text:
\section{Introduction}
We consider the fractional Korteweg-de Vries equation with cubic nonlinearity:
\begin{align}\label{eq:main}
	\partial_t u-|D|^{\alpha} \partial_x u=-u^2\partial_x u, \quad -1<\alpha<0,
\end{align}
where \(u\) maps \(\mathbb{R}_t\times\mathbb{R}_x\) to \(\mathbb{R}\) and \(|D|^{\alpha}\) is the usual Fourier multiplier operator with the symbol \(|\xi|^\alpha\).  The initial data is given by
\begin{align}\label{eq:initial}
	u(0,x)=u_0(x).
\end{align}

Although it does not seem to appear in a physical context, this equation, which will be from now on referred to as the modified fractional KdV equation (modified fKdV),  is a good toy model to understand the influence of a weak dispersion on the dynamics of a scalar conservation law such as the modified Burgers equation. When $\alpha=-\frac{1}{2},$ it is reminiscent for large frequencies, of a modified Whitham equation

\begin{equation}\label {modWhit}
	\partial_t u+\mathcal L \partial_xu=-u^2\partial_xu,
\end{equation}
where the Fourier multiplier operator $\mathcal L$ with symbol $\left(\frac{\tanh \xi}{\xi}\right)^{1/2}.$

We refer for instance to \cite{KLPS} for various issues and results on the usual (quadratic) Whitham equation, in particular on its KdV, long wave limit. It is straightforward to check that Theorem 1 in \cite{KLPS} extends to the modified Whitham equation, proving its long wave limit to the modified KdV equation.
More precisely we consider the rescaled modified Whitham equation
\begin{equation}\label{mWhit}
	u_t+\mathcal L_\epsilon u_x+\epsilon u^2u_x=0,
\end{equation}
where the non local operator $\mathcal L_\epsilon$ is related to the dispersion relation of the (linearized) water waves system and is defined by
$$\mathcal L_\epsilon =l(\sqrt{\epsilon}D):=\left(\frac{\tanh \sqrt \epsilon |D|}{\sqrt \epsilon |D|}\right)^{1/2} \quad \text{and} \quad D=-i\nabla=-i\frac{\partial}{\partial x}, $$
that we want to compare to the modified KdV equation

\begin{equation}\label{mKdV}
	v_t+v_x+\epsilon v^2v_x+\epsilon v_{xxx}=0.
\end{equation}
It is obvious that for any initial data $\phi\in H^s(\mathbb{R}), s>\frac{3}{2}$ \eqref {mWhit} admits a unique solution u in
$C([0,T_\epsilon);H^s(\mathbb{R}))$ where $T_\epsilon = O(1/\epsilon).$
Denoting $v$ the solution of\eqref{mKdV} with the same initial data $\phi$, one obtains, proceeding as in the proof of Theorem 1 in \cite{KLPS} which considered the Whitham equation \eqref{Whit}:

\begin{theorem}
	Let $\phi \in H^{\infty}(\mathbb R)$. Then, for all $j \in \mathbb N$, $j \ge 0$, there exists
	$M_j=M_j(\|\phi\|_{H^{j+8}})>0$  such that
	\begin{equation} \label{maintheo.1}
		\|(u-v)(t)\|_{H^j_x} \le M_j \epsilon^2t,
	\end{equation}
	for all $0 \le t \lesssim \epsilon^{-1}$.
\end{theorem}

It is well known that the modified Burgers equation undergoes shock formation, even for arbitrary small  smooth  initial data, provided  the square of the initial has a negative slope at some point. In particular no global solutions exist for arbitrary small smooth initial data in Sobolev spaces, such as a small gaussian.

The question we address here is whether this property is still true when adding a weakly dispersive term as in \eqref{eq:main}.  In fact we prove that adding this dispersive term allows the existence and (modified) scattering of small solutions.

Throughout the paper, we will always use the notation \(f(t)=e^{-t|D|^{\alpha} \partial_x}u(t)\) to denote the profile of \(u\). By time reversibility we  need only to consider the existence for positive time.

\vspace{0.3cm}
Our main result can be stated precisely as follows:
\begin{theorem}\label{th:main} Let \(\alpha\in(-1,0)\) and define the \(Z\)-norm
\[\|g\|_Z=\|(1+|\xi|)^{10}\widehat{g}(\xi)\|_{L^\infty_\xi}.\]
 Assume that \(N_0=100,\ p_0\in (0,1/1000]\cap (0,-\frac{\alpha}{100}] \) are fixed, and \(u_0\in H^{N_0}(\mathbb{R})\) satisfies
\begin{align}\label{1}
\|u_0\|_{H^{N_0}}+\|u_0\|_{H^{1,1}}+\|u_0\|_Z=\varepsilon_0\leq \bar{\varepsilon},
\end{align}
for some constant \(\bar{\varepsilon}\) sufficiently small (depending only on \(\alpha\) and \(p_0\)). Then the Cauchy problem \eqref{eq:main}-\eqref{eq:initial} admits a unique
 global solution \(u\in C((\mathbb{R}):H^{N_0}(\mathbb{R}))\) satisfying the following uniform bounds for \(t\geq 1\)
\begin{align}\label{2}
t^{-p_0}\|u\|_{H^{N_0}}+t^{-p_0}\|f\|_{H^{1,1}}+\|f\|_Z\lesssim \varepsilon_0.
\end{align}
Moreover,  there exists \(w_\infty\in L^\infty(\mathbb{R})\) such that  for \(t\geq 1\)
\begin{align}\label{3}
t^{p_0}\left\|\exp\left(\frac{3\mathrm{i}\xi|\xi|^{1-\alpha}}{\alpha(\alpha+1)}\int_1^t|\widehat{f}(s,\xi)|^2\frac{\mathrm{d} s}{s}\right)(1+|\xi|)^{10}\widehat{f}(\xi)-w_\infty(\xi)\right\|_{L^\infty_\xi}\lesssim \varepsilon_0.
\end{align}
\end{theorem}

The local well-posedness on the time interval \([0,1]\) for \eqref{eq:main}-\eqref{eq:initial} is standard provided $||u_0||_{H^2}$ is small enough, in particular  under the smallness assumption \eqref{1}. Then the existence and uniqueness of global solutions may be constructed by a bootstrap argument which allows  to extend the local solutions. More precisely,
assume that the following \(X\)-norm is a priori small:
\begin{equation}\label{4}
	\begin{aligned}
		\|u\|_{X}&=\sup_{t\geq 1}\bigg(t^{-p_0}\|u\|_{H^{N_0}}+t^{-p_0}\|f\|_{H^{1,1}}+\|f\|_Z\bigg)\leq \varepsilon_1
	\end{aligned}
\end{equation}
with \(\varepsilon_1=\varepsilon_0^{1/3}\),
we then aim to show that the above a priori assumption may be improved to
\begin{align}\label{5}
	\|u\|_{X}\leq C(\varepsilon_0+\varepsilon_1^3),
\end{align}
for some absolute constant \(C>1\).

The litterature dealing with the problem of global existence and scattering of small solutions to nonlinear dispersive equations is vast and we refer to the Introduction of \cite{MR3121725} for a useful survey.

The present work is close to \cite{MR3961297,MR3552008,MR3121725} in methodology.
In comparison with \cite{MR3121725}, the presence of a derivative
on the nonlinearity in the equation \eqref{eq:main} plays a crucial role in our proof in the sense that on the one hand it eliminates part of resonances in low frequencies and on the other hand it avoids using \(\partial_s\widehat{f}_l\) in \(L^\infty\)-norm in some frequency regimes, which allows us to extend the estimates to all \(\alpha\in(-1,0)\).

To end this section, we list the notations frequently used throughout the paper.

We denote by \(\mathcal{F}(g)\) or \(\widehat{g}\) the Fourier transform of a Schwartz function \(g\) whose formula is given by
\begin{equation*}
	\begin{aligned}
		\mathcal{F}(g)(\xi)=\widehat{g}(\xi):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}g(x)e^{-\mathrm{i}x\xi}\,\mathrm{d} x
	\end{aligned}
\end{equation*}
with inverse
\begin{equation*}
	\begin{aligned}
		\mathcal{F}^{-1}(g)(x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}g(\xi)e^{\mathrm{i}x\xi}\,\mathrm{d} \xi,
	\end{aligned}
\end{equation*}
and by \(m(\partial_x)\) the Fourier multiplier with symbol \(m\) via the relation
\begin{equation*}
	\begin{aligned}
		\mathcal{F}\big(m(\partial_x)g\big)(\xi)=m(\mathrm{i}\xi)\widehat{g}(\xi).
	\end{aligned}
\end{equation*}

Let \(\varphi\) be a smooth function satisfying
\begin{equation*}
	\varphi(\xi)=
	\left\{
	\begin{aligned}
		1,\quad &|\xi|\leq 1\\
		0,\quad &|\xi|>2.
	\end{aligned}
	\right.
\end{equation*}
Set
\begin{equation*}
	\begin{aligned}
		\psi(\xi)=\varphi(\xi)-\varphi(2\xi),\quad \psi_j(\xi)=\psi(2^{-j}\xi),\quad \varphi_j(\xi)=\varphi(2^{-j}\xi),
	\end{aligned}
\end{equation*}
we then may define the
Littlewood-Paley projections \(P_j,P_{\leq j},P_{> j}\)  via
\begin{equation*}
	\begin{aligned}
		\widehat{P_jg}(\xi)=\psi_j(\xi)\widehat{g}(\xi),\quad \widehat{P_{\leq j}g}(\xi)=\varphi_j(\xi)\widehat{g}(\xi),\quad P_{> j}=1-P_{\leq j},
	\end{aligned}
\end{equation*}
and also \(P_{\sim},P_{\lesssim j},P_{\ll j}\) by
\begin{equation*}
	\begin{aligned}
		P_{\sim j}=\sum_{2^k\sim 2^j}P_k, \quad P_{\lesssim j}=\sum_{2^k\leq 2^{j+C}}P_k,\quad P_{\ll j}=\sum_{2^k\ll 2^j}P_k,
	\end{aligned}
\end{equation*}
and the obvious notation for \(P_{[a,b]}\).
We will also denote \(g_j=P_jg, g_{\lesssim j}=P_{\lesssim j}g\), and so on, for convenience.

The notation \(C\)  always denotes a nonnegative universal constant which may be different from line to line but is
independent of the parameters involved. Otherwise, we will specify it by  the notation \(C(a,b,\dots)\).
We write \(g\lesssim h\) (\(g\gtrsim h\)) when \(g\leq  Ch\) (\(g\geq  Ch\)), and \(g \sim h\) when \(g \lesssim h \lesssim g\).
We also write \(\sqrt{1+x^2}=\langle x\rangle\) for \(x\in\mathbb{R}\) and \(\|g\|_{H^{1,1}}=\|\langle x\rangle g\|_{H^1}\), and \(P_{[k-2,k+2]}:=P_k^\prime\) and for simplicity.

\section{Decay Estimates}
This section is devoted to presenting some decay estimates of the solution of the equation \eqref{eq:main}. We first recall a dispersive linear estimate from \cite[Lemma 2.3]{MR3961297}:
\begin{lemma} For any \(t\geq 1\),  the following linear dispersive estimates hold:
	\begin{align}\label{6}
	\left\|e^{t|D|^{\alpha}\partial_x}P_kg\right\|_{L^\infty}
\lesssim t^{-\frac{1}{2}}2^{\frac{1-\alpha}{2}k}\|\widehat{g}\|_{L^\infty}
+t^{-\frac{3}{4}}2^{-\frac{1+3\alpha}{4}k}(\|\widehat{g}\|_{L^2}+2^k\|\partial\widehat{g}\|_{L^2}),
	\end{align}
and
	\begin{align}\label{6.5}
	\left\|e^{t|D|^{\alpha}\partial_x}P_kg\right\|_{L^\infty}
	\lesssim t^{-\frac{1}{2}}2^{\frac{1-\alpha}{2}k}\|g\|_{L^1},
	\end{align}
for \(\alpha\in (-1,1)\setminus\{0\}\).
\end{lemma}

We next show the following decay estimates for the solution:
\begin{lemma}
Let \(\alpha\in (-1,0)\) and \(u\) be the solution to \eqref{eq:main}. Assume that
\begin{align*}
	t^{-p_0}\|u\|_{H^{N_0}}+(1+t)^{-p_0}\|f\|_{H^{1,1}}+\|f\|_Z\leq 1,
	\end{align*}
for any \(t\geq 1\), then we have
	\begin{align}\label{7}
	\|u\|_{L^\infty}+\|\partial_xu\|_{L^\infty}\lesssim t^{-1/2},
	\end{align}
for any \(t\geq 1\).
\end{lemma}
\begin{proof}  {\bf{Case  \(\alpha\in (-1,-\frac{1}{3}]\).}}
Observe  \(\frac{1-\alpha}{2}\in[\frac{2}{3},1)\) and \(-\frac{1+3\alpha}{4}\in[0,\frac{1}{2})\).  To prove  that \(\|u\|_{L^\infty}\) satisfy \eqref{7}, we first show
\begin{equation}\label{8}
	\begin{aligned}
	\quad \sup_{k\in\mathbb{Z}}(2^{2 k}\left\|P_ku\right\|_{L^\infty})
	\lesssim t^{-\frac{1}{2}}.
	\end{aligned}
	\end{equation}
 In the frequency regime \(2^k\geq t^{(1-4p_0)/10}\),  we use \eqref{6.5} and
 \eqref{75} to deduce that
	\begin{equation}\label{8.5}
	\begin{aligned}
	&\quad 2^{2 k}\left\|P_ku\right\|_{L^\infty}\lesssim t^{-\frac{1}{2}}2^{2 k}2^{\frac{1-\alpha}{2}k}\|P_k^\prime f\|_{L^1}\\
	&\lesssim t^{-\frac{1}{2}}2^{2 k}2^{\frac{1-\alpha}{2}k}2^{-\frac{k}{2}}\|\widehat{P_k^\prime f}\|_{L^2}^{1/2}(\|\widehat{P_k^\prime f}\|_{L^2}+2^k\|\partial\widehat{P_k^\prime f}\|_{L^2})^{1/2}\\
	&\lesssim t^{-\frac{1}{2}}2^{2 k}2^{\frac{1-\alpha}{2}k}2^{-\frac{k}{2}}2^{-\frac{N_0k}{2}}\|P_k^\prime f\|_{H^{N_0}}^{1/2}(\|\widehat{P_k^\prime f}\|_{L^2}+2^k\|\partial\widehat{P_k^\prime f}\|_{L^2})^{1/2}\\
	&\lesssim t^{-\frac{1}{2}}t^{\frac{1-4p_0}{10}(2+\frac{1-\alpha}{2}-\frac{1}{2}-\frac{N_0}{2})} t^{p_0},
	\end{aligned}
	\end{equation}
which gives a stronger bound than what we need.
We then consider the frequency regime \(2^k\leq t^{(1-4p_0)/10}\).
It follows from \eqref{6} that
 	\begin{equation}\label{9}
	\begin{aligned}
	\left\|P_ku\right\|_{L^\infty}
\lesssim t^{-\frac{1}{2}}2^{\frac{1-\alpha}{2}k}\big\|\widehat{P_k^\prime f}\big\|_{L^\infty}
+t^{-\frac{3}{4}}2^{-\frac{1+3\alpha}{4}k}\big(\big\|\widehat{P_k^\prime f}\big\|_{L^2}+2^k\big\|\partial\widehat{P_k^\prime f}\big\|_{L^2}\big).
	\end{aligned}
	\end{equation}
Now the desired bound \eqref{8} is a consequence of  \eqref{9} and of  the following facts
	\begin{equation*}
	\begin{aligned}
	\frac{2^{(\frac{1-\alpha}{2}+2)k}}{1+2^{10k}}\lesssim 1,\quad 2^{(-\frac{1+3\alpha}{4}+2)k} \lesssim t^{\frac{1}{4}-p_0},
	\end{aligned}
	\end{equation*}
	and
	\begin{equation*}
	\begin{aligned}
	(1+2^{10k})\big\|\widehat{P_k^\prime f}\big\|_{L^\infty}\lesssim 1,\quad \|\widehat{P_k^\prime f}\|_{L^2}+2^k\|\partial\widehat{P_k^\prime f}\|_{L^2}\lesssim t^{p_0}.
	\end{aligned}
	\end{equation*}

	We are ready to show \(\|u\|_{L^\infty}\) satisfy \eqref{7}.
       It first follows from \eqref{8} that
	 \begin{equation*}
	 \begin{aligned}
	  \sum_{0\leq k\in\mathbb{Z}}\|P_ku\|_{L^\infty}= \sum_{0\leq k\in\mathbb{Z}}2^{-2 k}(2^{2 k}\left\|P_k u\right\|_{L^\infty})
	 \lesssim t^{-\frac{1}{2}} \sum_{0\leq k\in\mathbb{Z}}2^{-2 k}\lesssim t^{-\frac{1}{2}} .
	 \end{aligned}
	 \end{equation*}
	Thus it  remains to estimate \( \sum_{0> k\in\mathbb{Z}}\|P_ku\|_{L^\infty}\).  When \(2^k\leq t^{-1}\), the desired bound follows from the following inequality after summation in \(k\)
	 \begin{equation*}
	 \begin{aligned}
	 \|P_k u\|_{L^\infty}
	 \lesssim 2^{k/2}\|u\|_{L^2}.
	 \end{aligned}
	 \end{equation*}
where we have used Bernstein inequality.
For  \(1\geq 2^k\geq t^{-1}\),  it follows from \eqref{9} that
	 \begin{equation}
	 \begin{aligned}
	 \sum_{1\geq 2^k\geq t^{-1}}\|P_k u\|_{L^\infty}
	 &\lesssim t^{-\frac{1}{2}}\sum_{1\geq 2^k\geq t^{-1}}\frac{2^{\frac{1-\alpha}{2}k}}{1+2^{10k}}
	 \big\|(1+2^{10k})\widehat{P_k^\prime f}\big\|_{L^\infty}\\
	 &\quad+t^{-\frac{3}{4}}t^{p_0}\sum_{1\geq 2^k\geq t^{-1}}2^{-\frac{1+3\alpha}{4}k}\\
	 &\lesssim t^{-\frac{1}{2}}
	 +t^{-\frac{3}{4}}t^{p_0}\log t \lesssim t^{-\frac{1}{2}}.
	 \end{aligned}
	 \end{equation}

We finally prove the bound for \(\|\partial_xu\|_{L^\infty}\).  One has
	\begin{equation*}
	\begin{aligned}
	&\quad\|\partial_xu\|_{L^\infty}\leq \sum_{k\in\mathbb{Z}}\|P_k\partial_xu\|_{L^\infty}\lesssim\sum_{k\in\mathbb{Z}}2^k\|P_ku\|_{L^\infty}\\
	&=\sum_{0\geq k\in\mathbb{Z}}2^k\|P_ku\|_{L^\infty}+\sum_{0\leq k\in \mathbb{Z}}2^{-k}(2^{2 k}\|P_ku\|_{L^\infty})\\
	&\lesssim \|u\|_{L^\infty}\sum_{0\geq k\in\mathbb{Z}}2^k+t^{-\frac{1}{2}}\sum_{0\leq k\in \mathbb{Z}}2^{-k}
\lesssim t^{-\frac{1}{2}}.
	\end{aligned}
	\end{equation*}

\noindent{\bf{Case  \(\alpha\in (-\frac{1}{3},0)\).}}
Observe that   \(\frac{1-\alpha}{2}\in(\frac{1}{2},\frac{2}{3})\) and \(-\frac{1+3\alpha}{4}\in(-\frac{1}{4},0)\). We now split the frequencies into \(2^k\geq t^{(1-4p_0)/8}\)
and \(2^k\leq t^{(1-4p_0)/8}\). In a similar fashion as before, one still has
\begin{equation*}
\begin{aligned}
\quad \sup_{k\in\mathbb{Z}}(2^{2 k}\left\|P_ku\right\|_{L^\infty})
\lesssim t^{-\frac{1}{2}},
\end{aligned}
\end{equation*}
and furthermore shows
\begin{equation*}
\begin{aligned}
\sum_{0\leq k\in\mathbb{Z}}\|P_ku\|_{L^\infty}\lesssim t^{-\frac{1}{2}} .
\end{aligned}
\end{equation*}
The desired bound for \( \sum_{2^k\leq t^{-1}}\|P_k u\|_{L^\infty}\) may be obtained as before. We now consider the frequency regime  \(1\geq 2^k\geq t^{-1}\).  It follows from \eqref{9} that
  	\begin{equation*}
  	\begin{aligned}
  	\sum_{1\geq 2^k\geq t^{-1}}\left\|P_ku\right\|_{L^\infty}
  	&\lesssim t^{-\frac{1}{2}}\sum_{1\geq 2^k\geq t^{-1}}\frac{2^{\frac{1-\alpha}{2}k}}{1+2^{10k}}
  	\big\|(1+2^{10k})\widehat{P_k^\prime f}\big\|_{L^\infty}\\
  	&\quad+t^{-\frac{3}{4}}t^{p_0}\sum_{1\geq 2^k\geq t^{-1}}2^{-\frac{1+3\alpha}{4}k}\\
  	&\lesssim t^{-\frac{1}{2}}+t^{-\frac{3}{4}}t^{\frac{1+3\alpha}{4}}t^{p_0}\lesssim t^{-\frac{1}{2}},
  	\end{aligned}
  	\end{equation*}
where we have used the assumption \(p_0\leq-\frac{\alpha}{100}\) in the last inequality.

The desired bound for \(\|\partial_xu\|_{L^\infty}\) can be estimated as before.

\end{proof}

\section{Estimates on \(\|u\|_{H^{N_0}}\) and \(\|f\|_{H^{1,1}}\)}
In this section we will prove uniform bounds for the energy part in \eqref{4}.  More precisely :
\begin{theorem}\label{pr:1} Let \(u\) be a solution of \eqref{eq:main}-\eqref{eq:initial} satisfying the a priori bounds \eqref{4}. Then the following estimates hold true:
	\begin{align}\label{10}
	\|u(t,\cdot)\|_{H^{N_0}}\leq C\varepsilon_0\langle t \rangle^{C\varepsilon_1^2},
	\end{align}
and
	\begin{align}\label{11}
	\|f(t,\cdot)\|_{H^{1,1}}\leq C(\varepsilon_0+ \varepsilon_1^3)\langle t \rangle^{C\varepsilon_1^2}.
	\end{align}
\end{theorem}

\begin{proof} Let
	\begin{align*}
	\mathcal{L}=\partial_t-|D|^\alpha \partial_x,\  \mathcal{J}=x+(\alpha+1)t|D|^\alpha,
\  \Lambda=\partial_x^{-1}\big((\alpha+1)t\partial_t+x\partial_x+1\big).
	\end{align*}
Since \([\mathcal{L},\partial_x]=0\), the solution \(u\) of  \eqref{eq:main}-\eqref{eq:initial} satisfies
\begin{align*}
\mathcal{L}\partial_x^ku=-\partial_x^k(u^2\partial_xu),\quad k=1,2,\dots,N_0.
\end{align*}
This gives the energy identity
\begin{align*}
\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d} t}\int|\partial_x^ku|^2\,\mathrm{d} x
=-\int\big(\partial_x^k(u^2\partial_xu)-u^2\partial_x^{k+1}u\big)\partial_x^ku\,\mathrm{d} x-\int u^2\partial_x^{k+1}u\partial_x^ku\,\mathrm{d} x.
\end{align*}
Notice that
\begin{align*}
\|\partial_x^k(u^2\partial_xu)-u^2\partial_x^{k+1}u\|_{L^2}
&\lesssim \|\partial_x(u^2)\|_{L^\infty}\|\partial_x^ku\|_{L^2}+\|\partial_xu\|_{L^\infty}\|\partial_x^k(u^2)\|_{L^2}\\
&\lesssim \|u\|_{L^\infty}\|\partial_xu\|_{L^\infty}\|\partial_x^ku\|_{L^2},
\end{align*}
and
\begin{align*}
\int u^2\partial_x^{k+1}u\partial_x^ku\,\mathrm{d} x
=-\frac{1}{2}\int \partial_x(u^2)|\partial_x^ku|^2\,\mathrm{d} x
\lesssim \|u\|_{L^\infty}\|\partial_xu\|_{L^\infty}\|\partial_x^ku\|_{L^2}^2.
\end{align*}
Hence
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d} t}\|\partial_x^ku\|_{L^2}^2
\lesssim \|u\|_{L^\infty}\|\partial_xu\|_{L^\infty}\|\partial_x^ku\|_{L^2}^2
\lesssim \varepsilon_1^2\langle t \rangle^{-1}\|\partial_x^ku\|_{L^2}^2,
\end{align*}
where we have used \eqref{7}, which entails \eqref{10}.

A small calculation shows
\begin{align*}
\mathcal{L}\Lambda u=u^2\partial_x\Lambda u+(\alpha-2)u^3/3.
\end{align*}
It follows that
\begin{equation*}
\begin{aligned}
\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d} t}\|\Lambda u\|_{L^2}^2&=\int u^2\partial_x\Lambda u \Lambda u\,\mathrm{d} x+\frac{\alpha-2}{3}\int u^3\Lambda u\,\mathrm{d} x\\
&\lesssim \|u\|_{L^\infty}\|\partial_x u\|_{L^\infty}\|\Lambda u\|_{L^2}^2+\|u\|_{L^\infty}^2\|u\|_{L^2}\|\Lambda u\|_{L^2},
\end{aligned}
\end{equation*}
which combined with  \eqref{7}  lead to
\begin{align}\label{12}
\|\Lambda u\|_{L^2}\lesssim  \varepsilon_0 \langle t \rangle^{C\varepsilon_1^2}.
\end{align}
Notice that
\begin{align}\label{13}
\mathcal{J}u=\Lambda u+(\alpha+1)t u^3/3,
\end{align}
then we employ \eqref{7}, \eqref{12} and \eqref{13} to deduce
\begin{equation}\label{13.5}
\begin{aligned}
\|x f\|_{L^2}=\|\mathcal{J}u\|_{L^2}\lesssim  \varepsilon_0 \langle t \rangle^{C\varepsilon_1^2}+t\|u\|_{L^\infty}^2\|u\|_{L^2}\lesssim (\varepsilon_0+ \varepsilon_1^3) \langle t \rangle^{C\varepsilon_1^2}.
\end{aligned}
\end{equation}

Using \([\mathcal{L},\partial_x]=0\) and \([\Lambda,\partial_x]=-\mathrm{id}\), one calculates that
\begin{align*}
\mathcal{L}\Lambda\partial_x u
=u^2\partial_x\Lambda \partial_x u+2u\partial_x u\Lambda \partial_x u+(\alpha+2) u^2\partial_x u.
\end{align*}
This results into the estimate
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d} t}\|\Lambda \partial_xu\|_{L^2}^2
\lesssim \|u\|_{L^\infty}\|\partial_xu\|_{L^\infty}\|\Lambda \partial_xu\|_{L^2}^2+\|u\|_{L^\infty}\|\partial_xu\|_{L^\infty}\|u\|_{L^2}\|\Lambda \partial_xu\|_{L^2},
\end{align*}
which yields
\begin{align}\label{14}
\|\Lambda \partial_x u\|_{L^2}\lesssim \varepsilon_0 \langle t \rangle^{C\varepsilon_1^2}.
\end{align}
We use \([\mathcal{J},\partial_x]=-\mathrm{id}\) to calculate
\begin{align}\label{15}
	\mathcal{J}\partial_x u
	=-\Lambda u-(\alpha+1)t u^3/3+(\Lambda\partial_x+1) u+(\alpha+1)t u^2\partial_x u.
\end{align}
Therefore the estimates \eqref{7}, \eqref{10}, \eqref{12} and \eqref{14} yield
\begin{align}\label{16}
\|x\partial_x f\|_{L^2}=\|\mathcal{J} \partial_x u\|_{L^2}\lesssim (\varepsilon_0+ \varepsilon_1^3) \langle t \rangle^{C\varepsilon_1^2}.
\end{align}

The estimate \eqref{11} is a consequence of \eqref{13.5} and \eqref{16}.

\end{proof}

\section{Estimates on \(\|f\|_Z\)}
Let
\begin{equation*}
\begin{aligned}
\Phi(\xi,\eta,\sigma)=|\xi|^{\alpha}\xi-|\xi-\eta-\sigma|^{\alpha}(\xi-\eta-\sigma)-|\eta|^{\alpha}\eta-|\sigma|^{\alpha}\sigma.
\end{aligned}
\end{equation*}
Taking the Fourier transform of \eqref{eq:main} gives
\begin{equation}\label{17}
\begin{aligned}
\partial_t\widehat{f}(\xi,t)&=-\mathrm{i}(2\pi)^{-1}\xi\int_{\mathbb{R}^2}e^{\mathrm{i}t\Phi(\xi,\eta,\sigma)}
\widehat{f}(\xi-\eta-\sigma,t)\widehat{f}(\eta,t)\widehat{f}(\sigma,t)\,\mathrm{d}\eta \mathrm{d} \sigma\\
&=\colon-\mathrm{i}(2\pi)^{-1}I(\xi,t).
\end{aligned}
\end{equation}
Set
\begin{align*}
	H(\xi,t):=\frac{3\xi|\xi|^{1-\alpha}}{\alpha(\alpha+1)}\int_1^t|\widehat{f}(s,\xi)|^2\frac{\mathrm{d} s}{s},
\end{align*}
and
\begin{align*}
g(\xi,t):=e^{\mathrm{i}H(\xi,t)}\widehat{f}(\xi,t).
\end{align*}
Then \(g\) satisfies the following evolutionary equation
\begin{align}\label{18}
\partial_t g(\xi,t)=-\mathrm{i}(2\pi)^{-1}e^{\mathrm{i}H(\xi,t)}\big( I(\xi,t)-3\tilde{c}t^{-1}\xi|\xi|^{1-\alpha}|\widehat{f}(\xi,t)|^2\widehat{f}(\xi,t)\big),
\end{align}
where \(\tilde{c}:=2\pi/[\alpha(\alpha+1)]\).

This section is  aimed to show the following theorem:
\begin{theorem}\label{pr:2}
It holds that
\begin{align}\label{19}
t_1^{p_0}\left\|(1+|\xi|)^{10}\big(g(\xi,t_2)-g(\xi,t_1)\big)\right\|_{L^\infty_\xi}\lesssim\epsilon_0,
\end{align}
for any \(t_1\leq t_2\in[1,T]\).
\end{theorem}
\noindent  The \(Z\)-norm part of \eqref{5} is an immediate consequence of the estimate \eqref{19} which also entails \eqref{3}.

\subsection{Proof of Theorem \ref{pr:2}}
We decompose in frequencies
\begin{equation}\label{20}
\begin{aligned}
I(\xi,t)=\sum_{k_1,k_2,k_3\in \mathbb{Z}}I_{k_1,k_2,k_3}(\xi,t),
\end{aligned}
\end{equation}
in which
\begin{equation}\label{21}
\begin{aligned}
I_{k_1,k_2,k_3}(\xi,t):=\xi\int_{\mathbb{R}^2}e^{\mathrm{i}t\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}}(\xi-\eta-\sigma,t)\widehat{f_{k_2}}(\eta,t)\widehat{f_{k_3}}(\sigma,t)\,\mathrm{d}\eta \mathrm{d} \sigma.
\end{aligned}
\end{equation}

For \eqref{19}, it suffices to show that if \(t_1\leq t_2\in [2^m-1,2^{m+1}]\cap[1,T]\), for some \(m\in \{1,2,\dots\}\), then
\begin{align}\label{22}
\left\|(1+|\xi|)^{10}\big(g(\xi,t_2)-g(\xi,t_1)\big)\right\|_{L^\infty_\xi}\lesssim\epsilon_02^{-p_0m}.
\end{align}

Let \(k\in\mathbb{Z}\) and \(|\xi|\in[2^k,2^{k+1}]\) and \(s\in[2^m-1,2^{m+1}]\cap[1,T]\). Using the interpolation \eqref{75}, similar to \eqref{8.5}, it is easy to see that
\begin{equation}\label{24}
	\begin{aligned}
		\big|(1+|\xi|^{10})\widehat{f_k}(\xi)\big|\lesssim \epsilon_02^{-p_0m},
	\end{aligned}
\end{equation}
for  \(k\in [p_0m,\infty)\cap \mathbb{Z}\), which entails \eqref{22} in this frequency regime.

By the assumptions \eqref{4}, for \(s\in[2^m-1,2^{m+1}]\cap[1,T]\) and any \(l\in\mathbb{Z}\), we have
\begin{equation}\label{23}
\begin{aligned}
&\|\widehat{f_l}(s)\|_{L^2}\lesssim \epsilon_12^{p_0m}2^{-N_0l_+},\\
&\|\partial\widehat{f_l}(s)\|_{L^2}\lesssim \epsilon_12^{p_0m}2^{-l},\\
&\|\widehat{f_l}(s)\|_{L^\infty}\lesssim \epsilon_12^{-10l_+},\\
&\|e^{\mp\mathrm{i}s\Lambda}f_l(s)\|_{L^\infty}\lesssim \epsilon_12^{-m/2},
\end{aligned}
\end{equation}
where \(l_+=\max(l,0)\).

It remains to consider \eqref{22} for \(|\xi|\in[2^k,2^{k+1}]\) with \(k\in (-\infty,p_0m]\).  By the equation \eqref{18} and the decomposition \eqref{20}-\eqref{21}, it suffices to show that
\begin{equation}\label{25}
\begin{aligned}
&\sum_{k_1,k_2,k_3\in \mathbb{Z}}\bigg|\int_{t_1}^{t_2}e^{\mathrm{i}H(\xi,s)}\big( I_{k_1,k_2,k_3}(\xi,s)-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(-\xi,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad-\tilde{c}
s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(-\xi,s)\widehat{f_{k_3}}(\xi,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(-\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(\xi,s)\big)\,\mathrm{d} s\bigg|\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\lesssim
\epsilon_1^32^{-p_0m}2^{-10k_+},
\end{aligned}
\end{equation}
for \(t_1\leq t_2\in [2^m-1,2^{m+1}]\cap [1,T]\).

Using \eqref{23}, it is straightforward to show
\begin{equation}\label{26}
\begin{aligned}
|I_{k_1,k_2,k_3}(\xi,s)|&\lesssim \epsilon_1^32^{3p_0m}2^k2^{\min(k_1,k_2,k_3)/2}2^{-N_0({k_1}_++{k_2}_++{k_3}_+)},
\end{aligned}
\end{equation}
and
\begin{equation}\label{27}
\begin{aligned}
|I_{k_1,k_2,k_3}(\xi,s)|\lesssim \epsilon_1^32^k2^{\min(k_1,k_2,k_3)}2^{\text{med}(k_1,k_2,k_3)}2^{-10\max({k_1}_+,{k_2}_+,{k_3}_+)},
\end{aligned}
\end{equation}
and
\begin{equation}\label{28}
\begin{aligned}
&\quad s^{-1}|\xi|^{2-\alpha}\big(\big|\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(-\xi,s)\big|+\big|\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(-\xi,s)\widehat{f_{k_3}}(\xi,s)\big|\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\big|\widehat{f_{k_1}}(-\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(\xi,s)\big|\big)\\
&\lesssim \epsilon_1^32^{-m}2^{(2-\alpha)k}2^{-30k_+}{\bf{1}}_{[0,4]}\big(\max(|k_1-k|,|k_2-k|,|k_3-k|)\big).
\end{aligned}
\end{equation}

With the bounds \eqref{26}-\eqref{28} at hand, one easily verifies \eqref{25} if one of the following conditions holds
\begin{equation*}
\begin{aligned}
&\min(k_1,k_2,k_3)\leq -4m,\\
&\max(k_1,k_2,k_3)\geq p_0m/10,\\
&\min(k_1,k_2,k_3)+\mathrm{med}(k_1,k_2,k_3)\leq -(1+10p_0)m.
\end{aligned}
\end{equation*}

Therefore, to complete the proof \eqref{25},  we are  reduced to show
\begin{theorem}\label{th:reduce}
Assume that \(k,k_1,k_2,k_3\in\mathbb{Z}\), \(m\in\mathbb{Z}\cap[100,\infty)\), \(|\xi|\in[2^k,2^{k+1}]\), and \(t_1\leq t_2\in [2^m-1,2^{m+1}]\cap [1,T]\). If
\begin{equation}\label{29}
\begin{aligned}
&k\in (-\infty,p_0m],\\
&k_1,k_2,k_3\in[-4m,p_0m/10],\\
&\min(k_1,k_2,k_3)+\mathrm{med}(k_1,k_2,k_3)\geq -(1+10p_0)m,
\end{aligned}
\end{equation}
then
\begin{equation}\label{30}
\begin{aligned}
&\bigg|\int_{t_1}^{t_2}e^{\mathrm{i}H(\xi,s)}\big( I_{k_1,k_2,k_3}(\xi,s)-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(-\xi,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad-\tilde{c}
s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(-\xi,s)\widehat{f_{k_3}}(\xi,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(-\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(\xi,s)\big)\,\mathrm{d} s\bigg|\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\lesssim\epsilon_1^32^{-2p_0m}2^{-10k_+}.
\end{aligned}
\end{equation}
\end{theorem}

\subsection{Proof of Theorem \ref{th:reduce}}
We divide the proof into several propositions.
\begin{proposition}\label{le:1}
 Let \(k,k_1,k_2,k_3\in\mathbb{Z}\) and \(k_1,k_2,k_3\in[k-20,k+20]\), then the bound \eqref{30} holds.
\end{proposition}
 It is easy to check \eqref{30} if \(k\leq -m/2\). Hence we will assume \(k\geq -m/2\) in the following.
Without loss of generality, we may assume that \(\xi>0\) and \(\xi\in[2^k,2^{k+1}]\). We split the integral \(I_{k_1,k_2,k_3}\) as follows:
\begin{align*}
I_{k_1,k_2,k_3}(\xi,s)=\sum_{\iota_1,\iota_2,\iota_3\in\{+,-\}}I_{k_1,k_2,k_3}^{\iota_1,\iota_2,\iota_3}(\xi,s),
\end{align*}
with
\begin{align*}
I_{k_1,k_2,k_3}^{\iota_1,\iota_2,\iota_3}(\xi,s)=\xi\int_{\mathbb{R}^2}e^{it\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^{\iota_2}}(\eta,s)\widehat{f_{k_3}^{\iota_3}}(\sigma,s)\,\mathrm{d}\eta\mathrm{d}\sigma,
\end{align*}
where \(\widehat{f_{l}^{\iota}}(\mu):=\widehat{f_{l}}(\mu)1_{\iota}(\mu),1_+:=1_{[0,\infty)},1_-:=1_{(-\infty,0]}\). We first observe that \(I_{k_1,k_2,k_3}^{-,-,-}(\xi,s)=0\), so that  to finish the proof of \eqref{30} under the assumption of Proposition \ref{le:1}, we are then left to show the following two lemmas:
\begin{lemma}\label{le:2} It holds that
	\begin{equation}\label{31}
	\begin{aligned}
	&\quad\bigg|I_{k_1,k_2,k_3}^{+,+,-}(\xi,s)-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(-\xi,s)\bigg|\\
	&\quad+\bigg|I_{k_1,k_2,k_3}^{+,-,+}(\xi,s)-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(\xi,s)\widehat{f_{k_2}}(-\xi,s)\widehat{f_{k_3}}(\xi,s)\bigg|\\
	&\quad+\bigg|I_{k_1,k_2,k_3}^{-,+,+}(\xi,s)-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}}(-\xi,s)\widehat{f_{k_2}}(\xi,s)\widehat{f_{k_3}}(\xi,s)\bigg|\\
	&\lesssim \varepsilon_1^32^{-m}2^{-3p_0m}2^{-10k_+}.
	\end{aligned}
	\end{equation}
\end{lemma}
\begin{lemma}\label{le:3}
	We have
	\begin{align}\label{32}
	\bigg|\int_{t_1}^{t_2}e^{\mathrm{i}H(s,\xi)}I_{k_1,k_2,k_3}^{\iota_1,\iota_2,\iota_3}(\xi,s)\,\mathrm{d} s\bigg|
	\lesssim \varepsilon_1^32^{-2p_0m}2^{-10k_+},
	\end{align}
	for \((\iota_1,\iota_2,\iota_3)\in \{(+,+,+),(+,-,-),(-,+,-),(-,-,+)\}\).
\end{lemma}

\bigskip
We start by showing \eqref{31}.
\begin{proof}[Proof of Lemma \ref{le:2}] We only prove that the third term in LHS of \eqref{31} is bounded by \(\varepsilon_1^32^{-3p_1m}2^{-10k_+}\) and the other two terms may be handled similarly.

Let \(\bar{l}\) be the smallest integer with the property that
\begin{align*}
2^{\bar{l}}\geq 2^{(1-\alpha)k/2}2^{-49m/100}.
\end{align*}
Since \(k\geq -m/2\), one has \(\bar{l}\leq k-20\).  We may decompose
\begin{align*}
I_{k_1,k_2,k_3}^{-,+,+}(\xi,s)=\sum_{l_1,l_2=\bar{l}}^{k+20}J_{l_1,l_2}(\xi,s),
\end{align*}
with
\begin{equation*}
\begin{aligned}
J_{l_1,l_2}(\xi,s)&=\xi\int_{\mathbb{R}^2}e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s)\\
&\quad\times\varphi_{l_1}^{(\bar{l})}(\xi-\eta)\varphi_{l_2}^{(\bar{l})}(\xi-\sigma)\,\mathrm{d}\eta \mathrm{d} \sigma,
\end{aligned}
\end{equation*}
for any \(l_1,l_2\geq \bar{l}\), and where
\begin{equation*}
	\begin{aligned}
		\varphi_{k}^{(l)}(x):=\varphi(x/2^k),\  \text{if}\  k=l,
	\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\varphi_{k}^{(l)}(x):=\varphi(x/2^k)-\varphi(x/2^{k-1}),\  \text{if}\  k\geq l+1.
\end{aligned}
\end{equation*}

\bigskip
{\bf{Step 1: \(l_2\geq \max(l_1,\bar{l}+1)\) or \(l_1\geq \max(l_2,\bar{l}+1)\).}} We only consider the case \(l_2\geq \max(l_1,\bar{l}+1)\), a similar argument may apply to the other case.

We will show
\begin{align*}
|J_{l_1,l_2}(\xi,s)|\lesssim \epsilon_1^32^{-m}2^{-3p_0m}2^{-10k_{+}}.
\end{align*}
On the support of the integral, one has \(|\xi-\eta-\sigma|\approx|\eta|\approx 2^k\) and \(|\xi-\sigma|\approx 2^{l_2}\), and then one finds
\begin{equation}\label{33}
\begin{aligned}
|\partial_\eta\Phi(\xi,\eta,\sigma)|
=(\alpha+1)\big||\xi-\eta-\sigma|^{\alpha}-|\eta|^{\alpha}\big|
\gtrsim 2^{l_2}2^{(\alpha-1)k}.
\end{aligned}
\end{equation}
Using the identity
\begin{equation*}
	\begin{aligned}
	e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}=\frac{\partial_\eta e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}}{\mathrm{i}s\partial_\eta\Phi(\xi,\eta,\sigma)},
	\end{aligned}
	\end{equation*}
we use integration by parts in \(\eta\) to obtain
\begin{align*}
|J_{l_1,l_2}(\xi,s)|\leq |J_{l_1,l_2,1}(\xi,s)|+|F_{l_1,l_2,1}(\xi,s)|+|G_{l_1,l_2,1}(\xi,s)|,
\end{align*}
where
\begin{equation*}
\begin{aligned}
J_{l_1,l_2,1}(\xi,s)&=\xi\int_{\mathbb{R}^2}\partial_\eta m_1(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s)\, \mathrm{d} \eta\mathrm{d} \sigma,\\
F_{l_1,l_1,1}(\xi,s)&=\xi\int_{\mathbb{R}^2}m_1(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\partial_\eta\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s) \, \mathrm{d} \eta\mathrm{d} \sigma,\\
G_{l_1,l_1,1}(\xi,s)&=\xi\int_{\mathbb{R}^2}m_1(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\partial_\eta\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s) \, \mathrm{d} \eta\mathrm{d} \sigma,
\end{aligned}
\end{equation*}
with
\begin{align}\label{34}
m_1(\eta,\sigma):=\frac{\varphi_{l_1}^{(\bar{l})}(\xi-\eta)\varphi_{l_2}(\xi-\sigma)}{s\partial_\eta\Phi(\xi,\eta,\sigma)}
\varphi_{k_1}^\prime(\xi-\eta-\sigma)\varphi_{k_2}^\prime(\eta)\varphi_{k_3}^\prime(\sigma).
\end{align}

Following \eqref{33} and \eqref{34}, a straightforward calculation shows that
\begin{equation*}
\begin{aligned}
|\partial_\eta^a\partial_\sigma^bm_1(\eta,\sigma)|&\lesssim (2^{-m}2^{-l_2}2^{(1-\alpha)k})(2^{-al_1}2^{-bl_2})\\
&\quad \times{\bf{1}}_{[0,2^{l_1+4}]}(|\xi-\eta|){\bf{1}}_{[2^{l_2-4},2^{l_2+4}]}(|\xi-\sigma|),
\end{aligned}
\end{equation*}
for \(a,b\in[0,20]\cap\mathbb{Z}\). Hence
\begin{equation*}
\begin{aligned}
\|\mathcal{F}^{-1}(m_1)\|_{L^1}\lesssim 2^{-m}2^{-l_2}2^{(1-\alpha)k}.
\end{aligned}
\end{equation*}

We first estimate the term \(F_{l_1,l_1,1}\).
Fix \(\xi\) and \(s\), let
\begin{equation*}
\begin{aligned}
&\widehat{f}(\theta):=e^{-\mathrm{i}s|\xi-\theta|^\alpha(\xi-\theta)}\partial_\eta\widehat{f_{k_1}^-}(\xi-\theta,s),\\
&\widehat{g}(\eta):=e^{-\mathrm{i}s|\eta|^\alpha\eta}\widehat{f_{k_2}^+}(\eta,s),\\
&\widehat{h}(\sigma):=e^{-\mathrm{i}s|\sigma|^\alpha\sigma}\widehat{f_{k_3}^+}(\sigma,s).
\end{aligned}
\end{equation*}
which  in light of \eqref{23} gives
\begin{equation*}
\begin{aligned}
&\|f\|_{L^2}\lesssim \epsilon_12^{-k}2^{p_0m},\\
&\|g\|_{L^\infty}\lesssim \epsilon_12^{-m/2},\\
&\|h\|_{L^2}\lesssim \epsilon_12^{p_0m}2^{-N_0k_+}.
\end{aligned}
\end{equation*}
It then follows from Lemma \ref{symbol estimate} that
\begin{equation*}
\begin{aligned}
|F_{l_1,l_1,1}(\xi,s)|&\lesssim 2^k\|\mathcal{F}^{-1}(m_1)\|_{L^1}\|f\|_{L^2}\|g\|_{L^\infty}\|h\|_{L^2}\\
&\lesssim \epsilon_1^32^{-3m/2+2p_0m}2^{-l_2}2^{(1-\alpha)k}2^{-N_0k_+}\\
&\lesssim \epsilon_1^32^{-3m/2+2p_0m+49m/100}2^{(1-\alpha)k/2}2^{-N_0k_+}\\
&\lesssim \epsilon_1^32^{-m}2^{-m/200}2^{-10k_+},
\end{aligned}
\end{equation*}
which is stronger than what we need. A similar argument yields
\begin{align*}
|G_{l_1,l_1,1}(\xi,s)|\lesssim \epsilon_1^32^{-m}2^{-m/200}2^{-10k_+}.
\end{align*}

To estimate the term \(J_{l_1,l_1,1}\), we integrate by parts in \(\eta\) again to deduce
\begin{align*}
|J_{l_1,l_2,1}(\xi,s)|\leq |J_{l_1,l_2,2}(\xi,s)|+|F_{l_1,l_2,2}(\xi,s)|+|G_{l_1,l_2,2}(\xi,s)|,
\end{align*}
in which
\begin{equation*}
\begin{aligned}
J_{l_1,l_2,2}(\xi,s)&=\xi\int_{\mathbb{R}^2}\partial_\eta m_2(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s)\, \mathrm{d} \eta\mathrm{d} \sigma,\\
F_{l_1,l_2,2}(\xi,s)&=\xi\int_{\mathbb{R}^2}m_2(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\partial_\eta\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s) \, \mathrm{d} \eta\mathrm{d} \sigma,\\
G_{l_1,l_2,2}(\xi,s)&=\xi\int_{\mathbb{R}^2}m_2(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\partial_\eta\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s) \, \mathrm{d} \eta\mathrm{d} \sigma,
\end{aligned}
\end{equation*}
with
\begin{align}\label{35}
m_2(\eta,\sigma):=\frac{\partial_\eta m_1(\eta,\sigma)}{s\partial_\eta\Phi(\xi,\eta,\sigma)}.
\end{align}

It follows from \eqref{33} and \eqref{35} that \(m_2\) satisfies the following stronger estimate
\begin{equation*}
\begin{aligned}
|\partial_\eta^a\partial_\sigma^bm_2(\eta,\sigma)|&\lesssim (2^{-m}2^{-l_1-l_2}2^{(1-\alpha)k})(2^{-m}2^{-l_2}2^{(1-\alpha)k})(2^{-al_1}2^{-bl_2})\\
&\quad\times{\bf{1}}_{[0,2^{l_1+4}]}(|\xi-\eta|){\bf{1}}_{[2^{l_2-4},2^{l_2+4}]}(|\xi-\sigma|),
\end{aligned}
\end{equation*}
for \(a,b\in[0,19]\cap\mathbb{Z}\).
In a similar fashion as \(F_{l_1,l_1,1}(\xi,s)\) and \(G_{l_1,l_1,1}(\xi,s)\), we employ Lemma \ref{symbol estimate} to obtain
\begin{equation*}
\begin{aligned}
|F_{l_1,l_1,2}(\xi,s)|+|G_{l_1,l_1,2}(\xi,s)|
\lesssim \epsilon_1^32^{-m}2^{-m/200}2^{-10k_+}.
\end{aligned}
\end{equation*}
We finally estimate the left term \(J_{l_1,l_1,2}(\xi,s)\) as follows:
 \begin{equation*}
 \begin{aligned}
 |J_{l_1,l_1,2}(\xi,s)|&\lesssim  2^k2^{l_1}2^{l_2}|\partial_\eta m_2|\big\|\widehat{f_{k_1}^-}\big\|_{L^\infty}\big\|\widehat{f_{k_2}^+}\big\|_{L^\infty}\big\|\widehat{f_{k_3}^+}\big\|_{L^\infty}\\
 &\lesssim \epsilon^32^{-m}2^{-l_1-l_2}2^{(1-\alpha)k}2^{-m}2^{(2-\alpha)k}2^{-30k_+}\\
 &\lesssim \epsilon^32^{-m}2^{-m/200}2^{-10k_+}.
 \end{aligned}
 \end{equation*}

\bigskip
{\bf{Step 2: \(l_1=l_2=\bar{l}\).}} In this case, it suffices to prove that
\begin{align}\label{36}
\big|J_{\bar{l},\bar{l}}(\xi,s)-\tilde{c}s^{-1}\xi|\xi|^{1-\alpha}\widehat{f_{k_1}^-}(-\xi,s)\widehat{f_{k_2}^+}(\xi,s)\widehat{f_{k_3}^+}(\xi,s)\big|
\lesssim \epsilon_1^32^{-m}2^{-2p_0m}2^{-10k_+}.
\end{align}
Define
\begin{equation*}
\begin{aligned}
\tilde{J}_{\bar{l},\bar{l}}(\xi,s)&=\xi\int_{\mathbb{R}^2}e^{\mathrm{i}s\alpha(\alpha+1)(\xi-\eta)(\xi-\sigma)/|\xi|^{1-\alpha}}\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s)\\
&\quad\times\varphi\big(2^{-\bar{l}}(\xi-\eta)\big)\varphi\big(2^{-\bar{l}}(\xi-\sigma)\big)\,\mathrm{d}\eta \mathrm{d} \sigma,
\end{aligned}
\end{equation*}
and observe
\begin{align*}
\Phi(\xi,\eta,\sigma)=\alpha(\alpha+1)|\xi|^{\alpha-1}(\xi-\eta)(\xi-\sigma)
+2^{(\alpha-2)k}\mathcal{O}\big[(|\xi-\eta|+|\xi-\sigma|)^3\big],
\end{align*}
when \(|\xi-\eta|+|\xi-\sigma|\leq 2^{k-5}\),
we then estimate
\begin{equation}\label{38}
\begin{aligned}
|J_{\bar{l},\bar{l}}(\xi,s)-\tilde{J}_{\bar{l},\bar{l}}(\xi,s)|
\lesssim \epsilon^32^m2^{(\alpha-1)k}2^{5\bar{l}}2^{-30k_+}
\lesssim \epsilon^32^{-m}2^{-2m/5}2^{-10k_+}.
\end{aligned}
\end{equation}
Notice that
\begin{equation*}
\begin{aligned}
\big|\widehat{f_{k_1}^+}(\xi+r,s)-\widehat{f_{k_1}^+}(\xi,s)\big|\lesssim 2^{\bar{l}/2}2^{-k}2^{p_0m},\quad \text{for}\ |r|\leq 2^{\bar{l}},
\end{aligned}
\end{equation*}
we thus obtain
\begin{equation*}
\begin{aligned}
&\big|\widehat{f_{k_1}^-}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)
\widehat{f_{k_3}^+}(\sigma,s)-\widehat{f_{k_1}^-}(-\xi,s)\widehat{f_{k_2}^+}(\xi,s)\widehat{f_{k_3}^+}(\xi,s)\big|\\
&\lesssim \epsilon^32^{\bar{l}/2}2^{p_0m}2^{-k}2^{-20k_+},\quad \text{for}\ |\xi-\eta|+|\xi-\sigma|\leq 2^{\bar{l}+4}.
\end{aligned}
\end{equation*}
It then follows that
\begin{equation}\label{39}
\begin{aligned}
&\bigg|\tilde{J}_{\bar{l},\bar{l}}(\xi,s)-\xi\int_{\mathbb{R}^2}e^{\mathrm{i}s\alpha(\alpha+1)(\xi-\eta)(\xi-\sigma)/|\xi|^{1-\alpha}}\widehat{f_{k_1}^-}(-\xi,s)\widehat{f_{k_2}^+}(\xi,s)\widehat{f_{k_3}^+}(\xi,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\varphi\big(2^{-\bar{l}}(\xi-\eta)\big)\varphi\big(2^{-\bar{l}}(\xi-\sigma)\big)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|\\
&\lesssim \epsilon^32^{\bar{l}/2}2^{p_0m}2^{2\bar{l}}2^{-20k_+}
\lesssim \epsilon^32^{-6m/5}2^{-10k_+}.
\end{aligned}
\end{equation}

One calculates
\begin{equation*}
\begin{aligned}
\int_{\mathbb{R}^2} e^{-\mathrm{i}xy}e^{-x^2/N^2}e^{-y^2/N^2}\,\mathrm{d} x \mathrm{d} y
=\frac{2\pi N}{\sqrt{4N^{-2}+N^2}}=2\pi+\mathcal{O}(N^{-1}),
\end{aligned}
\end{equation*}
in which we have used the formula
\begin{equation*}
\begin{aligned}
\int_\mathbb{R} e^{-ax^2-bx}\,\mathrm{d} x=e^{b^2/(4a)}\sqrt{\pi/a}, \quad \mathrm{for}\ a,b\in \mathcal{C}, \Re a>0.
\end{aligned}
\end{equation*}
It  follows that
\begin{equation*}
\begin{aligned}
\int_{\mathbb{R}^2} e^{-\mathrm{i}xy}\varphi(x/N)\varphi(y/N)\,\mathrm{d} x \mathrm{d} y
=2\pi+\mathcal{O}(N^{-1/2}),\quad \text{for}\ N\geq 1.
\end{aligned}
\end{equation*}
We then may estimate
\begin{equation*}
\begin{aligned}
&\left|\int_{\mathbb{R}^2}e^{\mathrm{i}s\alpha(\alpha+1)(\xi-\eta)(\xi-\sigma)/|\xi|^{1-\alpha}}\varphi\big(2^{-\bar{l}}(\xi-\eta)\big)\varphi\big(2^{-\bar{l}}(\xi-\sigma)\big)\,\mathrm{d}\eta \mathrm{d} \sigma+\frac{2\pi|\xi|^{1-\alpha}}{s\alpha(\alpha+1)}\right|\\
&\lesssim 2^{(1-\alpha)k}2^{-m}\big(2^{m/100}\big)^{-1/2}
\lesssim 2^{-m}2^{-m/300}.
\end{aligned}
\end{equation*}
Hence
\begin{equation}\label{40}
\begin{aligned}
&\bigg|\xi\int_{\mathbb{R}^2}e^{\mathrm{i}s\alpha(\alpha+1)(\xi-\eta)(\xi-\sigma)/|\xi|^{1-\alpha}}\widehat{f_{k_1}^-}(-\xi,s)\widehat{f_{k_2}^+}(\xi,s)\widehat{f_{k_3}^+}(\xi,s)\varphi\big(2^{-\bar{l}}(\xi-\eta)\big)\\
&\quad\times\varphi\big(2^{-\bar{l}}(\xi-\sigma)\big)\,\mathrm{d}\eta \mathrm{d} \sigma+\frac{2\pi \xi|\xi|^{1-\alpha}}{s\alpha(\alpha+1)}\widehat{f_{k_1}^-}(-\xi,s)\widehat{f_{k_2}^+}(\xi,s)\widehat{f_{k_3}^+}(\xi,s)\bigg|\\
&\lesssim \epsilon^32^k2^{-m}2^{-m/300}2^{-30k_+}
\lesssim \epsilon^32^{-m}2^{-m/300}2^{-10k_+}.
\end{aligned}
\end{equation}

We finally conclude \eqref{36} from  \eqref{38}-\eqref{40}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{le:3}]
	Since \(\alpha\in(-1,0)\), it is straightforward to check
	\begin{align}\label{41}
	a^{\alpha+1}+b^{\alpha+1}+c^{\alpha+1}-(a+b+c)^{\alpha+1}\gtrsim b^{\alpha+1},
	\end{align}
if \(a\geq b\geq c\in(0,\infty)\). We only present the proof of the case \((\iota_1,\iota_2,\iota_3)=(+,-,-)\), since the other cases can be handled in a similar fashion.
	Use \eqref{41} and recall \(k_1,k_2,k_3\in[k-20,k+20]\), we then have
	\begin{equation}\label{42}
	\begin{aligned}
	|\Phi(\xi,\eta,\sigma)|\gtrsim 2^{(\alpha+1)k}.
	\end{aligned}
	\end{equation}
	Observing
	\begin{equation*}
	\begin{aligned}
	e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}=\frac{\partial_se^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}}{\mathrm{i}\Phi(\xi,\eta,\sigma)},
	\end{aligned}
	\end{equation*}
	we integrate by parts in \(s\) to deduce
	\begin{equation}\label{43}
	\begin{aligned}
	&\quad\bigg|\xi\int_{t_1}^{t_2}\int_{\mathbb{R}^2}e^{\mathrm{i}H(\xi,s)}e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^{\iota_2}}(\eta,s)\widehat{f_{k_3}^{\iota_3}}(\sigma,s)
	\,\mathrm{d}\eta \mathrm{d} \sigma\mathrm{d} s\bigg|\\
	&\lesssim  \int_{t_1}^{t_2}\bigg|\xi\int_{\mathbb{R}^2}\frac{e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}}{\Phi(\xi,\eta,\sigma)}\frac{\mathrm{d}}{\mathrm{d} s}\bigg[e^{\mathrm{i}H(\xi,s)}\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^{\iota_2}}(\eta,s)\widehat{f_{k_3}^{\iota_3}}(\sigma,s)\bigg]\,\mathrm{d}\eta \mathrm{d} \sigma \bigg|\mathrm{d} s\\
	&\quad+ \sum_{j=1}^2\bigg|\xi\int_{\mathbb{R}^2}\frac{e^{\mathrm{i}H(\xi,s_j)}e^{\mathrm{i}s_j\Phi(\xi,\eta,\sigma)}}{\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}
		^{\iota_1}}(\xi-\eta-\sigma,s_j)\widehat{f_{k_2}^{\iota_2}}(\eta,s_j)\widehat{f_{k_3}^{\iota_3}}(\sigma,s_j)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|\\
	&=\colon A^0(\xi,s)+\sum_{j=1}^2A_j(\xi,s_j).
	\end{aligned}
	\end{equation}

	We first handle the two terms \(A_1(\xi,s)\) and \(A_2(\xi,s)\).
	Let
	\begin{equation}\label{44}
	\begin{aligned}
	m_3(\eta,\sigma):=\frac{1}{\Phi(\xi,\eta,\sigma)}\varphi_{k_1}^\prime(\xi-\eta-\sigma)\varphi_{k_2}^\prime(\eta)\varphi_{k_3}^\prime(\sigma).
	\end{aligned}
	\end{equation}
	It thus follows from \eqref{42} and \eqref{44} that
	\begin{equation*}
	\begin{aligned}
	\|\mathcal{F}^{-1}(m_3)\|_{L^1}\lesssim 2^{-(\alpha+1)k}.
	\end{aligned}
	\end{equation*}
	We therefore, in light of Lemma \ref{symbol estimate}, may estimate
	\begin{equation*}
	\begin{aligned}
	\sup_{s\in[t_1,t_2]}A_j(\xi,s_j)
	\lesssim \varepsilon_1^32^{-\alpha k}2^{(2p_0-1/2)m}2^{-2N_0k_+}
	\lesssim \epsilon_1^32^{-m/4}2^{-10k_+},
	\end{aligned}
	\end{equation*}
	for \(j=1,2\), where we have used \(\alpha\in(-1,0)\) in the last inequality.

	We now come to estimate the term \(A^0(\xi,s)\). For this, we expand \(\mathrm{d} /\mathrm{d} s\) to deduce
	\begin{equation}\label{45}
	\begin{aligned}
	A^0(\xi,s)\lesssim 2^m\sup_{s\in[t_1,t_2]}\big(A_0^0(\xi,s)+A_1^0(\xi,s)+A_2^0(\xi,s)+A_3^0(\xi,s)\big),
	\end{aligned}
	\end{equation}
	where
	\begin{equation}\label{46}
	\begin{aligned}
	A_0^0(\xi,s)&=\bigg|\xi\int_{\mathbb{R}^2}m_4(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^{\iota_2}}(\eta,s)\widehat{f_{k_3}^{\iota_3}}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,\\
	A_1^0(\xi,s)&= \bigg|\xi\int_{\mathbb{R}^2}m_3(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\partial_s\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^{\iota_2}}(\eta,s)\widehat{f_{k_3}^{\iota_3}}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,\\
	A_2^0(\xi,s)&= \bigg|\xi\int_{\mathbb{R}^2}m_3(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\partial_s\widehat{f_{k_2}^{\iota_2}}(\eta,s)\widehat{f_{k_3}^{\iota_3}}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,\\
	A_3^0(\xi,s)&= \bigg|\xi\int_{\mathbb{R}^2}m_3(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^{\iota_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^{\iota_2}}(\eta,s)\partial_s\widehat{f_{k_3}^{\iota_3}}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,
	\end{aligned}
	\end{equation}
	with
	\begin{equation}\label{47}
	\begin{aligned}
	m_4(\eta,\sigma):=\partial_s H(\xi,s)m_3(\eta,\sigma).
	\end{aligned}
	\end{equation}

	To control \(A_0^0(\xi,s)\), one observes
	\begin{equation}\label{48}
	\begin{aligned}
	|\partial_s H(\xi,s)|
	\lesssim \epsilon_1^22^{(2-\alpha)k}2^{-m}2^{-20k_+},
	\end{aligned}
	\end{equation}
	and thus obtains
	\begin{equation*}
	\begin{aligned}
	\|\mathcal{F}^{-1}(m_4)\|_{L^1}\lesssim \epsilon_1^22^{(1-2\alpha)k}2^{-m}2^{-20k_+}.
	\end{aligned}
	\end{equation*}
	We apply Lemma \ref{symbol estimate} again to obtain
	\begin{equation*}
	\begin{aligned}
	\sup_{s\in[t_1,t_2]}A_0^0(\xi,s)
	&\lesssim \epsilon_1^52^{(2-2\alpha)k}2^{(2p_0-3/2)m}2^{-2N_0k_+}2^{-20k_+}\\
	&\lesssim \epsilon_1^52^{-m}2^{-m/4}2^{-10k_+}.
	\end{aligned}
	\end{equation*}

	We next consider the term \(A_1^0(\xi,s)\). For this, in view of \eqref{17}, \eqref{20} and \eqref{23}, we first easily see that
 \begin{align}\label{49}
	\|\partial_s\widehat{f_l}(s)\|_{L^2}\lesssim \epsilon_1^32^{-m}2^{3p_0m}2^{-20l_+}.
	\end{align}
Let
	\begin{equation*}
	\begin{aligned}
	&\widehat{f}(\theta):=e^{-\mathrm{i}s|\xi-\theta|^\alpha(\xi-\theta)}\partial_s\widehat{f_{k_1}^{\iota_1}}(\xi-\theta,s),\\
	&\widehat{g}(\eta):=e^{-\mathrm{i}s|\eta|^\alpha\eta}\widehat{f_{k_2}^{\iota_2}}(\eta,s),\\
	&\widehat{h}(\sigma):=e^{-\mathrm{i}s|\sigma|^\alpha\sigma}\widehat{f_{k_3}^{\iota_3}}(\sigma,s),
	\end{aligned}
	\end{equation*}
we then use \eqref{23} and \eqref{49} to get
	\begin{equation*}
	\begin{aligned}
	&\|f\|_{L^2}\lesssim \epsilon_12^{3p_0m}2^{-20k_+}2^{-m},\\ &\|g\|_{L^2}
	\lesssim \epsilon_12^{-N_0k_+}2^{p_0m},\\
	&\|h\|_{L^\infty}\lesssim \epsilon_12^{-m/2}.
	\end{aligned}
	\end{equation*}
	Recall \(\alpha\in(-1,0)\), it then follows from Lemma \ref{symbol estimate} that	\begin{equation*}
	\begin{aligned}
	\sup_{s\in[t_1,t_2]}A_1^0(\xi,s)
	&\lesssim \varepsilon_1^32^{-\alpha k}2^{(4p_0-3/2)m}2^{-N_0k_+}2^{-20k_+}\\
	&\lesssim \epsilon_1^32^{-m}2^{-m/4}2^{-10k_+}.
	\end{aligned}
	\end{equation*}
	Similarly, one has
	\begin{equation*}
	\begin{aligned}
	\sup_{s\in[t_1,t_2]}\big(A_2^0(\xi,s)+A_3^0(\xi,s)\big)
	\lesssim \epsilon_1^32^{-m}2^{-m/4}2^{-10k_+}.
	\end{aligned}
	\end{equation*}

\end{proof}

\begin{proposition}  Let
	\begin{equation}\label{49.5}
	\begin{aligned}
	&\min(k_1,k_2,k_3)\geq -\frac{(1-20p_0)m}{2(\alpha+1)},\\
	&\max(|k_1-k|,|k_2-k|,|k_3-k|)\geq 21,
	\end{aligned}
	\end{equation}
then the bound \eqref{30} holds.
\end{proposition}

\begin{proof}  It suffices to show
	\begin{equation*}
	\begin{aligned}
	| I_{k_1,k_2,k_3}(\xi,s)|
	\lesssim \epsilon_1^32^{-m}2^{-3p_0m}2^{-10k_+}.
	\end{aligned}
	\end{equation*}

\bigskip
\noindent {\bf{Step 1: \(\max(|k_1-k_2|,|k_1-k_3|,|k_2-k_3|)\geq 5\).}}
Without loss of generality, we may assume that \(|k_1-k_2|\geq5\). On the support  \(|\xi-\eta-\sigma|\in [2^{k_1-2},2^{k_1+2}]\) and \(|\eta|\in [2^{k_2-2},2^{k_2+2}]\), one has
\begin{equation}\label{50}
\begin{aligned}
|\partial_\eta\Phi(\xi,\eta,\sigma)|
=(\alpha+1)\big||\xi-\eta-\sigma|^{\alpha}-|\eta|^{\alpha}\big|
\gtrsim 2^{\alpha\min(k_1,k_2)},
\end{aligned}
\end{equation}
and then integrate by parts in \(\eta\) to control
\begin{align}\label{51}
|I_{k_1,k_2,k_3}(\xi,s)|\leq |J_1(\xi,s)|+|F_1(\xi,s)|+|G_1(\xi,s)|,
\end{align}
where
\begin{equation}\label{52}
\begin{aligned}
J_1(\xi,s)&=\xi\int_{\mathbb{R}^2}\partial_\eta m_5(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}}(\eta,s)\widehat{f_{k_3}}(\sigma,s)\, \mathrm{d} \eta\mathrm{d} \sigma,\\
F_1(\xi,s)&=\xi\int_{\mathbb{R}^2}m_5(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\partial_\eta\widehat{f_{k_1}}(\xi-\eta-\sigma,s)\widehat{f_{k_2}}(\eta,s)\widehat{f_{k_3}}(\sigma,s) \, \mathrm{d} \eta\mathrm{d} \sigma,\\
G_1(\xi,s)&=\xi\int_{\mathbb{R}^2}m_5(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}}(\xi-\eta-\sigma,s)\partial_\eta\widehat{f_{k_2}}(\eta,s)\widehat{f_{k_3}}(\sigma,s) \, \mathrm{d} \eta\mathrm{d} \sigma,
\end{aligned}
\end{equation}
with
\begin{align}\label{53}
m_5(\eta,\sigma):=\frac{1}{s\partial_\eta\Phi(\xi,\eta,\sigma)}\varphi_{k_1}^\prime(\xi-\eta-\sigma)\varphi_{k_2}^\prime(\eta)\varphi_{k_3}^\prime(\sigma).
\end{align}

Using \eqref{50} and \eqref{53}, one finds that \(m_5\) satisfies the following estimates
\begin{equation}\label{54}
\begin{aligned}
\|\mathcal{F}^{-1}(m_5)\|_{L^1}\lesssim 2^{-m}2^{-\alpha\min(k_1,k_2)},
\end{aligned}
\end{equation}
and
\begin{equation}\label{55}
\begin{aligned}
\|\mathcal{F}^{-1}(\partial_\eta m_5)\|_{L^1}\lesssim 2^{-m}2^{-(\alpha+1)\min(k_1,k_2)}.
\end{aligned}
\end{equation}

Applying Lemma \ref{symbol estimate}, we use \eqref{23} and \eqref{55} to see
\begin{align*}
|J_1(\xi,s)|
\lesssim \epsilon_1^32^k2^{-(\alpha+1)\min(k_1,k_2)}2^{(2p_0-3/2)m}2^{-N_0\max({k_1}_+,{k_2}_+,{k_3}_+)},
\end{align*}
and instead use \eqref{23} and \eqref{54} to find
\begin{align*}
|F_1(\xi,s)|
\lesssim\epsilon_1^32^k2^{-k_1}2^{-\alpha\min(k_1,k_2)}2^{(2p_0-3/2)m}2^{-N_0\max({k_2}_+,{k_3}_+)},
\end{align*}
and
\begin{align*}
|G_1(\xi,s)|\lesssim\epsilon_1^32^k2^{-k_2}2^{-\alpha\min(k_1,k_2)}2^{(2p_0-3/2)m}2^{-N_0\max({k_1}_+,{k_3}_+)}.
\end{align*}
We finally conclude that
\begin{equation*}
\begin{aligned}
&\quad|J_1(\xi,s)|+|F_1(\xi,s)|+|G_1(\xi,s)|\\
&\lesssim \epsilon_1^32^k2^{-(\alpha+1)\min(k_1,k_2)}2^{(2p_0-3/2)m}2^{-10k_+}
(2^{10\max({k_1}_+,{k_2}_+,{k_3}_+)}+1)\\
&\lesssim\epsilon_1^32^{-m}2^{-5p_0m}2^{-10k_+},
\end{aligned}
\end{equation*}
where we have used \eqref{29} and \eqref{49.5} in the last inequality.

\bigskip
\noindent{\bf{Step 2: \(\max(|k_1-k_2|,|k_1-k_3|,|k_2-k_3|)\leq 4\).}}
In this case  \(\partial_\eta\Phi\neq0\) or \(\partial_\sigma\Phi\neq0\). Without loss of generality, we assume \(\partial_\eta\Phi\neq0\). On the support \(|\xi-\eta-\sigma|\in [2^{k_1-2},2^{k_1+2}]\), \(|\eta|\in [2^{k_2-2},2^{k_2+2}]\) and \(|\sigma|\approx 2^{k_3}\), recalling \(2^{k_1}\approx 2^{k_2}\approx 2^{k_3}\), we have
\begin{equation}\label{56}
\begin{aligned}
|\partial_\eta\Phi(\xi,\eta,\sigma)|
=(\alpha+1)\big||\xi-\eta-\sigma|^{\alpha}-|\eta|^{\alpha}\big|
\gtrsim 2^{\alpha k_2}.
\end{aligned}
\end{equation}
One may use integration by parts in \(\eta\) as \eqref{51}-\eqref{52} to control \(I_{k_1,k_2,k_3}\). Due to \eqref{56}, we instead have
\begin{equation}\label{57}
\begin{aligned}
\|\mathcal{F}^{-1}(m_5)\|_{L^1}\lesssim 2^{-m}2^{-\alpha k_2},
\end{aligned}
\end{equation}
and
\begin{equation}\label{58}
\begin{aligned}
\|\mathcal{F}^{-1}(\partial_\eta m_5)\|_{L^1}\lesssim 2^{-m}2^{-(\alpha+1)k_2}.
\end{aligned}
\end{equation}
Recall \(2^{k_1}\approx 2^{k_2}\approx 2^{k_3}\) and repeat the arguments of {\bf{Step 2}}  using instead  \eqref{57}-\eqref{58}, we finally conclude that
\begin{equation*}
\begin{aligned}
|J_2(\xi,s)|+|F_2(\xi,s)|+|G_2(\xi,s)|
&\lesssim\epsilon_1^32^k2^{-(\alpha+1)k_2}2^{(2p_0-3/2)m}2^{-N_0{k_2}_+}\\
&\lesssim \epsilon_1^32^{-m}2^{-5p_0m}2^{-10k_+},
\end{aligned}
\end{equation*}
in which we have used the assumption
\begin{equation*}
\begin{aligned}
-(\alpha+1) k_2\leq (\frac{1}{2}-10p_0)m.
\end{aligned}
\end{equation*}
\end{proof}

\begin{proposition} Let
	\begin{equation}\label{58.5}
	\begin{aligned}
	&\min(k_1,k_2,k_3)\leq -\frac{(1-20p_0)m}{2(\alpha+1)},\\
	&\max(|k_1-k|,|k_2-k|,|k_3-k|)\geq 21,
	\end{aligned}
	\end{equation}
then the bound \eqref{30} holds.
\end{proposition}
\begin{proof} It suffices to show
	\begin{align*}
	\bigg|\int_{t_1}^{t_2}e^{\mathrm{i}H(s,\xi)}I_{k_1,k_2,k_3}^{\iota_1,\iota_2,\iota_3}(\xi,s)\,\mathrm{d} s\bigg|
	\lesssim \varepsilon_1^32^{-2p_0m}2^{-10k_+}.
	\end{align*}

\bigskip
\noindent{\bf{Case 1: \((\iota_1,\iota_2,\iota_3)= (+,+,+)\).}} In this case,
one may estimate
\begin{equation*}
\begin{aligned}
-\Phi(\xi,\eta,\sigma)&=(\xi-\eta-\sigma)^{\alpha+1}+\eta^{\alpha+1}+\sigma^{\alpha+1}-\xi^{\alpha+1}\\
&\gtrsim 2^{(\alpha+1)\mathrm{med}(k_1,k_2,k_3)}.
\end{aligned}
\end{equation*}
It follows from \eqref{29} and \eqref{58.5} that
\begin{equation*}
\begin{aligned}
\mathrm{med}(k_1,k_2,k_3)\geq -(1+10p_0)m+\frac{(1-20p_0)m}{2(\alpha+1)}.
\end{aligned}
\end{equation*}
Therefore
\begin{equation}\label{59}
\begin{aligned}
-\Phi(\xi,\eta,\sigma)\gtrsim 2^{-(1+10p_0)(\alpha+1)m+(\frac{1}{2}-10p_0)m}\gtrsim 2^{-(\frac{1}{2}-20p_0)m},
\end{aligned}
\end{equation}
where we have used the assumption \(p_0\leq-\frac{\alpha}{100}\).

Due to \eqref{59}, we may integrate by parts in \(s\) as \eqref{43}-\eqref{46} and need to control the terms \(A_j(\xi,s),A_l^0(\xi,s),j=1,2;l=0,1,2,3\).
Using \eqref{59} and \eqref{44}, one calculates that
\begin{equation}\label{60}
\begin{aligned}
\|\mathcal{F}^{-1}(m_3)\|_{L^1}\lesssim 2^{(\frac{1}{2}-20p_0)m},
\end{aligned}
\end{equation}
and
\begin{equation}\label{61}
\begin{aligned}
\|\mathcal{F}^{-1}(m_4)\|_{L^1}\lesssim \epsilon_1^22^{(\frac{1}{2}-20p_0)m}2^{(2-\alpha)k}2^{-m}2^{-20k_+}.
\end{aligned}
\end{equation}
The symbol-type estimate \eqref{60} together with  Lemma \ref{symbol estimate} yields
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}A_j(\xi,s_j)
&\lesssim \epsilon_1^32^k2^{(\frac{1}{2}-20p_0)m}2^{(2p_0-1/2)m}2^{-N_0({k_1}_+,{k_2}_+,{k_3}_+)}\\
&\lesssim \epsilon_1^32^{-5p_0m}2^{-10k_+},\quad \text{for}\ j=1,2,
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}A_1^0(\xi,s)
&\lesssim \epsilon_1^32^k2^{(\frac{1}{2}-20p_0)m}2^{(4p_0-3/2)m}2^{-20{k_1}_+}2^{-N_0({k_2}_+,{k_3}_+)}\\
&\lesssim \epsilon_1^52^{-m}2^{-5p_0m}2^{-10k_+},
\end{aligned}
\end{equation*}
and the same bound for \(A_2^0(\xi,s)\) and \(A_3^0(\xi,s)\).
Applying Lemma \ref{symbol estimate} with \eqref{61}, we may estimate
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}A_0^0(\xi,s)
&\lesssim \epsilon_1^32^{(3-\alpha)k}2^{(\frac{1}{2}-20p_0)m}2^{(2p_0-3/2)m}2^{-N_0({k_1}_+,{k_2}_+,{k_3}_+)}2^{-20k_+}\\
&\lesssim \epsilon_1^52^{-m}2^{-5p_0m}2^{-10k_+}.
\end{aligned}
\end{equation*}

\bigskip
\noindent{\bf{Case 2: \((\iota_1,\iota_2,\iota_3)\in \{(+,-,-),(-,+,-),(-,-,+)\}\).}} Since they are similar, we only analyze the case \((\iota_1,\iota_2,\iota_3)= (+,-,-)\).
In this case, we have
\begin{equation*}
\begin{aligned}
\Phi(\xi,\eta,\sigma)
&=\xi^{\alpha+1}+(-\eta)^{\alpha+1}+(-\sigma)^{\alpha+1}-(\xi-\eta-\sigma)^{\alpha+1}\\
&\gtrsim 2^{(\alpha+1)\mathrm{med}(k,k_2,k_3)}.
\end{aligned}
\end{equation*}
There are three sub-cases to consider:\\
{\bf{(i)}} If \(\mathrm{med}(k,k_2,k_3)=k\), then
\begin{equation*}
\begin{aligned}
\Phi(\xi,\eta,\sigma)\gtrsim 2^{(\alpha+1)k}.
\end{aligned}
\end{equation*}
{\bf{(ii)}} If \(\mathrm{med}(k,k_2,k_3)=k_2\), then \(k\leq k_2\leq k_3 \) or \(k_3\leq k_2\leq k\).\\
When \(k\leq k_2\leq k_3 \), one has
\begin{equation*}
\begin{aligned}
\Phi(\xi,\eta,\sigma)\gtrsim 2^{(\alpha+1)k_2}\geq 2^{(\alpha+1)k}.
\end{aligned}
\end{equation*}
For \(k_3\leq k_2\leq k\), it holds
\begin{equation*}
\begin{aligned}
\Phi(\xi,\eta,\sigma)\gtrsim 2^{(\alpha+1)k_2}\geq 2^{(\alpha+1)\mathrm{med}(k_1,k_2,k_3)}.
\end{aligned}
\end{equation*}
{\bf{(iii)}} If \(\mathrm{med}(k,k_2,k_3)=k_3\), then,  similarly to (ii), one may show \(\Phi\) enjoys the same bounds as (ii).

We conclude from (i)-(iii) that
\begin{equation}\label{62}
\begin{aligned}
\Phi(\xi,\eta,\sigma)\gtrsim 2^{(\alpha+1)k},
\end{aligned}
\end{equation}
or
\begin{equation}\label{63}
\begin{aligned}
\Phi(\xi,\eta,\sigma)\gtrsim 2^{(\alpha+1)\mathrm{med}(k_1,k_2,k_3)}\gtrsim 2^{-(\frac{1}{2}-20p_0)m}.
\end{aligned}
\end{equation}

For the case of \eqref{62}, the Phase \(\Phi\) enjoys the same bound as \eqref{42}, thus the terms \(A_j(\xi,s_j),A_l^0(\xi,s),j=1,2;l=0,1,2,3\) can be estimated as it in Lemma \ref{le:3}.
The latter case \eqref{63} can be handled identically to {\bf{Case 1}}.

\bigskip
\noindent{\bf{Case 3: \((\iota_1,\iota_2,\iota_3)\in \{(+,+,-),(+,-,+),(-,+,+)\}\).}} We only consider the case \((\iota_1,\iota_2,\iota_3)=(+,+,-)\), and the other cases may be handled in a similar fashion.
In this case, one has
\begin{equation*}
\begin{aligned}
\Phi(\xi,\eta,\sigma)=\xi^{\alpha+1}-(\xi-\eta-\sigma)^{\alpha+1}-\eta^{\alpha+1}+(-\sigma)^{\alpha+1}.
\end{aligned}
\end{equation*}

We shall divide it into two sub-cases.\\
{\bf{(i) \(k_3=\min(k_1,k_2,k_3)\).}} Recalling the assumption
\(
k_1,k_2,k_3\in[-4m,p_0m/10],
\)
we have
\[
k_3\in \bigg[-(1+20p_0)m,-\frac{(1-20p_0)m}{2(\alpha+1)}\bigg],\quad k_1,k_2\in \bigg[-\frac{(1-40p_0)m}{2(\alpha+1)},\frac{p_0m}{10}\bigg].
\]
Notice that
\begin{align*}
a^{\alpha+1}+b^{\alpha+1}-(a+b)^{\alpha+1}\gtrsim b^{\alpha+1},
\end{align*}
if \(a\geq b\in(0,\infty)\).
We then may estimate
\begin{equation}\label{66}
\begin{aligned}
-\Phi(\xi,\eta,\sigma)
&=\big(-\xi^{\alpha+1}+(\xi-\eta-\sigma)^{\alpha+1}
+(\eta+\sigma)^{\alpha+1}\big)\\
&\quad+\big(\eta^{\alpha+1}-(\eta+\sigma)^{\alpha+1}\big)
-(-\sigma)^{\alpha+1}\\
&\gtrsim 2^{(\alpha+1)\min(k_1,k_2)}\geq 2^{-(\frac{1}{2}-20p_0)m},
\end{aligned}
\end{equation}
provided that \(|\xi-\eta-\sigma|\in [2^{k_1-2},2^{k_1+2}]\), \(|\eta|\in [2^{k_2-2},2^{k_2+2}]\) and \(|\sigma|\in [2^{k_3-2},2^{k_3+2}]\).

Proceed as \eqref{43}-\eqref{46} by integration by parts in \(s\). It follows from \eqref{66} and \eqref{44} that
\begin{equation}\label{67}
\begin{aligned}
\|\mathcal{F}^{-1}(m_3)\|_{L^1}\lesssim 2^{(\frac{1}{2}-20p_0)m},
\end{aligned}
\end{equation}
and
\begin{equation}\label{68}
\begin{aligned}
\|\mathcal{F}^{-1}(m_4)\|_{L^1}\lesssim \epsilon_1^22^{(2-\alpha)k}2^{(\frac{1}{2}-20p_0)m}2^{-m}2^{-20k_+}.
\end{aligned}
\end{equation}
With the symbol-type bounds \eqref{67} and \eqref{68} at hand, repeating the argument of {\bf{Case 1}}, one may estimate
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}A_j(\xi,s)
\lesssim \epsilon_1^32^{-5p_0m}2^{-10k_+},\quad \text{for}\ j=1,2,
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}\big(A_0^0(\xi,s)+A_1^0(\xi,s)+A_2^0(\xi,s)+A_3^0(\xi,s)\big)\lesssim \epsilon_1^32^{-m}2^{-5p_0m}2^{-10k_+}.
\end{aligned}
\end{equation*}

\bigskip
\noindent {\bf{(ii) \(k_3\neq \min(k_1,k_2,k_3)\).}} By symmetry we may assume
\(k_1=\min(k_1,k_2,k_3)\).
In this case, we have
\[
k_1\in \bigg[-(1+20p_0)m,-\frac{(1-20p_0)m}{2(\alpha+1)}\bigg],\quad k_2,k_3\in \bigg[-\frac{(1-40p_0)m}{2(\alpha+1)},\frac{p_0m}{10}\bigg].
\]

Define
\begin{equation*}
\chi_{k,m}(\eta,\sigma)=
\left\{
\begin{aligned}
&1,&& \text{if}\ |k-k_1|\geq 11,\\
&1-\varphi\big(2^{(1+20p_0)m}(\eta+\sigma)\big),&& \text{if}\ |k-k_1|\leq 10.
\end{aligned}
\right.
\end{equation*}
We first observe that
\begin{equation*}
\begin{aligned}
&\quad\bigg|\xi\int_{t_1}^{t_2}\int_{\mathbb{R}^2}\big(1-\chi_{k,m}(\eta,\sigma)\big)e^{\mathrm{i}H(\xi,s)}e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\mathrm{d} s\bigg|\\
&\leq 2^m2^k\bigg|\int_{\mathbb{R}^2}\varphi\big(2^{(1+20p_0)m}(\eta+\sigma)\big)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|\\
&\lesssim \epsilon_1^32^m2^k2^{-(\frac{1}{2}+10p_0)m}2^{-m/2}2^{p_0m}2^{-10k_+}\\
&\lesssim\epsilon_1^32^{-5p_0m}2^{-10k_+}.
\end{aligned}
\end{equation*}
To complete the proof it remains to show
\begin{equation}\label{69}
\begin{aligned}
&\quad\bigg|\xi\int_{t_1}^{t_2}\int_{\mathbb{R}^2}\chi_{k,m}(\eta,\sigma)e^{\mathrm{i}H(\xi,s)}e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\mathrm{d} s\bigg|\\
&\lesssim\epsilon_1^32^{-5p_0m}2^{-10k_+}.
\end{aligned}
\end{equation}

The crucial ingredient in showing \eqref{69}
is that the phase \(\Phi\) satisfies the following weakly elliptic bound
\begin{equation}\label{70}
|\Phi(\xi,\eta,\sigma)|\geq \lambda:=
\left\{
\begin{aligned}
&2^{(\alpha+1)k_1-100},&&\quad \text{if}\ k\leq k_1-11,\\
&2^{(\alpha+1)\min(k,k_2,k_3)-100},&&\quad \text{if}\ k\geq k_1+11,\\
&2^{-(1+20p_0)m}2^{\alpha k_1-100},&&\quad \text{if}\ |k-k_1|\leq 10,
\end{aligned}
\right.
\end{equation}
provided that \(|\xi-\eta-\sigma|\in [2^{k_1-2},2^{k_1+2}]\), \(|\eta|\in [2^{k_2-2},2^{k_2+2}]\), \(|\sigma|\in [2^{k_3-2},2^{k_3+2}]\) and \(\chi_{k,m}(\eta,\sigma)\neq 0\). We now show \eqref{70} as follows:

If \(k\leq k_1-11\), we then estimate
\begin{equation*}
\begin{aligned}
-\Phi(\xi,\eta,\sigma)
&\geq -\xi^{\alpha+1}+(\xi-\eta-\sigma)^{\alpha+1}-|\eta^{\alpha+1}-(-\sigma)^{\alpha+1}|\\
&\geq -2^{(\alpha+1)k+1}+2^{(\alpha+1)k_1-1}-2^{k_1+\alpha k_2+10}\\
&\geq 2^{(\alpha+1)k_1-2}.
\end{aligned}
\end{equation*}

If \(k\geq k_1+11\), we then deduce
\begin{equation*}
\begin{aligned}
\quad\Phi(\xi,\eta,\sigma)
&\geq \big(\xi^{\alpha+1}+(-\sigma)^{\alpha+1}-(\xi-\sigma)^{\alpha+1}\big)-(\xi-\eta-\sigma)^{\alpha+1}\\
&\quad-|\eta^{\alpha+1}-(\xi-\sigma)^{\alpha+1}|\\
&\geq 2^{(\alpha+1)\min(k,k_3)-1}-2^{(\alpha+1)k_1+1}-2^{k_1+\alpha k_2+10}\\
&\geq 2^{(\alpha+1)\min(k,k_3)-2}.
\end{aligned}
\end{equation*}

If \(|k-k_1|\leq 10\), first notice \(2^{-(1+20p_0)m}\leq |\eta+\sigma|\leq 2^{k_1+11}\), then one has
\begin{equation*}
\begin{aligned}
|\Phi(\xi,\eta,\sigma)|&=(\alpha+1)\big|\big(\mu\xi+(1-\mu)(\xi-\eta-\sigma)\big)^{\alpha}-\big(\theta\eta+(1-\theta)(-\sigma)\big)^{\alpha}\big||\eta+\sigma|\\
&\gtrsim 2^{\alpha k_1}2^{-(1+20p_0)m},
\end{aligned}
\end{equation*}
for some \(\mu,\theta\in (0,1)\).

To show \eqref{69} in view of \eqref{70},
we integrate by parts in \(s\) to get
\begin{equation*}
\begin{aligned}
&\quad\bigg|\xi\int_{t_1}^{t_2}\int_{\mathbb{R}^2}\chi_{k,m}(\eta,\sigma)e^{\mathrm{i}H(\xi,s)}e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)
\,\mathrm{d}\eta \mathrm{d} \sigma\mathrm{d} s\bigg|\\
&\lesssim  \int_{t_1}^{t_2}\bigg|\xi\int_{\mathbb{R}^2}\chi_{k,m}(\eta,\sigma)\frac{e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}}{\Phi(\xi,\eta,\sigma)}\frac{\mathrm{d}}{\mathrm{d} s}\bigg[e^{\mathrm{i}H(\xi,s)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\bigg]\,\mathrm{d}\eta \mathrm{d} \sigma \bigg|\mathrm{d} s\\
&\quad+ \sum_{j=1}^2\bigg|\xi\int_{\mathbb{R}^2}\chi_{k,m}(\eta,\sigma)\frac{e^{\mathrm{i}H(\xi,s_j)}e^{\mathrm{i}s_j\Phi(\xi,\eta,\sigma)}}{\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s_j)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\widehat{f_{k_2}^+}(\eta,s_j)\widehat{f_{k_3}^-}(\sigma,s_j)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|\\
&=\colon B^0(\xi,s)+\sum_{j=1}^2B_j(\xi,s_j),
\end{aligned}
\end{equation*}
and the term \(B^0(\xi,s)\) may be bounded by
\begin{equation*}
\begin{aligned}
B^0(\xi,s)\lesssim 2^m\sup_{s\in[t_1,t_2]}\big(B_0^0(\xi,s)+B_1^0(\xi,s)+B_2^0(\xi,s)+B_3^0(\xi,s)\big),
\end{aligned}
\end{equation*}
in which each term is defined by
\begin{equation*}
\begin{aligned}
B_0^0(\xi,s)&=\bigg|\xi\int_{\mathbb{R}^2}m_7(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,\\
B_1^0(\xi,s)&= \bigg|\xi\int_{\mathbb{R}^2}m_6(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\partial_s\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,\\
B_2^0(\xi,s)&= \bigg|\xi\int_{\mathbb{R}^2}m_6(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\partial_s\widehat{f_{k_2}^+}(\eta,s)\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,\\
B_3^0(\xi,s)&= \bigg|\xi\int_{\mathbb{R}^2}m_6(\eta,\sigma)e^{\mathrm{i}s\Phi(\xi,\eta,\sigma)}\widehat{f_{k_1}^+}(\xi-\eta-\sigma,s)\widehat{f_{k_2}^+}(\eta,s)\partial_s\widehat{f_{k_3}^-}(\sigma,s)\,\mathrm{d}\eta \mathrm{d} \sigma\bigg|,
\end{aligned}
\end{equation*}
with
\begin{equation}\label{71}
\begin{aligned}
m_6(\eta,\sigma):=\frac{\chi_{k,m}(\eta,\sigma)}{\Phi(\xi,\eta,\sigma)}\varphi_{k_1}^\prime(\xi-\eta-\sigma)\varphi_{k_2}^\prime(\eta)\varphi_{k_3}^\prime(\sigma),
\end{aligned}
\end{equation}
and
\begin{equation}\label{72}
\begin{aligned}
m_7(\eta,\sigma):=\chi_{k,m}(\eta,\sigma)\partial_s H(\xi,s)m_6(\eta,\sigma).
\end{aligned}
\end{equation}

It follows from \eqref{70} and \eqref{71} that
\begin{equation}\label{73}
\begin{aligned}
\|\mathcal{F}^{-1}(m_6)\|_{L^1}\lesssim \lambda^{-1},
\end{aligned}
\end{equation}
and from \eqref{70} and \eqref{72} that
\begin{equation}\label{74}
\begin{aligned}
\|\mathcal{F}^{-1}(m_7)\|_{L^1}\lesssim \epsilon_1^2\lambda^{-1}2^{(2-\alpha)k}2^{-m}2^{-20k_+}.
\end{aligned}
\end{equation}
Applying Lemma \ref{symbol estimate} with \eqref{73} yields
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}B_j(\xi,s_j)
\lesssim \epsilon_1^3\lambda^{-1}2^k2^{(2p_0-1/2)m}2^{-10k_+},\quad \text{for}\ j=1,2.
\end{aligned}
\end{equation*}
So we need to estimate \(\lambda^{-1}2^k\). If \(k\leq k_1-11\), then
\begin{equation*}
\begin{aligned}
\lambda^{-1}2^k
\lesssim 2^{-(\alpha+1)k_1}2^k
\lesssim 2^{-(\alpha+1)k}2^k
\lesssim 2^{-\alpha p_0m}.
\end{aligned}
\end{equation*}
If \(k\geq k_1+11\), then when \(k\leq\min(k_2,k_3)\)
\begin{equation*}
\begin{aligned}
\lambda^{-1}2^k
\lesssim 2^{-(\alpha+1)k}2^k
\lesssim 2^{-\alpha p_0m},
\end{aligned}
\end{equation*}
and when \(k\geq\min(k_2,k_3)\)
\begin{equation*}
\begin{aligned}
\lambda^{-1}2^k
\lesssim 2^{-(\alpha+1)\min(k_2,k_3)}2^k
\lesssim 2^{(\frac{1}{2}-20p_0)m}2^{p_0m}.
\end{aligned}
\end{equation*}
If \(|k-k_1|\leq 10\), then
\begin{equation*}
\begin{aligned}
\lambda^{-1}2^k
\lesssim 2^{(1+20p_0)m}2^{(1-\alpha)k_1}
\lesssim 2^{(1+20p_0)m}2^{-\frac{(1-\alpha)(1-20p_0)m}{2(\alpha+1)}}
\lesssim 2^{(\frac{1}{2}-20p_0)m},
\end{aligned}
\end{equation*}
where we have used the assumption \(p_0\leq-\frac{\alpha}{100}\).
We finally conclude
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}B_j(\xi,s_j)\lesssim \epsilon_1^32^{-5p_0m}2^{-10k_+},\quad \text{for}\ j=1,2.
\end{aligned}
\end{equation*}
Similarly, we obtain
\begin{equation}
\begin{aligned}
\sup_{s\in[t_1,t_2]}\big(B_1^0(\xi,s)+B_2^0(\xi,s)+B_3^0(\xi,s)\big)
&\lesssim \epsilon_1^3\lambda^{-1}2^k2^{(4p_0-3/2)m}2^{-10k_+}\\
&\lesssim \epsilon_1^32^{-m}2^{-5p_0m}2^{-10k_+}.
\end{aligned}
\end{equation}
We apply Lemma \ref{symbol estimate} instead of using \eqref{74} to obtain
\begin{equation*}
\begin{aligned}
\sup_{s\in[t_1,t_2]}B_0^0(\xi,s)
&\lesssim \epsilon_1^5\lambda^{-1}2^{(3-\alpha)k}2^{(2p_0-3/2)m}2^{-10k_+}\\
&\lesssim \epsilon_1^52^{-m}2^{-5p_0m}2^{-10k_+}.
\end{aligned}
\end{equation*}

\end{proof}

\section{Concluding remarks}

We conclude this paper by some related and open questions.  Some previous results on long time behavior or finite time blow-up for solutions to fractional KdV type equations concerned equations with quadratic nonlinearities, namely the fractional KdV equation (fKdV)

\begin{align}\label{fKdV}
	\partial_t u-|D|^{\alpha} \partial_x u=-u\partial_x u, \quad -1<\alpha<0,
\end{align}
or the Whitham equation
\begin{equation}\label {Whit}
	\partial_t u+\mathcal L \partial_xu=-u\partial_x u.
\end{equation}

It was proven in \cite{EW} that the lifespan of solutions to \eqref{fKdV} with $-1<\alpha<1, \; \alpha \neq 0$ and with initial data of size $O(\epsilon)$ in $H^N(\mathbb{R}), N\geq 3$ is $O(1/\epsilon^2).$

This result is specially striking when $-1<\alpha<0$ since for $\alpha>0$ one expects global existence of small solutions. Note that for the inviscid Burgers equation this lifespan is $O(1/\epsilon).$

Proving a global existence result of small solutions of the  fKdV equation in the range $-1<\alpha<0$ or for the Whitham equation (and for the modified Whitham equation) is a challenging open question.

On the other hand, a finite time blow-up by shock formation was observed via numerical approach and conjectured in \cite{MR3317254, KLPS}. This has been   now  proven in \cite{HT,Hur} both for the fKdV equation when $-1<\alpha<-\frac{1}{3}$ and for the Whitham equation.

\section{Appendix}
For the reader's  convenience, we list two technical lemmas proven in \cite{MR3552008} and \cite{MR3121725} respectively.
The first one is the following interpolation inequality:
\begin{lemma}[\cite{MR3552008}]\label{interpolation} It holds that
	\begin{align}\label{75}
	\|\widehat{P_kg}\|_{L^\infty}^2\lesssim \left\|P_kg\right\|_{L^1}^2\lesssim 2^{-k}\|\widehat{P_kg}\|_{L^2}(\|\widehat{P_kg}\|_{L^2}+2^k\|\partial\widehat{P_kg}\|_{L^2}).
	\end{align}
\end{lemma}

\bigskip
The other one is the bound on pseudo-product operators satisfying certain strong integrability conditions:
\begin{lemma}[\cite{MR3121725}]\label{symbol estimate}
	Assume that \(L^1(\mathbb{R}\times\mathbb{R})\) satisfies
	\begin{align*}
	\left\|\int_{\mathbb{R}^2}m(\eta,\sigma)e^{\mathrm{i}x\eta}e^{\mathrm{i}y\sigma}\,\mathrm{d} \eta\mathrm{d} \sigma\right\|_{L^1_{x,y}}\lesssim A,
	\end{align*}
	for some \(A\in(0,\infty)\). Then for any \((p,q,r)\in\{(2,2,\infty),(2,\infty,2),(\infty,2,2)\}\),
	\begin{align*}
	\left|\int_{\mathbb{R}^2}m(\eta,\sigma)\widehat{f}(\eta)\widehat{g}(\sigma)\widehat{h}(-\eta-\sigma)\,\mathrm{d} \eta\mathrm{d} \sigma\right|\lesssim A\|f\|_{L^p}\|g\|_{L^q}\|h\|_{L^r}.
	\end{align*}
\end{lemma}

\vspace{0.5cm}
\noindent {\bf Acknowledgments.} The work of both authors was partially  supported by the ANR project ANuI ( ANR-17-CE40-0035-02).

redpajama_set_name: RedPajamaArXiv


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What is the average cost for a good propane catalytic heater?
If you want to buy the best propane catalytic heater on the market, remember that it does not have to be the most expensive one. Of course, costly models are usually good, but there is always a possibility that you can find an option with better price/quality ratio. If you want to get the most value out of your propane catalytic heater, choose wisely and consider all its characteristics, not only the price.
What should be my main choice criteria when buying a propane catalytic heater?
As you might have guessed, the weight of a propane catalytic heater is one of the main factor to consider. Another important aspects is its material. If the propane catalytic heater is assembled well, it will serve you for many years.


The Valhalla state of the art 2400 Series Power Analyzers offer high performance in both single and three phase. Unlike other instruments at this price level the 2400 is designed to operate with extreme signals generated on frequency inverter drivers. These analyzers provide precise reliable measurements for any waveform. Large clear monitor lets you read displayed values from a distance of four meters.
Checking power to determine the pertinent power parameters of a frequency inverter driven system is simple. All values display large letters easily read even in dark rooms. The user menu makes operation easy.
The new 2400 power analyzer measures, computes, and displays all of your critical power variable to let you concentrate on more efficient reliable testing. It is available in single or three phase versions and combines a wattmeter, oscilloscope, and a power spectrum analyzer in a single compact package. From the simultaneous and precise voltage and current measurements, you can measure and monitor all of the power parameters you need. You can display them in the format that fits your application.


text:
2.5 years prison sentence for Brazilian mother guilty of importing cocaine
by Thiago
A young mother caught smuggling more than 1.5 kg of cocaine into her body has been sentenced to two and a half years in prison.
Hyonaru Moreno, 26, was stopped by Irish customs officers at Dublin airport after demonstrating what was described as suspicious behavior.
Upon questioning, the young mother read a false statement about her planned stay in Ireland as noted today during a hearing at the Dublin Criminal Court earlier this week.
It was found that the Brazilian carried on her body, including her genitals, cocaine with a total street value of over €37,000. Moreno, who has an address in Sao Paulo, pleaded guilty to possession of drugs for sale in June this year. The defendant has no criminal record.
Grace Daly, a Gardai member, told Joseph Barnes, prosecuting, that Moreno was stopped by customs officials who were analyzing profiles of suspects originating from Brazil.
Barry White, defending Morena, argued that his client was persuaded to go to Ireland. He pointed out that she was told she should smuggle the drugs just hours before the flight, and reports that Moreno is a single mother who was struggling.
"What we have here is a naive mule," White told the court. "She laments what she has done and is willing to receive the punishment that the court deems appropriate, just as she is looking forward to returning to her children. "
In giving the sentence, Judge Martin Nolan stated that Moreno was involved in "desperate-motivated escape" for a financial purpose, and stressed this was the second such case that week. "Unless someone is desperate, he would not be at risk," he said.
The judge also noted that a foreigner with limited knowledge of English a long prison sentence would be an even more difficult experience. He passed a sentence of 2.5 years, after which time Morena must return to her native country.
cocaine, drugs, mule, prison
Author: Thiago
I moved to Ireland 2012. I work as a business administrator and travel a lot to Europe with my work. I enjoy writing and athletics during my vacation and try to visit friends and family every year in Brazil.
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redpajama_set_name: RedPajamaCommonCrawl


Is your dog behaving badly at home?
Charlie is chewing up the furniture and barking at inappropriate hours.
I usually walk Charlie for 30 minutes twice a day. And to my surprise, 30 minutes twice a day is not enough for Puppies especially for Beagle Puppies. Puppies just have a seemingly endless amount of energy and you need to empty the tank by the end of the day to keep your puppy and yourself sane.
The trainer suggested that I'd either increase the frequency and length of each walk or continue my present routine and strap one of these on Charlie and load it down with a couple of water bottles.
The theory is to make Charlie work a little harder for a shorter period of time. So a 30 minute walk with the above apparatus equals 60 minute walk without.
Man… sometimes I wish I got a cat instead.
August 29, 2007 Posted by xraytech | How-To	| Comments Off on Is your dog behaving badly at home?
What do you think of these Transformer/Star Wars cross-over ideas? I can't decide whether to be giddy or nauseous. Be sure to check out the Millenium Falcon transformer, too.
You know those handy dandy keyless car entry systems? Well, the bad news is that since most manufacturers went with the same encryption system, KeeLoq, all these cars can be broken into in the same fashion. The good news is that the perp would have to monitor the transmission between the key and the lock for about an hour in order to do so.
I bet you've never heard them like this. Be sure to listen to the second solo.
Here's a new sport for you adrenaline junkies out there.
Here are my photos of August 28th's Lunar Eclipse. Shot from central coast, CA using 1000 mm f/5 Newtonian telescope and Canon Digital Rebel.


I have and virt-manager installed, but I can do a bit of virsh also. I want to add 2 physical SATA disks to my system without creating a raid. I want the RAID on the virtual machine so I can install Gentoo. This is so I can experiment with btrfs (butter filesystem to me) for the reasons of developing temporal database capabilities.
What's the problem? You can add those drives to the VM as separate VirtIO disks. And then you can create RAID with Gentoo on them.
It's weird, doesn't seem to have the option. Might I have to install an extra package after the 26 alpha? I went alpha for once is all.
I don't think so. Do you know where to look? Open virt-manager, then open the virtual machine (don't need to turn it on), then select the second button to open VM configuration. Then you can add new hardware to the VM.


Midshipmen are required to meet with their class advisor within the first few weeks of the semester (September or January). The following documents must be completed and brought with the Midshipman to the advising meeting.
Midshipmen are required to meet with their class advisor half-way through the academic semester (generally October or March). Midshipmen must complete the memo below to report their academic performance to their class advisor.


text:
HHS rescinds rule protecting faith-based foster care, adoption agencies
The U.S. Department of Health and Human Services Nov. 18 rescinded a Trump administration rule that ensured faith-based social service agencies that provide adoption and foster care would continue receiving federal funding for services that follow their religious beliefs.
Nation/World. July 5, 2019
HHS delays implementation of conscience protection rule for health workers
The U.S. Department of Health and Human Services agreed in a July 1 court filing to postpone implementation of a new conscience protection rule "because it is the most efficient way to adjudicate the final rule on the merits."
Nation/World. August 8, 2018
HHS urged to reissue rules to enforce required notice on abortion coverage
The U.S. Department of Health and Human Services must issue new regulations to enforce a requirement that consumers be notified if a subsidized health plan offered under the Affordable Care Act covers elective abortion and informed they must pay an extra amount for the plan's abortion coverage.
Nation/World. October 25, 2017
HHS looks into removing barriers to program funding for religious groups
The U.S. Department of Health and Human Services announced Oct. 25 that it is looking for ways to ensure that faith-based and religious groups are not blocked from receiving HHS funding for their programs.
Nation/World. August 30, 2017
Catholic Benefits Association presses Trump to end contraceptive mandate
Frustrated by federal court inaction and the Department of Justice blocking the way, the Catholic Benefits Association has called on President Trump to intervene directly in the legal battle over the Affordable Care Act's contraception mandate.
Nation/World. December 29, 2016
Benefits association, diocese file suit over HHS transgender regulation
The Catholic Benefits Association, the Diocese of Fargo and Catholic Charities North Dakota filed a lawsuit Dec. 28 in U.S. District Court in North Dakota against a federal regulation scheduled to take effect Jan. 1 that redefines "sex" for anti-discrimination purposes to include sexual orientation and gender identity.


August 20 is World Mosquito Day, a commemoration of the 1897 discovery by British doctor Ronald Ross that female anopheles mosquitoes transmit malarial parasites to humans. 121 years on, despite significant scientific progress in treating and preventing malaria, mosquitoes remain the world's deadliest animal. These tiny insects cause more than half a million deaths annually, not just from malaria, but also a range of other diseases such as yellow fever, dengue and chikungunya.
In most cases of human malaria, one of four species of Plasmodium parasite is transmitted directly between mosquitoes and people, but in some parts of rural south east Asia, another species, known as 'Monkey Malaria' is on the rise. In this rarer form of the disease, a fifth species of the parasite can be transmitted by mosquitoes between monkeys and people. In Malaysia, Bayer is partnering with the Ministry of Health on new research to help protect local populations from this little-known disease.
Dr. Rohani Ahmad is currently the Head of Medical Entomology Unit at the Institute for Medical Research (IMR), the research arm of the Ministry of Health of Malaysia. Since 2015, she has been studying the effectiveness of residual spraying of insecticide to prevent Monkey Malaria in six rural villages on the island of Borneo in eastern Malaysia. One of the first of its kind, her study is funded principally by the MOH of Malaysia and are supported by Bayer's Asia Pacific Vector Control team, led by the ES team in Malaysia and Say Piau Lim, Vector Control Market Segment Manager APAC.
"Cases of human malaria have dropped rapidly in Malaysia since 2000 and the Ministry of Health aims to eradicate the disease by 2020. But cases of monkey malaria in humans have been on the rise since the country's first reported cases in 2008. Indeed incidence has increased by over 170% to 3614 cases between 2016 and 2017.
"This form of malaria is endemic to monkeys living in the rain forests of Borneo and is caused by Plasmodium knowlesi. The parasite proliferates in Anopheles balabacensis mosquitoes that commonly bite infected monkeys. The mosquitoes then transmit it to humans, causing symptoms that are similar to human malaria, but with a shorter incubation period of under one week. Monkey malaria can be cured if treated rapidly, but there have been fatalities reported when treatment was delayed.
Dr. Rohani and her team are focusing on six villages deep in the rural areas around the town of Tenom in the eastern Malaysian state of Sabah. The project began in late 2015 with colonies of Anopheles balabacensis mosquitoes collected and studied in the laboratory before being returned and monitored in the field. As the mosquitoes mainly remain and bite outside dwellings, exterior walls were sprayed, with K-Othrine Polyzone, first in April 2016 and again in November/December 2016.
"The first results are very positive, especially in what we call outdoor covered areas, such as roofed terraces where local populations meet to eat or sit. After monitoring the number of mosquitoes that died following exposure to walls in these areas, we estimate that the product retains 90-100% efficacy after six months. The number of cases of monkey malaria reported also dropped dramatically in the six villages, compared to other local villages where monkey malaria is still increasing.
"Education of the local population is another important part of the project. Dr. Rohani and her team have involved the heads of all of the six villages, who help them to organize meetings to explain how the disease is transmitted, how the spraying is being organized and how it can help.


text:
Q: Jython : SyntaxError: Lexical error at line 29, column 32. Encountered: "$" (36), after : "" I am getting a syntax error in my code. Can anyone say what's wrong in the syntax? I am new to this language, don't have much of an idea.
Error Message:
WASX7017E: Exception received while running file "jdbcConn.jy"; exception information: com.ibm.bsf.BSFException: exception from Jython: Traceback (innermost last): (no code object) at line 0 File "", line 29 classpath = ["classpath" , ${ORACLE_JDBC_DRIVER_PATH}/ojdbc6.jar ] ^ SyntaxError: Lexical error at line 29, column 32. Encountered: "$" (36), after : ""
My Code :
import sys

## **JDBCProvider** ##

def OracleJDBC(cellName,serverName):
    name ="Oracle JDBC Driver"
    print " Name of JDBC Provider which will be created ---> " + name
    print "   ----------------------------------------------------------------------------------------- "
    # Gets the name of cell
    cell = AdminControl.getCell()
    print cell
    cellid = AdminConfig.getid('/Cell:'+ cell +'/')
    print cellid
    print "   ----------------------------------------------------------------------------------------- "
    ## Creating New JDBC Provider ##
    print " Creating New JDBC Provider :"+ name
    n1 = ["name" , "Oracle JDBC Driver" ]
    desc = ["description" , "Oracle JDBC Driver"]
    impn = ["implementationClassName" ,
            "oracle.jdbc.pool.OracleConnectionPoolDataSource"]
    classpath = ["classpath" , ${ORACLE_JDBC_DRIVER_PATH}/ojdbc6.jar ]
    attrs1 = [n1 , impn , desc , classpath]

    n1 = ["name" , "Oracle JDBC Driver" ]

    desc = ["description" , "Oracle JDBC Driver"]

    impn = ["implementationClassName" , "oracle.jdbc.pool.OracleConnectionPoolDataSource"]

    classpath = ["classpath" , "${ORACLE_JDBC_DRIVER_PATH}/ojdbc6.jar"]

    attrs1 = [n1 , impn , desc , classpath]

    Serverid = AdminConfig.getid("/Cell:" + cellName + "/ServerName:" + serverName +"/")

    jdbc = AdminConfig.create('JDBCProvider', Serverid, attrs1)

    print " New JDBC Provider created :" + name

    AdminConfig.save()

    print " Saving Configuraion "

    print "   ----------------------------------------------------------------------------------------- "

    ####################################################################################################################
    ####################################################################################################################

#main program starts here
if __name__ == '__main__':
    cellName = sys.argv[0]
    serverName = sys.argv[1]
    OracleJDBC(cellName,serverName)

A: Your problem is in this line:
classpath = ["classpath" , ${ORACLE_JDBC_DRIVER_PATH}/ojdbc6.jar ]

Instead, do something like
opath = os.getenv("ORACLE_JDBC_DRIVER_PATH")
classpath = ["classpath", "{}/ojdbc6.jar".format(opath)]

"${ORACLE_JDBC_DRIVER_PATH}" is shell syntax, not Python.

A: Change this line
classpath = ["classpath" , ${ORACLE_JDBC_DRIVER_PATH}/ojdbc6.jar ]

to this:
classpath = ["classpath" , "${ORACLE_JDBC_DRIVER_PATH}/ojdbc6.jar" ]

or better yet, just delete that line. Anyways, classpath is declared again later to the same value

redpajama_set_name: RedPajamaStackExchange


Parking spaces located on Highbank Drive in Liverpool. The spaces are suitable for vehicles up to the size of a Van. On-site there is Allocated Space and Arranged Transfers.
South Parkway Station providing access to the city of Liverpool and Manchester and many other cities in the North of England.
I am also a 10-minute drive from Liverpool John Lennon Airport. I offer Airport transfer when available for £5 per trip. If I am not available a local taxi number can be provided for you.
Reply quickly and easy to find.


Slate tiles may come in different thickness but no matter the thickness of a slate tile, cutting it is not a problem as long as you have the right materials. Having a general idea of how to cut the tile also helps.... Tile cutters also make it too easy to cut too much of the tile if you only need to remove one corner. Luckily there are methods for cutting tiles that don't require a tile cutter. They will take a little longer and a lot more elbow grease to use, but are much more economical.
Cutting width range: 35mm - 800mm. 800mm heavy duty all steel tile cutter with laser guide. Position-adjustable steel side support to prevent the tile from accidental moving. Extraordinarily long breaker bar saves you more energy.... Tiles of Solnhofen Natural Stone with hand-cut edges are an visually attractive alternative to the machined cut floor tiles. Additional the hand-cut tiles are provided with a very favourable price (30% under the price for machined cut floor tiles).
Hand sawing is time consuming but can achieve great results. Neat curves are possible with this method, for example. Don't use too much force or you risk breaking particularly lightweight tiles. You help to avoid this you can place some MDF or light wood under the tile to give it support. Cut through both the wood and the tile to reduce the risk of breakages. how to add a google+ brand page Cutting width range: 35mm - 800mm. 800mm heavy duty all steel tile cutter with laser guide. Position-adjustable steel side support to prevent the tile from accidental moving. Extraordinarily long breaker bar saves you more energy.
Hand sawing is time consuming but can achieve great results. Neat curves are possible with this method, for example. Don't use too much force or you risk breaking particularly lightweight tiles. You help to avoid this you can place some MDF or light wood under the tile to give it support. Cut through both the wood and the tile to reduce the risk of breakages.
Most score-and-snap tile cutters follow the same basic principle: You fit the tile into a cutting bed (sometimes with guides that help keep your cut straight), and then push or pull a rotating blade across the tile's glaze, to score it.
Tiles of Solnhofen Natural Stone with hand-cut edges are an visually attractive alternative to the machined cut floor tiles. Additional the hand-cut tiles are provided with a very favourable price (30% under the price for machined cut floor tiles).