The first upper bound ?? can be obtained by Proposition 2.2 in [20]. To prove ?? , we first observe that \[\begin{align} (Q(g,h),f)_{L^2_v}&=-\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*g\,\partial_ih\,\partial_jf\,dv+\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*g\,h\,\partial_if\,dv\\ &=\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*g\,h\,\partial_{ij}f\,dv+2\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*g\,h\,\partial_if\,dv:=Q_1+Q_2. \end{align}\] Since \(a(z)\lesssim|z|^{-1}\), the Cauchy–Schwarz inequality gives \[\begin{align} |Q_1|&\lesssim&\int_{|v-v_*|\le 1}\frac{1}{|v-v_*|}|g(v_*)\langle v\rangle^\omega h\langle v\rangle^{-\omega}\nabla^2f|\,dv_*dv\\ &&+\int_{|v-v_*|>1}\frac{1}{|v-v_*|}|g(v_*)\langle v\rangle^\omega h\langle v\rangle^{-\omega}\nabla^2f|\,dv_*dv\\ &\lesssim& \|g\|_{L^2_2}\|h\|_{L^2_\omega}\|\nabla^2f\|_{L^2_{-\omega}}\lesssim \|g\|_{L^2_2}\|h\|_{L^2_\omega}\|f\|_{H^2_{-\omega}}. \end{align}\]

For \(Q_2\), since \(b(z)\lesssim |z|^{-2}\), we have \[\begin{align} |Q_2|&\lesssim& \int_{|v-v_*|\le 1}\frac{1}{|v-v_*|^2}|g(v_*)\langle v\rangle^\omega h(v)\langle v\rangle^{-\omega}\nabla f|\,dv_*dv\\ &&+\int_{|v-v_*|>1}\frac{1}{|v-v_*|^2}|g(v_*)\langle v\rangle^\omega h(v)\langle v\rangle^{-\omega}\nabla f|\,dv_*dv\\ &\lesssim& \|g\|_{L^2}\|\langle v\rangle^\omega h\langle v\rangle^{-\omega}\nabla f\|_{L^{6/5}}+\|g\|_{L^2_2}\|h\|_{L^2_\omega}\|\nabla f\|_{L^2_{-\omega}}\\ &\lesssim& \|g\|_{L^2}\|h\|_{L^2_\omega}\|\nabla f\|_{L^3_{-\omega}}+\|g\|_{L^2_2}\|h\|_{L^2_\omega}\|\nabla f\|_{L^2_{-\omega}}\\ &\lesssim &\|g\|_{L^2_2}\|h\|_{L^2_\omega}\|f\|_{H^{3/2}_{-\omega}}. \end{align}\] In the three inequalities above, we have used the Hardy–Littlewood–Sobolev inequality, Hölder inequality \(\|fg\|_{L^{6/5}}\le \|f\|_{L^2}\|g\|_{L^3}\), and the Sobolev embedding \(H^{1/2}\hookrightarrow L^3\). Combining the estimates for \(Q_1\) and \(Q_2\) then yields ?? .

We complete the proof of this lemma.

We remark that ?? was established in Lemma 2.12 of [33], while the first inequality in ?? was proved in Lemma 2.3 of [26]. To establish the second inequality ?? , we rewrite it as \[\begin{align} &&\big(Q(g,\mu), f\langle v\rangle^{2k}\big)_{L^2_v}=-\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*g\partial_i\mu \partial_j(f\langle v\rangle^{2k})dv+\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*g\mu\partial_i(f\langle v\rangle^{2k})dv. \end{align}\] By integration by parts and 20 , we obtain that \[\begin{align} &&\big(Q(g,\mu), f\langle v\rangle^{2k}\big)_{L^2_v}=\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*g\partial_i\mu f\langle v\rangle^{2k}dv+\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*g\partial_{ij}\mu f\langle v\rangle^{2k}dv+8\pi\int_{\mathbb{R}^3}\mu gf\langle v\rangle^{2k}dv \nonumber\\ &&-\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*g\partial_i\mu f\langle v\rangle^{2k}dv=\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*g\partial_{ij}\mu f\langle v\rangle^{2k}dv+8\pi\int_{\mathbb{R}^3}\mu gf\langle v\rangle^{2k}dv. \label{Q46ac} \end{align}\tag{23}\] Since \(|a(z)|\lesssim|z|^{-1}\), we have \[\begin{align} &&\sum_{i,j=1}^3\int_{\mathbb{R}^3}|a_{ij}*g\partial_{ij}\mu f\langle v\rangle^{2k}|dv+8\pi\int_{\mathbb{R}^3}\mu |gf|\langle v\rangle^{2k}dv\\ &\leq& C_k \int_{\mathbb{R}^6}\frac{1}{|v-v_*|}|g(v_*)f(v)\langle v\rangle^{-2}|dv_*dv+C_k\int_{\mathbb{R}^3}|gf| dv\leq C_k\|g\|_{L^2_3}\|f\|_{L^2_v}. \end{align}\] We get the desired result.

Finally, we give the proof for ?? . we rewrite it as \[\begin{align} &&\big(Q(G,f), f\langle v\rangle^{2k}\big)_{L^2_v}=-\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*G\partial_if \partial_j(f\langle v\rangle^{2k})dv+\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*Gf\partial_i(f\langle v\rangle^{2k})dv\\ &&=-\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*G\partial_i(f\langle v\rangle^k) \partial_j(f\langle v\rangle^{k})dv-k^2\sum_{i,j=1}^3\int_{\mathbb{R}^3}a_{ij}*Gf^2\langle v\rangle^{2k-4}v_iv_jdv\\ &&+4\pi\int_{\mathbb{R}^3}G f^2\langle v\rangle^{2k}dv+k\sum_{i=1}^3\int_{\mathbb{R}^3}b_i*Gf^2\langle v\rangle^{2k-2}v_idv:=G_1+G_2+G_3+G_4. \end{align}\]

For \(G_1\), observing that \[\begin{align} \int_{\mathbb{R}^3}G|\log G| dv&=&\int_{G>1}G\log G dv+\int_{e^{-|v|^2}<G\leq1}G|\log G| dv+\int_{G\leq e^{-|v|^2}}G|\log G| dv\\ &\lesssim& \int_{\mathbb{R}^3}G^2dv+\int_{\mathbb{R}^3} G|v|^2 dv+\int_{G \leq e^{-|v|^2}}G^{\frac{1}{2}}dv\lesssim C_\lambda, \end{align}\] since \(\|G\|_{L^2_5}<\lambda\). Then by Proposition 2.1 in [20], we have \(G_1\leq -C_1\|f\|^2_{D_1(k)}\) with \(C_1\) depending on \(\delta\) and \(\lambda\). For \(G_2\), since \(a(z)\lesssim|z|^{-1}\), we have \[\begin{align} |G_2|&\leq& C_k \Big(\int_{|v-v_*|\leq 1}\frac{1}{|v-v_*|}G(v_*)f^2\langle v\rangle^{2k-2}dv_*dv+\int_{|v-v_*|> 1}\frac{1}{|v-v_*|}G(v_*)f^2\langle v\rangle^{2k-2}dv_*dv\Big)\\ &\leq& C_k\|G\|_{L^2_5}\|f\|^2_{L^2_{k-1}}. \end{align}\] For \(G_3\), we have \[\begin{align} |G_3|\leq C_k\|G\|_{L^2_5}\|f\|^2_{L^4_{k-2}}\leq C_k\|G\|_{L^2_5}\|f\|^2_{H^{3/4}_{k-2}}\leq \varepsilon\|f\|^2_{H^1_{k-\frac{3}{2}}}+C_\varepsilon\|G\|^4_{L^2_5}\|f\|^2_{L^2_{k-2}},\quad \forall \varepsilon>0. \end{align}\] In the three inequalities above, we have used Hölder inequality, Sobolev embedding \(H^{3/4}_{k-2}\hookrightarrow L^4_{k-2}\), the interpolation inequality and Young’s inequality \(\|f\|_{H^{3/4}_{k-4}}\lesssim\|f\|^{3/4}_{H^{1}_{k-4}}\|f\|^{1/4}_{L^{2}_{k-4}}\lesssim\varepsilon\|f\|_{H^{1}_{k-4}}+C_\varepsilon\|f\|_{L^{2}_{k-4}}\), respectively. For \(G_4\), noticing that \(b(z)\lesssim|z|^{-2}\), then by Hardy-Littlewood-Sobolev inequality, we have \[\begin{align} |G_4|&\lesssim& C_k \Big(\int_{|v-v_*|\leq 1}\frac{1}{|v-v_*|^2}G(v_*)f^2\langle v\rangle^{2k-1}dv_*dv+\int_{|v-v_*|> 1}\frac{1}{|v-v_*|^2}G(v_*)f^2\langle v\rangle^{2k-1}dv_*dv\Big)\\ &\lesssim&C_k\|G\|_{L^2_5}(\|f\|^2_{L^{12/5}_{k-2}}+\|f\|^2_{k-\frac{1}{2}})\lesssim\varepsilon\|f\|^2_{H^1_{k-\frac{3}{2}}}+C_{k,\varepsilon}(1+\|G\|^4_{L^2_5})\|f\|^2_{L^2_{k-\frac{1}{2}}}, \end{align}\] where we also use interpolation and the fact \(\langle v\rangle\sim \langle v_*\rangle\) when \(|v-v_*|\leq 1\).

Combining these estimates and noticing that \(\|f\|_{H^1_{k-\frac{3}{2}}}\leq \|f\|_{D_1(k)}\), we can get the desired result by choosing properly small \(\varepsilon>0\).

We complete the proof of the lemma.

From (2.6) in [26], we have \[\begin{align} \big({\mathcal{L}}(f),f\big)_{L^2_v}\lesssim-\|\langle v\rangle^{-\frac{1}{2}}(\mathbf{I-P})f\|^2_{L^2_v}-\|\langle v\rangle^{-\frac{3}{2}}\tilde{\nabla}_v (\mathbf{I-P})f\|^2_{L^2_v}, \end{align}\] where the anisotropic gradient \(\tilde{\nabla}_v f\) of a function \(f\) defined by \[\begin{align} \tilde{\nabla}_v f=P_v\nabla_v f+\langle v\rangle(\mathbf{I}-P_v)\nabla_v f \end{align}\] with the projection operator \(P_v\) on the \(v-\)direction defined by \(P_v\xi=\Big(\xi\cdot \frac{v}{|v|}\Big)\frac{v}{|v|},\forall \xi\in \mathbb{R}^3.\) A direct calculation yields \[\begin{align} \|\langle v\rangle^{-\frac{3}{2}}\tilde{\nabla}_v f\|^2_{L^2_v}&\gtrsim& \|\nabla_v f\|^2_{L^2_{-\frac{3}{2}}}+\|(\mathbf{I}-P_v)\nabla_v f\|^2_{L^2_{-\frac{1}{2}}}\\ &\gtrsim&\|\nabla_v f\|^2_{L^2_{-\frac{3}{2}}}+\|v\times \nabla_v f\|^2_{L^2_{-\frac{3}{2}}}\gtrsim\|\nabla_v f\|^2_{L^2_{-\frac{3}{2}}}+\|(-\Delta_{\mathbb{S}^2})^{\frac{1}{2}} f\|^2_{L^2_{-\frac{3}{2}}}. \end{align}\] Thus ?? holds true.

Next, we provide the proofs of ?? and ?? separately. By the definition of linearized Landau operator \({\mathcal{L}}\), we have that \[\begin{align} \big({\mathcal{L}}(f),f\langle v\rangle^{2k}\big)_{L^2_v}=\big(Q(\mu,\mu^{\frac{1}{2}}f),f\mu^{-\frac{1}{2}}\langle v\rangle^{2k}\big)_{L^2_v}+\big(Q(\mu^{\frac{1}{2}}f,\mu),f\mu^{-\frac{1}{2}}\langle v\rangle^{2k}\big)_{L^2_v}:={\mathcal{L}}_1+{\mathcal{L}}_2. \end{align}\]

For \({\mathcal{L}}_2\), by the same argument used in the proof of the second inequality in ?? , we obtain \[\begin{align} \label{cL2} {\mathcal{L}}_2\leq C_k\|f\|^2_{L^2_v}. \end{align}\tag{24}\]

For \({\mathcal{L}}_1\), we rewrite much as in 23 as \[\begin{align} {\mathcal{L}}_1=\sum_{i,j=1}^3\int_{\mathbb{R}^3}(a_{ij}*\mu)\partial_{ij}(\mu^{\frac{1}{2}}f) f\mu^{-\frac{1}{2}}\langle v\rangle^{2k}dv+8\pi \int_{\mathbb{R}^3}\mu f^2\langle v\rangle^{2k}dv. \end{align}\] Let \(g=f\langle v\rangle^k\), integration by parts yields that \[\begin{align} {\mathcal{L}}_1&=&-\sum_{i,j=1}^3\int_{\mathbb{R}^3}(a_{ij}*\mu)\partial_ig\partial_jgdv+\Big(8\pi\int_{\mathbb{R}^3}\mu g^2dv-\sum_{j=1}^3\int_{\mathbb{R}^3}(b_j*\mu)g\partial_j gdv\Big)\\ &&-\int_{\mathbb{R}^3}\Big(\sum_{i,j=1}^3(a_{ij}*\mu)\partial_i(\mu^{\frac{1}{2}}\langle v\rangle^{-k})\partial_j(\mu^{-\frac{1}{2}}\langle v\rangle^k)+\sum_{i=1}^3(b_i*\mu)\partial_i(\mu^{\frac{1}{2}}\langle v\rangle^{-k})\mu^{-\frac{1}{2}}\langle v\rangle^k\Big)g^2dv\\ &:=&{\mathcal{L}}_{1,1}+{\mathcal{L}}_{1,2}+{\mathcal{L}}_{1,3}. \end{align}\] For \({\mathcal{L}}_{1,1}\), from the estimate for \(G_1\) in Lemma 3, there exists a constant \(\lambda_0>0\) such that \[\begin{align} \label{cL11} {\mathcal{L}}_{1,1}\leq -\lambda_0 (\|\nabla g\|^2_{L^2_{-\frac{3}{2}}}+\|(-\Delta_{\mathbb{S}^2})^{\frac{1}{2}}g\|^2_{L^2_{-\frac{3}{2}}}). \end{align}\tag{25}\] For \({\mathcal{L}}_{1,2}\) and \({\mathcal{L}}_{1,3}\), thanks to 20 and integration by parts, we have \[\begin{align} {\mathcal{L}}_{1,2}=8\pi\int_{\mathbb{R}^3}\mu g^2dv+\frac{1}{2}\int_{\mathbb{R}^3}(c*\mu) g^2dv=4\pi\int_{\mathbb{R}^3}\mu g^2 dv. \end{align}\] By 22 , one may check that \[\begin{align} \label{cL1213} \notag{\mathcal{L}}_{1,2}+{\mathcal{L}}_{1,3}&=&\int_{\mathbb{R}^3}\Big(\sum_{i,j=1}^3(a_{ij}*\mu)v_iv_j(\frac{1}{2}+k\langle v\rangle^{-2})^2+\sum_{i=1}^3(b_i*\mu)v_i(\frac{1}{2}+k\langle v\rangle^{-2})+4\pi\mu\Big)g^2dv\\ \notag&=&\int_{\mathbb{R}^3}\Big((\frac{1}{2}+k\langle v\rangle^{-2})^2|v|^2l_1(v)-(\frac{1}{2}+k\langle v\rangle^{-2})|v|^2l_1(v)+4\pi\mu\Big)g^2dv\\ &\lesssim&-\frac{1}{4}\int_{\mathbb{R}^3}\langle v\rangle^{-1}g^2 dv+C_k\int_{\mathbb{R}^3}\langle v\rangle^{-3}g^2dv. \end{align}\tag{26}\] Now patching together estimates 24 , 25 and 26 and the fact that \(\|g\|^2_{L^2_{-\frac{3}{2}}}\leq \varepsilon\|g\|^2_{L^2_{-\frac{1}{2}}}+C_\varepsilon\|g\|^2_{L^2_{-k}}\), we complete the proof of ?? .

Finally, we handle with ?? . Let \(\mu^{\frac{1}{2}} g=\tilde{g},\mu^{\frac{1}{2}} h=\tilde{h}\) and \(\langle v\rangle^k f=\tilde{f}\), we rewrite it as \[\begin{align} &&\big(\Gamma(g,h),f\langle v\rangle^{2k}\big)_{L^2_v}=\big(Q(\tilde{g},\tilde{h}),\tilde{f}\mu^{-\frac{1}{2}}\langle v\rangle^{k}\big)_{L^2_v}\\ &=&\big(Q(\tilde{g},\mu^{-\frac{1}{2}}\langle v\rangle^{k}\tilde{h}),\tilde{f}\big)_{L^2_v}+\big(\mu^{-\frac{1}{2}}\langle v\rangle^{k}Q(\tilde{g},\tilde{h})-Q(\tilde{g},\mu^{-\frac{1}{2}}\langle v\rangle^{k}\tilde{h}),\tilde{f}\big)_{L^2_v}:=\Gamma_1+\Gamma_2. \end{align}\]

For \(\Gamma_1\), by Lemma 2, we have \[\begin{align} \label{Ga1} |\Gamma_1|\lesssim\|\tilde{g}\|_{L^2_5}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{D_2(0)}\|\tilde{f}\|_{D_2(0)}\leq C_k \|g\|_{L^2_v}\|h\|_{D_2(k)}\|f\|_{D_2(k)}. \end{align}\tag{27}\]

For \(\Gamma_2\), by integration by parts, we derive that \[\begin{align} \Gamma_2&=&2\sum_{i=1}^3\int_{\mathbb{R}^3}(b_i*\tilde{g})\partial_i(\mu^{-\frac{1}{2}}\langle v\rangle^k)\tilde{h}\tilde{f}dv+\sum_{i,j=1}^3\int_{\mathbb{R}^3}(a_{ij}*\tilde{g})\partial_{ij}(\mu^{-\frac{1}{2}}\langle v\rangle^k)\tilde{h}\tilde{f}dv\\ &&+2\sum_{i,j=1}^3\int_{\mathbb{R}^3}(a_{ij}*\tilde{g})\partial_i(\mu^{-\frac{1}{2}}\langle v\rangle^k)\partial_j\tilde{f}\tilde{h}dv:=\Gamma_{2,1}+\Gamma_{2,2}+\Gamma_{2,3}. \end{align}\] We first have \[\begin{align} \label{Ga21} \notag|\Gamma_{2,1}|&\lesssim&\Big|\int_{\mathbb{R}^3}\frac{(v-v_*)v}{|v-v_*|^3}\tilde{g}(v_*)(\frac{1}{2}+k\langle v\rangle^{-2})\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\tilde{f}(\mathbf{1}_{|v-v_*|\leq1}+\mathbf{1}_{|v-v_*|>1})dv_*dv\Big|\\ &\lesssim&C_k\|\tilde{g}\|_{L^2_6}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-\frac{1}{2}}}\big(\|\tilde{f}\|_{H^1_{-\frac{3}{2}}}+\|\tilde{f}\|_{L^2_{-\frac{1}{2}}}\big). \end{align}\tag{28}\] Here we use Hardy-Littlewood-Sobolev inequality, \(|\frac{1}{2}+k\langle v\rangle^{-2}|\leq C_k\) and the facts that \(\langle v_*\rangle\sim\langle v\rangle\) when \(|v-v_*|\leq 1\) and \(|v-v_*|\geq \langle v_*\rangle^{-1}\langle v\rangle\) when \(|v-v_*|>1\).

For \(\Gamma_{2,2}\), using the short notation \(\tilde{g}_\ast = \tilde{g}(v_\ast)\), the relation \(\sum_{i,j=1}^3a_{ij}(v-v_*)v_iv_j=\sum_{i,j=1}^3a_{ij}(v-v_*)(v_*)_i(v_*)_j\) and \((\mathbf{I} - z \otimes z/|z|^2) : (x \otimes y) = |z/|z| \times x|\cdot|z/|z| \times y|\), we have \[\begin{align} \label{Ga22} \notag|\Gamma_{2,2}|&\leq& C_k\big|\int_{\mathbb{R}^6}\frac{|(v-v_*)\times v_*|^2}{|v-v_*|^3}\tilde{g}_*\tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^k \tilde{f}dv_*dv\big|+C_k\big|\int_{\mathbb{R}^6}|v-v_*|^{-1}\langle v_*\rangle^2\tilde{g}_* \tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^k \tilde{f} dv_*dv\big|\\ \notag&\leq& C_k\int_{\mathbb{R}^6}|v-v_*|^{-1}(\mathbf{1}_{|v-v_*|\leq1}+\mathbf{1}_{|v-v_*|\geq1})\langle v_* \rangle^2|\tilde{g}_*||\tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^k||\tilde{f}|dv_*dv \\ &\leq&C_k \|\tilde{g}\|_{L^2_5}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-1/2}}\|\tilde{f}\|_{L^2_{-1/2}}. \end{align}\tag{29}\]

For \(\Gamma_{2,3}\), a direct computation gives \[\begin{align} |\Gamma_{2,3}|&\leq& C_k\bigg[\int_{\mathbb{R}^6}\frac{|(v-v_*)\times \nabla\tilde{f}|\cdot|(v-v_*)\times v_*|}{|v-v_*|^3}\mathbf{1}_{|v-v_*|<1}\tilde{g}_*\tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^kdv_*dv\\ &&+ \int_{\mathbb{R}^6}\frac{|(v-v_*)\times \nabla\tilde{f}|\cdot|(v-v_*)\times v_*|}{|v-v_*|^3}\mathbf{1}_{\substack{|v-v_*|\geq1\\|v-v_*|\leq\frac{1}{2}|v|} }\tilde{g}_*\tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^kdv_*dv\\ && + \int_{\mathbb{R}^6}\frac{|(v-v_*)\times \nabla\tilde{f}|\cdot|(v-v_*)\times v_*|}{|v-v_*|^3}\mathbf{1}_{\substack{|v-v_*|\geq1\\|v-v_*|>\frac{1}{2}|v|} }\tilde{g}_*\tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^kdv_*dv\bigg]\\ &:=&\Gamma_{2,3,1}+\Gamma_{2,3,2}+\Gamma_{2,3,3}. \end{align}\] One may easily check that \[\begin{align} |\Gamma_{2,3,1}| \leq C_k \iint_{\mathbb{R}^6}|v-v_*|^{-1}\mathbf{1}_{|v-v_*|< 1}|v_*||\tilde{g}_*||\nabla\tilde{f}||\tilde{h}\mu^{-\frac{1}{2}}\langle v\rangle^k| dv_*dv \leq C_k \|\tilde{g}\|_{L^2_5}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-1/2}}\|\nabla\tilde{f}\|_{L^2_{-3/2}}. \end{align}\] In the regime \(|v-v_*|\leq\frac{1}{2}|v|\), we have \(|v|\sim |v_*|\), which implies that \[\begin{align} |\Gamma_{2,3,2}|\leq C_k\iint_{\mathbb{R}^6}|v-v_*|^{-1}\mathbf{1}_{\substack{|v-v_*|\geq1\\|v-v_*|\leq\frac{1}{2}|v|}}|v_*|\tilde{g}_*|\nabla\tilde{f}||\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}|dv_*dv \leq C_k \|\tilde{g}\|_{L^2_5}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-1/2}}\|\nabla\tilde{f}\|_{L^2_{-3/2}}. \end{align}\] While in the regime \(|v-v_*|\geq1, |v-v_*|> \frac{1}{2}|v|\), we have \(|v-v_*|^{-1}\lesssim\langle v\rangle^{-1}\). Then we get that \[\begin{align} |\Gamma_{2,3,3}| &\leq &C_k\iint_{\mathbb{R}^6}\frac{|v\times\nabla\tilde{f}|}{\langle v\rangle^2}|v_*||\tilde{g}_*||\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}|dv_*dv+\iint_{\mathbb{R}^6}\frac{|\nabla f||v_*|^2}{\langle v\rangle^2}|\tilde{g}_*|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}|dv_*dv\\ &\leq&C_k \|\tilde{g}\|_{L^2_5}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-1/2}}\big(\|(\Delta_{\mathbb{S}^2})^{\frac{1}{2}}\tilde{f}\|_{L^2_{-3/2}}+\|\nabla\tilde{f}\|_{L^2_{-3/2}}\big). \end{align}\] Thus we obtain that \[\begin{align} \label{Ga23} |\Gamma_{2,3}|\leq C_k\|\tilde{g}\|_{L^2_5}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-1/2}}\big(\|(\Delta_{\mathbb{S}^2})^{\frac{1}{2}}\tilde{f}\|_{L^2_{-3/2}}+\|\nabla\tilde{f}\|_{L^2_{-3/2}}\big). \end{align}\tag{30}\] Together with 28 and 29 , this implies that \[\begin{align} \label{Ga2} |\Gamma_2|&\leq& C_k\|\tilde{g}\|_{L^2_6}\|\mu^{-\frac{1}{2}}\langle v\rangle^k\tilde{h}\|_{L^2_{-1/2}}\big(\|(\Delta_{\mathbb{S}^2})^{\frac{1}{2}}\tilde{f}\|_{L^2_{-3/2}}+\|\nabla\tilde{f}\|_{L^2_{-3/2}}+\|\tilde{f}\|_{L^2_{-\frac{1}{2}}}\big)\\ \notag&\leq& C_k\|g\|_{L^2_v}\|h\|_{D_2(k)}\|f\|_{D_2(k)}. \end{align}\tag{31}\] Therefore, we get the desired result ?? by combining 31 and 27 .

We complete the proof of this lemma.

Thanks to the definition of \(L\) in 14 , ?? and ?? in Lemma 3, the left-hand side of ?? has the upper bound \[\begin{align} -\lambda_0\|f\|^2_{D_1(k)}+C_k\|g\|_{L^2_{7}}\|f\|^2_{H^1_{k-\frac{3}{2}}}+C_k\|f\|^2_{L^2_3}-A\int_{\mathbb{R}^3}\chi_M f^2\langle v\rangle^{2k}dv. \end{align}\] Since \(k \geq 5\) and \(\langle v\rangle^k\geq 1\), we have \[\begin{align} C_k\|f\|^2_{L^2_3}-A\int_{\mathbb{R}^3}\chi_M f^2\langle v\rangle^{2k}dv\leq C_k\int_{|v|>M}f^2\langle v\rangle^{6}dv+(C_kM^6-A)\int_{|v|\leq M}f^2dv&\\ \leq\frac{C_k}{M}\|f\|^2_{L^2_{k-\frac{3}{2}}}+(C_kM^6-A)\int_{|v|\leq M}f^2dv\leq \frac{\lambda_0}{2}\|f\|^2_{D_1(k)}&, \end{align}\] if we choose \(M>2C_k/\lambda_0\) and \(A=C_kM^6\). It ends the proof of this lemma.

We begin with the proof of ?? . For the case \(|\alpha_1|=|\alpha_2|=1\), by ?? in Lemma 3, we have that \[\begin{align} &&\int_{\mathbb{T}^3}\big(Q(\partial^{\alpha_1}_xg,\partial^{\alpha_2}_xh),\partial^\alpha_xf\langle v\rangle^{2(k-4|\alpha|)}\big)_{L^2_v}dx \leq C_k \int_{\mathbb{T}^3}\|\partial^{\alpha_1}_x g\|_{L^2_7}\|\partial^{\alpha_2}_xh\|_{D_1(k-7)}\|\partial^\alpha_x f\|_{D_1(k-8)}dx. \end{align}\] By Hölder inequality, Sobolev embedding \(H^1_x(\mathbb{T}^3)\hookrightarrow L^6_x(\mathbb{T}^3),H^{1/2}_x(\mathbb{T}^3)\hookrightarrow L^3_x(\mathbb{T}^3)\), the right-hand side can be bounded by \[\begin{align} &&C_k\|\partial^{\alpha_1}_x g\|_{L^6_xL^2_7}\|\partial^{\alpha_2}_xh\|_{L^3_xD_1(k-7)}\|\partial^\alpha_x f\|_{L^2_xD_1(k-8)}\\ &\leq& C_k\|\partial^{\alpha_1}_x g\|_{H^1_xL^2_7}\|\partial^{\alpha_2}_xh\|_{H^{1/2}_xD_1(k-7)}\|\partial^\alpha_x f\|_{L^2_xD_1(k-8)}\\ &\leq& C_k\|g\|_{X_{15}}\|h\|_{Y_k}\|f\|_{Y_k}, \end{align}\] where the last step invokes Lemma 6 to obtain \[\begin{align} \|\partial^{\alpha_2}_xh\|_{H^{1/2}_xD_1(k-7)}\lesssim\|\partial^{\alpha_2}_xh\|_{H^{1/2}_xD_1(k-6)}\leq C_k\|h\|^{1/4}_{L^2_xD_1(k)}\|h\|^{3/4}_{H^{2}_xD_1(k-8)}\leq C_k\|h\|_{Y_k}. \end{align}\]

For \(|\alpha_1|=2\) we use the Sobolev embedding \(H^{3/2+\delta}_x(\mathbb{T}^3)\hookrightarrow L^\infty_x(\mathbb{T}^3)\) (\(\delta>0\)) together with the interpolation estimate \[\begin{align} \|h\|_{H^{7/4}_xD_1(k-7)}\leq C_k\|h\|^{1/8}_{L^2_xD_1(k)}\|h\|^{7/8}_{H^2_xD_1(k-8)}\leq C_k\|h\|_{Y_k} \end{align}\] supplied by Lemma 6. In fact, we have \[\begin{align} \int_{\mathbb{T}^3}\big(Q(\partial^{\alpha}_xg,h),\partial^\alpha_xf\langle v\rangle^{2(k-4|\alpha|)}\big)_{L^2_v}dx \leq C_k \int_{\mathbb{T}^3}\|\partial^{\alpha}_x g\|_{L^2_7}\|h\|_{D_1(k-7)}\|\partial^\alpha_x f\|_{D_1(k-8)}dx&\\ \leq C_k\|\partial^{\alpha}_x g\|_{L^2_xL^2_7}\|h\|_{H^{7/4}_xD_1(k-7)}\|\partial^\alpha_x f\|_{{L_x^2} D_1(k-8)}\leq C_k\|g\|_{X_{15}}\|h\|_{Y_k}\|f\|_{Y_k}&. \end{align}\] This completes the proof of ?? .

A similar argument yields ?? , this time invoking ?? instead of ?? . We omit the details and regard the lemma as proved.

The proof of the first assertion can be found in Lemma 1.68 in [35]. We give the proof for the second assertion. Suppose \(\{f_j\},j\in\mathbb{N}\) is a bounded sequence in space \(H^a_b\). By Lemma 4.15 in [34], we have \[\begin{align} \|f_{i}\|^2_{H^a_b}\sim \sum_{k=-1}^\infty \|\langle v\rangle^b{\mathcal{P}}_k f_i\|^2_{H^a}, \end{align}\] where the dyadic operator \({\mathcal{P}}_k\) is defined as \[\begin{align} {\mathcal{P}}_{-1} f=\psi f,\quad {\mathcal{P}}_kf=\varphi(2^{-k}\cdot)f,~~k\geq0, \end{align}\] with nonnegative functions \(\psi,\varphi\in C^\infty_c(\mathbb{R}^3)\) satisfying \(\psi+\sum_{k\geq0}\varphi(2^{-k}\cdot)\equiv1\). Then by the first assertion and diagonal argument, there exists a subsequence, still denoted by \(\{f_{i}\}\), such that \(\langle v\rangle^b{\mathcal{P}}_k f_i\) converges in \(L^2(\mathbb{R}^3)\) for any \(k\geq-1\). Since \[\begin{align} \|f_i-f_j\|^2_{L^2}&\sim& \sum_{k=-1}^\infty 2^{-2bk}\|\langle v\rangle^{b}{\mathcal{P}}_k(f_i-f_j)\|^2_{L^2}\\ &=&\sum_{k< M}2^{-2bk}\|\langle v\rangle^{b}{\mathcal{P}}_k(f_i-f_j)\|^2_{L^2}+\sum_{k\geq M}2^{-2bk}\|\langle v\rangle^{b}{\mathcal{P}}_k(f_i-f_j)\|^2_{L^2}\\ &\leq& M\sup_{k< M}\|\langle v\rangle^b{\mathcal{P}}_k(f_i-f_j)\|^2_{L^2}+C_b 2^{-2bM}\sup_{i}\|f_i\|^2_{H^a_b}. \end{align}\] For any \(\varepsilon>0\), we first choose large \(M\) such that the second term less that \(\varepsilon/2\). Noticing that \(\{\langle v\rangle^b{\mathcal{P}}_k f_i\}\) is a Cauchy sequence for any \(k\geq-1\), thus there exists \(N_M>0\) such that for \(i,j>N_M\), \[\begin{align} \sup_{k< M}\|\langle v\rangle^b{\mathcal{P}}_k(f_i-f_j)\|^2_{L^2}<\frac{\varepsilon{2M}}{,} \end{align}\]which implies that \(\{f_i\}\) is a Cauchy sequence in \(L^2(\mathbb{R}^3)\).

It is straightforward to extend the above results to the case \(\mathbb{T}^3\times\mathbb{R}^3\) and we end the proof of this lemma.

Since, for any given \(h\), the pair \((V_h,T_h)\) forms a self-contained system, we begin by deriving estimates for \(V_h(t)\), \(T_h(t)\), and \(R_h(t)\) on a short time interval. As \(T(0)>0\), continuity allows us to assume that \(T_h(t)>0\) for sufficiently small times. We first estimate \(R_h(t)\) in two different ways. On one hand, since \[\begin{align} \sup\limits_{t\in[0,{\mathcal{T}}]}\|\mu+h(t)\|_{L^2_{x}L^2_2\cap L^1_{x,v}}< 3, \end{align}\] we have \[\begin{align} \label{R146new} \notag |R_h(t)|&\leq&\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{|V_h(t)+T_h(t)^{\frac{1}{2}}v|}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle|V_h(t)+T_h(t)^{\frac{1}{2}}v|^2}|\mu+h|dvdx\\ \notag&\leq& \Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle^2|V_h(t)+T_h(t)^{\frac{1}{2}}v|^2}dvdx\Big)^{\frac{1}{2}}\Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}(\mu+h)^2dvdx\Big)^{\frac{1}{2}} \\ &\leq& 20T_h(t)^{-\frac{3}{4}}, \end{align}\tag{32}\] by a change of variables since \(v \mapsto \langle v \rangle^{-2} |v|^{-2}\) is integrable. On the other hand, we split the integral regime into \(|T_h(t)^{\frac{1}{2}}v|\leq \frac{1}{2} |V_h(t)|\) and \(|T_h(t)^{\frac{1}{2}}v|>\frac{1}{2} |V_h(t)|\), i.e., \[\begin{align} R_h(t)&=&\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{V_h(t)+T_h(t)^{\frac{1}{2}}v}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle|V_h(t)+T_h(t)^{\frac{1}{2}}v|^2}(\mu+h)dvdx\\ &=&\int_{|T_h(t)^{\frac{1}{2}}v|\leq \frac{1}{2} |V_h(t)|}+\int_{|T_h(t)^{\frac{1}{2}}v|>\frac{1}{2} |V_h(t)|}. \end{align}\] In the first regime, it holds that \(|V_h(t)+T_h(t)^{\frac{1}{2}}v|\geq \frac{1}{2}|V_h(t)|\) and in the second regime, we have \(|v|>\frac{1}{2}T_h(t)^{-\frac{1}{2}}|V_h(t)|\). Thus we have \[\begin{align} \label{R246new} \notag|R_h(t)|&\leq& 4\langle V_h(t)\rangle^{-1}|V_h(t)|^{-1}\|\mu+h\|_{L^1_{x,v}}+ \Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{|V_h(t)+T_h(t)^{\frac{1}{2}}v|^2\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle^2}dvdx\Big)^{\frac{1}{2}}\\ \notag&&\times2^k|V_h(t)|^{-k}T_h(t)^{k/2}\|\mu+h\|_{L^2_xL^2_k}\leq 5\times2^k(\langle V_h(t)\rangle^{-1}|V_h(t)|^{-1} + |V_h(t)|^{-k} T_h(t)^{k/2-3/4})\\ \end{align}\tag{33}\] for any \(k>0\). Using that \(a \leq b_1\) and \(a \leq b_2\) (for \(a \geq 0\)) implies \(a \leq \sqrt{b_1 b_2}\), we get (choosing \(k=2\)) \[\begin{align} \label{T4} |T_h'(t)|=\frac{4}{3} |R_h(t)\cdot V_h(t)|\leq 40T_h(t)^{-3/8}(1 + T_h(t)^{1/8}). \end{align}\tag{34}\] This suffices to show that there exists \(0<t_0<1\), depending only on \(T(0)\) (in fact, on the lower bound of \(T(0)\)), such that \[\begin{align} \label{T3} T(0)/2\leq T_h(t)\leq 3T(0)/2,\quad t\in[0,t_0]. \end{align}\tag{35}\] For example, we can choose \[\begin{align} \label{t0} t_0=\frac{1}{160}\min\{1,T(0)^{\frac{3}{2}}\}. \end{align}\tag{36}\] Then 35 implies that \[\begin{align} \label{Rupperbound} |R_h(t)|\leq 40 T(0)^{-\frac{3}{4}},\quad t\in[0,t_0]. \end{align}\tag{37}\] Furthermore, since \[\begin{align} V_h(t)=V(0)+Et-2\int_0^{t}R_h(\tau)d\tau, \end{align}\] together with 37 , this implies \[\begin{align} \label{Vupperbound} |V_h(t)|\leq |V(0)+Et|+80tT(0)^{-\frac{3}{4}},\quad t\in[0,t_0]. \end{align}\tag{38}\]

With the short-time behavior of \(T_h(t),R_h(t)\) and \(V_h(t)\), we are now in the position to handle the equation for \(g\) within \(t\in[0,{\mathcal{T}}],{\mathcal{T}}<t_0\). To be rigorous, we regularize the equation, and first consider the following system \[\label{linearkappa} \partial_tg_\kappa+T_h(t)^{\frac{1}{2}}v\cdot\nabla_x g_\kappa=T_h(t)^{-\frac{3}{2}}Q(\mu+h,g_\kappa)+T_h(t)^{-\frac{3}{2}}Q(h,\mu)+B_{1,h}(g_\kappa)+B_{2,h}(\mu)-T_h(t)^{-\frac{3}{2}}\kappa\langle v\rangle^6\big((\Lambda\mathbf{I}-\Delta_v)^2\big)g_k\tag{39}\] with \(\kappa>0\) and initial data \[\begin{align} (g_\kappa(0),V_h(0),T_h(0))=(g_0,V(0),T(0)). \end{align}\] Here, \(\Lambda>0\) is large enough such that for any \(\psi\), we have \[\begin{align} \label{D2} (\langle v\rangle^6\big((\Lambda \mathbf{I}-\Delta_v)^2\big)\psi,\psi)_{X_k}\geq \sum_{|\alpha|\leq 2}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x\psi\|^2_{L^2_xH^2_v}. \end{align}\tag{40}\] Let \(\mathcal{Q}\) be the linear operator given by \[\begin{align} {\mathcal{Q}}=-\partial_t+(T_h(t)^{\frac{1}{2}}v\cdot\nabla_x-T_h(t)^{-\frac{3}{2}}Q^*(\mu+h,\cdot))-B^*_{1,h}(\cdot)+T_h(t)^{-\frac{3}{2}}\kappa\langle v\rangle^6\big((\Lambda \mathbf{I}-\Delta_v)^2\big)^*, \end{align}\] where the adjoint operator \((\cdot)^*\) is taken with respect to the scalar product in \(X_k\). Then, for all \(z\in C^\infty([0,{\mathcal{T}}]\times\mathbb{T}^3,{\mathcal{S}}(\mathbb{R}^3))\) with \(z({\mathcal{T}})=0\) and \(0\leq t\leq {\mathcal{T}}\), we have \[\begin{align} (z(t),{\mathcal{Q}}z(t))_{X_k}&=&-\frac{1}{2}\frac{d}{dt}\|z\|^2_{X_k}+T_h(t)^{\frac{1}{2}}(v\cdot\nabla_xz,z)_{X_k}-T_h(t)^{-\frac{3}{2}}(Q(\mu+h,z),z)_{X_k}\\ &&-(B_{1,h}(z),z)_{X_k}+T_h(t)^{-\frac{3}{2}}\kappa(\langle v\rangle^6(\Lambda \mathbf{I}-\Delta_v^2)z,z)_{X_k}. \end{align}\] Denote the third and forth term on the right-hand side by \(I_1\) and \(J_1\), then we have \[\begin{align} I_1=-T_h(t)^{-\frac{3}{2}}\Big(\sum_{|\alpha|=0,2}(Q(\mu+h,\partial^\alpha_xz),\partial^\alpha_xz\langle v\rangle^{2(k-4|\alpha|)})_{L^2_{x,v}}-\sum_{|\alpha_1|\geq 1}(Q(\partial^{\alpha_1}_xh,\partial^{\alpha_2}_xz),\partial^\alpha_xz\langle v\rangle^{2(k-4|\alpha|)})_{L^2_{x,v}}\Big). \end{align}\] Due to Lemma 3 and Lemma 7, we can derive that \[\begin{align} \label{I1} I_1\geq T_h(t)^{-\frac{3}{2}}(\lambda_0-C_k\|h\|_{X_{15}})\|z\|^2_{Y_k}-C_kT_h(t)^{-\frac{3}{2}}\|z\|^2_{X_{15}}. \end{align}\tag{41}\] For \(J_1\), the direct computation yields that \[\begin{align} J_1=&\sum_{|\alpha|=0,2}\Big[\Big((-\frac{1}{2} T_h(t)^{-1}T_h'(t)v+2T_h(t)^{-\frac{1}{2}}R_h(t))\cdot\nabla_v \partial^\alpha_xz,\partial^\alpha_xz\langle v\rangle^{2(k-4|\alpha|)}\Big)_{L^2_{x,v}}\\ -&\frac{3}{2}\Big( T_h(t)^{-1} T_h'(t) \partial^\alpha_x z,\partial^\alpha_x z\langle v\rangle^{2(k-4|\alpha|)}\Big)_{L^2_{x,v}}+T_h(t)^{-1}\Big( \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \partial^\alpha_xz\Big),\nabla_v(\partial^\alpha_xz\langle v\rangle^{2(k-4|\alpha|)})\Big)_{L^2_{x,v}}\Big]\\ =&J_{1,1}+J_{1,2}+J_{1,3}. \end{align}\] Since \[\begin{align} &&\int_{\mathbb{T}^3\times\mathbb{R}^3}v\cdot\nabla_v \partial^\alpha_xz \langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_xz dvdx=-3\int_{\mathbb{T}^3\times\mathbb{R}^3}(\partial^\alpha_xz)^2\langle v\rangle^{2(k-4|\alpha|)}dvdx\\ &&-\int_{\mathbb{T}^3\times\mathbb{R}^3}v\cdot\nabla_v \partial^\alpha_x z \langle v\rangle^{2(k-4|\alpha|)} \partial^\alpha_x zdvdx-2(k-4|\alpha|)\int_{\mathbb{T}^3\times\mathbb{R}^3}\langle v\rangle^{2(k-4|\alpha|)-2}|v|^2 (\partial^\alpha_x z)^2dvdx, \end{align}\] which implies that \[\begin{align} \int_{\mathbb{T}^3\times\mathbb{R}^3}v\cdot\nabla_v \partial^\alpha_xz \langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_xzdvdx=&-\frac{3}{2} \|\partial^\alpha_xz\|^2_{L^2_xL^2_{k-4|\alpha|}}-(k-4|\alpha|)\int_{\mathbb{T}^3\times \mathbb{R}^3}\langle v\rangle^{2(k-4|\alpha|)-2}|v|^2 (\partial^\alpha_x z)^2dvdx\\ =&-(k-4|\alpha|+\frac{3}{2})\|\partial^\alpha_xz\|^2_{L^2_xL^2_{k-4|\alpha|}}+(k-4|\alpha|)\|\partial^\alpha_x z\|^2_{L^2_xL^2_{k-4|\alpha|-1}}. \end{align}\] Similarly, we have \[\begin{align} \int_{\mathbb{T}^3\times\mathbb{R}^3}\nabla_v \partial^\alpha_xz\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_xzdv=-(k-4|\alpha|)\int_{\mathbb{T}^3\times\mathbb{R}^3}\langle v\rangle^{2(k-4|\alpha|)-2}v(\partial^\alpha_xz)^2dvdx. \end{align}\] Thus we can derive that \[\begin{align} J_{1,1}+J_{1,2}&=&\frac{1}{2}T_h(t)^{-1}T_h'(t)\sum_{|\alpha|=0,2}(k-4|\alpha|-\frac{3}{2})\|\partial^\alpha_xz\|^2_{L^2_xL^2_{k-4|\alpha|}}+\frac{1}{2}T_h(t)^{-1}T_h'(t)\sum_{|\alpha|=0,2}(k-4|\alpha|)\|\partial^\alpha_xz\|^2_{L^2_xL^2_{k-4|\alpha|-1}}\\ &&-2T_h(t)^{-\frac{1}{2}}R_h(t)(k-4|\alpha|)\int_{\mathbb{T}^3\times\mathbb{R}^3}\langle v\rangle^{2(k-4|\alpha|)-2}v(\partial^\alpha_xz)^2dvdx. \end{align}\] For \(J_{1,3}\), we have \[\begin{align} J_{1,3}&=&\sum_{|\alpha|=0,2}T_h(t)^{-1}\Big( \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \partial^\alpha_xz\Big),\nabla_v(\partial^\alpha_xz)\langle v\rangle^{2(k-4|\alpha|)}\Big)_{L^2_{x,v}}\\ &&+\sum_{|\alpha|=0,2}2(k-4|\alpha|)T_h(t)^{-1}\Big( \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \partial^\alpha_xz\Big),\partial^\alpha_xz\langle v\rangle^{2(k-4|\alpha|-1)}v\Big)_{L^2_{x,v}}\\ &\geq&\sum_{|\alpha|=0,2}\frac{1}{2} T_h(t)^{-1}\Big( \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \partial^\alpha_xz\Big),\nabla_v(\partial^\alpha_xz)\langle v\rangle^{2(k-4|\alpha|)}\Big)_{L^2_{x,v}}\\ &&-C_kT_h(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}|\partial^\alpha_xz|^2\langle v\rangle^{2(k-4|\alpha|-1)}dvdx. \end{align}\] For the second term on the right-hand side, splitting the integral regime into \(|T_h(t)^{\frac{1}{2}}v|\leq \frac{1}{2} |V_h(t)|\) and \(|T_h(t)^{\frac{1}{2}}v|>\frac{1}{2} |V_h(t)|\), we can bound it by \[\begin{align} C_k\big(T_h(t)^{-1}\langle V_h(t)\rangle^{-1}+T_h(t)^{-1}\langle T_h(t)^{-\frac{1}{2}}|V_h(t)|\rangle^{-1}\big)\|z\|^2_{X_k}. \end{align}\]

Patching together the estimates of \(J_{1,1},J_{1,2}\) and \(J_{1,3}\), we have \[\begin{align} J_1\geq -C_kS_1(t)\|z\|^2_{X_k} \end{align}\] with \[\begin{align} \label{at} S_1(t)=T_h^{-1}(t)|T'_h(t)|+T_h(t)^{-\frac{1}{2}}|R_h(t)|+T_h(t)^{-1}\langle V_h(t)\rangle^{-1}+T_h(t)^{-1}\langle T_h(t)^{-\frac{1}{2}}|V_h(t)|\rangle^{-1}. \end{align}\tag{42}\] Therefore, combining with 41 and 40 , we obtain \[\begin{align} (z(t),{\mathcal{Q}}z(t))_{X_k} &\geq& -\frac{1}{2}\frac{d}{dt}\|z(t)\|^2_{X_k}+T_h(t)^{-\frac{3}{2}}(\lambda_0-C_k\|h\|_{X_{15}})\|z\|^2_{Y_k}-C_kT_h(t)^{-\frac{3}{2}}\|z\|^2_{X_{15}}\\ &&-C_kS_1(t)\|z\|^2_{X_k}+T_h(t)^{-\frac{3}{2}}\kappa\sum_{|\alpha|=0, 2}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x z\|^2_{L^2_xH^2_v}. \end{align}\] If \(\varepsilon_0<\lambda_0/2C_k\), we get that \[\begin{align} -\frac{d}{dt}(e^{W(t)}\|z\|^2_{X_k})+\lambda_0e^{W(t)}T_h(t)^{-\frac{3}{2}}\|z\|^2_{Y_k}\leq 2e^{W(t)}|(z,{\mathcal{Q}}z)_{X_k}|-2\kappa e^{W(t)}T_h(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x z\|^2_{L^2_xH^2_v}, \end{align}\] where \(W(t)=2C_k\int_t^{{\mathcal{T}}}(S_1(\tau)+T_h^{-\frac{3}{2}}(t))d\tau\). Note that \(T_h^{-1}(t)\), \(|T'_h(t)|\) and \(|R_h(t)|\) are all bounded from above by some constant \(C_{T(0)}\) which only depend on the lower bound of \(T(0)\) on \([0,t_0]\) thanks to 35 , 34 and 37 . Then from 42 , we have \[\begin{align} |S_1(t)+T_h^{-\frac{3}{2}}(t)|\lesssim T_h^{-1}(t)|T'_h(t)|+T_h^{-\frac{1}{2}}(t)|R_h(t)|+T_h^{-1}(t)+T_h^{-\frac{3}{2}}(t)\lesssim C_{T(0)},\quad t\in[0,{\mathcal{T}}] \end{align}\] is bounded and hence integrable. Since \(z({\mathcal{T}})=0\), for all \(t\in[0,{\mathcal{T}}]\), we have \[\begin{align} \label{zt} &&\notag\|z(t)\|^2_{X_k}+\kappa\int_t^{{\mathcal{T}}}T(\tau)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x z\|^2_{L^2_xH^2_v}d\tau\\ &\leq& 2\int_t^{{\mathcal{T}}}e^{W(\tau)}|(z,{\mathcal{Q}}z)_{X_k}|d\tau\leq 2e^{W(0)}\int_0^{{\mathcal{T}}}|(z,{\mathcal{Q}}z)_{X_k}|d\tau. \end{align}\tag{43}\] We estimates the right-hand side as follows: \[\begin{align} \int_0^{{\mathcal{T}}}|(z,{\mathcal{Q}}z)_{X_k}|d\tau \leq \int_0^{{\mathcal{T}}} \|z\|_{X_k}\|{\mathcal{Q}}z\|_{X_k}d\tau\leq \|z\|_{L^\infty([0,{\mathcal{T}}],X_k)}\|{\mathcal{Q}}z\|_{L^1([0,{\mathcal{T}}],X_k)}, \end{align}\] which implies that \[\begin{align} \label{injection} \|z\|_{L^\infty([0,{\mathcal{T}}],X_k)}\leq 2e^{W(0)}\|{\mathcal{Q}}z\|_{L^1([0,{\mathcal{T}}],X_k)} \end{align}\tag{44}\] and \[\begin{align} \kappa\int_0^{{\mathcal{T}}}T(\tau)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x z\|^2_{L^2_xH^2_v}d\tau\leq 4e^{2W(0)}\|{\mathcal{Q}}z\|_{L^1([0,{\mathcal{T}}],X_k)}. \end{align}\]

Consider the vector subspace \[\begin{align} {\mathcal{W}}=\{w={\mathcal{Q}}z: z\in C^\infty([0,{\mathcal{T}}]\times\mathbb{T}^3,{\mathcal{S}}(\mathbb{R}^3)),z({\mathcal{T}})=0\}\subset L^1([0,{\mathcal{T}}],X_k). \end{align}\] Since \(g_0\in X_k\), we define a linear functional \[\begin{align} \mathfrak{G}:{\mathcal{W}}\rightarrow \mathop{\mathbb{C}\kern 0pt}\nolimits,\quad w={\mathcal{Q}}z \mapsto (g_0,z(0))_{X_k}-(T_h(t)^{-\frac{3}{2}}Q(h,\mu),z)_{L^2([0,{\mathcal{T}}],X_k)}-(B_{2,h}(\mu),z)_{L^2([0,{\mathcal{T}}],X_k)}, \end{align}\] where \(z\in C^\infty([0,{\mathcal{T}}]\times\mathbb{T}^3,{\mathcal{S}}(\mathbb{R}^3))\) with \(z({\mathcal{T}})=0\). According 44 , the operator \({\mathcal{Q}}\) is injective. The functional \(\mathfrak{G}\) is therefore well defined.

Now let \[\begin{align} I_2=T_h(t)^{-\frac{3}{2}}(Q(h,\mu),z)_{X_k},\quad J_2=(B_{2,h}(\mu),z)_{X_k}. \end{align}\] Since \[\begin{align} (Q(h,\mu),z)_{X_k}=\sum_{|\alpha|=0,2}(Q(\partial^{\alpha}_xh,\mu),\partial^\alpha_x z\langle v\rangle^{2(k-4|\alpha|})_{L^2_{x,v}}, \end{align}\] due to Lemma 3, we can obtain that \[\begin{align} |I_2|\leq C_k T_h(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\|\partial^\alpha_xh\|_{L^2_xL^2_3}\|\partial^\alpha_xz\|_{L^2_xL^2_3}\leq C_k T_h(t)^{-\frac{3}{2}}\|h\|_{X_k}\|z\|_{X_k}. \end{align}\] For \(J_2\), it follows immediately from the definition of \(B_{2,h}\) that \[\begin{align} |J_2|&\leq& C_kS_1(t)\|z\|_{X_{k}}+T_h(t)^{-1}\Big(\mathrm{div} \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \mu\Big),z\Big)_{X_k}\\ &\lesssim& C_kS_1(t)\|z\|_{X_{k}}+T_h(t)^{-1}\Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\nabla^2_v \mu,z\Big)_{L^2_xL^2_k}+T_h(t)^{-\frac{1}{2}}\Big(\frac{1}{|V_h(t)+T_h(t)^{\frac{1}{2}}v|}|\nabla_v \mu|,|z|\Big)_{L^2_xL^2_k}, \end{align}\] where \(S_1(t)\) is defined in 42 . Using the same argument as 32 and 33 , we can bounded the last two terms on the right-hand side by \[\begin{align} \min\{T_h(t)^{-1}+T_h(t)^{-\frac{5}{4}},T_h(t)^{-1}\langle V_h(t)\rangle^{-1}+T_h(t)^{-1}\langle T_h(t)^{-\frac{1}{2}}|V_h(t)|\rangle^{-1}+T_h(t)^{-\frac{1}{2}}|V_h(t)|^{-1}+|V_h(t)|^{-1}\}. \end{align}\] Thus, we obtain \[\begin{align} |J_2|\leq C_kS(t)\|z\|_{X_{k}}\leq C_{k,T(0)}\|z\|_{X_{k}},\quad t\in[0,{\mathcal{T}}], \end{align}\] with \[\begin{align} \label{St} &&S(t) :=S_1(t)+\min\big\{T_h(t)^{-1}+T_h(t)^{-\frac{5}{4}},\\ \notag&&T_h(t)^{-1}\langle V_h(t)\rangle^{-1}+T_h(t)^{-1}\langle T_h(t)^{-\frac{1}{2}}|V_h(t)|\rangle^{-1}+T_h(t)^{-\frac{1}{2}}|V_h(t)|^{-1}+|V_h(t)|^{-1}\big\}. \end{align}\tag{45}\] Using 44 , it holds that \[\begin{align} |\mathfrak{G}(w)|&\leq&\|g_0\|_{X_k}\|z(0)\|_{X_k}+C_kT(0)^{-\frac{3}{2}}\|h\|_{L^2([0,{\mathcal{T}}],X_k)}\|z\|_{L^2([0,{\mathcal{T}}],X_k)}+C_{k,T(0)}\|z\|_{L^2([0,{\mathcal{T}}],X_k)}\\ &\leq& {C_{k,h,T(0)}\|z\|_{L^\infty([0,{\mathcal{T}}],X_k)}}\leq C\|{\mathcal{Q}}z\|_{L^1([0,{\mathcal{T}}],X_k)}=C\|w\|_{L^1([0,{\mathcal{T}}],X_k)} \end{align}\] with \(C\) depending on \(k,h\) and \(T(0)\). Using Hahn-Banach theorem, \(\mathfrak{G}\) may be extend as a continuous linear form on \(L^1([0,{\mathcal{T}}],X_k)\) with a norm smaller than \(C\). It follows that there exists \(g_\kappa\in L^\infty([0,{\mathcal{T}}],X_k)\) such that \[\begin{align} \label{a01} \mathfrak{G}(w)=\int_0^{\mathcal{T}}(g_\kappa(t),w(t))_{X_k}dt,\quadfor all\quad w\in L^1([0,{\mathcal{T}}],X_k). \end{align}\tag{46}\] Hence, for all \(h\), we have \[\label{weaksense} \mathfrak{G}({\mathcal{Q}}z)=\int_0^{\mathcal{T}}(g_\kappa,{\mathcal{Q}}z)_{X_k}dt=(g_0,z(0))_{X_k}-\int_0^{\mathcal{T}}T_{h}^{-\frac{3}{2}}(t)(Q(h(t),\mu),z(t))_{X_k}dt-\int_0^{\mathcal{T}}(B_{2,h}(\mu),z(t))_{X_k}dt.\tag{47}\] This shows that \(g_\kappa\) is a weak solution of the Cauchy problem ?? because \(\sum_{|\alpha|=0,2}\langle v\rangle^{2(k-4|\alpha|)}\partial^{2\alpha}_x\) is bijective in \(C^\infty_0((-\infty,{\mathcal{T}}],{\mathcal{S}}(\mathbb{T}^3\times\mathbb{R}^3))\).

Furthermore, since \[\begin{align} 2e^{W(0)}\int_0^{\mathcal{T}}|(z,{\mathcal{Q}}z)_{X_k}|d\tau &\leq&\frac{\kappa}{2}\int_0^{\mathcal{T}}T_h(\tau)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x z\|^2_{L^2_xH^2_v}d\tau\\ &&+\frac{1}{2\kappa}e^{2W(0)}\int_0^{\mathcal{T}}T_h(\tau)^{\frac{3}{2}}\sum_{|\alpha|=0,2}\|\langle v\rangle^{k-4|\alpha|-3}\partial^\alpha_x{\mathcal{Q}}z\|^2_{L^2_xH^{-2}_v}d\tau, \end{align}\] from 43 , we have \[\begin{align} \|z(t)\|^2_{X_k}+\frac{\kappa}{2} \int_0^{\mathcal{T}}T(\tau)^{-\frac{3}{2}}\sum_{\alpha}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_x z\|^2_{L^2_xH^2_v}d\tau\leq\frac{1}{2\kappa}e^{2W(0)}\int_0^{\mathcal{T}}T_h(\tau)^{\frac{3}{2}}\|\langle v\rangle^{k-4|\alpha|-3}\partial^\alpha_x{\mathcal{Q}}z\|^2_{L^2_xH^{-2}_v}d\tau, \end{align}\] which implies that \(\|z\|_{L^\infty([0,{\mathcal{T}}],X_k)}\lesssim\|{\mathcal{Q}}z\|_{L^2([0,{\mathcal{T}}],\tilde{Y}_k)}\), where we denote by \(L^2([0,{\mathcal{T}}],\tilde{Y}_k)\) the space such that \(\omega\in L^2([0,{\mathcal{T}}],\tilde{Y}_k)\) if and only if \[\begin{align} \sum_{|\alpha|=0,2}\int_0^{\mathcal{T}}\|\langle v\rangle^{k-4|\alpha|-3}\partial^\alpha_x{\mathcal{Q}}z\|^2_{L^2_xH^{-2}_v}d\tau<\infty. \end{align}\] Therefore, \[\begin{align} |\mathfrak{G}(w)|\leq {C\|z\|_{L^\infty([0,{\mathcal{T}}],X_k)}}\leq C\|{\mathcal{Q}}z\|_{L^2([0,{\mathcal{T}}],\tilde{Y}_k)}=C\|w\|_{L^2([0,{\mathcal{T}}],\tilde{Y}_k)}. \end{align}\] Together with 46 , we have that \(g_\kappa\in L^2([0,{\mathcal{T}}],\tilde{Y}_k^*),\) where \(\tilde{Y}_k^*\) is the dual space of \(\tilde{Y}_k\) and it leads to \[\begin{align} \sum_{|\alpha|=0,2}\int_0^{\mathcal{T}}\|\langle v\rangle^{k-4|\alpha|+3}\partial^\alpha_xg_\kappa\|^2_{L^2_xH^{2}_v}d\tau<\infty. \end{align}\] In particular, this implies \(\|g_\kappa\|_{L^2([0,{\mathcal{T}}],Y_{k+3})}<\infty\). Equipped with this regularity of \(g_\kappa\), we are ready to show the energy bound. Using the basic energy method and the same argument used to estimate \(I_1,I_2\) and \(J_1,J_2\), we obtain \[\begin{align} &&\frac{1}{2}\frac{d}{dt}\|g_\kappa\|^2_{X_k}+T_h(t)^{-\frac{3}{2}}(\lambda_0-C_k\|h\|_{X_k})\|g_\kappa\|^2_{Y_k}\\ &\leq& C_{k}(S_1(t)+T_h(t)^{-\frac{3}{2}}+1)\|g_\kappa\|^2_{X_k}+C_kT_h(t)^{-\frac{3}{2}}\|h\|^2_{X_k}+C_kS^2(t). \end{align}\] Rigorously speaking, due to the transport term \(v\cdot\nabla_x g_\kappa\), one should also regularize \(g_\kappa\) in \(x\) to justify the above inequality, we omit this here and refer the reader to the proof of Theorem 1.1 in [37] for details. This leads to \[\begin{align} &&\Big(\frac{1}{2}-C_k\int_0^{\mathcal{T}}(S_1(t)+T_h(t)^{-\frac{3}{2}}+1)dt\Big)\|g_\kappa\|^2_{L^\infty([0,{\mathcal{T}}],X_k)}+(\lambda_0-C_k\varepsilon_0)\|T_h^{-\frac{3}{2}}g_\kappa\|^2_{L^2([0,{\mathcal{T}}],Y_k)}\\ &\leq&\frac{1}{2}\|g_0\|_{X_k}^2+C_k\int_0^{\mathcal{T}}T_h(t)^{-\frac{3}{2}}dt\|h\|^2_{L^\infty([0,{\mathcal{T}}],X_k)}+C_k\int_0^{\mathcal{T}}S(t)^2dt. \end{align}\] Since \(S_1(t)+T_h(t)^{-\frac{3}{2}}+1\leq C_{T(0)}\) and \(T_h(t)^{-\frac{3}{2}}\leq 2^{\frac{3}{2}}T_h(0)^{-\frac{3}{2}}\) for \(t\in[0,t_0]\), which yields ?? under the assumption that \[\begin{align} \varepsilon_0<\lambda_0/(2C_k),\quad {\mathcal{T}}<\min\{1/(4C_kC_{T(0)}),(2^{-\frac{3}{2}}T(0)^{\frac{3}{2}}/(2C_k))^2\}. \end{align}\] We can define \[\begin{align} \label{T0} \varepsilon_0=\lambda_0/(4C_k),\quad {\mathcal{T}}_0=\frac{1}{2}\min\{1/(4C_kC_{T(0)}),(2^{-\frac{3}{2}}T(0)^{\frac{3}{2}}/(2C_k))^2,t_0\}. \end{align}\tag{48}\]

The existence of a weak solution of ?? is then, obtained by using uniform estimates and passing to the limit \(\kappa\rightarrow0\). Indeed, by 47 or 39 , it holds that \[\begin{align} \|\partial_t g_\kappa\|_{H^{-1}_xH^{-4}_{-6}}&\leq& T_h^{\frac{1}{2}}\|v\cdot\nabla_x g_\kappa\|_{H^{-1}_xH^{-4}_{-6}}+\|B_{1,h}(g_\kappa)+B_{2,h}(\mu)\|_{H^{-1}_xH^{-4}_{-6}}\\ &&+T_h^{-\frac{3}{2}}\|Q(\mu+h,g_\kappa)+Q(h,\mu)+\kappa\langle v\rangle^6\big((\lambda\mathbf{I}-\Delta_v)^2\big)g_\kappa\|_{H^{-1}_xH^{-4}_{-6}}. \end{align}\] We first have \[\begin{align} \|v\cdot\nabla_x g_\kappa\|_{H^{-1}_xH^{-4}_{-6}}+\|\kappa\langle v\rangle^6\big((\lambda\mathbf{I}-\Delta_v)^2\big)g_\kappa\|_{H^{-1}_xH^{-4}_{-6}}\lesssim\|g_\kappa\|_{X_k}. \end{align}\] From the definitions of \(B_{1,h}\) and \(B_{2,h}\) it is straightforward to obtain \[\begin{align} \|B_{1,h}(g_\kappa)+B_{2,h}(\mu)\|_{H^{-1}_xH^{-4}_{-6}}\lesssim(T_h^{-1}|T'_h|+T_h^{-\frac{1}{2}}|R_h|+T_h^{-1})(\|g_\kappa\|_{X_k}+1). \end{align}\] By duality and Lemma 2, we have \[\begin{align} \|Q(\mu+h,g_\kappa)+Q(h,\mu)\|_{H^{-1}_xH^{-4}_{-6}}\lesssim\|\mu+h\|_{L^2_xL^2_7}\|g_\kappa\|_{X_k}+\|h\|_{X_k}. \end{align}\] Thus, we obtain \[\begin{align} \label{patgkappa} \notag\|\partial_t g_\kappa\|_{L^2([0,{\mathcal{T}}], H^{-1}_xH^{-4}_{-6})}&\lesssim& {\mathcal{T}}^{\frac{1}{2}}\sup_{t\in[0,{\mathcal{T}}]}(T_h^{\frac{1}{2}}+T_h^{-\frac{3}{2}}+T_h^{-1}|T'_h|+T_h^{-\frac{1}{2}}|R_h|+T_h^{-1})(t)\\ &\times&(1+\|h\|_{L^\infty([0,{\mathcal{T}}],X_k)})(1+\|g_\kappa\|_{L^\infty([0,{\mathcal{T}}],X_k)}). \end{align}\tag{49}\] Noticing that the coefficients are bounded. Thus, \(\{g_\kappa\}\) is uniformly bounded in \(L^\infty([0,{\mathcal{T}}], X_k)\cap L^2([0,{\mathcal{T}}], Y_k)\cap W^{1,2}([0,{\mathcal{T}}],H^{-1}_xH^{-4}_{-6}),k\ge 17\). By Lemma 9, there exists a subsequence, denoted by \(g_{\kappa_i}\), converging to \(g\) in \(L^2([0,{\mathcal{T}}], H^1_xL^2_7)\) as \(\kappa_i\rightarrow 0\). We replace \(g_\kappa\) by \(g_{\kappa_i}\) in 39 and need to verify that 39 converges to ?? in weak sense as \(\kappa_i\to 0\). We only give the details for \[\begin{align} Q(\mu+h,g_{\kappa_i})\rightarrow Q(\mu+g,g),&&\mathrm{div} \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v g_{\kappa_i}\Big)\rightarrow \mathrm{div} \Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v g\Big)\\ && in\quad{\mathcal{D}}'([0,{\mathcal{T}}]\times\mathbb{T}^3\times\mathbb{R}^3), \end{align}\] and the remaining terms are handled similarly. To this end, let \(\phi\in C_c^\infty([0,{\mathcal{T}}]\times\mathbb{T}^3\times\mathbb{R}^3)\), by ?? in Lemma 2, we have \[\begin{align} &&\Big|\int_0^{\mathcal{T}}\int_{\mathbb{T}^3\times\mathbb{R}^3}(Q(\mu+h,g_{\kappa_i})-Q(\mu+h,g))\phi dvdxdt\Big|\\ &\leq&\Big|\int_0^{\mathcal{T}}\int_{\mathbb{T}^3\times\mathbb{R}^3}Q(\mu+h,g_{\kappa_i}-g)\phi dvdxdt\Big| \leq C_\phi\|\mu+h\|_{L^\infty([0,{\mathcal{T}}],L^2_xL^2_7)}\|g_{\kappa_i}-g\|_{L^2([0,{\mathcal{T}}],L^2_xL^2_7)}\rightarrow 0, \end{align}\] as \(\kappa_i\rightarrow 0\). We also have \[\begin{align} &&\Big|\int_0^{\mathcal{T}}\Big(\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v (g_{\kappa_i}-g),\nabla_v\phi\Big)_{L^2_{x,v}}dt\Big|\\ &\lesssim& \Big|\int_0^{\mathcal{T}}\Big(g_{\kappa_i}-g,\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}:\nabla_v^2\phi\Big)_{L^2_{x,v}}dt\Big|+\Big|\int_0^{\mathcal{T}}\Big(g_{\kappa_i}-g,\nabla_v\frac{\Pi(V_h(t)+T_h(t)^{\frac{1}{2}}v)}{\langle V_h(t)+T_h(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v\phi\Big)_{L^2_{x,v}}dt\Big|\\ &\lesssim& C_\phi {\mathcal{T}}^{\frac{1}{2}}\|g_{\kappa_i}-g\|_{L^2([0,{\mathcal{T}}],L^2_xL^2_7)}+C_\phi \sup_{t\in[0,{\mathcal{T}}]}(T_h^{\frac{1}{2}}+T_h^{-\frac{1}{4}})(t)\|g_{\kappa_i}-g\|_{L^2([0,{\mathcal{T}}],L^2_xL^2_7)}\rightarrow 0,\quadas\quad \kappa_i\rightarrow 0. \end{align}\] Here we use the fact \(\sup_{t\in[0,{\mathcal{T}}]}(T_h^{\frac{1}{2}}+T_h^{-\frac{1}{4}})(t)\) is bounded above. We thereby complete the proof of the existence of a weak solution \(g\) with the desired bound.

We now give the proof for non-negativity. We come back to the original equation, i.e., \[\begin{align} \partial_t F+v\cdot\nabla_x F+E\cdot\nabla_v F=Q(J,F)+div_v \bigg(\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}} \nabla_v F\bigg), \end{align}\] where \(F\) and \(J\) satisfy \(\mu+g=T_h(t)^{\frac{3}{2}}F(t,x+H_h(t),V_h(t)+T_h(t)^{\frac{1}{2}}v)\) and \(\mu+h=T_h(t)^{\frac{3}{2}}J(t,x+H_h(t),V_h(t)+T_h(t)^{\frac{1}{2}}v)\) with \(H_h(t)=\int_0^tV_h(\tau)d\tau\). Thus we need to prove \(F\geq 0\) under the condition that \(F(0)\geq0\) and \(J\geq0\).

In fact, let \(F_\pm=\pm\max\{\pm F,0\}\) then \[F_\pm\in L^\infty([0,T],X_k)\cap L^2([0,T],Y_k),\quad \nabla F_+=\left\{\begin{align} &\nabla F,\quad F>0;\\ &0,\quad F\leq0. \end{align}\right. \quad and\quad \nabla F_-=\left\{\begin{align} &\nabla F,\quad F< 0;\\ &0,\quad F\geq0. \end{align}\right.\] Multiply the above equation by \(F_-\) and integrate over \([0,t]\times\mathbb{T}^3\times\mathbb{R}^3\). Then, in view of \(F_-(0)=0\) and \[\begin{align} \int_0^t {\int_{\mathbb{T}^3\times\mathbb{R}^3}} v\cdot\nabla_x F F_-dxdvd\tau=\int_0^t {\int_{\mathbb{T}^3\times\mathbb{R}^3}} E\cdot\nabla_v F F_-dxdvd\tau=0, \end{align}\] we have \[\begin{align} {\frac{1}{2}}\int_{\mathbb{T}^3\times\mathbb{R}^3} |F_-(t)|^2dxdv=&\int_{0}^t\int_{\mathbb{T}^3\times\mathbb{R}^3}Q(J,F)F_-dvdxd\tau-\int_{0}^t\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}} \nabla_v F\nabla_vF_-dvdxd\tau\\ =&\int_{0}^t\int_{\mathbb{T}^3\times\mathbb{R}^3}Q(J,F_-)F_-dvdxd\tau-\int_{0}^t\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}} \nabla_v F_-\nabla_vF_-dvdxd\tau\leq 0, \end{align}\] which implies that \(F_-(t)=0\), that is \(F(t)\geq0\).

Thus we complete the proof of this lemma.

Consider the sequence of approximate solutions defined by \(g^0=0\) and \[\label{linear2} \left\{\begin{align} &\partial_tg^{n+1} +T_{g^n}(t)^{\frac{1}{2}}v\cdot\nabla_x g^{n+1}=T_{g_n}(t)^{-\frac{3}{2}}Q(\mu+g^n,g^{n+1})+T_{g_n}(t)^{-\frac{3}{2}}Q(g^n,\mu)+B_{1,g^n}(g^{n+1})+B_{2,g^n}(\mu),\\ &V_{g_n}'(t)=E-2R_{g^n}(t),\quad T_{g^n}'(t)=\frac{4}{3}V_{g^n}(t)\cdot R_{g^n}(t) \end{align}\right.\tag{50}\] with initial data \((g^{n+1}(0),V_{g^n}(0),T_{g^n}(0))=(g_0,V(0),T(0))\). Using Lemma 10 with \(g=g^{n+1}, h=g^{n}\), it follows from ?? that \[\label{boundofg} \|g^{n+1}\|^2_{L^\infty([0,{\mathcal{T}}],X_k)}+\lambda_0\|T_{g^n}^{-\frac{3}{2}}g^{n+1}\|^2_{L^2([0,{\mathcal{T}}],Y_k)} \leq2\big(\|g_0\|^2_{X_k}+{\mathcal{T}}^{\frac{1}{2}}\|g^n\|^2_{L^\infty([0,{\mathcal{T}}],X_k)}\big)+C_k\int_0^{\mathcal{T}}S^2(t)dt\leq\varepsilon_1^2<\varepsilon_0^2,\tag{51}\] inductively, if \({\mathcal{T}}\) and \(\varepsilon_1\) are taken such that \[\begin{align} \label{Tvep0} 2(\varepsilon_1^4+{\mathcal{T}}^{\frac{1}{2}}\varepsilon_1^2)+C_k\int_0^{\mathcal{T}}S^2(t)dt\leq \varepsilon_1^2. \end{align}\tag{52}\] It is sufficient for 52 to hold is by choosing \[\begin{align} \varepsilon_1<1/(2\sqrt{2}),\quad {\mathcal{T}}<1/64\quad and\quad C_k\int_0^{\mathcal{T}}S^2(t)dt<\varepsilon_1^2/4. \end{align}\] Noticing that when \({\mathcal{T}}<t_0\), \(|S(t)|\leq C_{T(0)}\) which only depend on the lower bound of \(T(0)\), thus we can choose \[\begin{align} \label{T02} \varepsilon_1<\min\{\varepsilon_0,1/(2\sqrt{2})\},\quad {\mathcal{T}}<\min\{{\mathcal{T}}_0,1/64,\varepsilon_1^2/(4C_{k,T(0)})\}. \end{align}\tag{53}\] Thanks to the estimate 49 , we also have that \(\{g^{n}\}\) is uniformly bounded in \(L^\infty([0,{\mathcal{T}}], X_k)\cap L^2([0,{\mathcal{T}}], Y_k)\cap W^{1,2}([0,{\mathcal{T}}],H^{-1}_xH^{-4}_{-6})\), then Lemma 9 implies that \(g^n\rightarrow g\) in \(L^2([0,{\mathcal{T}}], H^1_xL^2_7)\). To complete the proof of local existence for the nonlinear equation, we need to verify that 50 remains valid in weak sense when both \(g^{n+1}\) and \(g^n\) are replaced by the limit \(g\).

We first prove convergence for the last two equations. Setting \(\omega^n=g^{n}-g,(\Delta T)^n=T_{g^{n}}-T_{g}\) and \((\Delta V)^n=V_{g^{n}}-V_{g}\). Since \[\begin{align} &&R_{g^n}(t)-R_{g}(t)=\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{V_{g^n}(t)+T_{g^n}^{\frac{1}{2}}v}{\langle V_{g^n}(t)+T_{g^n}^{\frac{1}{2}}v\rangle|V_{g^n}(t)+T_{g^n}^{\frac{1}{2}}v|^2}\omega^{n}dvdx\\ &&+\int_{\mathbb{T}^3\times\mathbb{R}^3}\Big(\frac{V_{g^n}(t)+T_{g^n}^{\frac{1}{2}}v}{\langle V_{g^n}(t)+T_{g^n}^{\frac{1}{2}}v\rangle|V_{g^n}(t)+T_{g^n}^{\frac{1}{2}}v|^2}-\frac{V_{g}(t)+T_{g}^{\frac{1}{2}}v}{\langle V_{g}(t)+T_{g}^{\frac{1}{2}}v\rangle|V_{g}(t)+T_{g}^{\frac{1}{2}}v|^2}\Big)(\mu+g)dvdx\\ &&:=O_1+O_2. \end{align}\] Use the same argument as in the estimates of \(R_h(t)\) in 32 and 33 (choosing \(k=3/2\)), we can derive that \[\begin{align} |O_1|\lesssim\min\big\{T_{g^n}(t)^{-\frac{3}{4}},|V_{g^n}(t)|^{-2}+|V_{g^n}(t)|^{-\frac{3}{2}}\big\}\|\omega^{n}\|_{L^2_xL^2_7}. \end{align}\] For \(O_2\), denoting \(V_{g^n}(t)+T_{g^n}^{1/2}v\) and \(V_g(t)+T_g^{1/2}v\) by \(W_n\) and \(W\), respectively, we have \[\begin{align} |W_n-W|\leq |V_{g^n}-V_{g}|+|T^{\frac{1}{2}}_{g^n}-T^{\frac{1}{2}}_{g}||v|\leq \big(1+(T^{\frac{1}{2}}_{g^n}+T^{\frac{1}{2}}_{g})^{-1}\big) \big(|(\Delta V)^{n}|+|(\Delta T)^{n}|\big)\langle v\rangle, \end{align}\] and \[\begin{align} \frac{W_n}{\langle W_n\rangle|W_n|^2}-\frac{W}{\langle W\rangle|W|^2}&=&\frac{W_n-W}{\langle W_n\rangle|W_n|^2}+W\frac{\langle W\rangle-\langle W_n\rangle}{\langle W_n\rangle|W_n|^2\langle W\rangle}+W\frac{|W|^2-|W_n|^2}{|W_n|^2\langle W\rangle|W|^2}\\ &\lesssim&|W_n-W|\Big(\frac{1}{\langle W_n\rangle|W_{n}|^2}+\frac{1}{\langle W\rangle|W|^2}\Big). \end{align}\] Then \[\label{O2} \begin{align} |O_2|\leq& \big(1+(T^{\frac{1}{2}}_{g^n}+T^{\frac{1}{2}}_{g})^{-1}\big)\big( |(\Delta V)^{n}|+|(\Delta T)^{n}|\big)\\ &\times\int_{\mathbb{T}^3\times\mathbb{R}^3}\Big(\frac{1}{\langle W_n\rangle|W_{n}|^2}+\frac{1}{\langle W\rangle|W|^2}\Big)\langle v\rangle|\mu+g|dvdx. \end{align}\tag{54}\] We take the first term in the integral as an example. By the same arguments as in 32 and 33 , we have on the one hand \[\begin{align} \int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle W_n\rangle|W_{n}|^2}\langle v\rangle|\mu+g|dvdx&\lesssim& \Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle W_n\rangle^{\frac{6}{5}}|W_{n}|^{\frac{12}{5}}}dvdx\Big)^{\frac{5}{6}}\Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}\langle v\rangle^6|\mu+g|^6dvdx\Big)^{\frac{1}{6}}\\ &\lesssim&T_{g^n}^{-\frac{5}{4}}\|\mu+g\|_{L^2_xH^1_1}, \end{align}\] where we have used the fact that \(v\mapsto \langle v\rangle^{-6/5}|v|^{-12/5}\) is integrable, together with the Sobolev embedding \(H^1_1(\mathbb{R}^3)\hookrightarrow L^6_1(\mathbb{R}^3)\). On the other hand, we have \[\begin{align} &&\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle W_n\rangle|W_{n}|^2}\langle v\rangle|\mu+g|dvdx=\int_{|T_{g^n}(t)^{\frac{1}{2}}v|\leq \frac{1}{2} |V_{g^n}(t)|}+\int_{|T_{g^n}(t)^{\frac{1}{2}}v|>\frac{1}{2} |V_{g^n}(t)|}\\ &\lesssim& |T_{g^n}|^{-2}\|\mu+g\|_{L^2_{x}L^2_3}+|V_{g^n}|^{-2}T_{g^n}^{-\frac{1}{4}}\|\mu+g\|_{L^2_xH^1_{3}}. \end{align}\] Plugging these into 54 , we obtain \[\begin{align} |O_2|&\lesssim& \big(1+(T^{\frac{1}{2}}_{g^n}+T^{\frac{1}{2}}_{g})^{-1}\big)\big( |(\Delta V)^{n}|+|(\Delta T)^{n}|\big)\\ &\times&\min\Big\{1+T_{g^n}^{-\frac{5}{4}}+T_{g}^{-\frac{5}{4}},|V_{g^n}|^{-2}+|V_{g^n}|^{-2}T_{g^n}^{-\frac{1}{4}}+|V_{g}|^{-2}+|V_{g}|^{-2}T_{g}^{-\frac{1}{4}}\Big\}\|\mu+g\|_{L^2_xH^1_3}. \end{align}\] Combining the estimates of \(O_1\) and \(O_2\), we conclude that \[\begin{align} \label{0001} \notag|\big((\Delta V)^{n}\big)'|\lesssim|R_{g^n}(t)-R_{g}(t)|\lesssim C_1(t)(\|\omega^{n}\|_{L^2_xL^2_7}+\big( |(\Delta V)^{n}|+|(\Delta T)^{n}|\big)\|\mu+g\|_{L^2_xH^1_3})\\ \end{align}\tag{55}\] with \[\begin{align} \label{C1t} C_1(t)&:=&\min\big\{T_{g^n}(t)^{-\frac{3}{4}},|V_{g^n}(t)|^{-2}+|V_{g^n}(t)|^{-\frac{3}{2}}\big\}+\big(1+(T^{\frac{1}{2}}_{g^n}+T^{\frac{1}{2}}_{g})^{-1}\big)\\ \notag&&\times\min\Big\{1+T_{g^n}^{-\frac{5}{4}}+T_{g}^{-\frac{5}{4}},|V_{g^n}|^{-2}+|V_{g^n}|^{-2}T_{g^n}^{-\frac{1}{4}}+|V_{g}|^{-2}+|V_{g}|^{-2}T_{g}^{-\frac{1}{4}}\Big\}. \end{align}\tag{56}\]

Similarly, we have \[\begin{align} &&|V_{g^n}(t)\cdot R_{g^n}(t)-V_{g}(t)\cdot R_{g}(t)|=(\Delta V)^{n}\cdot R_{g}(t)+V_{g^{n}}(t)\cdot(R_{g^n}(t)-R_{g}(t))\\ &\lesssim& \min\big\{T_{g}(t)^{-\frac{3}{4}},|V_{g}(t)|^{-2}+|V_{g}(t)|^{-\frac{3}{2}}\big\}|(\Delta V)^{n}|\\ &&+C_1(t)|V_{g^n}(t)|\Big(\|\omega^{n}\|_{L^2_xL^2_7}+|\big( |(\Delta V)^{n}|+|(\Delta T)^{n}|\big)\|\mu+g\|_{L^2_xH^1_5}\Big), \end{align}\] where we use 32 and 33 (choosing \(k=3/2\)) to bound \(R_{g}\). We denote coefficient by \[\begin{align} \label{C2t} C_2(t):=\min\big\{T_{g}(t)^{-\frac{3}{4}},|V_{g}(t)|^{-2}+|V_{g}(t)|^{-\frac{3}{2}}\big\}+C_1(t)(1+|V_{g^n}(t)|) \end{align}\tag{57}\] and conclude that \[\begin{align} \label{0002} |\big((\Delta T)^{n}\big)'|&\lesssim& |V_{g^n}(t)\cdot R_{g^n}(t)-V_{g}(t)\cdot R_{g}(t)|\\ \notag&\lesssim& C_2(t)\Big(\|\omega^{n}\|_{L^2_xL^2_7}+\big( |(\Delta V)^{n}|+|(\Delta T)^{n}|\big)(1+\|T_{g}^{-\frac{3}{4}}g\|_{L^2_xH^1_5})\Big). \end{align}\tag{58}\]

Combining 55 and 58 and noting that \(((\Delta V)^{n}(0),(\Delta T)^{n}(0))=(0,0)\), we obtain \[\begin{align} &&|(\Delta V)^{n}(t)|+|(\Delta T)^{n}(t)| \leq \sup_{\tau\in[0,{\mathcal{T}}]}C_2(\tau)\Big( \|\omega^{n}\|_{L^1([0,{\mathcal{T}}],L^2_xL^2_7)}\\ &&+\int_0^t(\|\mu+g(\tau)\|_{L^2_xH^1_5})\big(|(\Delta V)^{n-1}(\tau)|+|(\Delta T)^{n-1}(\tau)|\big)d\tau\Big),\quad for any \quad t\in(0,{\mathcal{T}}]. \end{align}\] Noticing that \(\|\mu+g\|_{L^2([0,{\mathcal{T}}],L^2_xH^1_5)}\lesssim 1+\|g\|_{L^2([0,{\mathcal{T}}],Y_k)}\lesssim 1\) in view of 51 , then by Grönwall’s inequality, we can derive that \[\begin{align} \label{DVDT} |(\Delta V)^{n}(t)|+|(\Delta T)^{n}(t)|\lesssim\sup_{\tau\in[0,{\mathcal{T}}]}C_2(\tau) \|\omega^{n}\|_{L^1([0,{\mathcal{T}}],L^2_xL^2_7)},\quad for any \quad t\in(0,{\mathcal{T}}]. \end{align}\tag{59}\] It is easy to see that \(\sup_{\tau\in[0,{\mathcal{T}}]}C_2(\tau)<\infty\), we thus have \(V_{g_n}\to V_g\) and \(T_{g_n}\to T_g\) as \(n\to\infty\) since \(\omega^n=g^n-g\rightarrow 0\) in \(L^2([0,{\mathcal{T}}], H^1_xL^2_7)\) as \(n\to\infty\).

For the first equation in 50 , in virtue of the convergence of \(V_{g^n}\) and \(T_{g^n}\), we only give the details for \[\begin{align} Q(\mu+g^n,g^{n+1})\rightarrow Q(\mu+g,g)\quad in\quad{\mathcal{D}}'([0,{\mathcal{T}}]\times\mathbb{T}^3\times\mathbb{R}^3), \end{align}\] and the remaining terms are handled similarly. To this end, let \(\phi\in C_c^\infty([0,{\mathcal{T}}]\times\mathbb{T}^3\times\mathbb{R}^3)\), by ?? in Lemma 2, we have \[\begin{align} &&\Big|\int_0^{\mathcal{T}}\int_{\mathbb{T}^3\times\mathbb{R}^3}(Q(\mu+g^n,g^{n+1})-Q(\mu+g,g))\phi dvdxdt\Big|\\ &\leq&\Big|\int_0^{\mathcal{T}}\int_{\mathbb{T}^3\times\mathbb{R}^3}Q(g^n-g,g^{n+1})\phi dvdxdt\Big|+\Big|\int_0^{\mathcal{T}}\int_{\mathbb{T}^3\times\mathbb{R}^3}Q(\mu+g,g^{n+1}-g)\phi dvdxdt\Big|\\ &\leq&C_\phi\|g^n-g\|_{L^1([0,{\mathcal{T}}],L^2_xL^2_7)}\|g^{n+1}\|_{L^\infty([0,{\mathcal{T}}],L^2_xL^2_7)}+C_\phi\|\mu+g\|_{L^\infty([0,{\mathcal{T}}],L^2_xL^2_7)}\|g^{n+1}-g\|_{L^1([0,{\mathcal{T}}],L^2_xL^2_7)}\rightarrow 0, \end{align}\] as \(n\rightarrow \infty\). We thereby complete the proof of the existence of a weak solution \(g\) with the desired bound.

Finally, we go back to the original equation for \(F\) and give the proof of uniqueness. By the change of variables between \(F\) and \(G\) (see 12 ), together with the boundedness of \(|V(t)|\), \(T(t)\) and \(T^{-1}(t)\), we obtain \[\begin{align} &&\int_{\mathbb{T}^3\times\mathbb{R}^3}|\partial^\alpha_x F(t,x,v)|^2\langle v\rangle^{16}dvdx=T(t)^{-3}\int_{\mathbb{T}^3\times\mathbb{R}^3}|\partial^\alpha_x G(t,x-H(t),T(t)^{-\frac{1}{2}}(v-V(t)))|^2\langle v\rangle^{16}dvdx\\ &\lesssim&T(t)^{-\frac{3}{2}}\int_{\mathbb{T}^3\times\mathbb{R}^3}|\partial^\alpha_x G(t,x,v)|^2\langle V(t)+T(t)^{\frac{1}{2}}v\rangle^{16}dvdx\lesssim T(t)^{-\frac{3}{2}}(1+\langle V(t)\rangle^{16}+T(t)^{8})\|G\|^2_{X_k},\quad k\geq 16, \end{align}\] which implies that \(\|F\|_{H^2_xL^2_8}\lesssim C_{\mathcal{T}}\|G\|_{X_k}\). Indeed, we can get that \[\begin{align} \label{FG} &&\|F\|_{H^2_xL^2_8}\lesssim C_{\mathcal{T}}\|G\|_{X_k}\in L^\infty[0,{\mathcal{T}}],\quad \|F\|_{H^2_xD_1(8)}\lesssim C_{\mathcal{T}}\|G\|_{Y_k}\in L^2[0,{\mathcal{T}}],\\ \notag &&and\quad \|F(t,x,\cdot)\|_{L^1_v}\geq\inf_{x\in\mathbb{T}^3}\|(\mu+g)(t,x,\cdot)\|_{L^1_v}\geq \|\mu\|_{L^1_v}-\|g\|_{X_k}\geq\frac{1}{2},\quad \forall t\in[0,{\mathcal{T}}]. \end{align}\tag{60}\] Suppose \(F_1\) and \(F_2\) are two solutions of 6 with the same initial data \(F_0\) and satisfy 60 . Let \(\tilde{F}=F_1-F_2\), then we have \(\tilde{F}(0)=0\) and \[\begin{align} \partial_t \tilde{F} + v\cdot\nabla_x \tilde{F} +E \cdot \nabla_v \tilde{F} = Q(F_1,\tilde{F})+Q(\tilde{F},F_2)+div\bigg(\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}} \nabla_v \tilde{F}\bigg). \end{align}\] Consider the energy in space \(L^2_xL^2_7\), we have \[\begin{align} \frac{1}{2}\frac{d}{dt}\|\tilde{F}\|^2_{L^2_xL^2_7}=(Q(F_1,\tilde{F}),\tilde{F}\langle v\rangle^{14})_{L^2_{x,v}}+(Q(\tilde{F}, F_2),\tilde{F}\langle v\rangle^{14})_{L^2_{x,v}}-\Big(\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}}\nabla_v\tilde{F},\nabla_v(\tilde{F}\langle v\rangle^{14})\Big)_{L^2_{x,v}}. \end{align}\] Since \(\Pi(v)\nabla_v\langle v\rangle^{14}=14\langle v\rangle^{12}\Pi(v)v=0\), then \[\begin{align} \Big(\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}}\nabla_v\tilde{F},\nabla_v(\tilde{F}\langle v\rangle^{14})\Big)_{L^2_{x,v}}=\Big(\frac{\Pi(v)}{(1+|v|^2)^{\frac{1}{2}}}\nabla_v\tilde{F},\nabla_v\tilde{F}\langle v\rangle^{14}\Big)_{L^2_{x,v}}\geq 0. \end{align}\] Thanks to 60 and ?? , ?? in Lemma 3, we obtain that \[\begin{align} \frac{1}{2}\frac{d}{dt}\|\tilde{F}\|^2_{L^2_xL^2_7}+\lambda_0\|\tilde{F}\|^2_{L^2_xD_1(7)}&\lesssim& \|\tilde{F}\|_{L^2_xL^2_7}\|F_2\|_{H^2_xD_1(8)}\|\tilde{F}\|_{L^2_xD_1(7)}+\|\tilde{F}\|^2_{L^2_xL^2_7}\\ &\lesssim& C_\varepsilon\|\tilde{F}\|^2_{L^2_xL^2_7}\|F_2\|^2_{H^2_xD_1(8)}+\varepsilon\|\tilde{F}\|^2_{L^2_xD_1(7)}+\|\tilde{F}\|^2_{L^2_xL^2_7}, \end{align}\] where \(\lambda_0\) depend on the lower bound of \(\|F(t,x,\cdot)\|_{L^1_v}\) and \(\|F\|_{L^\infty([0,{\mathcal{T}}],H^2_xL^2_8)}\). Choosing small \(\varepsilon>0\) implies that \[\begin{align} \frac{d}{dt}\|\tilde{F}\|^2_{L^2_xL^2_7}\lesssim(1+\|F_2\|^2_{H^2_xD_1(8)})\|\tilde{F}\|^2_{L^2_xL^2_7}. \end{align}\] Since \(\|F_2\|^2_{H^2_xD_1(8)}\) is integrable by 60 , we have \(\|\tilde{F}(t)\|^2_{L^2_xL^2_7}\lesssim\|\tilde{F}(0)\|^2_{L^2_xL^2_7}=0\) for \(t\in[0,{\mathcal{T}}]\), i.e., \(F_1=F_2\). From 3 and the change of variables 12 between \(F\) and \(G\), it follows that the system \((g,V,T)\) is also unique.

We complete the proof of this theorem.

Lemma 10 gives the behavior for the short interval \(t\in[0,t_0]\) with \(0<t_0<1\) depending on the lower bound of \(T(0)\). We now derive the estimates for \(t> t_0\). Firstly, thanks to 32 , 33 and 35 , we have \(|R(t)|\leq 20\times2^{\frac{3}{4}}T(0)^{-\frac{3}{4}},t\leq t_0\) and \(|R(t)|\leq5\times 2^{3/2}(|V(t)|^{-2}+|V(t)|^{-\frac{3}{2}}),t>t_0\)(choosing \(k=3/2\)). Then by ?? , we have \[\begin{align} \label{Vttleqt0} |V(t)|\geq |V(0)+Et|-80T(0)^{-\frac{3}{4}},\quad t\leq t_0, \end{align}\tag{61}\] and \[\begin{align} |V(t)|\geq |V(0)+Et|-80T(0)^{-\frac{3}{4}}-40\int_{t_0}^t\Big(\frac{1}{|V(\tau)|^2}+\frac{1}{|V(\tau)|^{\frac{3}{2}}}\Big)d\tau,\quad t>t_0. \end{align}\] There exists a large constant \(C_{T(0),V(0)}\) such that if \(|E|>C_{T(0),V(0)}\), then \[\begin{align} \label{con1} \frac{1}{2}|V(0)+Et|>80T(0)^{-\frac{3}{4}}\quad and\quad\frac{1}{3}|V(0)+Et|\geq \max\Big\{1+\frac{1}{4}|E|t,\frac{8\times 720}{|E|}\Big\}\quadfor\quad t> t_0, \end{align}\tag{62}\] which leads that \[\begin{align} |V(t)|\geq \frac{1}{2} |V(0)+Et|-40\int_{t_0}^t\Big(\frac{1}{|V(\tau)|^2}+\frac{1}{|V(\tau)|^{\frac{3}{2}}}\Big)d\tau,\quad t> t_0. \end{align}\] Furthermore, \[\begin{align} \label{Vttgeqt0} |V(t)|\geq \frac{1}{3}|V(0)+Et|\geq 1+\frac{1}{4}|E|t,\quad t> t_0. \end{align}\tag{63}\] This is the lower bound for \(V(t)\). We can also get the upper bound of \(|V(t)|\) as follows: \[\begin{align} \label{upperboundV} |V(t)-V(0)-Et|\leq 80T(0)^{-\frac{3}{4}}+80\int_{t_0}^t\frac{8}{(4+|E|\tau)^{\frac{3}{2}}}d\tau\leq 80T(0)^{-\frac{3}{4}}+2000/|E|,\quad t\geq 0. \end{align}\tag{64}\]

We now aim to prove that \(T'(t)\geq 0\) for any \(t> t_0\). Leaving \(\tilde{v}=V(t)+T(t)^{\frac{1}{2}}v\), we recompute \[\begin{align} R(t)\cdot V(t)&=&\int_{\mathbb{T}^3\times \mathbb{R}^3}\frac{V(t)+T(t)^{\frac{1}{2}}v}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\cdot \int_{\mathbb{T}^3\times\mathbb{R}^3}(V(t)+T(t)^{\frac{1}{2}}v)(\mu+g)dvdx\\ &=&\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}\frac{V(t)+T(t)^{\frac{1}{2}}v}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\cdot \int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(V(t)+T(t)^{\frac{1}{2}}v)(\mu+g)dvdx\\ &&+\int_{|v|> \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}\frac{V(t)+T(t)^{\frac{1}{2}}v}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\cdot \int_{\mathbb{T}^3\times\mathbb{R}^3}(V(t)+T(t)^{\frac{1}{2}}v)(\mu+g)dvdx\\ &&+\int_{|v|\leq \frac{1}{2} T(t)^{-\frac{1}{2}}|V(t)|}\frac{V(t)+T(t)^{\frac{1}{2}}v}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\cdot \int_{|v|> \frac{1}{2} T(t)^{-\frac{1}{2}}|V(t)|}(V(t)+T(t)^{\frac{1}{2}}v)(\mu+g)dvdx\\ &=&\sum_{i=1}^3D_i. \end{align}\] Since \(\frac{1}{2}|V(t)|\leq|\tilde{v}|\leq 2|V(t)|\) when \(|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|\), we have \[\begin{align} D_1&\geq& |V(t)|^2\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}\frac{1}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\Big)\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(\mu+g)dvdx\Big)\\ &&-|V(t)|T(t)^{\frac{1}{2}}\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}\frac{|v|}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\Big)\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}|v|(\mu+g)dvdx\Big)\\ &&-T(t)\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}\frac{|v|}{\langle\tilde{v}\rangle|\tilde{v}|^2}(\mu+g)dvdx\Big)\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}|v|(\mu+g)dvdx\Big)\\ &\gtrsim& |V(t)|^{-1}\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(\mu+g)dvdx\Big)^2\\ &&-(|V(t)|^{-2}T(t)^{\frac{1}{2}}+|V(t)|^{-3}T(t))\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(\mu+g)\langle v\rangle dvdx\Big)^2. \end{align}\] Recall from 34 that \[\begin{align} \label{T5} |T'(t)|\leq 40T(t)^{-3/8}(1 + T(t)^{1/8}), \end{align}\tag{65}\] whether \(T(t)\ge 1\) or \(T(t)\le 1\), we have \(T(t)\leq 1+T(0)+80t\). Thus for some large constant \(C_{T(0),V(0)}\), if \(|E|>C_{T(0),V(0)}\) and \(t>t_0\), we can obtain from 63 that \[\begin{align} T(t)^{-\frac{1}{2}}|V(t)|\geq \frac{1}{3}(1+T(0)+80t)^{-\frac{1}{2}}|V(0)+Et|>100, \end{align}\] which implies \[\begin{align} \int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(\mu+g)dvdx>\int_{|v|\leq 100}\mu dvdx-\int_{\mathbb{T}^3\times\mathbb{R}^3}|g|dvdx>\sqrt{3c_0} \end{align}\] since \(\|g(t)\|_{L^\infty([0,\infty),L^2_xL^2_3)}<\eta\). We can also get that \[\begin{align} &&(|V(t)|^{-2}T(t)^{\frac{1}{2}}+|V(t)|^{-3}T(t))\Big(\int_{|v|\leq \frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(\mu+g)\langle v\rangle dvdx\Big)^2\\ &\leq&|V(t)|^{-1}\big((1+T(0)+80t)^{\frac{1}{2}}|V(0)+Et|^{-1}+(1+T(0)+80t)|V(0)+Et|^{-2}\big)\|\mu+g\|^2_{L^\infty([0,\infty),L^2_xL^2_3)}\\ &\leq &c_0|V(t)|^{-1},\quad t>t_0. \end{align}\] Thus, we have \(D_1\geq 2c_0|V(t)|^{-1}\) for \(t>t_0\).

For \(D_2\) and \(D_3\), using the fact that \(|v|>\frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|\) and Cauchy-Schwarz inequality, we can obtain \[\begin{align} |D_2|&\leq& \Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle\tilde{v}\rangle^2|\tilde{v}|^2}dxdv\Big)^{\frac{1}{2}}\Big(\int_{|v|>\frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}(\mu+g)^2\langle v\rangle^{2k}\langle v\rangle^{-2k}dxdv\Big)^{\frac{1}{2}}\\ &&\times(|V(t)|+T(t)^{\frac{1}{2}})\int_{\mathbb{T}^3\times\mathbb{R}^3}|\mu+g|\langle v\rangle dxdv\\ &\lesssim& |V(t)|^{-k}T(t)^{\frac{k}{2}-\frac{3}{4}} (|V(t)|+T(t)^{\frac{1}{2}}) \end{align}\] and \[\begin{align} |D_3|&\leq& |V(t)|^{-2}\int_{\mathbb{T}^3\times\mathbb{R}^3}|\mu+g|dxdv \times (|V(t)|+T(t)^{\frac{1}{2}})\int_{|v|>\frac{1}{2}T(t)^{-\frac{1}{2}}|V(t)|}|\mu+g|\langle v\rangle^{1+k}\langle v\rangle^{-k}dxdv\\ &\lesssim& |V(t)|^{-2-k}(|V(t)|+T(t)^{\frac{1}{2}})T(t)^{\frac{k}{2}}. \end{align}\] Choosing \(k=3\), for \(|E|>C_{T(0),V(0)}\), we have that \[\begin{align} |D_2|+|D_3|\lesssim|V(t)|^{-2}\leq c_0|V(t)|^{-1},\quad t>t_0. \end{align}\] Therefore, we conclude that \[\begin{align} \label{T39tgeq0} T'(t)=\frac{4}{3}R(t)\cdot V(t)>D_1-|D_2|-|D_3|\geq {c_0}|V(t)|^{-1}>0\quad for \quad t> t_0, \end{align}\tag{66}\] which imoleis that for all \(t\geq0\), \(T(t)\) has the lower bound \(T(0)/2\).

Furthermore, thanks to 64 , we have \[\begin{align} |V(t)|\leq |V(0)|+|E|t+80T(0)^{-\frac{3}{4}}+2000/|E|\leq 1+\frac{3}{2}|E|t,\quad t>t_0, \end{align}\] if \(|E|>C_{T(0),V(0)}\). Then \[\begin{align} \label{lowerboundofT} T(t)\geq T(t_0)+c_0\int_{t_0}^t\frac{1}{|V(t)|}d\tau\geq \frac{T(0)}{2}+\frac{2c_0}{3|E|}(\ln(1+\frac{3}{2}|E|t)-\ln(1+\frac{3}{2}|E|t_0)),\quad t>t_0. \end{align}\tag{67}\] As for the upper bound of \(T(t)\), choosing \(k=3\) in 33 , we can derive that \[\begin{align} \label{upperboundofR} |R(t)|\leq 40(\langle V(t)\rangle^{-1}|V(t)|^{-1}+|V(t)|^{-3}T(t)^{\frac{3}{4}})\lesssim|V(t)|^{-2}. \end{align}\tag{68}\] On one hand, since \(|V(t)|^{-1} \lesssim(1+\frac{1}{4}|E|t)^{-1}\) for \(t>t_0\), we have \[\begin{align} \label{Rgeqt0} |R(t)|\leq c_1(1+\frac{1}{4}|E|t)^{-2},\quad t>t_0. \end{align}\tag{69}\] On the other hand, together with 63 , it leads to \[\begin{align} \label{T39geqt0} |T'(t)|\leq \frac{4}{3} |R(t)||V(t)|\lesssim|V(t)|^{-1}, \end{align}\tag{70}\] thus we have \[\begin{align} T(t)-T(t_0)\lesssim\int_{t_0}^t(1+\frac{1}{4}|E|\tau)^{-1} d\tau\lesssim\frac{1}{|E|}\ln(1+|E|t), \end{align}\] which yields that \[\begin{align} \label{upperboundofT} T(t)\leq \frac{3}{2} T(0)+\frac{c_2}{|E|}\ln(1+|E|t),\quad for some \quad c_2>0. \end{align}\tag{71}\] Therefore, we can get ?? by combining 67 , 71 , 69 and 64 .

Finally, we give the proof of ?? . We recall from 45 and 42 that \[\begin{align} &&S(t)\leq T^{-1}(t)|T'(t)|+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-1}\langle V(t)\rangle^{-1}+T(t)^{-1}\langle T(t)^{-\frac{1}{2}}|V(t)|\rangle^{-1}\\ &&+\min\big\{T(t)^{-1}+T(t)^{-\frac{5}{4}},\\ \notag&&T(t)^{-1}\langle V(t)\rangle^{-1}+T(t)^{-1}\langle T(t)^{-\frac{1}{2}}|V(t)|\rangle^{-1}+T(t)^{-\frac{1}{2}}|V(t)|^{-1}+|V(t)|^{-1}\big\}. \end{align}\] For the first on the right-hand side, by choosing \(k=\frac{5}{4}\) in 33 , we have \[\begin{align} |T'(t)|\lesssim|R(t)||V(t)|\lesssim\langle V(t)\rangle^{-1}+|V(t)|^{-\frac{1}{4}}T(t)^{-\frac{1}{8}},\quad t\in(0,t_0]. \end{align}\] Noticing \(T(t)>T(0)/2\) for any \(t\in[0,t_0]\), by 61 , we have \[\begin{align} &&\int_0^{t_0} \big(T^{-1}(t)|T'(t)|\big)^2dt\lesssim C_{T(0)} \int_{0}^{t_0}\Big(\frac{1}{\langle V(t)\rangle^2}+\frac{1}{|V(t)|^{\frac{1}{2}}}\Big)dt\\ &\lesssim& C_{T(0)}\Big(\int_0^{t_0} \frac{1}{1+(|V(0)+Et|-80T(0)^{-\frac{3}{4}})^2}dt+\int_0^{t_0}\frac{1}{\big||V(0)+Et|-80T(0)^{-\frac{3}{4}}\big|^{\frac{1}{2}}}dt\Big)\\ &:=&A_1+A_2. \end{align}\] We only give the detailed computation for \(A_2\). By an orthogonal transformation, we may assume without loss of generality that \(E=(|E|,0,0)\) and \(V(0)=(v_1,v_2,v_3)\) and denote \(80T(0)^{-\frac{3}{4}}=T_0\), then \[\begin{align} &&\int_0^{t_0}\frac{1}{\big||V(0)+Et|-80T(0)^{-\frac{3}{4}}\big|^{\frac{1}{2}}}dt=\int_0^{t_0}\frac{1}{\big||(|E|t+v_1,v_2,v_3)|-T_0\big|^{\frac{1}{2}}}dt\\ &=&\frac{1}{|E|}\int_{v_1}^{|E|t_0+v_1}\frac{1}{\big||(t^2+v^2_2+v^2_3)^{\frac{1}{2}}-T_0\big|^{\frac{1}{2}}}dt \leq \frac{2}{|E|}\int_0^{|E|t_0+|v_1|}\frac{1}{\big||(t^2+v^2_2+v^2_3)^{\frac{1}{2}}-T_0\big|^{\frac{1}{2}}}dt. \end{align}\] Consider separately the cases \((t^2+v_2^2+v_3^2)^{1/2}-T_0\ge 0\) and \((t^2+v_2^2+v_3^2)^{1/2}-T_0<0\), we can deduce that \[\begin{align} \frac{2}{|E|}\int_0^{|E|t_0+|v_1|}\frac{1}{\big||(t^2+v^2_2+v^2_3)^{\frac{1}{2}}-T_0\big|^{\frac{1}{2}}}dt\leq \frac{4}{|E|}\int_0^{((|E|t_0+|v_1|)^2+v_2^2+v_3^2)^{\frac{1}{2}}+T_0}\frac{1}{t^{\frac{1}{2}}}dt\lesssim C_{T(0),V(0)}|E|^{-\frac{1}{2}}. \end{align}\] Similar argument can be applied to the estimate of \(A_1\) and we can get that \[\begin{align} \int_0^{t_0} \big(T^{-1}(t)|T'(t)|\big)^2dt\lesssim A_1+A_2\lesssim C_{T(0),V(0)}(|E|^{-1}+|E|^{-\frac{1}{2}}). \end{align}\]

For the second term, using that \(a\leq b_1\) and \(a\leq b_2\)(for \(a\geq0\)) implies \(a\leq b_1^{\frac{3}{4}} b_2^{\frac{1}{4}}\), we get from 32 and 33 that(choosing \(k=1\)) \[\begin{align} |R(t)|\lesssim T(t)^{-\frac{9}{16}}(\langle V(t)\rangle^{-1}+T(t)^{-\frac{1}{4}})^{-\frac{1}{4}}|V(t)|^{-\frac{1}{4}}\lesssim C_{T(0)}|V(t)|^{-\frac{1}{4}},\quad t\in[0,t_0], \end{align}\] which implies \(T(t)^{-\frac{1}{2}}|R(t)|\lesssim C_{T(0)}|V(t)|^{-\frac{1}{4}}\) and we can obtain the same upper bound as above after integration. For the rest term, it is easy to get that \[\begin{align} T(t)^{-1}\langle V(t)\rangle^{-1}+T(t)^{-1}\langle T(t)^{-\frac{1}{2}}|V(t)|\rangle^{-1}\lesssim C_{T(0)}\langle V(t)\rangle^{-\frac{1}{4}} \end{align}\] and \[\begin{align} \min\big\{T(t)^{-1}+T(t)^{-\frac{5}{4}},T(t)^{-1}\langle V(t)\rangle^{-1}+T(t)^{-1}\langle T(t)^{-\frac{1}{2}}|V(t)|\rangle^{-1}+T(t)^{-\frac{1}{2}}|V(t)|^{-1}+|V(t)|^{-1}\big\}\lesssim C_{T(0)}|V(t)|^{-\frac{1}{4}} \end{align}\] by using that \(a\leq b_1\) and \(a\leq b_2\) implies \(a\leq b_1^{\frac{3}{4}} b_2^{\frac{1}{4}}\). Therefore, we can conclude that \[\begin{align} \int_0^{t_0} S^2(t)dt\lesssim C_{T(0),V(0)}(|E|^{-1}+|E|^{-\frac{1}{2}}), \end{align}\] thus we can get the desired result ?? for large \(E\). It ends the proof of this lemma.

To extend the solution up to time \(t_0\), we need to verify two facts: (i) the smallness of the solution is preserved, and (ii) starting from any time as the new initial instant, the solution can be extended over a fixed interval. Recalling the dependence of \({\mathcal{T}}\) in Lemma 11 and Theorem 3, i.e., 48 and 53 , we know that it depends on the lower bound of \(T(t)\). Since \(T(t)\geq T(0)/2\) when \(t\in[0,t_0]\), it suffices to ensure the smallness of \(g\). Using the energy method exactly as in the proof of ?? , we obtain \[\begin{align} \frac{1}{2}\frac{d}{dt}\|g\|^2_{X_k}+T(t)^{-\frac{3}{2}}(\lambda_0-C_k\|g\|_{X_k})\|g\|^2_{Y_k} \leq C_{k}T(t)^{-\frac{3}{2}}\|g\|^2_{X_k}+C_kS^2(t). \end{align}\] Multiply \(e^{-C_k\int_0^tT(\tau)^{-\frac{3}{2}}d\tau}\) on both sides and integrate with respect to \(t\in[0,t_0]\), we have \[\begin{align} \|g\|^2_{L^\infty([0,t_0],X_k)}+2(\lambda_0-C_k\|g\|_{L^\infty([0,t_0],X_k)})\|T(t)^{-\frac{3}{4}}g\|^2_{L^2([0,t_0],Y_k)}\leq e^{C_k\int_0^{t_0}T(\tau)^{-\frac{3}{2}}d\tau}\big(\|g_0\|^2_{X_k}+C_k\int_0^{t_0}S^2(t)dt\big). \end{align}\] Under the a priori assumption that \(\|g\|_{L^\infty([0,t_0],X_k)}<\lambda_0/(2C_k)\) and the fact \(T(t)\geq T(0)/2\), we can derive that \[\begin{align} \|g\|^2_{L^\infty([0,t_0],X_k)}+\lambda_0\|T(t)^{-\frac{3}{4}}g\|^2_{L^2([0,t_0],Y_k)}&\leq& e^{C_kt_0T(0)^{-\frac{3}{2}}}\big(\|g_0\|^2_{X_k}+C_k\int_0^{t_0}S^2(t)dt\big)\\ &\leq&e^{C_k}\big(\|g_0\|^2_{X_k}+C_k\int_0^{t_0}S^2(t)dt\big), \end{align}\] since \(t_0< T(0)^{\frac{3}{2}}\). Therefore, ?? holds ture if \(e^{C_k}\varepsilon_1^2<\frac{1}{4}\) and \(e^{C_k}C_k\int_0^{t_0} S(t)^2dt<\frac{1}{4} \varepsilon_1^2\). The latter is ensured by Lemma 11. It ends the proof of this corollary.

For any \(|\alpha|=0,2\), the equation for \(\partial^\alpha_x g_1\) is \[\begin{align} \label{paxg1} &&\partial_t \partial^\alpha_xg_1(t,x,v)+T(t)^{\frac{1}{2}}v\cdot\nabla_x \partial^\alpha_xg_1+(-\frac{1}{2} T(t)^{-1}T'(t)v+2T(t)^{-\frac{1}{2}}R(t))\cdot\nabla_v \partial^\alpha_xg_1(t,x,v)\\ &=&\notag T(t)^{-\frac{3}{2}}(L-A\chi_M)\partial^\alpha_xg_1+T(t)^{-\frac{3}{2}}\partial^\alpha_xQ(g_1,g_1)+\frac{3}{2} T(t)^{-1} T'(t) \partial^\alpha_xg_1+\partial^\alpha_x{\mathcal{H}}(\mu)+T(t)^{-1}\\ \notag&&\timesdiv\Big(\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v (\partial^\alpha_xg_1+\mu^{\frac{1}{2}}\partial^\alpha_xg_2)\Big)+T(t)^{-\frac{3}{2}}\big(\partial^\alpha_xQ(\mu^\frac{1}{2} g_2,g_1)+\partial^\alpha_xQ(g_1,\mu^{\frac{1}{2}}g_2)\big). \end{align}\tag{76}\] Multiply \(\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1,k\geq 17\) to 76 , integrate in \((x,v)\in\mathbb{T}^3\times\mathbb{R}^3\) and sum over \(|\alpha|=0,2\), the first two terms on the left-hand side gives \[\begin{align} \frac{1}{2} \frac{d}{dt}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}|\partial^\alpha_x g_1|^2\langle v\rangle^{2(k-4|\alpha|)}dvdx=\frac{1}{2}\frac{d}{dt}\|g_1\|^2_{X_k}, \end{align}\] recalling the energy functional \(X_k\) in 19 .

We denote the rest terms including the third term on the left-hand side by \(I_1\) to \(I_7\) and estimate them one by one. For \(I_1\) and \(I_4\), by the estimates in Lemma 10, we have \[\begin{align} I_1&=&\frac{1}{2}T(t)^{-1}T'(t)\sum_{|\alpha|=0,2}(k-4|\alpha|+\frac{3}{2})\|\partial^\alpha_xg_1\|^2_{L^2_xL^2_{k-4|\alpha|}}+\frac{1}{2}T(t)^{-1}T'(t)\sum_{|\alpha|=0,2}(k-4|\alpha|)\|\partial^\alpha_xg_1\|^2_{L^2_xL^2_{k-4|\alpha|-1}}\\ &&-2T(t)^{-\frac{1}{2}}R(t)(k-4|\alpha|)\int_{\mathbb{T}^3\times\mathbb{R}^3}\langle v\rangle^{2(k-4|\alpha|)-2}v(\partial^\alpha_xg_1)^2dvdx. \end{align}\] A direct computation of \(I_4\) gives \[\begin{align} I_4=\frac{3}{2} T(t)^{-1}T'(t)\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}(\partial^\alpha_xg_1)^2\langle v\rangle^{2(k-4|\alpha|)}dvdx=\frac{3}{2} T(t)^{-1}T'(t)\sum_{|\alpha|=0,2}\|\partial^\alpha_xg_1\|^2_{L^2_xL^2_{k-4|\alpha|}}. \end{align}\] Thanks to 72 and the fact that \(k\geq 17\), it holds that \[\begin{align} \label{I14} I_1-I_4\geq 2T(t)^{-1}T'(t)\|g_1\|^2_{X_k}-C_k\big(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|\big)\|g_1\|^2_{X_{k-\frac{1}{2}}}. \end{align}\tag{77}\]

For \(I_2\) and \(I_3\), we have that \[\begin{align} I_2=T(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\Big(\int_{\mathbb{T}^3\times\mathbb{R}^3}L(\partial^\alpha_xg_1)\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1dvdx-\int_{\mathbb{T}^3\times\mathbb{R}^3}A\chi_M\partial^\alpha_x g_1\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1dvdx\Big). \end{align}\] and \[\begin{align} I_3&=&T(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3}\big(Q(g_1,\partial^\alpha_xg_1),\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1\big)_{L^2_v}dx\\ &&+T(t)^{-\frac{3}{2}}\sum_{|\alpha_1|\geq1}\int_{\mathbb{T}^3}\big(Q(\partial^{\alpha_1}_xg_1,\partial^{\alpha_2}_xg_1),\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1\big)_{L^2_v}dx. \end{align}\] Thanks to Lemma 5 and ?? in Lemma 7, \[\begin{align} \label{I23} I_2+I_3\leq T(t)^{-\frac{3}{2}}\Big(-\frac{\lambda_0}{2} \|g_1\|^2_{Y_k}+C_k\|g_1\|_{X_{17}}\|g_1\|^2_{Y_k}\Big). \end{align}\tag{78}\]

For \(I_5\), since \(\partial^\alpha_x {\mathcal{H}}(\mu)=0\) for \(|\alpha|=2\), thus we have that \[\begin{align} I_5&=&\int_{\mathbb{T}^3\times\mathbb{R}^3}{\mathcal{H}}(\mu)g_1\langle v\rangle^{2k}dvdx=\int_{\mathbb{T}^3\times\mathbb{R}^3}\Big(\big(\frac{1}{2} T(t)^{-1}T'(t)v-2T(t)^{-\frac{1}{2}}R(t)\big)\cdot\nabla_v \mu+\frac{3}{2} T(t)^{-1} T'(t) \mu\Big)g_1\langle v\rangle^{2k}dvdx\\ &&+\int_{\mathbb{T}^3\times\mathbb{R}^3}T(t)^{-1}div\Big(\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \mu\Big)g_1\langle v\rangle^{2k}dvdx:=I_{5,1}+I_{5,2}. \end{align}\] Using the exponential decay of \(\mu\), it is direct to show that \[\begin{align} \label{I51} I_{5,1}\leq C_k\big(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|\big)\|g_1\|_{L^2_{x,v}}. \end{align}\tag{79}\] For \(I_{5,2}\), since \[\begin{align} \left|T(t)^{-1}div\Big(\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v \mu\Big)\right|\lesssim\Big(\frac{T(t)^{-\frac{1}{2}}}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle|V(t)+T(t)^{\frac{1}{2}}v|}+\frac{T(t)^{-1}}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\Big)\mu^{\frac{1}{2}}. \end{align}\] It yields that \[\begin{align} |I_{5,2}|\lesssim\int_{\mathbb{T}^3\times\mathbb{R}^3}\Big(\frac{T(t)^{-\frac{1}{2}}}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle|V(t)+T(t)^{\frac{1}{2}}v|}+\frac{T(t)^{-1}}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\Big)\mu^{\frac{1}{2}} g_1\langle v\rangle^{2k}dvdx:=I_{5,2,1}+I_{5,2,2}. \end{align}\] For \(I_{5,2,1}\), we split the integration domain \(\mathbb{R}^3\) into two parts: \(|T(t)^{\frac{1}{2}}v|\leq \frac{1}{2}|V(t)|\) and \(|T(t)^{\frac{1}{2}}v|> \frac{1}{2}|V(t)|\). For these two cases we have, respectively, \[\begin{align} \langle V(t)+T(t)^{\frac{1}{2}}v\rangle^{-1}|V(t)+T(t)^{\frac{1}{2}}v|^{-1}\lesssim|V(t)|^{-2},\quad and\quad \mu^{\frac{1}{4}}\lesssim|v|^{-2}\lesssim T(t)|V(t)|^{-2}. \end{align}\] Then it holds that \[\begin{align} I_{5,2,1}&\lesssim& T(t)^{-\frac{1}{2}}|V(t)|^{-2}\int_{\mathbb{T}^3\times\mathbb{R}^3} \mu^{\frac{1}{2}}|g_1|\langle v\rangle^{2k}dvdx+T(t)^{\frac{1}{2}}|V(t)|^{-2}\int_{\mathbb{T}^3\times\mathbb{R}^3} \frac{1}{|V(t)+T(t)^{\frac{1}{2}}v|}\mu^{\frac{1}{4}}|g_1|\langle v\rangle^{2k}dvdx\\ &\lesssim& C_k\big(T(t)^{-\frac{1}{2}}|V(t)|^{-2}+T(t)^{\frac{1}{2}}|V(t)|^{-2}+T(t)^{-\frac{1}{4}}|V(t)|^{-2}\big)\|g_1\|_{L^2_{x,v}}. \end{align}\] Similarly, we can derive that \[\begin{align} I_{5,2,2}\lesssim C_k\big(T(t)^{-1}|V(t)|^{-1}+|V(t)|^{-2}\big)\|g_1\|_{L^2_{x,v}}. \end{align}\] Then we obtain that \[\label{I52} |I_{5,2}|\leq C_k\big(T(t)^{-\frac{1}{2}}|V(t)|^{-2}+T(t)^{\frac{1}{2}}|V(t)|^{-2}+T(t)^{-\frac{1}{4}}|V(t)|^{-2}+T(t)^{-1}|V(t)|^{-1}+|V(t)|^{-2}\big)\|g_1\|_{L^2_{x,v}}.\tag{80}\] Combining 79 , 80 and using 73 , we conclude that \[\begin{align} \label{I955} |I_5| \leq C_kT(t)^{-1}|V(t)|^{-1}\|g_1\|_{L^2_{x,v}}. \end{align}\tag{81}\]

For \(I_6\), we write it as \[\begin{align} I_6&=&-T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\Big(\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\cdot \nabla_v (\partial^\alpha_xg_1+\mu^{\frac{1}{2}}\partial^\alpha_xg_2)\Big)\nabla_v(\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_xg_1)dvdx\\ &=&-T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\nabla_v\partial^\alpha_xg_1\big(\nabla_v\partial^\alpha_xg_1\langle v\rangle^{2(k-4|\alpha|)}+2(k-4|\alpha|)\langle v\rangle^{2(k-4|\alpha|-1)}v\partial^\alpha_xg_1\big)dvdx\\ &&-T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\nabla_v(\mu^{\frac{1}{2}}\partial^\alpha_xg_2)\nabla_v(\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_xg_1)dvdx:=I_{6,1}+I_{6,2}. \end{align}\] For \(I_{6,1}\), since \(\Pi(V(t)+T(t)^{\frac{1}{2}}v)\) is positive, by Cauchy-Schwarz inequality, we have that \[\begin{align} \label{I61} I_{6,1}&\leq&-\frac{3}{4}T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\nabla_v\partial^\alpha_xg_1\cdot \nabla_v\partial^\alpha_xg_1\langle v\rangle^{2(k-4|\alpha|)}dvdx\\ \notag &&+C_kT(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{1}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}(\partial^\alpha_xg_1)^2\langle v\rangle^{2(k-4|\alpha|-1)}dvdx\\ \notag&\leq&-\frac{3}{4}T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\nabla_v\partial^\alpha_xg_1\cdot \nabla_v\partial^\alpha_xg_1\langle v\rangle^{2(k-4|\alpha|)}dvdx\\ \notag&&+C_k\big(T(t)^{-1}|V(t)|^{-1}\|g_1\|^2_{X_{k-1}}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\|g_1\|^2_{X_{k-\frac{1}{4}}}\big). \end{align}\tag{82}\] In the last step we used the same splitting of the integration domain that was employed to estimate \(I_{5,2,1}\).

For \(I_{6,2}\), due to the exponential decay of \(\mu\) and Cauchy-Schwarz inequality, we can derive that \[\begin{align} \label{I62} \notag|I_{6,2}|&\leq& \frac{1}{4}T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\nabla_v\partial^\alpha_xg_1\cdot \nabla_v\partial^\alpha_xg_1\langle v\rangle^{2(k-4|\alpha|)}dvdx\\ &&+C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\big(\|g_1\|^2_{X_{k-1}}+\|g_2\|^2_{{\mathcal{D}}_0}\big). \end{align}\tag{83}\] Combining 82 and 83 , we conclude that \[\begin{align} \label{I6} \notag I_6&\leq& -\frac{1}{2}T(t)^{-1}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}\frac{\Pi(V(t)+T(t)^{\frac{1}{2}}v)}{\langle V(t)+T(t)^{\frac{1}{2}}v\rangle}\nabla_v\partial^\alpha_xg_1\cdot \nabla_v\partial^\alpha_xg_1\langle v\rangle^{2(k-4|\alpha|)}dvdx\\ &&+C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\big(\|g_1\|^2_{X_{k-\frac{1}{4}}}+\|g_2\|^2_{{\mathcal{D}}_0}\big). \end{align}\tag{84}\]

For \(I_7\), we first write it as \[\begin{align} I_7&=&T(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\sum_{\alpha_1+\alpha_2=\alpha}\Big(\int_{\mathbb{T}^3}\big(Q(\mu^{\frac{1}{2}}\partial^{\alpha_1}_xg_2,\partial^{\alpha_2}_xg_1),\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1\big)_{L^2_v}dx\\ &&+\int_{\mathbb{T}^3}\big(Q(\partial^{\alpha_1}_xg_1,\mu^{\frac{1}{2}}\partial^{\alpha_2}_xg_2),\langle v\rangle^{2(k-4|\alpha|)}\partial^\alpha_x g_1\big)_{L^2_v}dx\Big). \end{align}\] Due to ?? , ?? , we have that \[\begin{align} \label{I7} \notag|I_7|&\leq&C_kT(t)^{-\frac{3}{2}}\big(\|g_2\|_{{\mathcal{E}}_0}\|g_1\|^2_{Y_k}+\|g_1\|_{X_{17}}\|g_2\|_{{\mathcal{D}}_0}\|g_1\|_{Y_k}\big)\\ &\leq&C_kT(t)^{-\frac{3}{2}}\big(\|g_2\|_{{\mathcal{E}}_0}\|g_1\|^2_{Y_k}+\|g_1\|_{X_{17}}\|g_1\|^2_{Y_k}+\|g_1\|_{X_{17}}\|g_2\|^2_{{\mathcal{D}}_0}\big). \end{align}\tag{85}\]

Patching together the estimates of \(I_1\) to \(I_7\), i.e., 77 , 78 , 81 ,84 and 85 , we can conclude the desired result ?? .

For any \(|\alpha|=0,2\), the equation for \(\partial^\alpha_xg_2\) is \[\begin{align} \label{paxg2} &&\partial_t \partial^\alpha_xg_2(t,x,v)+T(t)^{\frac{1}{2}}v\cdot\nabla_x \partial^\alpha_xg_2=T(t)^{-\frac{3}{2}}\mathcal{L}\partial^\alpha_xg_2+T(t)^{-\frac{3}{2}}\mu^{-\frac{1}{2}}A\chi_M(v)\partial^\alpha_xg_1\\ &&\notag+T(t)^{-\frac{3}{2}}\partial^\alpha_x\Gamma(g_2,g_2)-(-\frac{1}{2} T(t)^{-1}T'(t)v+2T(t)^{-\frac{1}{2}}R(t))\cdot(\nabla_v\partial^\alpha_xg_2-\frac{v}{2} \partial^\alpha_xg_2)+\frac{3}{2}T(t)^{-1}T'(t)\partial^\alpha_xg_2. \end{align}\tag{86}\] We first prove the estimates ?? . Multiply \(\langle v\rangle^{2k}\partial^\alpha_x g_2,k\geq17\) to 86 , integrate in \((x,v)\in\mathbb{T}^3\times\mathbb{R}^3\) and sum over \(|\alpha|=0,2\). Then the first two terms on the left-hand side give \[\begin{align} \frac{1}{2} \frac{d}{dt}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3}|\partial^\alpha_x g_2|^2\langle v\rangle^{2k}dvdx=\frac{1}{2}\frac{d}{dt}\|g_2\|^2_{{\mathcal{E}}_k}. \end{align}\]

We denote the rest terms by \(J_{1,k}\) to \(J_{5,k}\) and estimate them one by one. For \(J_{1,k}\), by ?? in Lemma 4, we have that \[\begin{align} \label{J1} J_{1,k}=T(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3}\big({\mathcal{L}}\partial^\alpha_xg_2,\partial^\alpha_xg_2\langle v\rangle^{2k}\big)_{L^2_v}dx\leq T(t)^{-\frac{3}{2}}\Big( -\lambda_0\|g_2\|^2_{{\mathcal{D}}_k}+C_k\|g_2\|^2_{{\mathcal{D}}_0}\Big). \end{align}\tag{87}\]

For \(J_{2,k}\), we write it as \[\begin{align} J_{2,k}=T(t)^{-\frac{3}{2}}\sum_{|\alpha|=0,2}\int_{\mathbb{T}^3\times\mathbb{R}^3} \mu^{-\frac{1}{2}}A\chi_M(v)\partial^\alpha_xg_1\langle v\rangle^{2k}\partial^\alpha_xg_2dvdx. \end{align}\] Noticing that \(\chi_M\) has compact support, thus it has the upper bound \[\begin{align} \label{J2} J_{2,k}\leq \varepsilon T(t)^{-\frac{3}{2}}\|g_2\|^2_{{\mathcal{D}}_0}+ C_{\varepsilon,k}T(t)^{-\frac{3}{2}}\|g_1\|^2_{Y_k} \end{align}\tag{88}\] for any \(\varepsilon>0\). In the below, we choose \(\varepsilon=\lambda_0/2\).

For \(J_{3,k}\), we write it as \[\begin{align} J_{3,k}=T(t)^{-\frac{3}{2}}\sum_{\alpha_1+\alpha_2=\alpha}\int_{\mathbb{T}^3}\big(\Gamma(\partial^{\alpha_1}_xg_2,\partial^{\alpha_1}_xg_2),\langle v\rangle^{2k}\partial^\alpha_xg_2\big)_{L^2_v}dx. \end{align}\] Due to ?? in Lemma 7, we have that \[\begin{align} \label{J3} J_{3,k}\leq C_kT(t)^{-\frac{3}{2}} \|g_2\|_{{\mathcal{E}}_{17}}\|g_2\|^2_{{\mathcal{D}}_k}. \end{align}\tag{89}\]

For \(J_{4,k}\) and \(J_{5,k}\), similar to the estimate of \(I_1\) and \(I_4\) in Proposition 2, we can derive that \[\begin{align} \label{J45} J_{4,k}+J_{5,k}&\leq& -\frac{1}{4}T(t)^{-1}T'(t)\|g_2\|^2_{{\mathcal{E}}_{k+1}}+C_k\big(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|\big)\|g_2\|^2_{{\mathcal{E}}_{k+\frac{1}{2}}}. \end{align}\tag{90}\] Combining the estimates 87 , 88 , 89 and 90 , we obtain the desired result ?? .

The similar argument can be applied to obtain ?? , the only difference is the estimate for \(J_{1,0}\), where we appeal to ?? instead of ?? . We omit the details and list the resulting bounds for each term as follows. \[\begin{align} &&J_{1,0}\leq -T(t)^{-\frac{3}{2}}\lambda_0\|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0},~ J_{2,0}\lesssim T(t)^{-\frac{3}{2}}\|g_2\|_{{\mathcal{D}}_0}\|g_1\|_{Y_0},\\ &&J_{3,0}\leq C_0T(t)^{-\frac{3}{2}} \|g_2\|_{{\mathcal{E}}_0}\|g_2\|^2_{{\mathcal{D}}_0},~ J_{4,0}+J_{5,0}\leq -\frac{1}{4}T(t)^{-1}T'(t)\|g_2\|^2_{{\mathcal{E}}_{1}}+C\big(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|\big)\|g_2\|^2_{{\mathcal{E}}_{\frac{1}{2}}}. \end{align}\] Decomposing \(g_2=\mathbf{P} g_2+(\mathbf{I}-\mathbf{P})g_2\) and noticing that \(\|\mathbf{P}g_2\|_{{\mathcal{D}}_0}\lesssim\|\mathbf{P}g_2\|_{H^2_{x}L^2_v}\), we arrive at ?? and the proof of the lemma is complete.

The proof proceeds in a standard way. Splitting \[\begin{align} g_2=\mathbf{P}g_2+(\mathbf{I-P})g_2=[a^{g_2}+b^{g_2}\cdot v+c^{g_2}(|v|^2-3)]\mu^{\frac{1}{2}}+(\mathbf{I-P})g_2. \end{align}\] We rewrite ?? as \[\begin{align} \label{macro1} \notag&&(\partial_ta^{g_2}+\partial_t b^{g_2}\cdot v+\partial_t c^{g_2}(|v|^2-3))\mu^{\frac{1}{2}}+T(t)^{\frac{1}{2}}v\cdot\nabla_x(a^{g_2}+b^{g_2}\cdot v+c^{g_2}(|v|^2-3))\mu^{\frac{1}{2}}\\ &=&-(\partial_t+T(t)^{\frac{1}{2}}v\cdot\nabla_x)(\mathbf{I-P})g_2+T(t)^{-\frac{3}{2}}{\mathcal{L}}(\mathbf{I-P})g_2+r(g_1,g_2). \end{align}\tag{91}\] It is noteworthy that \[\begin{align} \label{macro4} \int_{\mathbb{R}^3}(1,v_i^2,v_i^2v_j^2,v_i^4,|v|^2,|v|^4)\mu dv=(1,1,1,3,3,15),\quad i,j=1,2,3,i\neq j, \end{align}\tag{92}\] and \[\begin{align} \label{macro3} &&\big((\mathbf{I-P})g_2,\psi(v)\big)_{L^2_v}=0,~\big({\mathcal{L}}(\mathbf{I-P})g_2,\psi(v)\big)_{L^2_v}=0,\\ \notag&&\big(\partial_t(\mathbf{I-P})g_2,\psi(v)\big)_{L^2_v}=0,~\big(v\cdot\nabla_x(\mathbf{I-P})g_2,\mu^{\frac{1}{2}}\big)_{L^2_v} = 0,\quadfor\quad\psi=\mu^{\frac{1}{2}}(1,v,|v|^2). \end{align}\tag{93}\] Let us prove the first equation in ?? as a typical case. Taking inner product of 91 with \(\mu^{\frac{1}{2}}\) over \(v\in\mathbb{R}^3\) and by 92 , we have \[\begin{align} \notag\partial_t a^{g_2}+T(t)^{\frac{1}{2}}\nabla_x\cdot b^{g_2}=&\big(-(\partial_t+T(t)^{\frac{1}{2}}v\cdot\nabla_x)(\mathbf{I-P})g_2+T(t)^{-\frac{3}{2}}{\mathcal{L}}(\mathbf{I-P})g_2+r(g_1,g_2),\mu^{\frac{1}{2}}\big)\\ =&(r(g_1,g_2),\mu^{\frac{1}{2}}):=\mathsf{T}_{11}, \end{align}\] where the last step follows from 93 .

Analogously, taking inner product of 91 with \(\mu^{\frac{1}{2}}v_iv_j,i\neq j\) and \(\mu^{\frac{1}{2}}(v^2_i-1)/2,i=1,2,3\) over \(v\in\mathbb{R}^3\), then with \(\mu^{\frac{1}{2}}v,\mu^{\frac{1}{2}}\frac{|v|^2-3}{6}\) and finally with \(\mu^{\frac{1}{2}}\frac{v(|v|^2-5)}{10}\). By subsequently applying 92 and 93 , we can obtain the remaining equations outlined in ?? .

We only give the proof of \(\mathsf{T}_{51}\) with \(|\alpha|\leq 1\) since the similar argument can be applied for the rest terms. Noticing that \[\begin{align} \partial^\alpha_x\mathsf{T}_{51}=\big(-T(t)^{\frac{1}{2}}v\cdot\nabla_x\partial^\alpha_x(\mathbf{I-P})g_2+T(t)^{-\frac{3}{2}}{\mathcal{L}}(\mathbf{I-P})\partial^\alpha_xg_2+\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}\frac{v(|v|^2-5)}{10}\big)_{L^2_v}, \end{align}\] and we have \[\begin{align} &&\big|\big(v\cdot\nabla_x\partial^\alpha_x(\mathbf{I-P})g_2,\mu^{\frac{1}{2}}\frac{v(|v|^2-5)}{10}\big)_{L^2_v}\big|\lesssim\|\langle v\rangle^{-2}\nabla_x\partial^\alpha_x(\mathbf{I-P})g_2\|_{L^2_{v}},\\ &&\big|\big({\mathcal{L}}(\partial^\alpha_x(\mathbf{I-P})g_2),\mu^{\frac{1}{2}}\frac{v(|v|^2-5)}{10}\big)_{L^2_v}\big|\lesssim\|\langle v\rangle^{-2}\partial^\alpha_x(\mathbf{I-P})g_2\|_{L^2_{v}}, \end{align}\] where we used ?? and ?? in the second inequality. Together with the fact \[\begin{align} \int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}\frac{v(|v|^2-5)}{10}\big)^2_{L^2_v}dx\lesssim\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx, \end{align}\] we obtain the desired results.

We will complete the proof in several steps.

Step 1: Control of \(\nabla_x\partial^\alpha_xc^{g_2}\). Applying \(\partial^\alpha_x,|\alpha|\leq 1\) on both sides of the last equation in ?? and taking inner product with \(\nabla_x\partial^\alpha_x c^{g_2}\) over \(x\in\mathbb{T}^3\), we deduce that \[\begin{align} \label{a001} T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}=\frac{d}{dt}\int_{\mathbb{T}^3}\partial^\alpha_x\mathsf{T}_{52}\nabla_x\partial^\alpha_x c^{g_2}dx-\int_{\mathbb{T}^3}\nabla_x\partial^\alpha_x\mathsf{T}_{52}\partial^\alpha_x \partial_t c^{g_2}dx+\int_{\mathbb{T}^3}\partial^\alpha_x\mathsf{T}_{51}\nabla_x\partial^\alpha_xc^{g_2}dx. \end{align}\tag{94}\] For the term \(\partial_t\partial^\alpha_xc^{g_2}\) in 94 , it follows from the third equation in ?? that \[\begin{align} \partial_t \partial^\alpha_xc^{g_2}=-\frac{1}{3}T(t)^{\frac{1}{2}}\nabla_x\cdot\partial^\alpha_xb^{g_2}+\partial^\alpha_x\mathsf{T}_{31}. \end{align}\] By Cauchy-Schwarz inequality, for any small \(\varepsilon>0\), we have \[\begin{align} -\int_{\mathbb{T}^3}\nabla_x\partial^\alpha_x\mathsf{T}_{52}\partial^\alpha_x \partial_t c^{g_2}dx&=&\frac{1}{3}T(t)^{\frac{1}{2}}\int_{\mathbb{T}^3}\nabla_x\partial^\alpha_x\mathsf{T}_{52}\nabla_x\cdot\partial^\alpha_x b^{g_2}dx-\int_{\mathbb{T}^3}\nabla_x\partial^\alpha_x\mathsf{T}_{52}\partial^\alpha_x \mathsf{T}_{31}dx\\ &\leq&\varepsilon T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}+C_\varepsilon(T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x\mathsf{T}_{52}\|^2_{L^2_x}+T(t)^{-\frac{1}{2}}\|\partial^\alpha_x\mathsf{T}_{31}\|^2_{L^2_x}). \end{align}\] For the third term on the right-hand side of 94 , also by Cauchy-Schwarz inequality, we have \[\begin{align} \int_{\mathbb{T}^3}\partial^\alpha_x\mathsf{T}_{51}\nabla_x\partial^\alpha_xc^{g_2}dx\leq \frac{1}{2} T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_xc^{g_2}\|^2_{L^2_x}+2 T(t)^{-\frac{1}{2}}\|\partial^\alpha_x \mathsf{T}_{51}\|^2_{L^2_x}. \end{align}\] Finally, by ?? , we can conclude that \[\begin{align} \label{c} \notag&&\frac{1}{2}T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\leq \frac{d}{dt}\int_{\mathbb{T}^3}\partial^\alpha_x\mathsf{T}_{52}\nabla_x\partial^\alpha_x c^{g_2}dx+\varepsilon T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}\\ &&+ C_{\alpha,\varepsilon}\Big( T(t)^{\frac{1}{2}} \|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0}+T(t)^{-\frac{1}{2}}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx\Big), \end{align}\tag{95}\] where we note that \(\|\langle v\rangle^{-2}(\mathbf{I-P})g_2\|^2_{H^2_xL^2_2}\lesssim\|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0}\).

Step 2: Control of \(\nabla_x\partial^\alpha_xb^{g_2}\). Next, we use the forth and fifth equations in ?? to compute that \[\begin{align} &&T(t)^{\frac{1}{2}}\Delta_x b_i^{g_2}=T(t)^{\frac{1}{2}}\sum_{j\neq i}\partial_{jj}b_i^{g_2}+T(t)^{\frac{1}{2}}\partial_{ii}b_i^{g_2}\\ &=&\sum_{j\neq i}\big(T(t)^{\frac{1}{2}}(-\partial_{ij}b_j^{g_2})+\partial_j(\mathsf{T}_{41})_{ij}+\partial_t\partial_j(\mathsf{T}_{42})_{ij}\big)+\partial_i(\mathsf{T}_{41})_{ii}+\partial_t\partial_i(\mathsf{T}_{42})_{ii}-\partial_t\partial_i c^{g_2}\\ &=&\sum_{j\neq i}\big(\partial_t\partial_ic^{g_2}-\partial_i(\mathsf{T}_{41})_{jj}-\partial_t\partial_i(\mathsf{T}_{42})_{jj}\big)+\sum_{j\neq i}\big(\partial_j(\mathsf{T}_{41})_{ij}+\partial_t\partial_j(\mathsf{T}_{42})_{ij}\big)+\partial_i(\mathsf{T}_{41})_{ii}+\partial_t\partial_i(\mathsf{T}_{42})_{ii}-\partial_t\partial_i c^{g_2}\\ &=&\sum_{j\neq i}\big(\partial_j(\mathsf{T}_{41})_{ij}+\partial_t\partial_j(\mathsf{T}_{42})_{ij}-\partial_i(\mathsf{T}_{41})_{jj}-\partial_t\partial_i(\mathsf{T}_{42})_{jj}\big)-T(t)^{\frac{1}{2}}\partial_{ii} b_i^{g_2}+2\big(\partial_i(\mathsf{T}_{41})_{ii}+\partial_t\partial_i(\mathsf{T}_{42})_{ii}\big). \end{align}\] Applying \(\partial^\alpha_x\) on both sides and taking inner product with \(\partial^\alpha_xb_i^{g_2}\) over \(x\in\mathbb{T}^3\), and summation over \(i=1,2,3\), we deduce that \[\begin{align} &&T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}=\sum_{i=1}^3\Big(-\int_{\mathbb{T}^3}\sum_{j\neq i}\partial^\alpha_x\big(\partial_j(\mathsf{T}_{41})_{ij}+\partial_t\partial_j(\mathsf{T}_{42})_{ij}-\partial_i(\mathsf{T}_{41})_{jj}-\partial_t\partial_i(\mathsf{T}_{42})_{jj}\big)\partial^\alpha_x b_i^{g_2}dx\Big)\\ &&-\sum_{i=1}^3\Big(T(t)^{\frac{1}{2}}\int_{\mathbb{T}^3}|\partial_i\partial^\alpha_x b_i^{g_2}|^2dx-2\int_{\mathbb{T}^3}\partial^\alpha_x\big(\partial_i(\mathsf{T}_{41})_{ii}+\partial_t\partial_i(\mathsf{T}_{42})_{ii}\big)\partial^\alpha_x b_i^{g_2}dx\Big)\\ &\leq& \sum_{j\neq i}\frac{d}{dt}\int_{\mathbb{T}^3}\partial^\alpha_x\big(-\partial_j(\mathsf{T}_{42})_{ij}+\partial_i(\mathsf{T}_{42})_{jj}\big)\partial^\alpha_xb_i^{g_2}dx+2\sum_{i=1}^3\frac{d}{dt}\int_{\mathbb{T}^3}\partial^\alpha_x\partial_i(\mathsf{T}_{42})_{ii}\partial^\alpha_xb_i^{g_2}dx\\ &&+\sum_{i=1}^3\Big(-\int_{\mathbb{T}^3}\sum_{j\neq i}\partial^\alpha_x\big(\partial_j(\mathsf{T}_{41})_{ij}-\partial_i(\mathsf{T}_{41})_{jj}\big)\partial^\alpha_x b_i^{g_2}dx\Big)-\sum_{i=1}^3\Big(2\int_{\mathbb{T}^3}\partial^\alpha_x\partial_i(\mathsf{T}_{41})_{ii}\partial^\alpha_x b_i^{g_2}dx\Big)\\ &&+\sum_{i=1}^3\Big(\int_{\mathbb{T}^3}\sum_{j\neq i}\partial^\alpha_x\big(\partial_j(\mathsf{T}_{42})_{ij}-\partial^\alpha_x\partial_i(\mathsf{T}_{42})_{jj}\big) \partial_t\partial^\alpha_xb_i^{g_2}dx\Big)+\sum_{i=1}^3\Big(2\int_{\mathbb{T}^3}\partial^\alpha_x\partial_i(\mathsf{T}_{42})_{ii} \partial_t\partial^\alpha_xb_i^{g_2}dx\Big). \end{align}\] For the term \(\partial_t\partial^\alpha_x b^{g_2}\), by the second equation in ?? , we have that \[\begin{align} \partial_t \partial^\alpha_xb^{g_2}+T(t)^{\frac{1}{2}}(\nabla_x \partial^\alpha_xa^{g_2}+2\nabla_x\partial^\alpha_xc^{g_2})=\partial^\alpha_x\mathsf{T}_{21}, \end{align}\] Then by ?? and Cauchy-Schwarz inequality, we conclude that \[\label{b} \begin{align} &\frac{1}{2}T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}\leq \sum_{j\neq i}\frac{d}{dt}\int_{\mathbb{T}^3}\partial^\alpha_x\big(-\partial_j(\mathsf{T}_{42})_{ij}+\partial_i(\mathsf{T}_{42})_{jj}\big)\partial^\alpha_xb_i^{g_2}dx+2\sum_{i=1}^3\frac{d}{dt}\int_{\mathbb{T}^3}\partial^\alpha_x\partial_i(\mathsf{T}_{42})_{ii}\partial^\alpha_xb_i^{g_2}dx\\ &+\varepsilon T(t)^{\frac{1}{2}}\big(\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}+\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\big)+C_{\varepsilon,\alpha}\Big( T(t)^{\frac{1}{2}} \|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0}+T(t)^{-\frac{1}{2}}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx\Big). \end{align}\tag{96}\]

Step 3: Control of \(\nabla_x\partial^\alpha_xa^{g_2}\). For \(\nabla_x\partial^\alpha_x a^{g_2}\), from the second and the sixth equations in ?? , we have that \[\begin{align} T(t)^{\frac{1}{2}}\nabla_x a^{g_2}=-\partial_tb^{g_2}+\mathsf{T}_{21}-2\mathsf{T}_{51}-2\partial_t\mathsf{T}_{52}. \end{align}\] Applying \(\partial^\alpha_x\) on both sides and taking inner product with \(\nabla_x\partial^\alpha_x a^{g_2}\) over \(\mathbb{T}^3\), we deduce that \[\begin{align} &T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}=-\int_{\mathbb{T}^3}\partial_t (\partial^\alpha_xb^{g_2}+2\partial^\alpha_x\mathsf{T}_{52})\nabla_x\partial^\alpha_x a^{g_2}dx+\int_{\mathbb{T}^3}(\partial^\alpha_x\mathsf{T}_{21}-2\partial^\alpha_x\mathsf{T}_{51})\nabla_x\partial^\alpha_x a^{g_2}dx\\ =&-\frac{d}{dt}\int_{\mathbb{T}^3}(\partial^\alpha_xb^{g_2}+2\partial^\alpha_x\mathsf{T}_{52})\nabla_x\partial^\alpha_x a^{g_2}dx+\int_{\mathbb{T}^3}\nabla_x(\partial^\alpha_xb^{g_2}+2\partial^\alpha_x\mathsf{T}_{52})\partial^\alpha_x \partial_ta^{g_2}dx+\int_{\mathbb{T}^3}(\partial^\alpha_x\mathsf{T}_{21}-2\partial^\alpha_x\mathsf{T}_{51})\nabla_x\partial^\alpha_x a^{g_2}dx. \end{align}\] In view of ?? , we have \[\begin{align} \partial_t \partial^\alpha_xa^{g_2}+T(t)^{\frac{1}{2}}\nabla_x\cdot \partial^\alpha_xb^{g_2}=\partial^\alpha_x\mathsf{T}_{11}. \end{align}\] Thus by Cauchy-Schwarz inequality and ?? , we can conclude that \[\begin{align} \label{a} \frac{1}{2}T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}&\leq& -\frac{d}{dt}\int_{\mathbb{T}^3}(\partial^\alpha_xb^{g_2}+2\partial^\alpha_x\mathsf{T}_{52})\nabla_x\partial^\alpha_x a^{g_2}dx+2T(t)^{\frac{1}{2}}\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}\\ \notag&&+C_\alpha\Big( T(t)^{\frac{1}{2}} \|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0}+T(t)^{-\frac{1}{2}}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx\Big). \end{align}\tag{97}\]

Step 4: Macroscopic estimates. Compute \(\eqref{c}+\kappa\times \eqref{b}+\eqref{a}\) for some constant \(\kappa\) and summation over \(|\alpha|\leq 1\), we have \[\begin{align} &&\frac{d}{dt}{\mathcal{G}}+T(t)^{\frac{1}{2}}\sum_{|\alpha|\leq 1}\Big((\frac{1}{2}-\kappa\varepsilon)\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}+(\frac{\kappa}{2}-2-\varepsilon)\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}+(\frac{1}{2}-\kappa\varepsilon)\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\Big)\\ &\leq&C_{\varepsilon,\alpha,\kappa}\Big( T(t)^{\frac{1}{2}} \|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0}+T(t)^{-\frac{1}{2}}\sum_{|\alpha|\leq 1}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx\Big). \end{align}\] with \[\begin{align} \label{cG} {\mathcal{G}}&=&\sum_{|\alpha|\leq 1}\Big(-\int_{\mathbb{T}^3}\partial^\alpha_x\mathsf{T}_{52}\nabla_x\partial^\alpha_x c^{g_2}dx+\int_{\mathbb{T}^3}(\partial^\alpha_xb^{g_2}+2\partial^\alpha_x\mathsf{T}_{52})\nabla_x\partial^\alpha_x a^{g_2}dx\\ \notag&&+\kappa\sum_{j\neq i}\int_{\mathbb{T}^3}\partial^\alpha_x\big(\partial_j(\mathsf{T}_{42})_{ij}-\partial_i(\mathsf{T}_{42})_{jj}\big)\partial^\alpha_xb_i^{g_2}dx+2\kappa\sum_{i=1}^3\int_{\mathbb{T}^3}\partial^\alpha_x\partial_i(\mathsf{T}_{42})_{ii}\partial^\alpha_xb_i^{g_2}dx\Big). \end{align}\tag{98}\] Thus we can choose \(\kappa=8\) and \(\varepsilon=1/32\) to get that \[\begin{align} \label{cG2} \notag &&\frac{d}{dt}{\mathcal{G}}+\frac{1}{4}T(t)^{\frac{1}{2}}\sum_{|\alpha|\leq 1}\Big(\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}+\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}+\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\Big)\\ &\lesssim& T(t)^{\frac{1}{2}} \|(\mathbf{I-P})g_2\|^2_{{\mathcal{D}}_0}+T(t)^{-\frac{1}{2}}\sum_{|\alpha|\leq 1}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx. \end{align}\tag{99}\]

On one hand, thanks to 98 and ?? , \({\mathcal{G}}\) has the upper bound \[\begin{align} \label{cGupperbound} |{\mathcal{G}}|\lesssim\|(\mathbf{I-P})g_2\|^2_{H^{2}_xL^2_{v}}+\sum_{|\alpha|\leq 1}\Big(\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}+\|\partial^\alpha_x b^{g_2}\|^2_{L^2_x}+\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\Big)\lesssim\|g_2\|^2_{H^2_xL^2_v}. \end{align}\tag{100}\] On the other hand, by 15 and the fact \(g=g_1+\mu^{\frac{1}{2}}g_2\), we can derive that \[\begin{align} \int_{\mathbb{T}^3}(a^{g_2}(x),b^{g_2}(x),c^{g_2}(x))dx=-\int_{\mathbb{T}^3\times\mathbb{R}^3}g_1(1,v,|v|^2-3)dvdx=:\big(\bar{a},\bar{b},\bar{c}). \end{align}\] Then by Poincaré inequality in \(\mathbb{T}^3\) and the bound \(|\bar{a}|^2+|\bar{b}|^2+|\bar{c}|^2\lesssim\|g_1\|^2_{L^2_{x}L^2_{4}}\), we have that \[\begin{align} &&\sum_{|\alpha|\leq 1}\Big(\|\nabla_x\partial^\alpha_x a^{g_2}\|^2_{L^2_x}+\|\nabla_x\partial^\alpha_x b^{g_2}\|^2_{L^2_x}+\|\nabla_x\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\Big)\\ &\gtrsim& \sum_{|\alpha|\leq 2}\Big(\|\partial^\alpha_x( a^{g_2}-\bar{a})\|^2_{L^2_x}+\|\partial^\alpha_x (b^{g_2}-\bar{b})\|^2_{L^2_x}+\|\partial^\alpha_x (c^{g_2}-\bar{c})\|^2_{L^2_x}\Big)\\ &\gtrsim& \sum_{|\alpha|\leq 2}\Big(\|\partial^\alpha_xa^{g_2}\|^2_{L^2_x}+\|\partial^\alpha_x b^{g_2}\|^2_{L^2_x}+\|\partial^\alpha_x c^{g_2}\|^2_{L^2_x}\Big)-\|g_1\|^2_{L^2_xL^2_4}. \end{align}\] Together with 99 and 100 , this yields the desired result ?? and completes the proof of the lemma.

Firstly, we compute \(\eqref{cE0cD0}+\frac{\kappa_1}{2}T(t)^{-2}\times\eqref{macroscopic}\) to get that \[\begin{align} \label{001} \notag&&\frac{1}{2}\frac{d}{dt}\big(\|g_2\|^2_{{\mathcal{E}}_0}+\kappa_1T(t)^{-2}{\mathcal{G}}\big)+T(t)^{-\frac{3}{2}}(\lambda_0-C_0\|g_2\|_{{\mathcal{E}}_0}-C\kappa_1)\|(\mathbf{I-P})g_2\|^2_{\mathcal{D}_0}+T(t)^{-\frac{3}{2}}(\kappa_1-C\|g_2\|_{{\mathcal{E}}_0})\\ \notag&&\times\|\mathbf{P}g_2\|^2_{H^2_xL^2_v}+\frac{1}{4}T(t)^{-1}T'(t)\|g_2\|^2_{{\mathcal{E}}_1}\lesssim T(t)^{-\frac{3}{2}}(\|g_2\|_{{\mathcal{D}}_0}\|g_1\|_{Y_0}+\|g_1\|^2_{L^2_xL^2_4})+(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|)\\ &&\times \|g_2\|^2_{{\mathcal{E}}_{\frac{1}{2}}}+\kappa_1T(t)^{-3}T'(t){\mathcal{G}}+\kappa_1T(t)^{-\frac{5}{2}}\sum_{|\alpha|\leq 1}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx. \end{align}\tag{101}\] On one hand, since \(|\mathcal{G}|\lesssim \|g_2\|_{H^2_xL^2_v}^2=\|g_2\|_{\mathcal{E}_0}^2\) and \(T(t)\geq T(0)/2\)(see 72 ), we can fix \(\kappa_1=\min\{T(0)^2,1\}\times\kappa_2\) with a sufficiently small \(\kappa_2>0\) such that \(C\kappa_1\leq C\kappa_2\le \lambda_0/4\) and the functional \[\|g_2\|_{\bar{\mathcal{E}}_0}^2:=\|g_2\|_{\mathcal{E}_0}^2+\kappa_1 T(t)^{-2}\mathcal{G}=\|g_2\|_{\mathcal{E}_0}^2+\kappa_2 \min\{T(0)^2,1\} T(t)^{-2}\mathcal{G}\] is equivalent to \(\|g_2\|_{\mathcal{E}_0}^2\). On the other hand, by the definition of \(r(g_1,g_2)\) in ?? and applying ?? to the term \(\Gamma(g_2,g_2)\), we obtain \[\begin{align} &&T(t)^{-\frac{5}{2}}\sum_{|\alpha|\leq 1}\sum_{i=1}^{13}\int_{\mathbb{T}^3}\big(\partial^\alpha_xr(g_1,g_2),\mu^{\frac{1}{2}}e_i\big)^2_{L^2_v}dx\\ &\lesssim& T(t)^{-\frac{5}{2}}\Big(T(t)^{-3}\|g_1\|^2_{H^2_xL^2_v}+T(t)^{-3}\|g_2\|^4_{{\mathcal{E}}_{0}}+(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|)^2\|g_2\|^2_{{\mathcal{E}}_{0}}\\ &\lesssim& T(t)^{-\frac{11}{2}}(\|g_1\|^2_{H^2_xL^2_v}+\|g_2\|^4_{{\mathcal{E}}_0})+(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|)\|g_2\|^2_{{\mathcal{E}}_{0}}. \end{align}\] In the last inequality we used 73 . Plug these into 101 , we have \[\begin{align} \notag&&\frac{1}{2}\frac{d}{dt}\|g_2\|_{\bar{\mathcal{E}}_0}^2+T(t)^{-\frac{3}{2}}(\frac{3}{4}\lambda_0-C_0\|g_2\|_{{\mathcal{E}}_0})\|(\mathbf{I-P})g_2\|^2_{\mathcal{D}_0}+T(t)^{-\frac{3}{2}}(\kappa_2\min\{T(0)^2,1\}-C\|g_2\|_{{\mathcal{E}}_0})\\ \notag&&\times\|\mathbf{P}g_2\|^2_{H^2_xL^2_v}\lesssim T(t)^{-\frac{3}{2}}(\|g_2\|_{{\mathcal{D}}_0}\|g_1\|_{Y_0}+\|g_1\|^2_{L^2_xL^2_4})+(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t))\\ &&\times \|g_2\|^2_{{\mathcal{E}}_{\frac{1}{2}}}+T(t)^{-\frac{11}{2}}(\|g_1\|^2_{H^2_xL^2_v}+\|g_2\|^4_{{\mathcal{E}}_{0}}). \end{align}\] Noticing \(\|(\mathbf{I-P})g_2\|^2_{\mathcal{D}_0}+\|\mathbf{P}g_2\|^2_{H^2_xL^2_v}\geq c_0 \|g_2\|^2_{\mathcal{D}_0}\) and \(\|g_2\|_{{\mathcal{D}}_0}\|g_1\|_{Y_0}\leq \varepsilon\|g_2\|^2_{{\mathcal{D}}_0}+\varepsilon^{-1}\|g_1\|^2_{Y_0}\) for any \(\varepsilon>0\), if we set \(2\lambda_1:=c_0\min\{\frac{3}{4}\lambda_0, \kappa_2T(0)^2,\kappa_2\}\) and choose \(\varepsilon=\lambda_1\), we can get \[\begin{align} \label{barcE0} &\frac{1}{2}\frac{d}{dt}\|g_2\|^2_{\bar{{\mathcal{E}}}_0}+T(t)^{-\frac{3}{2}}(\lambda_1-C\|g_2\|_{{\mathcal{E}}_0})\|g_2\|^2_{{\mathcal{D}}_0}\lesssim T(t)^{-\frac{3}{2}}(\lambda_1^{-1}\|g_1\|^2_{Y_0}+\|g_1\|^2_{L^2_xL^2_4})\\ &\notag+(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t)) \|g_2\|^2_{{\mathcal{E}}_{\frac{1}{2}}}+T(t)^{-\frac{11}{2}}(\|g_1\|^2_{H^2_xL^2_v}+\|g_2\|^4_{{\mathcal{E}}_{0}}) \end{align}\tag{102}\] with the positive constant \(1>\lambda_1\gtrsim\min\{T(0)^2,1\}\).

Next, we compute \(\eqref{barcE0}+\kappa_3\eqref{cEkcDk}\) with \(0<\kappa_3<1\) to get that \[\begin{align} &&\frac{1}{2}\frac{d}{dt}(\|g_2\|^2_{\bar{{\mathcal{E}}}_0}+\kappa_3\|g_2\|^2_{{\mathcal{E}}_k})+T(t)^{-\frac{3}{2}}(\lambda_1-C_0\|g_2\|_{{\mathcal{E}}_0}-\kappa_3C_k)\|g_2\|^2_{{\mathcal{D}}_0}+\kappa_3T(t)^{-\frac{3}{2}}(\frac{\lambda_0}{2}-C_k\|g_2\|_{{\mathcal{E}}_{17}})\|g_2\|^2_{\mathcal{D}_k}\\ &&+\frac{\kappa_3}{4}T(t)^{-1}T'(t)\|g_2\|^2_{{\mathcal{E}}_{k+1}}\leq C_k(T(t)^{-\frac{3}{2}}\lambda_1^{-1}+T(t)^{-\frac{11}{2}})\|g_1\|^2_{Y_k}\\ &&+C_k(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t))\|g_2\|^2_{{\mathcal{E}}_{k+\frac{1}{2}}}+C_kT(t)^{-\frac{11}{2}}\|g_2\|^4_{{\mathcal{E}}_0}, \end{align}\] where we use the fact \(\|g_1\|_{Y_0}+\|g_1\|_{L^2_xL^2_4}+\|g_1\|_{H^2_xL^2_v}\lesssim\|g_1\|_{Y_k},k\geq 17\). Fixing \(\kappa_3\) such that \(\kappa_3C_k=\lambda_1/2\), then \(\kappa_3\gtrsim_k \lambda_1\gtrsim\min\{T(0)^2,1\}\) and the functional \[\begin{align} \label{func1} \min\{T(0)^2,1\}\|g_2\|_{\mathcal{E}_k}^2\lesssim_k \kappa_3 \|g_2\|_{\mathcal{E}_k}^2 \leq\|g_2\|_{\bar{\mathcal{E}}_k}^2:=\|g_2\|_{\bar{\mathcal{E}}_0}^2+\kappa_3 \|g_2\|_{\mathcal{E}_k}^2\leq \|g_2\|_{\mathcal{E}_k}^2 \end{align}\tag{103}\] is equivalent to \(\|g_2\|_{\mathcal{E}_k}^2\). Rearranging terms, we can obtain \[\begin{align} &&\frac{1}{2}\frac{d}{dt}\|g_2\|^2_{\bar{{\mathcal{E}}}_k}+T(t)^{-\frac{3}{2}}(\lambda_2-C_k\|g_2\|_{\bar{{\mathcal{E}}}_{17}})\|g_2\|^2_{\mathcal{D}_k}+\lambda_2T(t)^{-1}T'(t)\|g_2\|^2_{{\mathcal{E}}_{k+1}}\leq C_k\Big((T(t)^{-\frac{3}{2}}\lambda_2^{-1}+T(t)^{-\frac{11}{2}})\\ &&\times\|g_1\|^2_{Y_k}+(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t))\|g_2\|^2_{{\mathcal{E}}_{k+\frac{1}{2}}}+T(t)^{-\frac{11}{2}}\|g_2\|^4_{{\mathcal{E}}_0}\Big) \end{align}\] with the positive constant \(\lambda_2\gtrsim_k \min\{T(0)^2,1\}\).

The proof of the proposition is complete.

Combining Proposition 2 and Proposition 3, we compute \(\eqref{evolutiong1}+\kappa_4\times \eqref{barcEk2}\) with \(0<\kappa_4<1\) to get that \[\begin{align} &&\frac{1}{2}\frac{d}{dt}(\|g_1\|^2_{X_k}+\kappa_4\|g_2\|^2_{\bar{{\mathcal{E}}}_k})+T(t)^{-\frac{3}{2}}\big(\frac{\lambda_0}{2}-C_k(\|g_1\|_{X_{17}}+\|g_2\|_{\bar{{\mathcal{E}}}_{17}})-C_k\kappa_4\lambda_2^{-1}-C_k \kappa_4T(t)^{-4}\big)\|g_1\|^2_{Y_k}\\ &&+T(t)^{-\frac{3}{2}}(\kappa_4\lambda_2-\kappa_4C_k\|g_2\|_{\bar{{\mathcal{E}}}_{17}}-C_k\|g_1\|_{X_{17}})\|g_2\|^2_{{\mathcal{D}}_{k}}+T(t)^{-1}T'(t)\big(\|g_1\|^2_{X_k}+\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{E}}_{k+1}}\big)\\ &\leq&C_k(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t))\|g_2\|^2_{{\mathcal{E}}_{k+\frac{1}{2}}}+C_kT(t)^{-\frac{11}{2}}\|g_2\|^4_{{\mathcal{E}}_0}\\ &&+C_k\big(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\|g_1\|^2_{X_{k-\frac{1}{4}}}\\ &&+C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\|g_2\|^2_{{\mathcal{D}}_0}+C_kT(t)^{-1}|V(t)|^{-1}\|g_1\|_{L^2_{x,v}}, \end{align}\] where we use the fact \(\|g_2\|_{{\mathcal{D}}_0}\leq \|g_2\|_{{\mathcal{D}}_k}\) and \(\|g_2\|_{{\mathcal{E}}_0}\leq \|g_2\|_{\bar{{\mathcal{E}}}_{17}}\). Choosing \(\kappa_4\) satisfies \(\kappa_4(C_k\lambda_2^{-1}+C_kT(t)^{-4})=\lambda_0/4\). Note that \(\lambda_2\gtrsim_k \min\{T(0)^2,1\}\) and \(T(t)\geq T(0)/2\), then \(\kappa_4\gtrsim_k \min\{ T(0)^4,1\}\) and by 103 , the functional \[\begin{align} \label{func2} \min\{T(0)^6,1\}(\|g_1\|^2_{X_k}+\|g_2\|^2_{{{\mathcal{E}}}_k})\lesssim_k \|(g_1,g_2)\|^2_{\mathbf{E}_k}:=\|g_1\|^2_{X_k}+\kappa_4\|g_2\|^2_{\bar{{\mathcal{E}}}_k}\lesssim\|g_1\|^2_{X_k}+\|g_2\|^2_{{{\mathcal{E}}}_k}. \end{align}\tag{104}\] Noticing that \[\begin{align} \|g_1\|_{X_{17}}+\|g_2\|_{\bar{{\mathcal{E}}}_{17}}\lesssim_k \kappa_4^{-\frac{1}{2}}\|(g_1,g_2)\|_{\mathbf{E}_{17}}, \quad \|g_1\|_{X_{17}}+\kappa_4\|g_2\|_{\bar{{\mathcal{E}}}_{17}}\lesssim_k \|(g_1,g_2)\|_{\mathbf{E}_{17}}, \end{align}\] we can obtain from above \[\begin{align} &&\frac{1}{2}\frac{d}{dt}\|(g_1,g_2)\|^2_{\mathbf{E}_k}+T(t)^{-\frac{3}{2}}\big(\frac{\lambda_0}{4}-C_k\kappa_4^{-\frac{1}{2}}\|(g_1,g_2)\|_{\mathbf{E}_{17}}\big)\|g_1\|^2_{Y_k}\\ &&+T(t)^{-\frac{3}{2}}(1-C_k\kappa_4^{-1}\lambda_2^{-1}\|(g_1,g_2)\|_{\mathbf{E}_{17}})\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{D}}_{k}}+T(t)^{-1}T'(t)\big(\|g_1\|^2_{X_k}+\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{E}}_{k+1}}\big)\\ &\leq&C_k(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t))\|g_2\|^2_{{\mathcal{E}}_{k+\frac{1}{2}}}+C_kT(t)^{-\frac{11}{2}}\|g_2\|^4_{{\mathcal{E}}_0}\\ &&+C_k\big(T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\|g_1\|^2_{X_{k-\frac{1}{4}}}\\ &&+C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\|g_2\|^2_{{\mathcal{D}}_0}+C_k T(t)^{-1}|V(t)|^{-1}\|g_1\|_{L^2_{x,v}}. \end{align}\] Finally, we define the dissipation norm \[\begin{align} \min\{T(0)^6,1\}(\|g_1\|^2_{Y_k}+\|g_2\|^2_{{\mathcal{D}}_k}) \lesssim\|(g_1,g_2)\|^2_{\mathbf{D}_k}:=\|g_1\|^2_{Y_k}+\kappa_4\lambda_2\|g_2\|_{{\mathcal{D}}_k}\lesssim\|g_1\|^2_{Y_k}+\|g_2\|^2_{{\mathcal{D}}_k}, \end{align}\] and since \(\kappa_4^{-1}\lambda_2^{-1}\geq\kappa_4^{-\frac{1}{2}}\), we can get the desired result.

We will use ?? to close the energy estimates. Denote the terms on the right-hand side by \(R_i,i=1,\cdots,5\) and we first handle \(R_4\). Since \[\begin{align} R_4&=&C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\|g_2\|^2_{{\mathcal{D}}_0}\leq C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)\|g_2\|^2_{{\mathcal{D}}_k}\\ &\leq& C_k\big(T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\big)(1+T(0)^{-6})\|(g_1,g_2)\|^2_{\mathbf{D}_k}\leq \frac{\lambda_0 }{5} T(t)^{-\frac{3}{2}}\|(g_1,g_2)\|^2_{\mathbf{D}_k}, \end{align}\] where we use ?? and 107 in the last two steps.

For \(R_1\), by interpolation inequality, we have \[\begin{align} \|g_2\|^2_{{\mathcal{E}}_{k+\frac{1}{2}}}\leq \|g_2\|^{\frac{8}{5}}_{{\mathcal{E}}_{k+1}}\|g_2\|^{\frac{2}{5}}_{{\mathcal{E}}_{k-\frac{3}{2}}}\leq\epsilon \|g_2\|^2_{{\mathcal{E}}_{k+1}}+C\epsilon^{-4} \|g_2\|^2_{{\mathcal{D}}_k},\quad \forall \epsilon>0. \end{align}\] Due to 105 and 106 , the coefficient \[\begin{align} &&T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-3}T'(t)\lesssim(1+T(0)^{-2})T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|V(t)|^{-2}\\ &\lesssim&(1+T(0)^{-2})T(t)^{-1}T'(t) \lesssim(1+T(0)^{-2})(\kappa_4\lambda_2)^{-1}T(t)^{-1}T'(t)\kappa_4\lambda_2\lesssim(1+T(0)^{-8})T(t)^{-1}T'(t)\kappa_4\lambda_2, \end{align}\] where we use the fact \(\kappa_4^{-1}\lambda_2^{-1}\lesssim_k 1+T(0)^{-6}\). Thus we let \(\epsilon=\min\{T(0)^8,1\}\epsilon_1/C_k,\epsilon_1>0\) to get that \[\begin{align} R_1&\leq& \epsilon_1T(t)^{-1}T'(t)\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{E}}_{k+1}}+C_k\epsilon_1^{-4}(1+T(0)^{-24})T(t)^{-1}T'(t)\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{D}}_k}\\ &\leq&\epsilon_1T(t)^{-1}T'(t)\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{E}}_{k+1}}+C_k\epsilon_1^{-4}(1+T(0)^{-30})T(t)^{-1}T'(t)\|(g_1,g_2)\|^2_{\mathbf{D}_k}\\ &\leq& \frac{1}{2} T(t)^{-1}T'(t)\kappa_4\lambda_2\|g_2\|^2_{{\mathcal{E}}_{k+1}}+\frac{\lambda_0}{5} T(t)^{-\frac{3}{2}}\|(g_1,g_2)\|^2_{\mathbf{D}_k}, \end{align}\] where we choose \(\varepsilon_1=\frac{1}{2}\) and use ?? and 107 in the last two step.

Similarly, for \(R_3\), we have the interpolation inequality \[\begin{align} \|g_1\|^2_{X_{k-\frac{1}{4}}}\leq \|g_1\|^{\frac{5}{3}}_{X_k}\|g_1\|^{\frac{1}{3}}_{X_{k-\frac{3}{2}}}\leq \epsilon \|g_1\|^2_{X_k}+C\epsilon^{-5}\|g_1\|_{Y_k}. \end{align}\] Due to 106 , the coefficient \[\begin{align} T(t)^{-1}T'(t)+T(t)^{-\frac{1}{2}}|R(t)|+T(t)^{-1}|V(t)|^{-1}+T(t)^{-\frac{1}{4}}|V(t)|^{-\frac{3}{2}}\lesssim T(t)^{-1}T'(t). \end{align}\] Then choose suitably small \(\epsilon>0\) and using ?? , 107 again, we have \[\begin{align} R_3&\leq& \frac{1}{2} T(t)^{-1}T'(t)\|g_1\|^2_{X_k}+C_kT(t)^{-1}T'(t)\|g_1\|^2_{Y_k}\\ &\lesssim& \frac{1}{2} T(t)^{-1}T'(t)\|g_1\|^2_{X_k}+C_k(1+T(0)^{-6})T(t)^{-1}T'(t)\|(g_1,g_2)\|^2_{\mathbf{D}_k}\\ &\leq& \frac{1}{2} T(t)^{-1}T'(t)\|g_1\|^2_{X_k} +\frac{\lambda_0}{5}T(t)^{-\frac{3}{2}}\|(g_1,g_2)\|^2_{\mathbf{D}_k}. \end{align}\]

For \(R_2\) and \(R_5\), noticing that \[\begin{align} &&C_kT(t)^{-\frac{11}{2}}\|g_2\|^4_{{\mathcal{E}}_0}\leq C_k(1+T(0)^{-16})\|(g_1,g_2)\|^2_{\mathbf{E}_{17}}T(t)^{-\frac{3}{2}}\|(g_1,g_2)\|^2_{\mathbf{D}_k},\\ &&and\quad T(t)^{-1} |V(t)|^{-1}\|g_1\|_{L^2_{x,v}}\leq \frac{\lambda_0}{5}T(t)^{-\frac{3}{2}}\|(g_1,g_2)\|^2_{\mathbf{D}_k}+ (1+T(0)^{-6})T(t)^{-\frac{1}{2}}|V(t)|^{-2}. \end{align}\] We can rewrite evolution of \((g_1,g_2)\) by combining the estimates of \(R_i,i=1\cdots,5\) as follows: \[\begin{align} \label{simpleenergyinequ} \notag&&\frac{1}{2}\frac{d}{dt}\|(g_1,g_2)\|^2_{\mathbf{E}_k}+T(t)^{-\frac{3}{2}}\Big(\frac{\lambda_0}{5}-C_k\kappa_4^{-1}\lambda_2^{-1}\|(g_1,g_2)\|_{\mathbf{E}_{17}}-C_k(1+T(0)^{-16})\|(g_1,g_2)\|^2_{\mathbf{E}_{17}}\Big)\\ &&\times\|(g_1,g_2)\|^2_{\mathbf{D}_k}\leq C_k(1+T(0)^{-6})T(t)^{-\frac{1}{2}}|V(t)|^{-2}. \end{align}\tag{109}\] Since \(\kappa_4^{-1}\lambda_2^{-1}\lesssim_k 1+T(0)^{-6}\) and recall that \(g_1(t_0)=g(t_0)\), \(g_2(t_0)=0\), if we make the a priori assumption that \[\begin{align} \label{aprioriassumption} C_k\sup_{t>t_0}\|(g_1,g_2)(t)\|^2_{\mathbf{E}_k}\lesssim\frac{\min\{\lambda_0,\lambda_0^2\}}{400}\min\{1,T(0)^{16}\}, \end{align}\tag{110}\] it leads that \[\begin{align} &&\|(g_1,g_2)(t)\|_{\mathbf{E}_{k}}^2\leq \|(g_1,g_2)(t_0)\|_{\mathbf{E}_{k}}^2+C_k(1+T(0)^{-6})\int_{t_0}^tT(t)^{-1}|V(t)|^{-2}d\tau\\ &\leq&\|g(t_0)\|_{X_{k}}^2+\frac{C_k(1+T(0)^{-6})}{|E|}\int_{t_0}^\infty (1+\tau)^{-2}d\tau\leq \varepsilon_1^2+C_k(1+T(0)^{-6})|E|^{-1},\quad t>t_0, \end{align}\] where we use 104 and 108 . Consequently, since \[\begin{align} \|g(t)\|^2_{X_k}&=&\|g_1+\mu g_2\|^2_{X_k}\leq C_k(\|g_1\|^2_{X_k}+\|g_2\|^2_{{\mathcal{E}}_k})\leq C_k(1+T(0)^{-6})\|(g_1,g_2)(t)\|^2_{\mathbf{E}_{k}}\\ &\leq& C_k(1+T(0)^{-6})\varepsilon_1^2+C_k(1+T(0)^{-6})(1+T(0)^{-6})|E|^{-1}, \end{align}\] then the a priori assumption 110 can be maintained, provided that \[\begin{align} \label{varepsilon1} \varepsilon_1\leq C_{k,\lambda_0}\min\{1,T(0)^{11}\}\eta,\quad |E|^{-1}\leq C_{k,\lambda_0}\min\{1,T(0)^{28}\}\eta \end{align}\tag{111}\] with small \(\eta>0\). Meanwhile, \(\|g(t)\|^2_{X_k}\lesssim\eta\) remains small and thus the argument for long-time behavior of \(T(t)\) and \(V(t)\) is closed, thanks to Lemma 11.

Finally, we give the decay estimates of the solution. We remark that the constants \(C_i\) below depend on \(k\), \(T(0)\) and \(V(0)\), however, since we are concerned with the regime \(t\gg 1\), we shall not indicate this dependence explicitly.

Choosing \(k=17\) and \(k=k_1>17\) separately, we obtain on the one hand that \(\sup_{t\in[t_0,\infty]}\|(g_1,g_2)(t)\|_{\mathbf{E}_{k_1}}\le C_{k_1}\) and that 110 holds for \(k=17\). On the other hand, by ?? and 108 , we have \[\begin{align} T(t)^{-\frac{3}{2}}\geq C_1\big(\ln (3+|E|t)\big)^{-\frac{3}{2}},\quad T(t)^{-\frac{1}{2}}|V(t)|^{-2}\leq C_2\langle Et\rangle^{-2},\quad t> t_0. \end{align}\] Then we obtain from 109 that \[\begin{align} \frac{d}{dt}\|(g_1,g_2)\|_{\mathbf{E}_{17}}^2+C(\ln (3+|E|t))^{-\frac{3}{2}}\|(g_1,g_2)\|_{\mathbf{D}_{17}}^2\leq C_2\langle Et\rangle^{-2},\quad t>t_0. \end{align}\] The interpolation inequality \[\begin{align} \|(g_1,g_2)\|_{\mathbf{E}_{17}}\leq \|(g_1,g_2)\|^\theta_{\mathbf{D}_{17}}\|(g_1,g_2)\|^{1-\theta}_{\mathbf{E}_{k_1}},\quad with\quad \theta=\frac{k_1-17}{k_1-31/2}<1 \end{align}\] implies that \[\begin{align} \|(g_1,g_2)\|_{\mathbf{D}_{17}}\geq \|(g_1,g_2)\|^{1-\frac{1}{\theta}}_{\mathbf{E}_{k_1}}\|(g_1,g_2)\|^{\frac{1}{\theta}}_{\mathbf{E}_{17}}\geq C_{k_1}\|(g_1,g_2)\|^{\frac{1}{\theta}}_{\mathbf{E}_{17}}. \end{align}\] Thus we have \[\begin{align} Z'(t)+C_1(\ln (3+|E|t))^{-\frac{3}{2}} Z(t)^{1+\theta_1}\leq C_{2}\langle Et\rangle^{-2},\quad with\quad \theta_1=\frac{1}{\theta}-1=\frac{3}{2(k_1-17)}, \end{align}\] where \(Z(t):=\|(g_1,g_2)(t)\|^2_{\mathbf{E}_{17}}\).

If the initial data satisfies \(C_1(\ln (3+|E|t_0))^{-\frac{3}{2}} Z(t_0)^{1+\theta_1}\leq 2C_2\langle Et_0\rangle^{-2}\), at this point, if for any \(t\geq t_0\), \(C_1(\ln (3+|E|t))^{-\frac{3}{2}} Z(t)^{1+\theta_1}\leq 2C_2\langle Et\rangle^{-2}\), we have \(Z(t)\leq (2C_2/C_1)^{\frac{1}{1+\theta_1}} ((\ln (3+|E|t))^{\frac{3}{2}}\langle Et\rangle^{-2})^{\frac{1}{1+\theta_1}}\). Otherwise, by continuity, there is a time interval \(t\in[t_1,t_2)\) such that \[\begin{align} \label{Z1} Z'(t)+\frac{C_1}{2}(\ln (3+|E|t))^{-\frac{3}{2}} Z(t)^{1+\theta_1}\leq 0,\quad t\in[t_1,t_2) \end{align}\tag{112}\] and \(C_1(\ln (3+|E|t_1))^{-\frac{3}{2}} Z(t_1)^{1+\theta_1}= 2C_2\langle Et_1\rangle^{-2}\) which yields that \[\begin{align} Z(t)&\leq& \Big(Z^{-\theta_1}(t_1)+\frac{C_1\theta_1}{2}\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}(t-t_1)\Big)^{-\frac{1}{\theta_1}}\\ &\leq&\Big(\big(\frac{C_1}{2C_2}\big)^{\frac{\theta_1}{1+\theta_1}}\big(\ln(3+|E|t_1)\big)^{-\frac{3}{2}\frac{\theta_1}{1+\theta_1}}\langle Et_1\rangle^{\frac{2\theta_1}{1+\theta_1}}+\frac{C_1\theta_1}{2}\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}(t-t_1)\Big)^{-\frac{1}{\theta_1}}. \end{align}\] Observing that \[\begin{align} \big(\frac{C_1}{2C_2}\big)^{\frac{\theta_1}{1+\theta_1}}\big(\ln(3+|E|t_1)\big)^{-\frac{3}{2}\frac{\theta_1}{1+\theta_1}}\langle Et_1\rangle^{\frac{2\theta_1}{1+\theta_1}}\geq \frac{C_1\theta_1}{2} \big(\ln(3+|E|t)\big)^{-\frac{3}{2}}t_1,~~t\geq t_1 \end{align}\] for large \(|E|\) depending on \(k_1\), \(T(0)\) and \(V(0)\), but independent of \(t_1\). Indeed, \[\begin{align} \big(\ln(3+|E|t_1)\big)^{-\frac{3}{2}\frac{\theta_1}{1+\theta_1}}\langle Et_1\rangle^{\frac{2\theta_1}{1+\theta_1}}\gtrsim\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}\frac{2|E|}{1+\theta_1}\theta_1t_1,~~t\geq t_1, \end{align}\] where we use \((1+x)^a\geq ax\) for any \(a,x>0\). Thus we can choose \(|E|\gtrsim C_1(1+\theta_1)\big(\frac{2C_2}{C_1}\big)^{\frac{\theta_1}{1+\theta_1}}\) to get that \[\begin{align} Z(t)\leq C_3\Big(\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}t\Big)^{-\frac{1}{\theta_1}},\quad t\in[t_1,t_2). \end{align}\] Therefore, we conclude that \[\begin{align} \label{Z2} Z(t)\leq C_4\max\Big\{(\ln (3+|E|t))^{\frac{3}{2}}\langle Et\rangle^{-2})^{1-\frac{3}{2k_1-31}},\Big(\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}t\Big)^{-\frac{2(k_1-17)}{3}}\Big\},\quad t>t_0. \end{align}\tag{113}\] Note that the constant \(C_4\) does not depend on \(t_0\).

If the initial data satisfies \(C_1(\ln (3+|E|t_0))^{-\frac{3}{2}} Z(t_0)^{1+\theta_1}> 2C_2\langle Et_0\rangle^{-2}\) and it holds for any \(t\geq t_0\), then by 112 , we deduce that \[\begin{align} Z(t)\leq\Big(Z^{-\theta_1}(t_0)+\frac{C_1\theta_1}{2}\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}(t-t_0)\Big)^{-\frac{1}{\theta_1}}\leq C_4 \Big(\big(\ln(3+|E|t)\big)^{-\frac{3}{2}}t\Big)^{-\frac{1}{\theta_1}},~~t\geq t_0. \end{align}\] Otherwise, there exists a smallest time \(t_3\) such that \(C_1(\ln (3+|E|t_3))^{-\frac{3}{2}} Z(t_3)^{1+\theta_1}\leq 2C_2\langle Et_3\rangle^{-2}\). Taking \(t_3\) as the initial time as \(t_0\) and by the previous argument, we can still obtain 113 .

Consequently, combining 111 (note that to obtain \(\|g(t_0)\|_{X_k}<\varepsilon_1\), we require \(\|g(0)\|_{X_k}\le \varepsilon_1^2\) according to Corollary 1), 113 and ?? , we complete the proof of Theorem 2. Theorem 1 follows immediately by the change of variables 12 .