{t}+\frac{1}{2}\psi _{xx}=0,\quad x\in \mathbb{R},\; t\in \mathbb{R},
\\
\psi (0,x)=\psi _{0}(x),\quad x\in \mathbb{R},
\end{gather*}
where $\psi _{0}(x)=(\exp (-2\int_{0}^{x}(
u_{0}(y)-s(0,y))dy)-1)\exp
(-2\int_{0}^{x}s(0,y)dy)$. Thus the solution of (
\ref{1.1 a}) have a form
\begin{equation*}
u(t,x)=s(t,x)-\frac{1}{2}\partial _{x}\log (
1+\psi (t,x)\exp (2\int_{0}^{x}s(t,y)dy)).
\end{equation*}

\section{\label{S10}Appendix 2, Existence of a self-similar solution}

Here we study a self-similar solution of the form $v(t,\xi )
=w(\xi \sqrt{t})$
\begin{equation*}
v_{t}(t,\xi )=\xi \int_{\mathbb{R}}e^{itS}\overline{v(
t,\eta -\xi )}\overline{v(t,-\eta )}d\eta ,
\end{equation*}
where
\begin{equation*}
S=\frac{1}{2}(\xi ^{2}+(\xi -\eta )^{2}+\eta ^{2})
\end{equation*}
and the solution $u(t,x)$ of \eqref{1.1} can be obtained from
$v(t,\xi )$ through the formula
\begin{equation*}
u(t)=\mathcal{U}(t)\mathcal{F}^{-1}v(t).
\end{equation*}
For $w(\xi \sqrt{t})$ we get
\begin{equation*}
w'(\xi )=2\int_{\mathbb{R}}e^{iS}\overline{w(
\eta -\xi )}\overline{w(-\eta )}d\eta .
\end{equation*}
We use the identity
$e^{iS}=(\partial _{\xi }\xi +\partial _{\eta }\eta )(Ke^{iS})+Fe^{iS}$,
where
\begin{equation*}
K=\frac{2+iS}{2(1+iS)^{2}},\quad
F=\frac{1-iS}{(1+iS)^{3}}.
\end{equation*}
Then we get
\begin{align*}
&\frac{d}{d\xi }(w(\xi )+2\int_{\mathbb{R}}\xi Ke^{iS}
\overline{w(\eta -\xi )}\overline{w(-\eta )}d\eta)\\
&= 2\int_{\mathbb{R}}Fe^{iS}\overline{w(\eta -\xi )}\overline{
w(-\eta )}d\eta
-\int_{\mathbb{R}}K\eta e^{iS}\frac{d}{d\eta }(\overline{w(
\eta -\xi )}\overline{w(-\eta )})d\eta .
\end{align*}
Denote
\begin{equation}
\phi (\xi )=w(\xi )+2\int_{\mathbb{R}}\xi K(
\xi ,\eta )e^{iS}\overline{w(\eta -\xi )}\overline{
w(-\eta )}d\eta ,  \label{c1}
\end{equation}
then in view of the symmetry $\eta -\zeta $ $\leftrightarrow $ $\zeta $ we
have
\begin{equation}
\frac{d\phi }{d\xi }=2\int_{\mathbb{R}}Fe^{iS}\overline{w(\eta -\xi
)}\overline{w(-\eta )}d\eta
-4\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(\eta -\xi )}
\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(y)dy
\label{c5}
\end{equation}
with
\begin{equation*}
Q=\frac{1}{2}(\eta ^{2}+(\eta -y)^{2}+y^{2}).
\end{equation*}
We will see later that the right-hand side of (\ref{c5}) has the
asymptotic
\begin{equation*}
\frac{16i\pi }{3\langle\xi \rangle }|\phi (\xi
)|^{2}\phi (\xi )+O(\langle\xi
\rangle ^{\gamma -2}).
\end{equation*}
To exclude the first divergent term we make a change $\phi =\varphi E$,
where
\begin{equation*}
E=\exp \Big(\frac{16}{3}i\int_{0}^{\xi }|\varphi (\xi )
|^{2}\frac{d\xi }{\langle\xi \rangle }\Big),
\end{equation*}
then we get from (\ref{c5}),
\begin{equation}
\begin{aligned}
\frac{d\varphi }{d\xi }
&= 2\overline{E}\int_{\mathbb{R}}Fe^{iS}\overline{
w(\eta -\xi )}\overline{w(-\eta )}d\eta -\frac{
16i\pi }{3\langle\xi \rangle }|\varphi (\xi )
|^{2}\varphi (\xi )  \\
&\quad -4\overline{E}\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(\eta
-\xi )}\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(y)dy.
\end{aligned} \label{c6}
\end{equation}
Integrating the above expression, we obtain
\begin{equation}
\begin{aligned}
\varphi (\xi )&= \varphi (0)+\int_{0}^{\xi }2
\overline{E}\int_{\mathbb{R}}Fe^{iS}\overline{w(\eta -\xi )}
\overline{w(-\eta )}d\eta d\xi\\
&\quad -\int_{0}^{\xi }\Big(\frac{16i\pi }{3\langle\xi \rangle }
|\varphi (\xi )|^{2}\varphi (\xi )
 \\
&\quad +4\overline{E}\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(
\eta -\xi )}\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(
y)dy\Big)d\xi .
\end{aligned} \label{c7}
\end{equation}
We solve integral equation (\ref{c7}) by the contraction mapping principle.
Define the transformation
\begin{equation}
\begin{aligned}
\mathcal{A}(\varphi )(\xi )
&= \varphi (0)+\int_{0}^{\xi }2\overline{E}
 \int_{\mathbb{R}}Fe^{iS}\overline{w(\eta -\xi )}
 \overline{w(-\eta )}d\eta d\xi
 \\
&\quad -\int_{0}^{\xi }\Big(\frac{16i\pi }{3\langle\xi \rangle }
|\varphi (\xi )|^{2}\varphi (\xi )
 \\
&\quad +4\overline{E}\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(
\eta -\xi )}\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(
y)dy\Big)d\xi .
\end{aligned}  \label{c8}
\end{equation}
We now find the inverse transformation to (\ref{c1}). Changing the variable
of integration $\eta =\eta '+\frac{\xi }{2}$ we rewrite (\ref{c1})
as (the prime we will omit)
\begin{equation}
w(\xi )=\phi (\xi )-e^{\frac{3}{4}i\xi ^{2}}\int_{
\mathbb{R}}e^{i\eta ^{2}}K_{1}(\xi ,\eta )\overline{w(
\eta -\frac{\xi }{2})}\overline{w(-\eta -\frac{\xi }{2})}
d\eta  \label{c3}
\end{equation}
where
\begin{equation*}
K_{1}(\xi ,\eta )=\xi K(\xi ,\eta +\frac{\xi }{2})=
\frac{\xi (2+iS_{1})}{(1+iS_{1})^{2}}, \quad
S_{1}=\frac{3}{4}\xi ^{2}+\eta ^{2}.
\end{equation*}
We substitute the representation
\begin{equation}
w(\xi )=\phi (\xi )+\psi _{1}(\xi )
-e^{\frac{3}{4}i\xi ^{2}}\psi _{2}(\xi )\label{c2}
\end{equation}
into (\ref{c3}). In view of the symmetry $\eta \to -\eta $, changing
in the second integral $\eta =\frac{3}{2}\xi -\eta '$ we find
\begin{align*}
\psi _{1}(\xi )-e^{\frac{3}{4}i\xi ^{2}}\psi _{2}(\xi)
&=-e^{\frac{3}{4}i\xi ^{2}}\int_{\mathbb{R}}e^{i\eta ^{2}}K_{1}(
\xi ,\eta )\overline{\psi _{3}(\eta -\frac{\xi }{2})}
\overline{\psi _{3}(-\eta -\frac{\xi }{2})}d\eta \\
&\quad +2\int_{\mathbb{R}}e^{\frac{1}{4}i\eta ^{2}}K_{2}(\xi ,\eta )
\overline{\psi _{3}(\xi -\eta )}\overline{\psi _{2}(\eta
-2\xi )}d\eta \\
&\quad -e^{\frac{3}{8}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{2}i\eta
^{2}}K_{1}(\xi ,\eta )\overline{\psi _{2}(-\eta -\frac{
\xi }{2})}\overline{\psi _{2}(-\eta -\frac{\xi }{2})}
d\eta ,
\end{align*}
where $\psi _{3}(\xi )=\phi (\xi )+\psi _{1}(\xi )$ and
$K_{2}(\xi ,\eta )=K_{1}(\xi ,\frac{3}{2}\xi -\eta )$.
Thus we obtain the system of integral equations for
the functions $\psi _{1}(\xi )$ and $\psi _{2}(\xi)$
\begin{equation}
\begin{gathered}
\begin{aligned}
\psi _{1}(\xi )&=2\int_{\mathbb{R}}e^{\frac{1}{4}i\eta
^{2}}K_{2}(\xi ,\eta )\overline{\psi _{2}(\eta -2\xi
)}\overline{\psi _{3}(\xi -\eta )}d\eta \\
&\quad +e^{-\frac{3}{8}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{2}i\eta
^{2}}K_{1}(\xi ,\eta )\overline{\psi _{2}(\eta -\frac{\xi
}{2})}\overline{\psi _{2}(-\eta -\frac{\xi }{2})}d\eta
\end{aligned},
\\
\psi _{2}(\xi )=\int_{\mathbb{R}}e^{i\eta ^{2}}K_{1}(\xi
,\eta )\overline{\psi _{3}(\eta -\frac{\xi }{2})}
\overline{\psi _{3}(-\eta -\frac{\xi }{2})}d\eta .
\end{gathered}  \label{c4}
\end{equation}
Denote the norm
\begin{equation*}
\|\psi \|_{\mathbf{Z}}\equiv \sup_{\xi \in \mathbb{R}}(
\langle\xi \rangle ^{1-\gamma }|\psi (\xi )
|+\langle\xi \rangle ^{1-2\gamma }|\psi '(\xi )|).
\end{equation*}

\begin{lemma}\label{Lemma A1}
Let $\phi \in \mathbf{C}^{1}(\mathbb{R})$
satisfy the estimate
\begin{equation}
\sup_{\xi \in \mathbb{R}}(|\phi (\xi )|
+\langle\xi \rangle ^{1-\gamma }|\phi '(\xi
)|)\leq \varepsilon ,  \label{s1}
\end{equation}
where $\varepsilon >0$ is sufficiently small and
$\gamma \in (0,1)$. Then there exist unique solutions
$\psi _{1},\psi _{2}\in \mathbf{C}^{1}(\mathbb{R})$ of a system of
integral equations (\ref{c4}) such that
\begin{equation}
\|\psi _{1}\|_{\mathbf{Z}}+\|\psi _{2}\|_{\mathbf{Z
}}\leq C\varepsilon ^{2}.  \label{r}
\end{equation}
\end{lemma}

\begin{proof}
We solve equations (\ref{c4}) by the contraction mapping principle in the
set
\begin{equation*}
\mathbf{Z}_{\varepsilon } = \big\{(\psi _{1},\psi _{2})
\in (\mathbf{C}^{1}(\mathbb{R}))^{2}:\|
\psi _{1}\|_{\mathbf{Z}}+\| \psi _{2}\|_{\mathbf{
Z}}\leq C\varepsilon ^{2}\big\}.
\end{equation*}
Define the transformation
\begin{gather*}
\begin{aligned}
\mathcal{M}_{1}(\psi _{1},\psi _{2})(\xi )
&=2\int_{\mathbb{R}}e^{\frac{1}{4}i\eta ^{2}}K_{2}(\xi ,\eta )\overline{
\psi _{2}(\eta -2\xi )}\overline{\psi _{3}(\xi -\eta
)}d\eta \\
&\quad +e^{-\frac{3}{8}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{2}i\eta
^{2}}K_{1}(\xi ,\eta )\overline{\psi _{2}(\eta -\frac{\xi
}{2})}\overline{\psi _{2}(-\eta -\frac{\xi }{2})}d\eta
\end{aligned},
\\
\mathcal{M}_{2}(\psi _{1},\psi _{2})(\xi )=\int_{
\mathbb{R}}e^{i\eta ^{2}}K_{1}(\xi ,\eta )\overline{\psi
_{3}(\eta -\frac{\xi }{2})}\overline{\psi _{3}(-\eta -
\frac{\xi }{2})}d\eta
\end{gather*}
for $(\psi _{1},\psi _{2})\in \mathbf{Z}_{\varepsilon }$. Via (
\ref{s1}) and by the fact that $\| \psi _{1}\|_{\mathbf{Z}
}+\| \psi _{2}\|_{\mathbf{Z}}\leq C\varepsilon ^{2}$ we
obtain the estimate
\begin{align*}
&|\mathcal{M}_{1}(\psi _{1},\psi _{2})(\xi)|\\
&\leq C\varepsilon ^{3}\int_{\mathbb{R}}(1+|\xi |+|\eta |)
^{-1}\big(\langle\eta -2\xi \rangle ^{\gamma -1}
 +\langle\eta -\frac{\xi }{2}\rangle ^{\gamma
-1}\langle\eta +\frac{\xi }{2}\rangle ^{\gamma -1}\big)d\eta\\
&\leq C\varepsilon ^{3}\langle\xi \rangle ^{\gamma -1}.
\end{align*}
Integrating by parts with respect to $\eta $ via the identity $e^{i\eta
^{2}}=A\partial _{\eta }(\eta e^{i\eta ^{2}})$ with $A=(
1+2i\eta ^{2})^{-1}$ we get
\begin{align*}
|\mathcal{M}_{2}(\psi _{1},\psi _{2})(\xi)|
&=\big|\int_{\mathbb{R}}e^{i\eta ^{2}}\overline{
\psi _{3}(\eta -\frac{\xi }{2})}\overline{\psi _{3}(-\eta
-\frac{\xi }{2})}\eta \partial _{\eta }(AK_{1})d\eta\\
 &\quad +2\int_{\mathbb{R}}e^{i\eta ^{2}}\eta AK_{1}\overline{\psi
_{3}(\eta -\frac{\xi }{2})}\overline{\psi _{3}'(
-\eta -\frac{\xi }{2})}d\eta \big|\\
&\leq C\varepsilon ^{2}\int_{\mathbb{R}}\Big(\langle\xi
\rangle ^{-1}\langle\eta \rangle ^{-2}+(1+|
\xi |+|\eta |)^{-2}\langle\eta
\rangle ^{-1}\\
&\quad +(1+|\xi |+|\eta |
)^{-1}\langle\eta \rangle ^{-1}\langle2\eta +\xi
\rangle ^{2\gamma -1}\Big)d\eta\\
&\leq C\varepsilon ^{2}\langle \xi \rangle ^{-1}.
\end{align*}
We now estimate the derivatives
\begin{align*}
&\frac{d}{d\xi }\mathcal{M}_{1}(\psi _{1},\psi _{2})(\xi)\\
&=2\int_{\mathbb{R}}e^{\frac{1}{4}i\eta ^{2}}\overline{\psi
_{2}(\eta -2\xi )}\overline{\psi _{3}(\xi -\eta )}
\partial _{\xi }K_{2}d\eta \\
&\quad +2\int_{\mathbb{R}}e^{\frac{1}{4}i\eta ^{2}}(\overline{\psi
_{2}(\eta -2\xi )}\overline{\psi _{3}'(\xi -\eta
)}-2\overline{\psi _{2}'(\eta -2\xi )}\overline{
\psi _{3}(\xi -\eta )})K_{2}d\eta \\
&\quad -e^{-\frac{3}{8}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{2}i\eta ^{2}}K_{1}
\overline{\psi _{2}'(\eta -\frac{\xi }{2})}\overline{
\psi _{2}(-\eta -\frac{\xi }{2})}d\eta \\
&\quad +e^{-\frac{3}{8}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{2}i\eta ^{2}}
\overline{\psi _{2}(\eta -\frac{\xi }{2})}\overline{\psi
_{2}(-\eta -\frac{\xi }{2})}(\partial _{\xi }K_{1}
-\frac{3}{4}i\xi K_{1})d\eta .
\end{align*}
and
\begin{align*}
\frac{d}{d\xi }\mathcal{M}_{2}(\psi _{1},\psi _{2})(\xi)
&=-\int_{\mathbb{R}}e^{i\eta ^{2}}K_{1}\overline{\psi _{3}'(\eta -\frac{\xi }{2})}\overline{\psi _{3}(-\eta -\frac{
\xi }{2})}d\eta \\
&\quad +\int_{\mathbb{R}}e^{i\eta ^{2}}\overline{\psi _{3}(\eta -\frac{\xi
}{2})}\overline{\psi _{3}(-\eta -\frac{\xi }{2})}
\partial _{\xi }K_{1}d\eta .
\end{align*}
Then we find the estimates
\begin{align*}
|\frac{d}{d\xi }\mathcal{M}_{1}(\psi _{1},\psi _{2})
(\xi )|
&\leq C\varepsilon ^{3}\int_{\mathbb{R}}(\langle\eta -2\xi
\rangle ^{2\gamma -1}+\langle\xi -\eta \rangle ^{2\gamma
-1})(1+|\xi |+|\eta |
)^{-1}d\eta \\
&\quad +C\varepsilon ^{4}\int_{\mathbb{R}}\langle\eta -\frac{\xi }{2}
\rangle ^{\gamma -1}\langle\eta +\frac{\xi }{2}\rangle
^{\gamma -1}d\eta \leq C\varepsilon ^{3}\langle\xi \rangle
^{2\gamma -1}
\end{align*}
and
\begin{align*}
|\frac{d}{d\xi }\mathcal{M}_{2}(\psi _{1},\psi _{2})(\xi )|
&\leq C\varepsilon ^{2}\int_{\mathbb{R}}(
1+|\xi |+|\eta |)
^{-1}\langle\eta -\frac{\xi }{2}\rangle ^{2\gamma -1}d\eta \\
&\quad +C\varepsilon ^{2}\int_{\mathbb{R}}(1+|\xi |
+|\eta |)^{-2}d\eta \leq C\varepsilon
^{2}\langle\xi \rangle ^{2\gamma -1}.
\end{align*}
Thus the mapping $(\mathcal{M}_{1},\mathcal{M}_{2})$ transforms
the set $\mathbf{Z}$ into itself. In the same manner we find
\begin{equation*}
\| \mathcal{M}_{j}(\psi _{1},\psi _{2})-\mathcal{M}
_{j}(\widetilde{\psi _{1}},\widetilde{\psi _{2}})\|_{
\mathbf{Z}}\leq \frac{1}{2}\| \psi _{1}-\widetilde{\psi _{1}}
\|_{\mathbf{Z}}+\frac{1}{2}\| \psi _{2}-\widetilde{\psi
_{2}}\|_{\mathbf{Z}}.
\end{equation*}
Therefore, $(\mathcal{M}_{1},\mathcal{M}_{2})$ is a contraction
mapping in $\mathbf{Z.}$ Hence there exist unique solutions $\psi _{1},\psi
_{2}\in \mathbf{Z}$ of a system of integral equations (\ref{c4}), which
satisfy estimate (\ref{r}). Lemma \ref{Lemma A1} is proved.
\end{proof}

We now evaluate the asymptotic form of the integral
\begin{equation*}
\mathcal{I}\equiv \int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )
^{2}}A(\xi ,\eta )\phi _{1}(\eta -\xi )\Phi d\eta ,
\end{equation*}
where
\begin{equation*}
\Phi =\int_{\mathbb{R}}e^{\beta iy^{2}}\phi _{2}(a_{1}\eta
-b_{1}y)\phi _{3}(a_{2}\eta -b_{2}y)dy,
\end{equation*}
with $\alpha ,\beta ,\mu ,a_{1},a_{2},b_{1},b_{2}\in \mathbf{R\backslash }
\{0\}$. Also we assume that $a_{1}b_{2}-a_{2}b_{1}\neq 0$.

\begin{lemma}\label{Lemma A2}
Suppose that
\begin{gather*}
\|\phi _{j}\|_{\mathbf{Y}}\equiv \sup_{\xi \in \mathbb{R}
}(|\phi _{j}(\xi )|+\langle\xi
\rangle ^{1-2\gamma }|\phi _{j}'(\xi )|)\leq C,
\\
|A(\xi ,\eta )|\leq C(1+|\xi |
+|\eta |)^{-1},
\\
|\partial _{\eta }A(\xi ,\eta )|\leq C(
1+|\xi |+|\eta |)^{-2}.
\end{gather*}
Then the asymptotic form is true
\begin{equation}
\mathcal{I}=\frac{i\pi }{\sqrt{\alpha \beta }}A(\xi ,\mu \xi )
\phi _{1}((\mu -1)\xi )\phi _{2}(a_{1}\mu
\xi )\phi _{3}(a_{2}\mu \xi )+O(\langle\xi
\rangle ^{\gamma -2}).  \label{AA}
\end{equation}
\end{lemma}

\begin{proof}
We first integrate by parts with respect to $y$ via identity
$e^{\beta iy^{2}}=B\partial _{y}(ye^{\beta iy^{2}})$ with $B=(
1+2i\beta y^{2})^{-1}$ we get $\Phi =\Phi _{1}+\Phi _{2}+\Phi _{3}$,
where
\begin{gather*}
\Phi _{1} = 2\int_{\mathbb{R}}e^{\beta iy^{2}}\phi _{2}(a_{1}\eta
-b_{1}y)\phi _{3}(a_{2}\eta -b_{2}y)B(B-1)
dy, \\
\Phi _{2} = b_{1}\int_{\mathbb{R}}e^{\beta iy^{2}}\phi _{2}'(
a_{1}\eta -b_{1}y)\phi _{3}(a_{2}\eta -b_{2}y)yB\,dy, \\
\Phi _{3} = b_{2}\int_{\mathbb{R}}e^{\beta iy^{2}}\phi _{2}(a_{1}\eta
-b_{1}y)\phi _{3}'(a_{2}\eta -b_{2}y)yB\,dy,
\end{gather*}
Now in the integral
\begin{equation*}
I=\int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )^{2}}A(\xi
,\eta )\phi _{1}(\eta -\xi )\Phi _{1}(\eta )
d\eta
\end{equation*}
we integrate by parts with respect to $\eta $ via identity
\begin{equation*}
e^{\alpha i(\eta -\mu \xi )^{2}}=H\partial _{\eta }(
(\eta -\mu \xi )e^{\alpha i(\eta -\mu \xi )
^{2}})
\end{equation*}
with $H=(1+2i\alpha (\eta -\mu \xi )^{2})^{-1}$ to obtain
 $I=I_{1}+\cdot \cdot \cdot +I_{5}$, where
\begin{align*}
I_{1} &= 2\phi _{1}((\mu -1)\xi )\Phi _{1}(
\mu \xi )A(\xi ,\mu \xi )\int_{\mathbb{R}}e^{\alpha
i(\eta -\mu \xi )^{2}}H(H-1)d\eta , \\
I_{2} &= \int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )^{2}}\phi
_{1}(\eta -\xi )\Phi _{1}(\eta )(\eta -\mu
\xi )H\partial _{\eta }Ad\eta \\
I_{3} &= \int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )
^{2}}\Big(\phi _{1}(\eta -\xi )\Phi _{1}(\eta )
A(\xi ,\eta ) \\
&\quad -\phi _{1}((\mu -1)\xi )\Phi _{1}(
\mu \xi )A(\xi ,\mu \xi )\Big)H(H-1)d\eta
\\
I_{4} &= \int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )^{2}}\phi
_{1}(\eta -\xi )\Phi _{1}'(\eta )A(
\eta -\mu \xi )Hd\eta \\
I_{5} &= \int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )^{2}}\phi
_{1}'(\eta -\xi )\Phi _{1}(\eta )A(
\eta -\mu \xi )Hd\eta
\end{align*}
We prove that the integral $I_{1}$ is the main term. Since
\begin{equation*}
2\int_{\mathbb{R}}e^{\beta iy^{2}}B(B-1)dy=\frac{\sqrt{i\pi }}{
\sqrt{\beta }},
\end{equation*}
and
$\langle y\rangle |\phi _{j}(a_{i}\mu \xi
-b_{i}y)-\phi _{j}(a_{i}\mu \xi )|\leq
C\langle\xi \rangle ^{\gamma -1}\langle y\rangle
^{-\gamma }$,
we have
\begin{align*}
\Phi _{1}(\mu \xi )&= 2\phi _{2}(a_{1}\mu \xi )
\phi _{3}(a_{2}\mu \xi )\int_{\mathbb{R}}e^{\beta
iy^{2}}B(B-1)dy \\
&\quad +2\int_{\mathbb{R}}e^{\beta iy^{2}}\Big(\phi _{2}(a_{1}\mu \xi
-b_{1}y)\phi _{3}(a_{2}\mu \xi -b_{2}y) \\
&\quad -\phi _{2}(a_{1}\mu \xi )\phi _{3}(a_{2}\mu \xi
)\Big)B(B-1)dy \\
&= \frac{\sqrt{i\pi }}{\sqrt{\beta }}\phi _{2}(a_{1}\mu \xi )
\phi _{3}(a_{2}\mu \xi )
+O\Big(\langle\xi \rangle
^{\gamma -1}\int_{\mathbb{R}}\langle y\rangle ^{-1-\gamma
}dy\Big).
\end{align*}
By a direct calculation,
\begin{equation*}
2\int_{\mathbb{R}}e^{\alpha i(\eta -\mu \xi )^{2}}H(
H-1)d\eta =\frac{1}{\sqrt{\alpha }}\int_{\mathbb{R}}e^{i\eta
^{2}}d\eta =\frac{\sqrt{i\pi }}{\sqrt{\alpha }}.
\end{equation*}
Therefore,
\begin{equation*}
I_{1}=\frac{i\pi }{\sqrt{\alpha \beta }}A(\xi ,\mu \xi )\phi
_{1}((\mu -1)\xi )\phi _{2}(a_{1}\mu \xi
)\phi _{3}(a_{2}\mu \xi )+O\big(\langle\xi
\rangle ^{\gamma -2}\big).
\end{equation*}
We have
\begin{align*}
|\Phi _{1}(\eta )|
&\leq C\int_{\mathbb{R}
}\langle a_{1}\eta -b_{1}y\rangle ^{-\gamma _{1}}\langle
a_{2}\eta -b_{2}y\rangle ^{-\gamma _{2}}\langle y\rangle
^{-2}dy\leq C\langle\eta \rangle ^{-\gamma _{1}-\gamma _{2}}, \\
|\Phi _{1}'(\eta )|
&\leq C\int_{\mathbb{R}}\langle a_{1}\eta -b_{1}y\rangle ^{\gamma
-1}\langle a_{2}\eta -b_{2}y\rangle ^{-\gamma _{2}}\langle
y\rangle ^{-2}dy \\
&\quad +C\int_{\mathbb{R}}\langle a_{1}\eta -b_{1}y\rangle ^{-\gamma
_{1}}\langle a_{2}\eta -b_{2}y\rangle ^{\gamma -1}\langle
y\rangle ^{-2}dy\leq C\langle\eta \rangle ^{\gamma -1}
\end{align*}
Hence
\begin{equation*}
|I_{2}|\leq C\langle\xi \rangle ^{\gamma
-2}\int_{\mathbb{R}}\langle\eta -\mu \xi \rangle
^{-1}\langle\eta \rangle ^{-\gamma }d\eta \leq C\langle\xi
\rangle ^{\gamma -2}.
\end{equation*}
Since
\begin{equation*}
\langle\eta -\mu \xi \rangle ^{-1}|\phi (\eta
)-\phi (\mu \xi )|\leq C\langle\xi
\rangle ^{\gamma -1}\langle\eta \rangle ^{-\gamma }
\end{equation*}
and
\begin{equation*}
\langle\eta -\mu \xi \rangle ^{-1}|A(\xi ,\eta
)-A(\xi ,\mu \xi )|\leq C\langle\xi
\rangle ^{\gamma -2}\langle\eta \rangle ^{-\gamma }
\end{equation*}
we find
\begin{equation*}
|I_{3}|\leq C\langle\xi \rangle ^{\gamma
-2}\int_{\mathbb{R}}\langle\eta \rangle ^{-\gamma }\langle
\eta -\mu \xi \rangle ^{-1}d\eta \leq C\langle\xi \rangle
^{\gamma -2}.
\end{equation*}
We also have
\begin{equation*}
|I_{4}|+|I_{5}|\leq C\langle
\xi \rangle ^{-1}\int_{\mathbb{R}}(\langle\eta
\rangle ^{\gamma -1}+\langle\eta -\xi \rangle ^{\gamma
-1})\langle\eta -\mu \xi \rangle ^{-1}d\eta \leq
C\langle\xi \rangle ^{\gamma -2}.
\end{equation*}
In the integral $\Phi _{2}$ we change $a_{1}\eta -b_{1}y=y'$ (the
prime we will omit) then with
\begin{gather*}
\widetilde{a_{2}} = \frac{1}{b_{1}}(a_{2}b_{1}-a_{1}b_{2})\neq
0, \quad
\widetilde{b_{2}}=\frac{b_{2}}{b_{1}}, \\
\widetilde{B}_{1} = (1+\frac{2\beta }{b_{1}^{2}}i(a_{1}\eta
-y)^{2})^{-1}, \quad
\beta _{1}=\frac{\beta }{b_{1}^{2}}
\end{gather*}
we have
\begin{equation*}
\Phi _{2}=C\int_{\mathbb{R}}e^{i\beta _{1}(a_{1}\eta -y)
^{2}}\phi _{2}'(y)\phi _{3}(\widetilde{a_{2}}
\eta -\widetilde{b_{2}}y)(a_{1}\eta -y)\widetilde{B}dy.
\end{equation*}
Then  with
$Q=\beta _{1}(a_{1}\eta -y)^{2}+\alpha (\eta -\mu \xi)^{2}$.
we define
\begin{equation*}
I_{6}=C\int_{\mathbb{R}}dy\phi _{2}'(y)\int_{\mathbb{R}
}A(\xi ,\eta )\phi _{1}(\eta -\xi )e^{iQ}\phi
_{3}(\widetilde{a_{2}}\eta -\widetilde{b_{2}}y)(a_{1}\eta
-y)\widetilde{B}d\eta\,.
\end{equation*}
Then we integrate by parts with respect to $\eta $ via identity
$e^{iQ}=H\partial _{\eta }(\widetilde{\eta }e^{iQ})$ with
$H=(1+i\widetilde{\eta }Q_{\eta })^{-1}$, where
\begin{equation*}
\widetilde{\eta }=\eta -\widetilde{b}y-\widetilde{a}\xi ,\quad
\widetilde{b}=\frac{
a_{1}\beta _{1}}{\alpha +\beta _{1}a_{1}^{2}},\quad
\widetilde{a}=\frac{\alpha \mu
}{\alpha +\beta _{1}a_{1}^{2}},
\end{equation*}
if $\alpha +\beta _{1}a_{1}^{2}\neq 0$ and $\widetilde{\eta }=\eta $ if
$\alpha +\beta _{1}a_{1}^{2}=0$ we get
\begin{equation*}
I_{6}=C\int_{\mathbb{R}}dy\phi _{2}'(y)\int_{\mathbb{R}
}e^{iQ}\widetilde{\eta }\partial _{\eta }HA(\xi ,\eta )\phi
_{1}(\eta -\xi )\phi _{3}(\widetilde{a_{2}}\eta -
\widetilde{b_{2}}y)(a_{1}\eta -y)\widetilde{B}d\eta
\end{equation*}
since $\langle\widetilde{\eta }Q_{\eta }\rangle ^{-1}\leq
C\langle\widetilde{\eta }\rangle ^{-2}$ if $\alpha +\beta
_{1}a_{1}^{2}\neq 0$ and
\begin{equation*}
\langle\widetilde{\eta }Q_{\eta }\rangle ^{-1}\leq C|
\widetilde{\eta }|^{\gamma -1}|a_{1}\beta _{1}y+\alpha
\mu \xi |^{\gamma -1}
\end{equation*}
if $\alpha +\beta _{1}a_{1}^{2}=0$ we obtain
\begin{align*}
|I_{6}|&\leq C\langle\xi \rangle
^{-1}\int_{\mathbb{R}}dy\langle y\rangle ^{\gamma -1}\int_{
\mathbb{R}}\langle\eta -\frac{y}{a_{1}}\rangle ^{-1}\langle
\eta -\widetilde{b}y-\widetilde{a}\xi \rangle ^{-1}d\eta \\
&\leq C\langle\xi \rangle ^{-1}\int_{\mathbb{R}}dy\langle
y\rangle ^{\gamma -1}\langle a_{3}y-\widetilde{a}\xi
\rangle ^{\gamma -1}\leq C\langle\xi \rangle ^{2\gamma -2}
\end{align*}
if $\alpha +\beta _{1}a_{1}^{2}\neq 0$, since $a_{1}$, $\widetilde{a}$,
$a_{3}\equiv \frac{1-a_{1}\widetilde{b}}{a_{1}}\neq 0$, and
\begin{align*}
|I_{6}|&\leq C\langle\xi \rangle ^{\gamma
-1}\int_{\mathbb{R}}dy\langle y\rangle ^{\gamma -1}|
a_{1}\beta _{1}y+\alpha \mu \xi |^{\gamma -1}\int_{\mathbb{R}
}\langle a_{1}\eta -y\rangle ^{-1}\Big(|\eta
|^{\gamma -1} \\
&\quad +\langle\eta -\xi \rangle ^{\gamma -1}+\langle
\widetilde{a_{2}}\eta -\widetilde{b_{2}}y\rangle ^{\gamma
-1}+\langle a_{1}\eta -y\rangle ^{-1}\Big)d\eta \leq
C\langle\xi \rangle ^{2\gamma -2}
\end{align*}
if $\alpha +\beta _{1}a_{1}^{2}=0$. In the same manner we estimate the
integral with $\Phi _{3}$. Thus we have the asymptotic form (\ref{AA}). Lemma
\ref{Lemma A2} is proved.
\end{proof}

Now we substitute (\ref{c2}) $w(\xi )=\psi _{3}(\xi
)-e^{\frac{3}{4}i\xi ^{2}}\psi _{2}(\xi )$ into the
second summand in the right-hand side of (\ref{c5}) as above changing the
variables of integration $y=y'+\frac{\eta }{2}$ or $y=2\eta
-y'$ we find
\begin{align*}
\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(y)dy
&=e^{-
\frac{3}{4}i\eta ^{2}}\int_{\mathbb{R}}e^{-iy^{2}}\psi _{3}(\frac{\eta
}{2}-y)\psi _{3}(\frac{\eta }{2}+y)dy \\
&\quad -2\int_{\mathbb{R}}e^{-\frac{1}{4}iy^{2}}\psi _{3}(y-\eta )
\psi _{2}(2\eta -y)dy \\
&\quad +e^{-\frac{3}{8}i\eta ^{2}}\int_{\mathbb{R}}e^{\frac{1}{2}iy^{2}}\psi
_{2}(\frac{\eta }{2}-y)\psi _{2}(\frac{\eta }{2}+y)
dy.
\end{align*}
So we have
\begin{align*}
&\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(\eta -\xi )}
\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(y)dy\\
&= \int_{\mathbb{R}}e^{\frac{1}{4}i(\eta -2\xi )^{2}}K\eta
\overline{\psi _{3}(\eta -\xi )}\Psi _{1}(\eta )
d\eta \\
&\quad +2e^{\frac{3}{4}i\xi ^{2}}\int_{\mathbb{R}}e^{i(\eta -\frac{1}{2}\xi
)^{2}}K\eta \overline{\psi _{3}(\eta -\xi )}\Psi
_{2}(\eta )d\eta \\
&\quad +e^{\frac{3}{5}i\xi ^{2}}\int_{\mathbb{R}}e^{\frac{5}{8}i(\eta -
\frac{4}{5}\xi )^{2}}K\eta \overline{\psi _{3}(\eta -\xi
)}\Psi _{3}(\eta )d\eta \\
&\quad +e^{\frac{3}{8}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{2}i(\eta -
\frac{1}{2}\xi )^{2}}K\eta \overline{\psi _{2}(\eta -\xi
)}\Psi _{1}(\eta )d\eta \\
&\quad +2\int_{\mathbb{R}}e^{\frac{1}{4}i(\eta +\xi )^{2}}K\eta
\overline{\psi _{2}(\eta -\xi )}\Psi _{2}(\eta )
d\eta \\
&\quad +e^{\frac{3}{4}i\xi ^{2}}\int_{\mathbb{R}}e^{-\frac{1}{8}i(\eta
-2\xi )^{2}}K\eta \overline{\psi _{2}(\eta -\xi )}\Psi
_{3}(\eta )d\eta
\end{align*}
where
\begin{gather*}
\Psi _{1}(\eta )= \int_{\mathbb{R}}e^{-iy^{2}}\psi _{3}(
\frac{\eta }{2}-y)\psi _{3}(\frac{\eta }{2}+y)dy, \\
\Psi _{2}(\eta )= \int_{\mathbb{R}}e^{-\frac{1}{4}iy^{2}}\psi
_{3}(y-\eta )\psi _{2}(2\eta -y)dy, \\
\Psi _{3}(\eta )= \int_{\mathbb{R}}e^{\frac{1}{2}iy^{2}}\psi
_{2}(\frac{\eta }{2}-y)\psi _{2}(\frac{\eta }{2}+y)
dy,
\end{gather*}
Applying Lemma \ref{Lemma A2} we obtain the asymptotic form
\begin{align*}
&\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(\eta -\xi )}
\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(y)dy \\
&= \int_{\mathbb{R}}e^{\frac{1}{4}i(\eta -2\xi )^{2}}K\eta
\overline{\psi _{3}(\eta -\xi )}\int_{\mathbb{R}
}e^{-iy^{2}}\psi _{3}(\frac{\eta }{2}-y)\psi _{3}(\frac{
\eta }{2}+y)dyd\eta
+O(\langle\xi \rangle ^{\gamma -2})\\
&= -\frac{4i\pi }{3\langle\xi \rangle }\overline{\psi _{3}(
\xi )}\psi _{3}(\xi )\psi _{3}(\xi )
+O(\langle\xi \rangle ^{\gamma -2})\\
&= -\frac{4i\pi }{3\langle\xi \rangle }|\phi (
\xi )|^{2}\phi (\xi )+O(\langle
\xi \rangle ^{\gamma -2}).
\end{align*}
Thus we can estimate the right-hand side of (\ref{c8})
\begin{align*}
\frac{d}{d\xi }\mathcal{A}(\varphi )(\xi )&= 2
\overline{E}\int_{\mathbb{R}}Fe^{iS}\overline{w(\eta -\xi )}
\overline{w(-\eta )}d\eta d\xi -\frac{16i\pi }{3\langle
\xi \rangle }|\varphi (\xi )|
^{2}\varphi (\xi )\\
&\quad -4\overline{E}\int_{\mathbb{R}}d\eta e^{iS}K\eta \overline{w(\eta
-\xi )}\int_{\mathbb{R}}e^{-iQ}w(\eta -y)w(
y)dy \\
&= O(\varepsilon ^{3}\langle\xi \rangle ^{\gamma-2}).
\end{align*}
We solve equation (\ref{c7}) by the contraction mapping principle in the set
\begin{equation*}
\mathbf{X}_{\varepsilon } = \{\varphi \in \mathbf{C}^{1}(
\mathbb{R}):\| \varphi \|_{\mathbf{X}}\leq
C\varepsilon ^{2}\}
\end{equation*}
where the norm
\begin{equation*}
\| \varphi \|_{\mathbf{X}}\equiv \sup_{\xi \in \mathbb{R}
}\big(|\varphi (\xi )|+\langle\xi
\rangle ^{2-2\gamma }|\varphi '(\xi )|\big).
\end{equation*}
When $\varphi \in \mathbf{X}_{\varepsilon }$, then
$\| \phi \|_{\mathbf{Y}}\equiv \sup_{\xi \in \mathbb{R}}(|
\phi (\xi )|+\langle\xi \rangle
^{1-2\gamma }|\phi '(\xi )|
)\leq C\varepsilon $ and by Lemma \ref{Lemma A1}
\begin{equation*}
\| \psi _{1}\|_{\mathbf{Z}}+\| \psi
_{2}\|_{\mathbf{Z}}+\| \psi _{3}\|_{\mathbf{Z}
}\leq C\varepsilon .
\end{equation*}
Then by Lemma \ref{Lemma A2}
\begin{equation*}
|\frac{d}{d\xi }\mathcal{A}(\varphi )(\xi
)|\leq C\varepsilon \langle\xi \rangle
^{2\gamma -2}.
\end{equation*}
And integrating we have
\begin{equation*}
|\mathcal{A}(\varphi )(\xi )|
\leq C\varepsilon \int_{0}^{\xi }\langle\xi \rangle ^{2\gamma
-2}d\xi +|\varphi (0)|\leq C\varepsilon .
\end{equation*}
In the same manner we can estimate the difference
\begin{equation*}
|\mathcal{A}(\varphi _{1})-\mathcal{A}(\varphi
_{2})|\leq C\varepsilon \| \varphi _{1}-\varphi
_{2}\|_{\mathbf{X}}.
\end{equation*}
Therefore, $\mathcal{A}$ is a contraction mapping in $\mathbf{X.}$ Hence
there exists a unique solution $\varphi \in \mathbf{X}$ of integral equation
(\ref{c7}).



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\end{document}