\documentclass[reqno]{amsart} \usepackage{amssymb} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 13, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/13\hfil Local well-posedness] {Local well-posedness for a higher order nonlinear Schr\"{o}dinger equation in Sobolev spaces of negative indices} \author[Xavier Carvajal\hfil EJDE-2004/13\hfilneg] {Xavier Carvajal} \address{Xavier Carvajal \hfill\break IMECC-UNICAMP, Caixa Postal:6065, 13083-859, Bar\~ao Geraldo, Campinas, SP, Brazil} \email{carvajal@ime.unicamp.br} \date{} \thanks{Submitted July 30, 2003. Published January 23, 2004.} \thanks{Partially supported by FAPESP, Brazil.} \subjclass[2000]{35Q58, 35Q60} \keywords{Schr\"{o}dinger equation, Korteweg-de Vries equation, \hfill\break\indent trilinear estimate, Bourgain spaces} \begin{abstract} We prove that the initial value problem associated with $$\partial_tu+i\alpha \partial^2_x u+\beta \partial^3_x u +i\gamma|u|^2u = 0, \quad x,t \in \mathbb{R},$$ is locally well-posed in $H^s$ for $s>-1/4$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}{Proposition} \section{Introduction} In this work, we study a particular case of the initial value problem (IVP) $$\label{0y0} \begin{gathered} \partial_tu+i\alpha \partial^2_x u+\beta \partial^3_x u+F(u) = 0, \quad x,t \in \mathbb{R},\\ u(x,0) = u_0(x)\,. \end{gathered}$$ Here $u$ is a complex valued function, $F(u)=i\gamma|u|^2u+\delta |u|^2\partial_x u+\epsilon u^2\partial_x \overline{u}$, $\gamma , \delta , \epsilon \in \mathbb{C}$ and $\alpha,\beta \in \mathbb{R}$ are constants. Hasegawa and Kodama \cite{[H-K], [Ko]} proposed (\ref{0y0}) as a model for propagation of pulse in optical fiber. We will study the IVP (\ref{0y0}) in Sobolev space $H^s(\mathbb{R})$ under the condition $\delta=\epsilon=0$, $\beta \neq 0$ (see case (iv) in Theorem \ref{Teo1} below). When $\gamma , \delta , \epsilon \in \mathbb{R}$, it was shown in \cite{[C2]} that the flow associated to the IVP (\ref{0y0}) leaves the following quantity $$I_1(v) = \int_{\mathbb{R}}|v|^2(x,t)\,dx, \label{1.2}$$ conserved in time. Also, when $\delta-\gamma=\epsilon \neq 0$ we have the following quantity conserved: $$I_2(v) = c_1\int_{\mathbb{R}}|\partial_x v|^2(x,t)\,dx +c_2\int_{\mathbb{R}}|v|^4(x,t)dx + c_3\int_{\mathbb{R}}v(x,t)\partial_x\overline{v(x,t)}dx, \, \label{1.3}$$ where $c_1=3\beta\,\epsilon$, $c_2=-\epsilon(\epsilon+\delta)/2$ and $c_3=i(\alpha(\epsilon+\delta)-3\beta\gamma)$. These quantities were used in \cite{[C2]} to establish global well-posedness for (\ref{0y0}) in $H^s(\mathbb{R})$, $s\geq 1$. Note that the quantity $i\int_{\mathbb{R}}v(x,t)\partial_x\overline{v(x,t)}dx$ in (\ref{1.3}) is real since $$\partial_t(\,i\int_{\mathbb{R}}v(x,t)\partial_x\overline{v(x,t)}dx)= 2\epsilon \, \mathop{\rm Im}(\int_{\mathbb{R}}[v(x,t)\partial_x\overline{v(x,t)}]^2dx).$$ We say that the IVP (\ref{0y0}) is locally well-posed in $X$ (Banach space) if the solution uniquely exists in certain time interval $[-T,T]$ (unique existence), the solution describes a continuous curve in $X$ in the interval $[-T,T]$ whenever initial data belongs to $X$ (persistence), and the solution varies continuously depending upon the initial data (continuous dependence) i.e. continuity of application $u_{0} \mapsto u(t)$ from $X$ to $\mathcal{C}([-T,T];X)$. We say that the IVP (\ref{0y0}) is globally well-posed in $X$ if the same properties hold for all time $T>0$. If some hypothesis in the definition of local well-posed fails, we say that the IVP is ill-posed. Particular cases of (\ref{0y0}) are the following: \noindent$\bullet$ Cubic nonlinear Schr\"odinger equation (NLS), ($\alpha=\mp 1$, $\beta=0$, $\gamma=-1$, $\delta=\epsilon=0$). $$\label{3y0} iu_t \pm u_{xx} + |u|^2u = 0, \quad x,t \in \mathbb{R}.$$ The best known local result for the IVP associated to (\ref{3y0}) is in $H^s(\mathbb{R})$, $s \geq 0$, obtained by Tsutsumi \cite{[Ts]}. \noindent$\bullet$ Nonlinear Schr\"odinger equation with derivative ($\alpha=-1$, $\beta=0$, $\gamma=0$, $\delta=2\epsilon$). $$\label{4y0} iu_t + u_{xx} + i\lambda(|u|^2u)_x =0, \quad x,t\in \mathbb{R}.$$ The best known local result for the IVP associated to (\ref{4y0}) is in $H^s(\mathbb{R})$, $s \geq 1/2$, obtained by Takaoka \cite{[T1]}. \noindent$\bullet$ Complex modified Korteweg-de Vries (mKdV) equation ($\alpha=0$, $\beta=1$, $\gamma=0$, $\delta=1$, $\epsilon=0$). $$\label{83y0} u_t+u_{xxx}+|u|^2u_x =0, \quad x,t \in \mathbb{R}.$$ If $u$ is real, (\ref{83y0}) is the usual mKdV equation and Kenig et al. \cite{[KPV1]} proved the IVP associated to it is locally well-posed in $H^s(\mathbb{R})$, $s \geq 1/4$. \noindent$\bullet$ When $\alpha \neq 0$ is real and $\beta=0$, we obtain a particular case of the well-known mixed nonlinear Schr\"odinger equation $$\label{6y0} u_t=i\alpha u_{xx}+\lambda(|u|^2)_x u+g(u),\quad x,t \in \mathbb{R},$$ where $g$ satisfies some appropriate conditions. Ozawa and Tsutsumi in \cite{[O-T]} proved that for any $\rho >0$, there is a positive constant $T(\rho)$ depending only on $\rho$ and $g$, such that the IVP (\ref{6y0}) is locally well-posed in $H^{1/2}(\mathbb{R})$, whenever the initial data satisfies $\|u_{0}\|_{\mathrm{H}^{1/2}} \leq \rho.$ There are other dispersive models similar to (\ref{0y0}). The interested readers can see the following works and the references therein \cite{[A],[CC],[P-Na],[P-S-S-M],[SS]}. Laurey \cite{[C1],[C2]} proved that the IVP associated to (\ref{0y0}) is locally well-posed in $H^s(\mathbb{R})$, $s > 3/4$. Staffilani \cite{[G]} improved this result by proving the IVP associated to (\ref{0y0}) is locally well-posed in $H^s(\mathbb{R})$, $s \geq 1/4$. When $\alpha, \beta$ are functions of\, $t$, we proved in \cite{C1, CL1} local well-posedness in $H^s(\mathbb{R})$, $s \geq 1/4$. Also we studied in \cite{C1, MP} the unique continuation property for the solution of (\ref{0y0}). Regarding the ill-posedness of the IVP (\ref{0y0}), we proved in \cite{CL0} the following theorem. \begin{theorem}\label{Teo1} The mapping data-solution $u_{0} \mapsto u(t)$ for the IVP (\ref{0y0}) is not $\mathcal{C}^3$ at origin in the following cases: \begin{itemize} \item[(i)] $\beta=0$, $\alpha \neq 0$, $\delta =\epsilon=0$, $\gamma \neq 0$ for $s<0$. \item[(ii)] $\beta=0$, $\alpha \neq 0$, $\delta \neq 0$ or $\epsilon \neq 0$ for $s<1/2$. \item[(iii)] $\beta \neq 0$, $\delta \neq 0$ or $\epsilon \neq 0$ for $s<1/4$. \item[(iv)] $\beta \neq 0$, $\delta = \epsilon=0$, $\gamma \neq 0$ for $s<-1/4$. \end{itemize} \end{theorem} In this work, we consider the case (iv) and prove the following result. \begin{theorem}\label{T2} Let $\beta \neq 0$ real and $\gamma \neq 0$ complex, then the following IVP $$\label{0y1} \begin{gathered} \partial_tu+i\alpha \partial^2_x u+\beta \partial^3_x u+i\gamma|u|^2u = 0, \quad x,t \in \mathbb{R},\\ u(x,0)= u_{0}, \end{gathered}$$ is locally well-posed in $H^s(\mathbb{R})$, $s > -1/4$. \end{theorem} The following trilinear estimate will be fundamental in the proof of Theorem \ref{T2}. \begin{theorem}\label{Tt3} Let $-1/4 < s \leq 0$, $b>7/12$, $b'0$, as it can be seen by combining $\langle\xi\rangle^s \leq \langle\xi-(\xi_{2}-\xi_{1})\rangle^s\langle\xi_{2}\rangle^s\langle\xi_{1}\rangle^s$ and the estimate (\ref{T3}) for $s=0$. \noindent$\bullet$ We will use the notation $\|u\|_{\{s,b\}}:=\|u\|_{X^{s,b}.}$ \noindent$\bullet$ When $\alpha =0, \beta=1$, we have the usual bilinear estimate due to Kenig et al. \cite{[KPV8]}, $\|(uv)_x \|_{\{-3/4+,-1/2+\}} \leq C \|u\|_{\{-3/4+,1/2+\}}\|v\|_{\{-3/4+,1/2+\}.}$ Also we have the 1/4 trilinear estimate due to Tao \cite{[Tao]}, $\|(uvw)_x \|_{\{1/4,-1/2+\}} \leq C \|u\|_{\{1/4,1/2+\}}\|v\|_{\{1/4,1/2+\}} \|w\|_{\{1/4,1/2+\}}.$ \section{Proofs of Main Result} \begin{proof}[Proof of Theorem \ref{T4}] As in \cite{[KPV8]} consider the set $B:=\{(\xi,\tau); N\leq \xi \leq N+N^{-1/2}, |\tau-\phi(\xi)| \leq 1\},$ where $\phi(\xi)=\alpha \xi^2+\beta \xi^3$. We have $|B| \thicksim N^{-1/2}$. Let us consider $\hat{v}=\chi_{B}$, it is not difficult to see that $\|v\|_{\{s,b\}}\leq N^s|B|^{1/2}$. Moreover $\mathcal{F}(|v|^2\overline{v}):=\chi_{B}\ast\chi_{B}\ast\chi_{-B}\gtrsim \frac{1}{N}\chi_{A},$ where $A$ is a rectangle contained in $B$ such that $|A|\thicksim N^{-1/2}$.\\ Therefore $\|\,|v|^2\overline{v}\|_{\{s,b'\}} =\|\langle \xi \rangle^s\langle \tau-\phi(\xi) \rangle^{b'} \mathcal{F}(|v|^2\overline{v})\|_{\mathrm{L_{\xi}^2L_{\tau}^2}} \gtrsim N^s\frac{1}{N}N^{-1/4}=N^{s-5/4}.$ As a consequence, for large $N$ the trilinear estimate fails if $3(s-1/4)1/2$, $a_{1}, a_{2} \in \mathbb{R}$ then $$\label{el1} \int_{\mathbb{R}}\frac{dx}{\langle x-a_{1}\rangle^{2b}\langle x-a_{2}\rangle^{2b}} \thicksim \frac{1}{\langle a_{1}-a_{2}\rangle^{2b}}.$$ \item If\, $0< c_{1}, c_{2}<1$, $c_{1}+c_{2}>1$, $a_{1}\neq a_{2}$, then $$\label{el2} \int_{\mathbb{R}}\frac{dx}{|x-a_{1}|^{c_{1}}|x-a_{2}|^{c_{2}}}\lesssim \frac{1}{|a_{1}- a_{2}|^{(c_{1}+c_{2}-1)}}.$$ \item Let $a \in \mathbb{R}$, $c_{1} \leq c_{2}$, then $$\label{el3} \frac{|x|^{c_{1}}}{\langle ax \rangle^{c_{2}}} \leq \frac{1}{|a|^{c_{1}}}.$$ \item Let $a, \eta \in \mathbb{R}$, $b>1/2$, then $$\label{el4} \int_{\mathbb{R}} \frac{dx}{\langle a(x^2-\eta^2)\rangle^{2b}} \lesssim \frac{1}{|a\eta|}.$$ \end{enumerate} \end{lemma} Now, let $f(\xi,\tau)=\,\langle\xi\rangle^s\,\langle\tau-\xi^3\rangle^{b}\hat{u}$, $g(\xi,\tau)=\,\langle\xi\rangle^s\,\langle\tau-\xi^3\rangle^{b}\hat{v}$, $h(\xi,\tau)=\langle\xi\rangle^s$, $\langle\tau-\xi^3\rangle^{b}\hat{w}$, $\eta=(\xi,\tau)$, $x=(\xi_{1},\tau_{1})$, $y=(\xi_{2},\tau_{2})$. We have \begin{align*} \|uv\overline{w}\|_{\{s,b'\}}= & \|\int_{\mathbb{R}^{4}} f(\eta+x-y)g(y)\overline{h}(x)K(\eta,x,y)dx dy \|_{\mathrm{L^2_{\eta}}}\\ \leq & \|K(\eta,x,y)\|_{\mathrm{L^{\infty}_{\eta}}\mathrm{L^2_{x,y}}} \|f\|_{\mathrm{L^2}}\|g\|_{\mathrm{L}^2}\|h\|_{\mathrm{L}^2}, \end{align*} where $K(\eta,x,y)=\frac{\langle\xi+\xi_{1}-\xi_{2}\rangle^{\rho} \langle\xi_{2}\rangle^{\rho}\langle\xi_{1}\rangle^{\rho}} {r(\xi,\tau)\langle\tau_{1}-\phi(\xi_{1})\rangle^{b}\langle\tau_{2} -\phi(\xi_{2})\rangle^{b}\langle\tau+ \tau_{1}-\tau_{2}-\phi(\xi+\xi_{1}-\xi_{2})\rangle^{b}}$ and $r(\xi,\tau)=\langle\xi\rangle^{\rho}\langle\tau-\phi(\xi)\rangle^{-b'}$, $\rho=-s$. Using (\ref{el1}) we obtain \begin{align*} I(\xi,\tau):=\|K\|_{\mathrm{L^2_{x,y}}}^2 \thicksim & \frac{1}{r(\xi,\tau)^2}\int_{\mathbb{R}^2} \frac{G_{\rho}(\xi, \xi_{1}, \xi_{2})\,\,d\xi_{1}d\xi_{2}} {\langle\tau-\phi(\xi+\xi_{1}-\xi_{2})-\phi(\xi_{2})+\phi(\xi_{1})\rangle^{2b}}\\ =& \frac{1}{r(\xi,\tau)^2}\int_{\mathbb{R}^2} \frac{G_{\rho}(\xi, \xi_{1}, \xi_{2})\,\,d\xi_{1}d\xi_{2}} {\langle\tau-\phi(\xi)+ g(\xi,\xi_{1},\xi_{2}) \rangle^{2b}} \end{align*} where $$\label{Gg} \begin{gathered} G_{\rho}(\xi,\xi_{1},\xi_{2}) :=\langle\xi+\xi_{1}-\xi_{2}\rangle^{2\rho}\langle\xi_{1}\rangle^{2\rho} \langle\xi_{2}\rangle^{2\rho},\\ g(\xi,\xi_{1},\xi_{2})= (\xi_{1}-\xi_{2})(\xi+\xi_{2})(2\alpha +3\beta (\xi-\xi_{1})). \end{gathered}$$ Assuming $y=\tau- \phi(\xi)$, to get Theorem \ref{Tt3} it is sufficient to prove the following lemma. \begin{lemma}\label{lpri} Let $0\leq\rho <1/4$, $b>7/12$, $b'< -\rho/3$. Then $I(\xi,y):=\frac{1}{\langle\xi\rangle^{2\rho}\langle y\rangle^{-2b'}}\int_{\mathbb{R}^2} \frac{G_{\rho}(\xi,-\xi_{1},-\xi_{2})d\xi_{1}d\xi_{2}}{\langle y+g(\xi,\xi_{1},\xi_{2})\rangle^{2b}} \leq C(\rho, b, b')< \infty,$ where $C(\rho, b, b')$ is a constant independent of $\xi$ and $y$. \end{lemma} To prove Lemma \ref{lpri} we need to prove the following lemmas. \begin{lemma}\label{lpri1} Let $\rho<1/4$. Then $I(0,0)= \int_{\mathbb{R}^2} \frac{G_{\rho}(0,-\xi_{1},-\xi_{2})d\xi_{1}d\xi_{2}}{\langle g(0,\xi_{1},\xi_{2})\rangle^{2b}} = \begin{cases} C(\rho, b)< \infty & \mbox{if \rho +1/3 7/12. Then \[ I(\xi,0)=\frac{1}{\langle\xi\rangle^{2\rho}}\int_{\mathbb{R}^2} \frac{G_{\rho}(\xi,-\xi_{1},-\xi_{2})d\xi_{1}d\xi_{2}}{\langle g(\xi,\xi_{1},\xi_{2})\rangle^{2b}} \leq C(\rho, b),$ where $C(\rho, b)$ is a constant independent of $\xi$. \end{lemma} For clarity in exposition, we consider the case $\alpha=0$, $\beta=1$, i.e. $\phi(\xi)=\xi^3$ (see the observation at the end of the proof of Lemma \ref{lpri}). In the definition of $I(\xi,y)$ if we make the change of variables $\xi-\xi_{1}:=\xi \xi_{1}$, $\xi+\xi_{2}:=\xi \xi_{2}$ and $y=\xi^3z$, then $I(\xi,y)$ becomes $$\label{intpri} I(\xi,z)=p(\xi,z)\int_{\mathbb{R}^2}\frac{H_{\rho}(\xi, \xi_{1}, \xi_{2})d\xi_{1}d\xi_{2}}{\langle\xi^3(z+F(\xi_{1},\xi_{2}))\rangle^{2b}},$$ where $p(\xi,z)=\xi^2\langle\xi^3z\rangle^{2b'}\langle\xi\rangle^{-2\rho}$,\, $F(\xi_{1},\xi_{2})=(2-(\xi_{1}+\xi_{2}))\xi_{1}\xi_{2}$\,\, and $$H_{\rho}(\xi,\xi_{1},\xi_{2})= \langle\xi(1-(\xi_{1}+\xi_{2}))\rangle^{2\rho}\langle\xi(1-\xi_{1})\rangle^{2\rho} \langle\xi(1-\xi_{2})\rangle^{2\rho}.$$ From here onwards we will suppose $z>0$, because if $z<0$ we can obtain the same result by symmetry (see Remark after Proposition \ref{prop2}). \begin{proof}[Proof of Lemma \ref{lpri1}] By symmetry it is sufficient to prove that the integrals \begin{gather*} I_{1}(0,0):=\int_{0}^{\infty}\int_{0}^{\infty}\frac{G_{\rho}(0,-\xi_{1},-\xi_{2}) d\xi_{1}d\xi_{2}}{\langle g(0,\xi_{1},\xi_{2})\rangle^{2b}} ,\\ I_{2}(0,0):=\int_{0}^{\infty}\int_{0}^{\infty}\frac{G_{\rho}(0,-\xi_{1},\xi_{2}) d\xi_{1}d\xi_{2}}{\langle g(0,\xi_{1},-\xi_{2})\rangle^{2b}} \end{gather*} are finite. We will prove that $I_{1}(0,0)$ is finite only; the same proof works for $I_{2}(0,0)$. Also, by symmetry we can suppose that $0\leq \xi_{2} \leq \xi_{1}$. We have \label{eql3} \begin{aligned} \int_{1}^{\infty}d\xi_{1}\int_{0}^{\xi_{1}}d\xi_{2} \frac{G_{\rho}(0,-\xi_{1},-\xi_{2})}{\langle g(0,\xi_{1},\xi_{2})\rangle^{2b}} =&\int_{1}^{\infty}d\xi_{1}\int_{0}^{\xi_{1}/2}d\xi_{2}+ \int_{1}^{\infty}d\xi_{1}\int_{\xi_{1}/2}^{\xi_{1}}d\xi_{2}\\ = &I_{1,1}+I_{1,2}. \end{aligned} Since $0\leq \xi_{2} \leq \xi_{1}$, we have $G_{\rho}(0,-\xi_{1},-\xi_{2}) \leq \langle\xi_{1}\rangle^{4\rho}\langle\xi_{2}\rangle^{2\rho}$. In $I_{1,1}$ we have $\xi_{1}/2< \xi_{1}-\xi_{2}<\xi_{1}$, therefore if $b>\rho+1/3$, \begin{align*} I_{1,1} & \lesssim \int_{1}^{\infty}\langle\xi_{1}\rangle^{4\rho}d\xi_{1}\int_{0}^{\xi_{1}/2}\frac{\langle\xi_{2}\rangle^{2\rho}d\xi_{2}} {\langle 3\xi_{1}^2\xi_{2}\rangle^{2b}}\\ \lesssim & \int_{1}^{\infty}\langle\xi_{1}\rangle^{4\rho}\Big(\frac{1}{\xi_{1}^2}+\frac{1}{\xi_{1}^{2+4\rho}}+ \frac{1}{\xi_{1}^{2+4\rho}} \int_{1}^{3\xi_{1}^3/2}\frac{x^{2\rho}dx}{(1+x)^{2b}}\Big)d\xi_{1} \\ =& C(\rho,b)< \infty. \end{align*} Analogously we can prove that $I_{1,1}=\infty$ if $b \leq \rho+1/3$. In $I_{1,2}$ we have $\xi_{1}/2\leq \xi_{2} \leq\xi_{1}$, so \begin{align*} I_{1,2} \lesssim & \int_{1}^{\infty}\langle\xi_{1}\rangle^{4\rho}d\xi_{1}\int_{\xi_{1}/2}^{\xi_{1}} \frac{\langle\xi_{1}-\xi_{2}\rangle^{2\rho}d\xi_{2}} {\langle(\xi_{1}-\xi_{2})\xi_{1}^2\rangle^{2b}}\\ =&\int_{1}^{\infty}\langle\xi_{1}\rangle^{4\rho}d\xi_{1}\int_{0}^{\xi_{1}/2} \frac{\langle x\rangle^{2\rho}dx} {\langle\xi_{1}^2x\rangle^{2b}} \\ \lesssim & \,C(\rho,b), \quad b>\rho+1/3. \end{align*} \end{proof} The propositions will be useful for proving Lemmas \ref{lpri} and \ref{lpri2}. \begin{proposition}\label{prop2} Let $0\leq\rho<1/4$, $b>1/3+2\rho/3$, then we have $J_{1}=\xi^{2+4\rho}\int_{\mathbb{R}^2}\frac{d\xi_{1}d\xi_{2}}{\langle\xi^3(z+F)\rangle^{2b}} \leq C,$ where $C$ is a constant independent of $\xi$ and $z$. \end{proposition} \begin{proof} If $\xi_{1}\leq 0$, $\xi_{2}\leq 0$, then $|z+F|\geq |\xi_{1}+\xi_{2}||\xi_{1}\xi_{2}|$. Therefore by Lemma \ref{lpri1} and by symmetry, it is enough to consider $\xi_{1} \geq 0$. We have $|z+F|=|\xi_{1}||(\xi_{2}+(\xi_{1}-2)/2)^2- (\xi_{1}-2)^2/4-z/\xi_{1}|$. Let $l^2=(\xi_{1}-2)^2/4+z/\xi_{1}$, $c(\rho)=(2+4\rho)/3$, then making change of variable $\eta=\xi_{2}+(\xi_{1}-2)/2$ and using (\ref{el2}) and (\ref{el3}) we have \begin{align*} J_{1} =& \xi^{2+4\rho}\int_{0}^{\infty}d\xi_{1} \int_{\mathbb{R}}\frac{d\eta}{\langle\xi^3\xi_{1}(\eta^2-l^2)\rangle^{2b}}\\ \lesssim &\int_{0}^{\infty}d\xi_{1}\int_{\mathbb{R}}\frac{l \,dx}{[|\xi_{1}|l^2|x^2-1|]^{c(\rho)}}\\ \lesssim &\int_{0}^{\infty} \frac{d\xi_{1}}{|\xi_{1}|^{c(\rho)}|\xi_{1}-2|^{(1+8\rho)/3}}\int_{\mathbb{R}} \frac{dx}{|x^2-1|^{c(\rho)}}\\ \lesssim & \,2^{-(20\rho+1)/3}, \quad 0<\rho<1/4. \end{align*} The case $\rho=0$ follows from the case $0<\rho<1/4$, taking the limit. \end{proof} \subsection*{Remark} When $z<0$, we make $\xi_1:=-\xi_1$, $\xi_2:=-\xi_2$ then $|z+F|=|\xi_{1}||(\xi_{2}+(\xi_{1}+2)/2)^2- (\xi_{1}+2)^2/4+z/\xi_{1}|$ and the proof is similar. \begin{proposition}\label{prop1} Let $|\xi|>1$, $b>1/2$, $0\leq\rho<1/4$. Then $J_{2}=\xi^{2+4\rho} \int_{0}^{\infty}\xi_{1}^{4\rho}d\xi_{1}\int_{\mathbb{R}}\frac{d\xi_{2}}{\langle\xi^3(z+F)\rangle^{2b}} \leq C,$ where $C$ is a constant independent of $\xi$ and $z$. \end{proposition} \begin{proof} By Proposition \ref{prop2} we can suppose $\xi_{1}>4$, so $(\xi_{1}-2)> \xi_{1}/2$. Using (\ref{el4}) and making change of variables as above, we have$J_{2}\lesssim \, \frac{\xi^{2+4\rho}}{|\xi|^3}\int_{4}^{\infty}\frac{\xi_{1}^{4\rho}}{\xi_{1}\,l} d\xi_{1} \leq C.$ \end{proof} \begin{proof}[Proof of Lemma \ref{lpri2}] \textbf{Case} $|\xi| \leq 1$. Let \begin{gather*} A_{1}=\{(\xi_{1},\xi_{2})/ |\xi_{1}|>2, |\xi_{2}|>2\},\quad A_{2}=\{(\xi_{1},\xi_{2})/|\xi_{1}|\leq 2, | \xi_{2}|\leq 2\},\\ A_{3}=\{(\xi_{1},\xi_{2})/ |\xi_{1}|\leq 2, | \xi_{2}|>2\},\quad A_{4}=\{(\xi_{1},\xi_{2})/ |\xi_{1}|>2, | \xi_{2}|\leq 2\}\,. \end{gather*} Consider $I(\xi,0)=\sum_{j=1}^{4}I_{j}(\xi,0)$, where $I_{j}(\xi,0)$ is defined in the region $A_{j}$. Obviously $I_{2}(\xi,0) \leq C$. In $A_{1}$ we have $|\xi-\xi_{1}|>|\xi_{1}|/2$ and $|\xi+\xi_{2}|>|\xi_{2}|/2$, therefore Lemma \ref{lpri1} gives $I_{1}(\xi,0) \leq C$. In $A_{3}$ we have $|\xi+\xi_{2}|>|\xi_{2}|/2$, and consequently \begin{align*} I_{3}(\xi,0)\lesssim & \frac{1}{\langle\xi\rangle^{2\rho}} \int_{A_{3}}\frac{\langle\xi_{2}\rangle^{4\rho}d\xi_{1}d\xi_{2}}{\langle(\xi_{1}-\xi_{2})\xi_{2}(\xi-\xi_{1})\rangle^{2b}}\\ =& \frac{1}{\langle\xi\rangle^{2\rho}}\int_{A_{3}\cap \{|\xi_{1}-\xi_{2}|>|\xi_{2}|\}}+ \frac{1}{\langle\xi\rangle^{2\rho}}\int_{A_{3}\cap \{|\xi_{1}-\xi_{2}|\leq |\xi_{2}|\}}\\ =&I_{3,1}(\xi,0)+I_{3,2}(\xi,0). \end{align*} In the first integral, for $\rho <1/4$, $b>1/2$ we have \begin{align*} I_{3,1}(\xi,0) &\lesssim \frac{1}{\langle\xi\rangle^{2\rho}}\int_{|\xi_{2}|>2}\langle\xi_{2} \rangle^{4\rho}d\xi_{2}\int_{|\xi_{1}|\leq2} \frac{d\xi_{1}}{\langle\xi_{2}^2(\xi-\xi_{1})\rangle^{2b}}\\ &\lesssim \frac{1}{\langle\xi\rangle^{2\rho}}\int_{|\xi_{2}|>2} \frac{\langle\xi_{2}\rangle^{4\rho}d\xi_{2}}{\xi_{2}^2} \leq C. \end{align*} To estimate $I_{3,2}(\xi,0)$ we make the change of variables $\eta_{2}=\xi_{1}-\xi_{2}$, $\eta_{1}=\xi_{1}$ and as $|\xi_{1}|\leq 2$ we obtain the same estimate as that for $I_{3,1}(\xi,0)$. By symmetry we can estimate $I_{4}(\xi,0)$ in the same manner as $I_{3}(\xi,0)$. \noindent \textbf{Case } $|\xi| >1$. Let us consider $I(\xi,0)$ in the form (\ref{intpri}) and let $B_{1}=\{ |\xi_{1}+ \xi_{2}|>4\}$ and $B_{2}=\{ |\xi_{1}+ \xi_{2}|\leq 4\}$, then $I(\xi,0)=I_{1}(\xi)+I_{2}(\xi)$, where $I_{j}(\xi)$ is defined in $B_{j}$. In $B_{1}$ we have $$\label{B1} |2-(\xi_{1}+\xi_{2})|>|\xi_{1}+\xi_{2}|/2,\quad |1-(\xi_{1}+\xi_{2})|\leq 5|\xi_{1}+\xi_{2}|/4,$$ moreover $B_{1} \subset \{|\xi_{1}| \geq 2\}\cup\{|\xi_{2}| \geq 2\}=:B_{1,1}\cup B_{1,2}$ and therefore $I_{1}(\xi)\leq I_{1,1}(\xi)+I_{1,2}(\xi)$, where $I_{1,j}(\xi)$ is defined in $B_{1,j}\cap B_{1}$. In $B_{1,1}$ we have $|\xi_{1}|/2 \leq |1-\xi_{1}|\leq 3|\xi_{1}|/2$, therefore using (\ref{B1}), we obtain that $I_{1,1}(\xi) \lesssim I(0,0) \leq C$ if $\rho <1/4$, $\rho+1/3 \rho +1/3$. \end{proof} \begin{proof}[Proof of Lemma \ref{lpri}] Let $0 \leq \rho <1/4$, $b>7/12$, $b'< -\rho/3$. Using symmetry and Lemma \ref{lpri2} it is sufficient to prove $J=p(\xi,z)\int_{0}^{\infty}\int_{\mathbb{R}}\frac{H_{\rho}(\xi, \xi_{1}, \xi_{2})d\xi_{1}d\xi_{2}}{\langle\xi^3(z+F(\xi_{1},\xi_{2})\rangle^{2b}}\leq C < \infty.$ By Lemma \ref{lpri2} we can suppose $|\xi|^3z \geq 1$; since if $|\xi|^3z < 1$, $$\langle\xi^3(z+F)\rangle^{-2b} \leq 2^{2b}\langle\xi^3F\rangle^{-2b}.$$ Also by symmetry we can suppose $|\xi_{2}| \leq |\xi_{1}|$. Therefore $%\label{H} H_{\rho}(\xi,\xi_{1},\xi_{2})\lesssim 1+|\xi|^{6\rho}+|\xi|^{6\rho}|\xi_{1}|^{6\rho}.$ Using Proposition \ref{prop2} we can suppose $|\xi_{1}|>4$ ($l^{-1} \leq |\xi_{1}|^{-1}$). \noindent\textbf{Case } $|\xi| |\xi_{1}| \leq 1$. We have $H_{\rho}\lesssim \langle\xi\rangle^{6\rho}$ and therefore $J \leq C<\infty$, by Proposition \ref{prop2}. \noindent\textbf{Case} $|\xi| |\xi_{1}| > 1$.\\ {\bf i)} If $|\xi_{1}|^3 \leq z$, $|\xi_{1}| \leq z^{1/3}$, we have $H_{\rho}(\xi,\xi_{1},\xi_{2})\lesssim 1+|\xi|^{6\rho}+|z|^{2\rho/3}|\xi|^{6\rho}|\xi_{1}|^{4\rho}$. Therefore using (\ref{el4}), in this region we have \begin{align*} \frac{\xi^{2+6\rho}|z|^{2\rho/3}}{\langle\xi^3z\rangle^{-2b'}}\int_{1/|\xi|}^{|z|^{1/3}} |\xi_{1}|^{4\rho}d\xi_{1} \int_{\mathbb{R}}\frac{d\eta}{\langle\xi^3\xi_{1}(\eta^2-l^2)\rangle^{2b}} \lesssim & \frac{\xi^{2+6\rho}|z|^{2\rho/3}}{\langle\xi^3z\rangle^{-2b'}|\xi|^3}\int_{1/|\xi|}^{\infty}\frac{|\xi_{1}|^{4\rho}d\xi_{1}}{|\xi_{1}|^2}\\ \lesssim & \frac{(|\xi|^3z)^{2\rho/3}}{\langle\xi^3z\rangle^{-2b'}}\\ \leq & C. \end{align*} {\bf ii)} If $|\xi_{1}|^3 \geq z$, $|\xi_{1}| \geq z^{1/3}$, we can proceed as follows. By Lemma \ref{lpri2} we can suppose $|z+F| \leq |F|/2$, so $|F| \leq 2z$, $|(2-(\xi_{1}+\xi_{2}))\xi_{1}\xi_{2}| \leq 2z$. This implies that $|1-\xi_{2}||1-(\xi_{1}+\xi_{2})|\lesssim 1+|\xi_{1}|+z^{2/3}$. Therefore \begin{align*} H_{\rho} \lesssim &(\langle\xi\rangle^{4\rho}+|\xi|^{6\rho})+ |\xi|^{4\rho}|\xi_{1}|^{4\rho}+|\xi|^{6\rho}|\xi_{1}|^{2\rho} +|\xi|^{4\rho}|\xi_{1}|^{2\rho} +|\xi|^{6\rho}|\xi_{1}|^{4\rho}\\ &+|\xi|^{4\rho}z^{4\rho/3}+|\xi|^{6\rho}z^{4\rho/3} +|\xi|^{6\rho}z^{4\rho/3}|\xi_{1}|^{2\rho}=: \sum_{j=1}^{8}l_{j}. \end{align*} We have, $$\label{4o} \frac{|\xi|^{6\rho}}{\langle\xi\rangle^{2\rho}} \leq |\xi|^{4\rho}.$$ To estimate the term that contains $l_{1}=\langle\xi\rangle^{4\rho}+|\xi|^{6\rho}$, we use (\ref{4o}) and Proposition \ref{prop2}. For terms $l_{j}$, $j=2, \ldots,5$, we use (\ref{4o}) and Propositions \ref{prop2} and \ref{prop1} if $|\xi|>1$. If $|\xi| <1$, we integrate in the region $\xi_{1}> 1/|\xi|$ as above. In $l_{6}=|\xi|^{4\rho}z^{4\rho/3}$, we have $\frac{|\xi|^2|\xi|^{4\rho}z^{4\rho/3}} {\langle\xi^3z\rangle^{-2b'}|\xi|^3\langle\xi\rangle^{2\rho}} \int_{z^{1/3}}^{\infty} \frac{d\xi_{1}}{\xi_{1}^2} \lesssim \frac{1}{(|\xi|^3z)^{(1-4\rho)/3}}\leq C.$ We estimate $l_{7}=|\xi|^{6\rho}z^{4\rho/3}$, as in $l_{6}$ using (\ref{4o}). Finally in $l_{8}=|\xi|^{6\rho}z^{4\rho/3}|\xi_{1}|^{2\rho}$, we have $\frac{|\xi|^{2+6\rho}z^{4\rho/3}}{\langle\xi^3z\rangle^{-2b'} \langle\xi\rangle^{2\rho}|\xi|^3}\int_{z^{1/3}}^{\infty} \frac{|\xi_{1}|^{2\rho}d\xi_{1}}{\xi_{1}^2} \lesssim \frac{(|\xi|^3z)^{(6\rho-1)/3}}{\langle\xi^3z\rangle^{-2b'}} \leq C.$ \end{proof} \subsection*{Remark} In the case $\alpha \neq 0$ under little modifications, the proofs of Propositions \ref{prop2} and \ref{prop1} and the proofs of Lemmas \ref{lpri}, \ref{lpri1} and \ref{lpri2} are similar to the case $\alpha = 0$. For example in order to prove Lemma \ref{lpri1} with $\alpha \neq 0$ we proceed as follows In (\ref{Gg}) we have $g(\xi,\xi_{1},\xi_{2})= (\xi_{1}-\xi_{2})\xi_{2}(2\alpha -3\beta \xi_{1}))$. In order to obtain symmetry in $\xi_{1}$ and $\xi_{2}$, we consider the change of variable $2\alpha -3\beta \xi_{1}:= 3\beta \xi_{1}$. In this way we have $$I(0,0) \lesssim C\left(\frac{\alpha}{\beta}\right)\int\limits_{\mathbb{R}^2}\frac{\textstyle \langle\xi_{1}+\xi_{2}\rangle^{2\rho} \langle\xi_{1}\rangle^{2\rho}\langle\xi_{2}\rangle^{2\rho}d\xi_{1}d\xi_{2}} {\textstyle\left\langle\beta\left(\frac{\textstyle 2\alpha}{\textstyle\beta}- (\xi_{1}+\xi_{2})\right)\xi_{1}\xi_{2}\right\rangle^{2\rho}}.$$ Now using symmetry, the rest of the proof is the same as that of Lemma \ref{lpri1}, if we replace the lower limit $1$ in the integrals in (\ref{eql3}) by $4\alpha/3\beta$. \section*{Proof of Theorem \ref{T2}} Consider a cut-off function $\psi \in \mathcal{C^{\infty}}$, such that $0\leq \psi \leq 1$, $\psi(t)=\begin{cases} 1 & \textrm{if } |t| \leq 1\\ 0 & \textrm{if } |t| \geq 2, \end{cases}$ and let $\psi_{T}(t):=\psi(t/T)$. To prove Theorem \ref{T2} we need the following result. \begin{proposition}\label{prop3} Let $-1/20$ and $X_{a}=\{v \in X^{s,b}; \|v\|_{\{s,b\}}\leq a\}.$ For $v \in X_{a}$ fixed, let us define $\Phi(v)=\psi_{1}(t)U(t)u_{0}- \psi_{T}(t) \int_{0}^{t}U(t-t')F(v)(t',\cdot)dt'.$ Let $\epsilon=1-b+b'\!>0$, \$b-1