In this section, by using Lemmas 2.1-2.2, we give the proof of Lemma 3.1.
Lemma 3.1
Let
\(u_{j}\in X_{s,\frac{1}{2}+\epsilon}\)
with
\(s\geq\frac{1}{4}\)
and
\(j=1,2,3\). Then we have
$$ \Biggl\Vert \partial_{x} \Biggl(\prod_{j=1}^{3}u_{j} \Biggr) \Biggr\Vert _{X_{s,-\frac{1}{2}+2\epsilon}}\leq C\prod_{j=1}^{3} \|u_{j}\|_{X_{s,\frac{1}{2}+\epsilon}}. $$
(3.1)
Proof
To prove (3.1), by duality, it suffices to prove that
$$ \int_{\mathbf {R}^{2}}\bar{u}(x,t)\partial_{x} \Biggl(\prod _{j=1}^{3}u_{j} \Biggr)\,dx\,dt\leq C \Biggl[\prod_{j=1}^{3}\|u_{j} \|_{X_{s,\frac{1}{2}+\epsilon}} \Biggr] \|u\|_{X_{-s,\frac{1}{2}-2\epsilon}}. $$
(3.2)
Let
$$\begin{gathered} F(\xi,\tau)=\langle\xi\rangle^{-s}\langle\sigma\rangle^{\frac {1}{2}-2\epsilon} \mathscr{F}u(\xi,\tau),\\ F_{j}(\xi_{j},\tau_{j})= \langle\xi_{j}\rangle^{s}\langle\sigma _{j} \rangle^{\frac{1}{2}+\epsilon} \mathscr{F}u_{j}(\xi_{j}, \tau_{j})\quad (j=1,2,3).\end{gathered} $$
(3.3)
To obtain (3.2), from (3.3), it suffices to prove that
$$\begin{aligned}[b] & \int_{\mathbf {R}^{2}} \underset{\tau=\tau_{1}+\tau_{2}+\tau_{3}}{\int_{ \xi=\xi_{1}+\xi_{2}+\xi_{3} }}\frac{|\xi|\langle\xi\rangle^{s} F(\xi,\tau)\prod _{j=1}^{3}F_{j}(\xi_{j},\tau_{j})}{\langle \sigma\rangle^{\frac{1}{2}-2\epsilon} \prod _{j=1}^{3}\langle\xi_{j}\rangle^{s}\langle\sigma_{j} \rangle^{\frac{1}{2}+\epsilon}}\,d\xi_{1}\,d \tau_{1}\,d\xi_{2}\,d\tau _{2}\,d\xi \,d\tau \\ &\quad\leq C \|F\|_{L_{\xi\tau}^{2}} \Biggl(\prod_{j=1}^{3} \|F_{j}\|_{L_{\xi\tau }^{2}} \Biggr).\end{aligned} $$
(3.4)
Without loss of generality, by using the symmetry, we assume that \(|\xi_{1}|\geq|\xi_{2}|\geq|\xi_{3}|\) and \(F(\xi,\tau)\geq 0\), \(F_{j}(\xi_{j},\tau_{j})\geq0\) (\(j=1,2\)). We define
$$\begin{gathered} \Omega_{1}=\Biggl\{ (\xi_{1}, \tau_{1},\xi_{2},\tau_{2},\xi ,\tau)\in{\mathrm{R}^{6}},\xi=\sum_{j=1}^{3} \xi_{j},\tau=\sum_{j=1}^{3} \tau_{j}, |\xi_{3}|\leq|\xi_{2}|\leq| \xi_{1}|\leq64 \Biggr\} , \\ \Omega_{2}=\Biggl\{ (\xi_{1}, \tau_{1},\xi_{2},\tau_{2},\xi ,\tau)\in{\mathrm{R}^{6}},\xi=\sum_{j=1}^{3} \xi_{j},\tau=\sum_{j=1}^{3} \tau_{j}, |\xi_{1}|\geq64, |\xi_{1}|\geq4| \xi_{2}|\Biggr\} , \\ \Omega_{3}=\Biggl\{ (\xi_{1}, \tau_{1},\xi_{2},\tau_{2},\xi ,\tau)\in{\mathrm{R}^{6}},\xi=\sum_{j=1}^{3} \xi_{j},\tau=\sum_{j=1}^{3} \tau_{j}, |\xi_{1}|\geq64, |\xi_{1}|\sim| \xi_{2}|,|\xi_{2}|\gg|\xi_{3}|\Biggr\} , \\ \Omega_{4}=\Biggl\{ (\xi_{1}, \tau_{1},\xi_{2},\tau_{2},\xi ,\tau)\in{\mathrm{R}^{6}},\xi=\sum_{j=1}^{3} \xi_{j},\tau=\sum_{j=1}^{3} \tau_{j}, |\xi_{1}|\geq64, |\xi_{1}|\sim| \xi_{2}|\sim|\xi_{3}| \Biggr\} .\end{gathered} $$
Obviously, \(\{(\xi_{1},\tau_{1},\xi_{2},\tau_{2},\xi,\tau)\in{\mathrm{ R}^{6}},\xi=\sum _{j=1}^{3}\xi_{j},\tau=\sum _{j=1}^{3}\tau_{j}, |\xi_{3}|\leq|\xi_{2}|\leq|\xi_{1}|\}\subset\bigcup _{j=1}^{4}\Omega_{j}\). Let
$$ K(\xi_{1},\tau_{1},\xi_{2}, \tau_{2},\xi,\tau)=\frac{|\xi |\langle\xi\rangle^{s}}{\langle\sigma\rangle^{\frac {1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle ^{\frac{1}{2}+\epsilon}} $$
(3.5)
and
$$ I= \int_{\mathbf {R}^{2}} \underset{\tau=\sum _{j=1}^{3}\tau_{j}}{\int_{ \xi=\sum _{j=1}^{3}\xi_{j} }}K(\xi_{1},\tau_{1}, \xi_{2},\tau_{2},\xi,\tau)F(\xi,\tau)\prod _{j=1}^{3}F_{j}(\xi_{j}, \tau_{j}) \,d\xi_{1}\,d\tau_{1}\,d\xi_{2}\,d \tau_{2}\,d\xi \,d\tau. $$
(1) \(\Omega_{1}\). In this subregion, we have
$$ K(\xi_{1},\tau_{1},\xi_{2},\tau_{2}, \xi,\tau)\leq\frac {C}{\langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}}. $$
(3.6)
By using (3.6) and the Cauchy-Schwartz inequality and the Plancherel identity and the Hölder inequality as well as (2.1), we have
$$\begin{aligned} I&\leq C \int_{\mathbf {R}^{2}} \underset{\tau=\sum _{j=1}^{3}\tau_{j}}{\int_{ \xi=\sum _{j=1}^{3}\xi_{j} }}\frac{ F(\xi,\tau)\prod _{j=1}^{3}F_{j}(\xi_{j},\tau_{j})}{\langle \sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3} \langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} \,d\xi_{1}\,d \tau_{1}\,d\xi_{2}\,d\tau_{2}\,d\xi \,d\tau \\ &\leq C \biggl\Vert \frac{F(\xi,\tau)}{\langle\sigma\rangle^{\frac {1}{2}-2\epsilon}} \biggr\Vert _{L_{\xi\tau}^{2}} \biggl\Vert \underset{\tau =\sum _{j=1}^{3}\tau_{j}}{\int_{ \xi=\sum _{j=1}^{3}\xi_{j} }}\frac{\prod _{j=1}^{3}F_{j}(\xi_{j},\tau_{j})}{\prod _{j=1}^{3} \langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}}\,d\xi_{1}\,d\tau _{1}\,d\xi_{2}\,d\tau_{2} \biggr\Vert _{L_{\xi\tau}^{2}} \\ &\leq C\|F\|_{L_{\xi\tau}^{2}} \Biggl(\prod _{j=1}^{3} \biggl\Vert \mathscr{F}^{-1} \biggl(\frac {F_{j}}{\langle\sigma_{j} \rangle^{\frac{1}{2}+\epsilon}} \biggr) \biggr\Vert _{L_{xt}^{6}} \Biggr) \\ &\leq C\|F\|_{L_{\xi\tau}^{2}} \Biggl(\prod _{j=1}^{3} \| F_{j}\|_{L_{\xi\tau}^{2}} \Biggr).\end{aligned} $$
(2) \(\Omega_{2}\). In this subregion, since \(\vert \phi^{\prime}(\xi_{1})-\phi^{\prime}(\xi_{2}) \vert =3|\xi_{1}^{2}-\xi_{2}^{2}| \vert 1+\frac{1}{3\xi_{1}^{2}\xi _{2}^{2}} \vert \geq3|\xi_{1}^{2}-\xi_{2}^{2}|\geq C|\xi|^{2}\) and \(|\xi|\sim|\xi _{1}|\), we have
$$\begin{aligned}[b] K(\xi_{1},\tau_{1},\xi_{2},\tau_{2}, \xi,\tau)&\leq\frac{C|\xi |}{\langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon }} \\ & \leq C\frac{C|\xi_{1}^{2}-\xi_{2}^{2}|^{\frac{1}{2}} \vert 1+\frac {1}{3\xi_{1}^{2}\xi_{2}^{2}} \vert ^{\frac{1}{2}}}{ \langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} = \frac{C|\phi^{\prime}(\xi_{1})-\phi^{\prime}(\xi_{2})|^{\frac{1}{2}}}{ \langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} .\end{aligned} $$
(3.7)
By using (3.7) and the Cauchy-Schwartz inequality and the Plancherel identity and the Hölder inequality as well as (2.3)-(2.4), since \(\frac{3}{4} (\frac{1}{2}+\epsilon )<\frac {1}{2}-2\epsilon\), we have
$$\begin{aligned} I\leq{}& C \int_{\mathbf {R}^{2}} \underset{\tau=\sum _{j=1}^{3}\tau_{j}}{\int_{ \xi=\sum _{j=1}^{3}\xi_{j} }}\frac{|\phi^{\prime}(\xi_{1})-\phi^{\prime}(\xi_{2})|^{\frac{1}{2}} F(\xi,\tau)\prod _{j=1}^{3}F_{j}(\xi_{j},\tau_{j})}{\langle \sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3} \langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} \,d\xi_{1}\,d \tau_{1}\,d\xi_{2}\,d\tau_{2}\,d\xi \,d\tau \\ \leq{}& C \biggl\Vert \mathscr{F}^{-1} \biggl(\frac{F}{\langle\sigma \rangle^{\frac{1}{2}-2\epsilon}} \biggr) \biggr\Vert _{L_{xt}^{4}} \biggl\Vert I^{\frac{1}{2}} \biggl( \mathscr{F}^{-1} \biggl(\frac {F_{1}}{\langle\sigma_{1} \rangle^{\frac{1}{2}+\epsilon}} \biggr),\mathscr{F}^{-1} \biggl(\frac{F_{1}}{\langle\sigma_{2} \rangle ^{\frac{1}{2}+\epsilon}} \biggr) \biggr) \biggr\Vert _{L_{xt}^{2}} \\ &\times \biggl\Vert \mathscr{F}^{-1} \biggl(\frac{F_{3}}{\langle\sigma _{3}\rangle^{\frac{1}{2}+\epsilon}} \biggr) \biggr\Vert _{L_{xt}^{4}} \\ \leq{}& C\|F\|_{L_{\xi\tau}^{2}} \Biggl(\prod_{j=1}^{3} \|F_{j}\| _{L_{\xi\tau}^{2}} \Biggr).\end{aligned} $$
(3) \(\Omega_{3}\). In this subregion, since \(\vert \phi^{\prime}(\xi_{2})-\phi^{\prime}(\xi_{3}) \vert =3|\xi_{2}^{2}-\xi_{3}^{2}| \vert 1+\frac{1}{3\xi_{2}^{2}\xi _{3}^{2}} \vert \geq3|\xi_{2}^{2}-\xi_{3}^{2}|\geq C|\xi_{1}|^{2}\), we have
$$\begin{aligned}[b] K(\xi_{1},\tau_{1},\xi_{2},\tau_{2}, \xi,\tau)&\leq\frac{C|\xi_{1}|}{ \langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3} \langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} \\ &\quad\leq C C\frac{C|\xi_{2}^{2}-\xi_{3}^{2}|^{\frac{1}{2}} \vert 1+\frac {1}{3\xi_{2}^{2}\xi_{3}^{2}} \vert ^{\frac{1}{2}}}{ \langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}}\leq \frac{C|\phi^{\prime}(\xi_{2})-\phi^{\prime}(\xi_{3})|^{\frac{1}{2}}}{ \langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} .\end{aligned} $$
(3.8)
By using (3.8) and the Cauchy-Schwartz inequality and the Plancherel identity and the Hölder inequality as well as (2.3)-(2.4), since \(\frac{3}{4} (\frac{1}{2}+\epsilon )<\frac {1}{2}-2\epsilon\), we have
$$\begin{aligned} I\leq{}& C \int_{\mathbf {R}^{2}} \underset{\tau=\sum _{j=1}^{3}\tau_{j}}{\int_{ \xi=\sum _{j=1}^{3}\xi_{j} }}\frac{|\phi^{\prime}(\xi_{2})-\phi^{\prime}(\xi_{3})|^{\frac{1}{2}} F(\xi,\tau)\prod _{j=1}^{3}F_{j}(\xi_{j},\tau_{j})}{\langle \sigma\rangle^{\frac{1}{2}-2\epsilon} \prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac {1}{2}+\epsilon}} \,d\xi_{1}\,d \tau_{1}\,d\xi_{2}\,d\tau_{2}\,d\xi \,d\tau \\ \leq{}& C \biggl\Vert \mathscr{F}^{-1} \biggl(\frac{F}{\langle\sigma \rangle^{\frac{1}{2}-2\epsilon}} \biggr) \biggr\Vert _{L_{xt}^{4}} \biggl\Vert I^{\frac{1}{2}} \biggl( \mathscr{F}^{-1} \biggl(\frac {F_{2}}{\langle\sigma_{2} \rangle^{\frac{1}{2}+\epsilon}} \biggr),\mathscr{F}^{-1} \biggl(\frac{F_{3}}{\langle\sigma_{3} \rangle ^{\frac{1}{2}+\epsilon}} \biggr) \biggr) \biggr\Vert _{L_{xt}^{2}} \\ & \times \biggl\Vert \mathscr{F}^{-1} \biggl(\frac{F_{1}}{\langle\sigma _{1}\rangle^{\frac{1}{2}+\epsilon}} \biggr) \biggr\Vert _{L_{xt}^{4}} \\ \leq{}& C\|F\|_{L_{\xi\tau}^{2}} \Biggl(\prod _{j=1}^{3} \| F_{j}\|_{L_{\xi\tau}^{2}} \Biggr).\end{aligned} $$
(4) \(\Omega_{4}\). In this subregion, since \(s\geq\frac {1}{4}\) and \(|\xi_{1}|\sim|\xi_{2}|\sim|\xi_{3}|\), we have
$$ K(\xi_{1},\tau_{1},\xi_{2},\tau_{2}, \xi,\tau)\leq\frac{C|\xi _{1}|^{1-2s}}{\langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon }}\leq\frac{C\prod _{j=1}^{3}|\xi_{j}|^{\frac{1}{6}}}{ \langle\sigma\rangle^{\frac{1}{2}-2\epsilon}\prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac{1}{2}+\epsilon}} . $$
(3.9)
By using (3.9) and the Cauchy-Schwartz inequality and the Plancherel identity and the Hölder inequality as well as (2.2), since \(\frac{3}{4} (\frac{1}{2}+\epsilon )<\frac {1}{2}-2\epsilon\), we have
$$\begin{aligned} I&\leq C \int_{\mathbf {R}^{2}} \underset{\tau=\sum _{j=1}^{3}\tau_{j}}{\int_{ \xi=\sum _{j=1}^{3}\xi_{j} }}\frac{ F(\xi,\tau)\prod _{j=1}^{3}|\xi_{j}|^{\frac{1}{6}}F_{j}(\xi _{j},\tau_{j})}{\langle\sigma\rangle^{\frac{1}{2}-2\epsilon} \prod _{j=1}^{3}\langle\sigma_{j}\rangle^{\frac {1}{2}+\epsilon}} \,d\xi_{1}\,d \tau_{1}\,d\xi_{2}\,d\tau_{2}\,d\xi \,d\tau \\ &\leq C \biggl\Vert \frac{F}{\langle\sigma\rangle^{\frac {1}{2}-2\epsilon}} \biggr\Vert _{L_{\xi\tau}^{2}} \Biggl( \prod_{j=1}^{3} \biggl\Vert D_{x}^{\frac{1}{6}}P^{2}\mathscr{F}^{-1} \biggl( \frac {F_{j}}{\langle\sigma_{j} \rangle^{\frac{1}{2}+\epsilon}} \biggr) \biggr\Vert _{L_{xt}^{6}} \Biggr) \\ &\leq C\|F\|_{L_{\xi\tau}^{2}} \Biggl(\prod _{j=1}^{3} \| F_{j}\|_{L_{\xi\tau}^{2}} \Biggr).\end{aligned} $$
This completes the proof of Lemma 3.1. □