Appendix
Our Appendix is about three problems: the derivation of the kernel of operator \(K_{{\mathbf {M}}}\) in (1.21), estimates of the kernels related to \(K_1\) and \(K_2\), and the construction and regularity estimates of the coefficients in the Hilbert expansion (1.8).
1.1 Appendix 1: Derivation of the kernel \(k_2\)
In this part, we derive the kernel \(k_2\) of the operator \(K_2\). For \(K_i, (i=1, 2)\) given in (1.21), their kernels are defined as
$$\begin{aligned} K_i(f)=\int _{{\mathbb {R}}^3}dq k_i(p,q)f(q),\qquad i=1,2. \end{aligned}$$
Following the derivation of \(k_2(p,q)\) for a global Maxwellian case in the Appendix of [37], we can obtain the following expression of \(k_2(p,q)\) for our case:
$$\begin{aligned} k_2(p,q)=\frac{C_1s^{\frac{3}{2}}}{gp^0q^0}U_1(p,q)\exp \{-U_2(p,q)\}, \end{aligned}$$
(8.1)
where \(C_1\) is some positive constant, \(U_2(p,q)\) and \(U_1(p,q)\) are smooth functions satisfying
$$\begin{aligned} U_2(p,q)&=[1+O(1)|u|]\frac{\sqrt{s}|p-q|}{2T_0g},\\ U_1(p,q)&=\frac{1}{U_2(p,q)}+\frac{(p^0+q^0)[1+O(1)|u|]}{2[U_2(p,q)]^2}+\frac{(p^0+q^0)[1+O(1)|u|]}{2[U_2(p,q)]^3}. \end{aligned}$$
The proof of (8.1) can be proceeded similarly as in the Appendix [37] except for the estimation of additional terms w.r.t. the velocity u, which arises due to the local Maxwellian \(\mathbf {M}\).
As in [37], we first write the operator \(K_2\) as
$$\begin{aligned} K_2(f)=\frac{1}{p^0}\int _{{\mathbb {R}}^3}\frac{dq}{q^0}\int _{{\mathbb {R}}^3}\frac{dq'}{q'^{0}}\int _{{\mathbb {R}}^3}\frac{dp'}{p'^{0}}W\sqrt{{\mathbf {M}}}(q)[\sqrt{{\mathbf {M}}}(q')f(p')+\sqrt{{\mathbf {M}}}(p')f(q')]. \end{aligned}$$
After the same exchanges of variables \(q, p', q'\) and changes of integration variables as in [37], we obtain the following form of kernel \(k_2\)
$$\begin{aligned} k_2(p,q)=\frac{C}{p^0q^{0}}\int _{{\mathbb {R}}^3}\frac{d{\bar{p}}}{{\bar{p}}^0}\delta ((q^{\mu }-p^{\mu }){\bar{p}}_{\mu }){\bar{s}} \exp \Big \{\frac{u^{\mu }{\bar{p}}_{\mu } }{2T_0}\Big \}, \end{aligned}$$
where the integration variable \({\bar{p}}_{\mu }=p'_{\mu }+q'_{\mu }\) and \({\bar{s}}={\bar{g}}^2+4\) with
$$\begin{aligned} {\bar{g}}^2=g^2+\frac{1}{2}(p^{\mu }+q^{\mu })(p_{\mu }+q_{\mu }-{\bar{p}}_{\mu }). \end{aligned}$$
Introduce a Lorentz transformation \(\Lambda \) which maps into the center-of-momentum system:
$$\begin{aligned} A^{\mu }=\Lambda ^{\mu }_{\nu }(p^{\nu }+q^{\nu })=(\sqrt{s},0,0,0), \quad B^{\mu }=-\Lambda ^{\mu }_{\nu }(p^{\nu }-q^{\nu })=(0,0,0,g). \end{aligned}$$
Here \(\Lambda \) is the following Lorentz transform derived in [34]:
$$\begin{aligned} \Lambda =(\Lambda ^{\mu }_{\nu })=\left( \begin{array}{cccc} \frac{p^0+q^0}{\sqrt{s}} &{} -\frac{p_1+q_1}{\sqrt{s}} &{} -\frac{p_2+q_2}{\sqrt{s}} &{}-\frac{p_3+q_3}{\sqrt{s}}\\ \Lambda ^{1}_0 &{} \Lambda ^{1}_{1} &{} \Lambda ^{1}_{2} &{} \Lambda ^{1}_3 \\ 0 &{} \frac{(p\times q)_1}{|p\times q|} &{} \frac{(p\times q)_2}{|p\times q|} &{} \frac{(p\times q)_3}{|p\times q|}\\ \frac{p^0-q^0}{g} &{} -\frac{p_1-q_1}{g} &{} -\frac{p_2-q_2}{g} &{}-\frac{p_3-q_3}{g} \end{array} \right) , \end{aligned}$$
where \(\Lambda ^{1}_0=\frac{2|p\times q|}{g\sqrt{s}}\) and
$$\begin{aligned} \Lambda ^{1}_i=\frac{2[p_i(p^0+q^0p^{\mu }q_{\mu })+q_i(q^0+p^0p^{\mu }q_{\mu })]}{g\sqrt{s}|p\times q|}, \quad i=1,2,3. \end{aligned}$$
Define \({\bar{u}}^\mu =\Lambda ^{\mu }_{\nu }u^{\nu }\):
$$\begin{aligned} {\bar{u}}^0&=\Lambda ^{0}_{\nu }u^{\nu }=\frac{(p^0+q^0)u^0}{\sqrt{s}} -\frac{(p+q)\cdot u}{\sqrt{s}},\\ {\bar{u}}^1&=\Lambda ^{1}_{\nu }u^{\nu }=\frac{2|p\times q|u^0}{g\sqrt{s}} +\frac{2[p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu })]\cdot u}{g\sqrt{s}|p\times q|},\\ {\bar{u}}^2&=\Lambda ^{2}_{\nu }u^{\nu }=\frac{(p\times q)\cdot u}{|p\times q|},\\ {\bar{u}}^3&=\Lambda ^{3}_{\nu }u^{\nu }=\frac{(p^0-q^0)u^0}{g} -\frac{(p-q)\cdot u}{g}. \end{aligned}$$
Then we have
$$\begin{aligned} \int _{{\mathbb {R}}^3}\frac{d{\bar{p}}}{{\bar{p}}^0}\delta ((q^{\mu }-p^{\mu }){\bar{p}}_{\mu }){\bar{s}} \exp \Big \{\frac{u^{\mu }{\bar{p}}_{\mu } }{2T_0}\Big \}=\int _{{\mathbb {R}}^3}\frac{d{\bar{p}}}{{\bar{p}}^0}\delta (B^{\mu }{\bar{p}}_{\mu }){\bar{s}}_{\Lambda } \exp \Big \{\frac{u^{\mu }{\bar{p}}_{\mu } }{2T_0}\Big \}, \end{aligned}$$
where \(B^{\mu }{\bar{p}}_{\mu }={\bar{p}}_3g,\)
$$\begin{aligned} {\bar{s}}_{\Lambda }&={\bar{g}}^2+4=g^2+\frac{1}{2}A^{\mu }(A_{\mu }-{\bar{p}}_{\mu })=g^2+\frac{1}{2}\sqrt{s}({\bar{p}}^0-\sqrt{s}),\\ {\bar{u}}^{\mu }{\bar{p}}_{\mu }=&{\bar{p}}^0\Big (-\frac{(p^0+q^0)u^0}{\sqrt{s}} +\frac{(p+q)\cdot u}{\sqrt{s}}\Big )\\&\quad +{\bar{p}}_1\Big (\frac{2|p\times q|u^0}{g\sqrt{s}} +\frac{2[p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu })]\cdot u}{g\sqrt{s}|p\times q|}\Big )\\&\quad +{\bar{p}}_2\frac{(p\times q)\cdot u}{|p\times q|}+{\bar{p}}_3\Big (\frac{(p^0-q^0)u^0}{g} -\frac{(p-q)\cdot u}{g}\Big ). \end{aligned}$$
Switch to polar coordinates
$$\begin{aligned} d{\bar{p}}=|{\bar{p}}|^2d|{\bar{p}}|\sin \phi d\phi d\theta ,\quad {\bar{p}}=|{\bar{p}}|(\sin \phi \cos \theta , \sin \phi \sin \theta , \cos \phi ), \end{aligned}$$
and use \(\cos \phi =0\) for \(\phi =\frac{\pi }{2}\) to rewrite \( k_2(p,q)\) as
$$\begin{aligned}&\frac{C}{gp^0q^0}\int _{0}^{2\pi }d\phi \int _0^{\infty }\frac{|{\bar{p}}|d{\bar{p}}}{{\bar{p}}^0}{\bar{s}}_{\Lambda } \exp \Big \{\frac{1}{2T_0}\Big [{\bar{p}}^0\Big (-\frac{(p^0+q^0)u^0}{\sqrt{s}} +\frac{(p+q)\cdot u}{\sqrt{s}}\Big )\\&\qquad +\Big (\frac{2|p\times q|u^0}{g\sqrt{s}} +\frac{2[p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu })]\cdot u}{g\sqrt{s}|p\times q|}\Big )|{\bar{p}}|\cos \phi \\&\qquad +\frac{(p\times q)\cdot u}{|p\times q|}|{\bar{p}}|\sin \phi \Big ] \Big \}\\&\quad =\frac{C}{gp^0q^0}\int _{0}^{\infty }\frac{|{\bar{p}}|d{\bar{p}}}{{\bar{p}}^0}{\bar{s}}_{\Lambda } \exp \Big \{\frac{{\bar{p}}^0}{2T_0}\Big (-\frac{(p^0+q^0)u^0}{\sqrt{s}} +\frac{(p+q)\cdot u}{\sqrt{s}}\Big )\Big \}\\&\qquad \times I_0\Big (\frac{\sqrt{{g^2s|(p\times q)\cdot u|^2+4[|p\times q|^2u^0+(p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu }))\cdot u]^2}}}{2T_0g\sqrt{s}|p\times q|}|{\bar{p}}|\Big ), \end{aligned}$$
where \(I_0\) is the first kind modified Bessel function of index zero:
$$\begin{aligned} I_0(y)=\frac{1}{2\pi }\int _0^{2\pi }e^{y\cos \phi }d\phi . \end{aligned}$$
By further changing variables of integration and applying some known integrals as in the Appendix [37], we obtain (8.1) with the following exact form of \(U_2(p,q)\):
$$\begin{aligned} 4T_0^2sU_2^2&=|(p^0+q^0)u^0-(p+q)\cdot u|^2-\frac{s|(p\times q)\cdot u|^2}{|p\times q|^2}\nonumber \\&\quad -\frac{4[|p\times q|^2u^0+(p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu }))\cdot u]^2}{g^2|p\times q|^2}\nonumber \\&=\frac{g^2(p^0+q^0)^2-4|p\times q|^2}{g^2}(u^0)^2+|(p+q)\cdot u|^2-\frac{s|(p\times q)\cdot u|^2}{|p\times q|^2}\nonumber \\&\quad -\frac{2u^0[g^2(p^0+q^0)(p+q)+4(p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu }))]\cdot u}{g^2}\nonumber \\&\quad -\frac{|(p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu }))\cdot u|^2}{g^2|p\times q|^2}. \end{aligned}$$
(8.2)
Now we estimate the terms in the right hand side of the second equality in (8.2). For the first term, we have
$$\begin{aligned} \begin{aligned}&g^2(p^0+q^0)^2-4|p\times q|^2\\&\quad =2[(2p^0q^0+2+|p|^2+|q|^2)(p^0q^0-p\cdot q-1)-4|p|^2|q|^2+4(p\cdot q)^2\\&\quad =4[(|p|^2+|q|^2)-(p^0q^0+1)p\cdot q]\\&\qquad +2(|p|^2+|q|^2)(p^0q^0-p\cdot q-1)+4(p\cdot q)^2\\&\quad =2(|p|^2+|q|^2)(p^0q^0-p\cdot q+1)-4p\cdot q(p^0q^0-p\cdot q+1)\\&\quad =s|p-q|^2. \end{aligned} \end{aligned}$$
(8.3)
It is straightforward to see that the upper bound of the second term is \(s|u|^2\). Noting
$$\begin{aligned}&g^2(p^0+q^0)(p+q)+4(p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu }))\\&\quad =2p[(p^0+q^0)(p^0q^0-p\cdot q-1)-2p^0|q|^2+2q^0p\cdot q)]\\&\qquad +2q[(p^0+q^0)(p^0q^0-p\cdot q-1)-2q^0|p|^2+2p^0p\cdot q)]\\&\quad =2p(p^0-q^0)(p^0q^0-p\cdot q+1)-2q(p^0-q^0)(p^0q^0-p\cdot q+1)\\&\quad =s(p^0-q^0)(p-q), \end{aligned}$$
the fourth term can be bounded by
$$\begin{aligned} \frac{2u^0|u|s|p-q|^2}{g^2}. \end{aligned}$$
For the corresponding fifth term, we rewrite its numerator as
$$\begin{aligned}&|p(p^0+q^0p^{\mu }q_{\mu })+q(q^0+p^0p^{\mu }q_{\mu })|^2\\&\quad =|p|^2[(p^0)^2|q|^4-2p^0q^0|q|^2p\cdot q+(q^0)^2(p\cdot q)^2]\\&\qquad +|q|^2[(q^0)^2|p|^4-2p^0q^0|p|^2p\cdot q+(p^0)^2(p\cdot q)^2]\\&\qquad +2p\cdot q[p^0q^0|p|^2|q|^2-((p^0)^2|q|^2+(q^0)^2|p|^2)p\cdot q+p^0q^0(p\cdot q)^2]\\&\quad =|p|^2|q|^2\left\{ [(p^0)^2|q|^2+(q^0)^2|p|^2+2p^0q^0p\cdot q]-4p^0q^0p\cdot q\right\} \\&\qquad +[|p|^2(q^0)^2+|q|^2(p^0)^2-2(|q|^2(p^0)^2+|p|^2(q^0)^2)](p\cdot q)^2+2p^0q^0(p\cdot q)^3\\&\quad =|p|^2|q|^2|p^0q-q^0p|^2-(p\cdot q)^2[(p^0)^2|q|^2+(q^0)^2|p|^2-2p^0q^0(p\cdot q)]\\&\quad =|p\times q|^2|p^0q-q^0p|^2. \end{aligned}$$
Then it can be bounded by
$$\begin{aligned} \frac{|u|^2|p\times q|^2|p^0q-q^0p|^2}{g^2|p\times q|^2}=\frac{|u|^2|p^0q-q^0p|^2}{g^2}. \end{aligned}$$
On the other hand, we have
$$\begin{aligned}&s|p- q|^2-|p^0q-q^0p|^2\\&\quad =2(p^0q^0-p\cdot q+1)(|p|^2+|q|^2-2p\cdot q)-|q|^2(p^0)^2-|p|^2(q^0)^2+2p^0q^0p\cdot q\\&\quad =4(p\cdot q)^2-2[(p^0)^2+(q^0)^2+p^0q^0]p\cdot q\\&\qquad +2(|p|^2+|q|^2)(p^0q^0+1)-[(p^0)^2|q|^2+(q^0)^2|p|^2]\\&\qquad \ge 4|p|^2|q|^2-(2|p|^2+2|q|^2+2p^0q^0+4)|p||q|\\&\qquad +2(|p|^2+|q|^2)p^0q^0+|p|^2+|q|^2-2|p|^2|q|^2\\&\quad =2(|p|^2+|q|^2-|p||q|)(p^0q^0-|p||q|)+(|p|^2+|q|^2-4|p||q|)\\&\quad \ge 3(|p|-|q|)^2 \end{aligned}$$
since \(p^0q^0-|p||q|\ge 1.\) We combine the above estimates in (8.2) to get
$$\begin{aligned} 4T_0^2U_2^2(p,q)=[1+O(1)|u|]^2\frac{s|p-q|^2}{g^2}. \end{aligned}$$
1.2 Appendix 2: Estimates of kernels
We first prove some important estimates for kernels \(k_1\) and \(k_2\). As a corollary, estimates for corresponding kernels of \({\bar{K}}(h^{\varepsilon })=\frac{w(|p|)\sqrt{\mathbf{M}}}{\sqrt{J_{M}}}K(\frac{\sqrt{J_{M}}}{w\sqrt{\mathbf{M}}}h^{\varepsilon })\) will also be established.
Lemma 8.1
For \(k_2\) given in (8.1) and \(i=1,2,3\), it holds that
$$\begin{aligned} k_2(p,q)\lesssim \frac{1}{p^0|p-q|}\exp \Big \{-\frac{\sqrt{s}|p-q|}{8T_0g}\Big \}, \end{aligned}$$
(8.4)
and
$$\begin{aligned}&\int _{{\mathbb {R}}^3}dq k_2(p,q)\lesssim \frac{1}{p^0},\nonumber \\&\quad \int _{{\mathbb {R}}^3}dq \partial _{x_i} k_2(p,q)\lesssim \frac{1}{p^0},\nonumber \\&\quad \int _{{\mathbb {R}}^3}dq \partial _{p_i} k_2(p,q)\lesssim \frac{1}{p^0}. \end{aligned}$$
(8.5)
It is straightforward to verify that \(k_1(p,q)\) also satisfies the same estimates.
Proof
We first prove (8.4). Note that, by Lemma 3.1 in [16], \(s=g^2+4\le 4p^0q^0+4\) and
$$\begin{aligned} \frac{|p-q|}{\sqrt{p^0q^0}}\le g\le |p-q|. \end{aligned}$$
We use the smallness assumption of |u| in (1.20) to have
$$\begin{aligned} U_2(p,q)\ge&\frac{\sqrt{s}|p-q|}{4T_0g},\qquad U_1(p,q)\lesssim \frac{p^0+q^0}{2[U_2(p,q)]^2}. \end{aligned}$$
Then we can further estimate the kernel \(k_2\) as
$$\begin{aligned} k_2(p,q)&\lesssim \frac{s^{\frac{3}{2}}}{gp^0q^0}(p^0+q^0)\times \frac{g^2}{s|p-q|^2}\exp \{-\frac{\sqrt{s}|p-q|}{4T_0g}\}\\&= \frac{(p^0+q^0)g^2}{p^0q^0|p-q|^3}\frac{\sqrt{s}|p-q|}{g}\exp \{-\frac{\sqrt{s}|p-q|}{4T_0g}\}\\&\lesssim \frac{1+|p-q|+q^0}{p^0q^0|p-q|}\exp \{-\frac{|p-q|}{6T_0}\}\\&\lesssim \frac{1}{p^0|p-q|}\exp \{-\frac{|p-q|}{8T_0}\}. \end{aligned}$$
Now we prove (8.5). Noting that
$$\begin{aligned} |p|^2\le 2|p-q|^2+2|q|^2,\qquad |q|^2\le 2|p-q|^2+2|p|^2, \end{aligned}$$
we get from (8.4) that
$$\begin{aligned} \int _{{\mathbb {R}}^3}dq k_2(p,q)\lesssim&\int _{{\mathbb {R}}^3}dq\frac{1}{p^0|p-q|}\exp \Big \{-\frac{|p-q|}{8T_0}\Big \}\lesssim \frac{1}{p^0}. \end{aligned}$$
For the second inequality in (8.5), again from (8.4), we obtain
$$\begin{aligned}&\int _{{\mathbb {R}}^3}dq \partial _{x_i} k_2(p,q)\\&\quad \lesssim \int _{{\mathbb {R}}^3}dq k_2(p,q) \frac{\sqrt{s}|p-q|}{g} \exp \{-\frac{\sqrt{s}|p-q|}{4T_0g}\}\\&\quad \lesssim \int _{{\mathbb {R}}^3}dq k_2(p,q) \exp \{-\frac{\sqrt{s}|p-q|}{8T_0g}\}\\&\quad \lesssim \frac{1}{p^0}. \end{aligned}$$
Now we prove the third inequality in (8.5). Note that
$$\begin{aligned} \partial _{p_i}U_2(p,q)&=[1+O(1)|u|]\partial _{p_i}\Big (\frac{\sqrt{s}|p-q|}{2T_0g}\Big )\\&=[1+O(1)|u|]\Big (\frac{|p-q|\partial _{p_i}s}{4T_0g\sqrt{s}}+\frac{\sqrt{s}(p_i-q_i)}{2T_0g|p-q|} -\frac{\sqrt{s}|p-q|\partial _{p_i}g}{2T_0g^2}\Big )\\&=[1+O(1)|u|]\Big [\frac{-4|p-q|}{2T_0g^2\sqrt{s}}\frac{q^0}{g}\Big (\frac{p_i}{p^0}-\frac{q_i}{q^0}\Big ) +\frac{\sqrt{s}(p_i-q_i)}{2T_0g|p-q|}\Big ]\\&\lesssim \frac{s^{\frac{3}{2}}|p-q|^3}{g^3} \frac{q^0}{s^2|p-q|\min \{p_0,q_0\}}+\frac{\sqrt{s}|p-q|}{2T_0g}\frac{1}{|p-q|}\\&\lesssim \frac{s^{\frac{3}{2}}|p-q|^3}{g^3} \frac{\min \{p_0,q_0\}+|p-q|}{|p-q|\min \{p_0,q_0\}}+\frac{\sqrt{s}|p-q|}{2T_0g}\frac{1}{|p-q|}, \end{aligned}$$
where we have used the following inequality:
$$\begin{aligned} \frac{p_i}{p^0}-\frac{q_i}{q^0}\le \frac{|p-q|}{\min \{p_0,q_0\}}. \end{aligned}$$
Then we use (8.4) to obtain
$$\begin{aligned}&\int _{{\mathbb {R}}^3}dq \partial _{p_i} k_2(p,q)\\&\quad \lesssim \int _{{\mathbb {R}}^3}dq \Big (\frac{s^{\frac{3}{2}}|p-q|^3}{g^3} \frac{\min \{p_0,q_0\}+|p-q|}{|p-q|\min \{p_0,q_0\}}+\frac{\sqrt{s}|p-q|}{2T_0g}\frac{1}{|p-q|}\Big )k_2(p,q)\\&\quad \lesssim \int _{{\mathbb {R}}^3}dq \frac{1}{p^0}\Big (\frac{s^{\frac{3}{2}}|p-q|^3}{g^3}+1\Big ) \Big (1+\frac{1}{|p-q|^2}\Big )\exp \Big \{-\frac{\sqrt{s}|p-q|}{8T_0g}\Big \}\\&\quad \lesssim \int _{{\mathbb {R}}^3}dq \frac{1}{p^0}\Big (1+\frac{1}{|p-q|^2}\Big )\exp \Big \{-\frac{\sqrt{s}|p-q|}{16T_0g}\Big \}\\&\quad \lesssim \frac{1}{p^0}. \end{aligned}$$
\(\square \)
According to the definition of \(h^{\varepsilon }\) in (1.23), the operator K corresponding to the equation satisfied by \(h^{\varepsilon }\) is \({\bar{K}}(h^{\varepsilon })=\frac{w(|p|)\sqrt{\mathbf{M}}}{\sqrt{J_{M}}}K(\frac{\sqrt{J_{M}}}{w\sqrt{\mathbf{M}}}h^{\varepsilon })\). Correspondingly, we can also define \({\bar{K}}_i, (i=1,2)\) and kernels \({\bar{k}}_i\) as follows:
$$\begin{aligned} {\bar{K}}_i(f)=\int _{{\mathbb {R}}^3}dq {\bar{k}}_i(p,q)f(q)=\int _{{\mathbb {R}}^3}dq \frac{w(|p|)\sqrt{J_{M}(q)}\sqrt{\mathbf{M}(p)}}{w(|q|)\sqrt{J_{M}(p)}\sqrt{\mathbf{M}(q)}}k_i(p,q)f(q),\qquad i=1,2. \end{aligned}$$
Namely, \({\bar{k}}_i(p,q)= \frac{w(|p|)\sqrt{J_{M}(q)}\sqrt{\mathbf{M}(p)}}{w(|q|)\sqrt{J_{M}(p)}\sqrt{\mathbf{M}(q)}}k_i(p,q), i=1, 2\). Then \({\bar{k}}_i(p,q)\) also satisfy the same estimates in Lemma 8.1.
Corollary 8.1
For \(i=1,2,3\), it holds that
$$\begin{aligned} {\bar{k}}_2(p,q)\lesssim \frac{1}{p^0|p-q|}\exp \Big \{-\frac{c_0\sqrt{s}|p-q|}{T_0g}\Big \}, \end{aligned}$$
(8.6)
for some small constant \(c_0\), and
$$\begin{aligned}&\int _{{\mathbb {R}}^3}dq {\bar{k}}_2(p,q)\lesssim \frac{1}{p^0},\nonumber \\&\quad \int _{{\mathbb {R}}^3}dq \partial _{x_i} {\bar{k}}_2(p,q)\lesssim \frac{1}{p^0},\nonumber \\&\quad \int _{{\mathbb {R}}^3}dq \partial _{p_i} {\bar{k}}_2(p,q)\lesssim \frac{1}{p^0}. \end{aligned}$$
(8.7)
\({\bar{k}}_1(p,q)\) also satisfies the same estimates.
Proof
In fact, we only need to note that
$$\begin{aligned} {\bar{k}}_2(p,q)\le&\frac{(1+|p|)^{\beta }}{(1+|q|)^{\beta }}\frac{s^{\frac{3}{2}}}{gp^0q^0}U_1(p,q)\\&\times \exp \Big \{-[1+O(1)|u|]\frac{\sqrt{s}|p-q|}{2T_0g}+\frac{(T_0-T_M)(p^0-q^0)}{2T_MT_0}\Big \}\\ \lesssim&\frac{s^{\frac{3}{2}}}{gp^0q^0}U_1(p,q) (1+|p-q|)^{\beta }\exp \Big \{[(T_0-2T_M)+O(1)|u|]\frac{\sqrt{s}|p-q|}{2T_0T_Mg}\}\\ \lesssim&\frac{s^{\frac{3}{2}}}{gp^0q^0}U_1(p,q) \exp \Big \{-c_0\frac{\sqrt{s}|p-q|}{T_0T_Mg}\} \end{aligned}$$
by (1.24). Then, similar to the proof of Lemma 8.1, we can obtain (8.6) and (8.7). \(\quad \square \)
1.3 Appendix 3: Construction and estimates of coefficients
We will present the existence of coefficients \(F_n, (1\le n\le 2k-1)\), and their momentum and time regularities estimates. For \(i\in [1, 2k-1]\), we decompose \(\frac{F_n}{\sqrt{{\mathbf {M}}}}\) as the sum of macroscopic and microscopic parts:
$$\begin{aligned} \begin{aligned} \frac{F_n}{\sqrt{{\mathbf {M}}}}&=\mathbf{P}\Big (\frac{F_n}{\sqrt{{\mathbf {M}}}}\Big )+{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_n}{\sqrt{{\mathbf {M}}}}\Big )\\&=[a_n(t,x)+b_n(t,x)\cdot p+c_n(t,x) p^0]\sqrt{{\mathbf {M}}}+{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_n}{\sqrt{{\mathbf {M}}}}\Big ). \end{aligned} \end{aligned}$$
To obtain the linear system satisfied by the abstract functions \(a_{n}(t,x), b_{n}(t,x), c_{n}(t,x)\) and \(E_{n}(t,x)\), \(B_{n}(t,x)\), we first derive the explicit expression of the third momentum \(T^{\alpha \beta \gamma }[{\mathbf {M}}]\), \((\alpha ,\beta ,\gamma \in \{0, 1, 2, 3\})\) :
$$\begin{aligned} T^{\alpha \beta \gamma }[{\mathbf {M}}] = \int _{{\mathbb {R}}^3} \frac{ p^\alpha p^\beta p^{\gamma } }{p^0} {\mathbf {M}}\, d p. \end{aligned}$$
We will first get the expression of \(T^{\alpha \beta \gamma }[\mathbf {M}]\) in the rest frame where \((u^0, u^1, u^2, u^3)=(1, 0, 0, 0)\). For convenience, we denote \(T^{\alpha \beta \gamma }[{\mathbf {M}}]\) as \({\bar{T}}^{\alpha \beta \gamma }\) in the rest frame, p as \({\bar{p}}\) in the rest frame, and \(-v\) as the velocity of general reference frame relative to the rest frame. Then, the corresponding boost matrix \({\bar{\Lambda }}\) is
$$\begin{aligned} {\bar{\Lambda }}=({\bar{\Lambda }}^{\mu }_{\nu })=\left( \begin{array}{cccc} \displaystyle {\tilde{r}} &{} {\tilde{r}}v_1 &{} {\tilde{r}}v_2 &{} {\tilde{r}}v_3\\ \displaystyle {\tilde{r}}v_1 &{} 1+({\tilde{r}}-1)\frac{v_1^2}{|v|^2} &{} ({\tilde{r}}-1)\frac{v_1v_2}{|v|^2} &{} ({\tilde{r}}-1)\frac{v_1v_3}{|v|^2}\\ \displaystyle {\tilde{r}}v_2 &{} ({\tilde{r}}-1)\frac{v_1v_2}{|v|^2} &{} 1+({\tilde{r}}-1)\frac{v_2^2}{|v|^2} &{} ({\tilde{r}}-1)\frac{v_2v_3}{|v|^2}\\ \displaystyle {\tilde{r}}v_3 &{} ({\tilde{r}}-1)\frac{v_1v_3}{|v|^2} &{} ({\tilde{r}}-1)\frac{v_2v_3}{|v|^2} &{}1+({\tilde{r}}-1)\frac{v_3^2}{|v|^2} \end{array} \right) , \end{aligned}$$
where \({\tilde{r}}=u^0, v_i=\frac{u_i}{u^0}\). Noting
$$\begin{aligned} p^{\mu }= {\bar{\Lambda }}^{\mu }_{\nu }{\bar{p}}^{\nu }, \qquad \frac{dp}{p^0}=\frac{d{\bar{p}}}{{\bar{p}}^0}, \end{aligned}$$
we have
$$\begin{aligned} T^{\alpha \beta \gamma }[{\mathbf {M}}]={\bar{\Lambda }}^{\alpha }_{\alpha '}{\bar{\Lambda }}^{\beta }_{\beta '}{\bar{\Lambda }}^{\gamma }_{\gamma '}{\bar{T}}^{\alpha ' \beta '\gamma '}. \end{aligned}$$
(8.8)
Then, we can obtain the expression of \(T^{\alpha \beta \gamma }[{\mathbf {M}}]\) by the expression of \({\bar{T}}^{\alpha \beta \gamma }\) and (8.8). Now we give the expression of \({\bar{T}}^{\alpha \beta \gamma }\) as follows:
Lemma 8.2
Let \(i, j, k\in \{1, 2, 3\}\). For the third momentum \({\bar{T}}^{\alpha \beta \gamma }\) which corresponds to \(T^{\alpha \beta \gamma }[{\mathbf {M}}]\) in the rest frame, we have
$$\begin{aligned}&{\bar{T}}^{000}= \frac{n_0[3K_3(\gamma )+\gamma K_2(\gamma )]}{\gamma K_2(\gamma )}, \end{aligned}$$
(8.9)
$$\begin{aligned}&{\bar{T}}^{0ii}={\bar{T}}^{ii0}={\bar{T}}^{i0i}= \frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}, \end{aligned}$$
(8.10)
$$\begin{aligned}&{\bar{T}}^{\alpha \beta \gamma }=0, \qquad \text{ if } ~(\alpha ,\beta ,\gamma )\ne (0,0,0), (0,i,i), (i,i,0), (i,0,i). \end{aligned}$$
(8.11)
Proof
It is straightforward to verify (8.11) by the symmetry of the variable of integration. Now we prove (8.9). It holds that
$$\begin{aligned}&{\bar{T}}^{000}= \int _{{\mathbb {R}}^3} (p^0)^2 \frac{n_0 \gamma }{4 \pi K_2(\gamma )} \exp \{- \gamma {\bar{p}}_{0} \} \, d p. \end{aligned}$$
As the proof of Proposition 3.3 in [33], we let \(y=\gamma {\bar{p}}_{0}\) to get \({\bar{p}}^0=\frac{y}{\gamma }\), \(|{\bar{p}}|=\frac{\sqrt{y^2-\gamma ^2}}{\gamma }\), and \(d|{\bar{p}}|=\frac{1 }{\gamma }\frac{y dy}{ \sqrt{y^2-\gamma ^2}}\). Then we have
$$\begin{aligned} {\bar{T}}^{000}&= \int _{\gamma }^{\infty } \frac{1}{\gamma ^2}y^2 \frac{n_0 \gamma }{ K_2(\gamma )} e^{- y} \frac{1}{\gamma ^2}(y^2-\gamma ^2)\frac{1 }{\gamma }\frac{y dy}{ \sqrt{y^2-\gamma ^2}}\\&= \frac{n_0}{\gamma ^4K_2(\gamma )}\int _{\gamma }^{\infty }[(y^2-\gamma ^2)^{\frac{3}{2}}+\gamma ^2\sqrt{y^2-\gamma ^2}]ye^{- y} dy\\&=\frac{n_0[3K_3(\gamma )+\gamma K_2(\gamma )]}{\gamma K_2(\gamma )}. \end{aligned}$$
Similarly, (8.10) can be proved as follows:
$$\begin{aligned} {\bar{T}}^{0ii}&=c \int _{{\mathbb {R}}^3} \frac{|p|^2}{3} \frac{n_0 \gamma }{4 \pi K_2(\gamma )} \exp \{- \gamma {\bar{p}}_{0} \} \, d p\\&= \int _{\gamma }^{\infty } \frac{1}{3\gamma ^2}(y^2-\gamma ^2) \frac{n_0 \gamma }{ K_2(\gamma )} e^{- y} \frac{1}{\gamma ^2}(y^2-\gamma ^2)\frac{1 }{\gamma }\frac{y dy}{ \sqrt{y^2-\gamma ^2}}\\&= \frac{n_0}{3\gamma ^4K_2(\gamma )}\int _{\gamma }^{\infty }(y^2-\gamma ^2)^{\frac{3}{2}}ye^{- y} dy\\&= \frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}. \end{aligned}$$
\(\square \)
Now we can use Lemma 8.2 and (8.8) to derive the explicit expression of \(T^{\alpha \beta \gamma }[{\mathbf {M}}]\).
Lemma 8.3
For \(i, j, k\in \{1, 2, 3\}\), we have
$$\begin{aligned} \begin{aligned}&T^{000}[{\mathbf {M}}]= \frac{n_0}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^3+3K_3(\gamma )u^0|u|^2],\\&T^{00i}[{\mathbf {M}}]= \frac{n_0}{\gamma K_2(\gamma )}\left[ (5K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2u_i+K_3(\gamma )|u|^2u_i\right] ,\\&T^{0ij}[{\mathbf {M}}]=\frac{n_0}{\gamma K_2(\gamma )}\left[ (6K_3(\gamma )+\gamma K_2(\gamma ))u^0u_iu_j+\delta _{ij} K_3(\gamma )u^0\right] ,\\&T^{ijk}[{\mathbf {M}}]=\frac{n_0}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u_iu_ju_k \\&\qquad +\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}(u_i\delta _{jk}+u_j\delta _{ik}+u_k\delta _{ij}). \end{aligned} \end{aligned}$$
(8.12)
Proof
We first prove \(T^{000}[{\mathbf {M}}]\) in (8.12). By Lemma 8.2 and (8.8), we have
$$\begin{aligned} T^{000}[{\mathbf {M}}]&= {\bar{\Lambda }}^0_{\alpha }{\bar{\Lambda }}^0_{\beta }{\bar{\Lambda }}^0_{\gamma }{\bar{T}}^{\alpha \beta \gamma }\\&=\sum _{\alpha =0}{\tilde{r}}\Big ({\tilde{r}}^2\frac{n_0[3K_3(\gamma )+\gamma K_2(\gamma )]}{\gamma K_2(\gamma )}+{\tilde{r}}^2|v|^2\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\Big )\\&\quad +\sum _{\alpha =1}^3{\tilde{r}}v_{\alpha }\times 2{\tilde{r}}^2v_{\alpha }\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\\&= \frac{n_0}{\gamma K_2(\gamma )}\Big ([3K_3(\gamma )+\gamma K_2(\gamma )](u^0)^3+K_3(\gamma )u^0|u|^2\Big )+\frac{2n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0|u|^2\\&=\frac{n_0}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^3+3K_3(\gamma )u^0|u|^2]. \end{aligned}$$
For \(T^{00i}[{\mathbf {M}}]\), we have
$$\begin{aligned} T^{00i}[{\mathbf {M}}]&= {\bar{\Lambda }}^0_{\alpha }{\bar{\Lambda }}^0_{\beta }{\bar{\Lambda }}^i_{\gamma }{\bar{T}}^{\alpha \beta \gamma }\\&=\sum _{\alpha =0}{\tilde{r}}\Big ({\tilde{r}}^2v_{i}\frac{n_0[3K_3(\gamma )+\gamma K_2(\gamma )]}{\gamma K_2(\gamma )}\\&\quad +\sum _{j=1}^3{\tilde{r}}v_{j}\Big (\frac{{\tilde{r}}-1}{|v|^2}v_iv_j+\delta _{ij}\Big )\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\Big )\\&\quad +\sum _{\alpha =1}^3{\tilde{r}}v_{\alpha }\Big [{\tilde{r}}^2v_{\alpha }v_i +{\tilde{r}}\Big (\frac{{\tilde{r}}-1}{|v|^2}v_iv_{\alpha }+\delta _{i\alpha }\Big ) \Big ]\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\\&= \frac{n_0}{\gamma K_2(\gamma )}[4K_3(\gamma )+\gamma K_2(\gamma )](u^0)^2u_i+\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_i[|u|^2+(u^0)^2]\\&=\frac{n_0}{\gamma K_2(\gamma )}\left[ (5K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2u_i+K_3(\gamma )|u|^2u_i\right] . \end{aligned}$$
Similarly, we obtain
$$\begin{aligned} T^{0ij}[{\mathbf {M}}]&= {\bar{\Lambda }}^0_{\alpha }{\bar{\Lambda }}^i_{\beta }{\bar{\Lambda }}^j_{\gamma }{\bar{T}}^{\alpha \beta \gamma }\\ =&\sum _{\alpha =0}{\tilde{r}}\Big ({\tilde{r}}^2v_{i}v_{j}\frac{n_0[3K_3(\gamma )+\gamma K_2(\gamma )]}{\gamma K_2(\gamma )}\\&+\sum _{k=1}^3\Big (\frac{{\tilde{r}}-1}{|v|^2}v_iv_k+\delta _{ik}\Big )\Big (\frac{{\tilde{r}}-1}{|v|^2}v_jv_k+\delta _{jk}\Big )\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\Big )\\&+\sum _{\alpha =1}^3{\tilde{r}}v_{\alpha }\Big [{\tilde{r}}\frac{v_{i}}{c}\Big (\frac{{\tilde{r}}-1}{|v|^2}v_jv_{\alpha }+\delta _{j\alpha }\Big ) +{\tilde{r}}v_{j}\Big (\frac{{\tilde{r}}-1}{|v|^2}v_iv_{\alpha }+\delta _{i\alpha }\Big )\Big ]\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\\&= \frac{n_0}{\gamma K_2(\gamma )}\{[4K_3(\gamma )+\gamma K_2(\gamma )]u^0u_iu_j+\delta _{ij}K_3(\gamma )u^0\}+2\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0u_iu_j\\&=\frac{n_0}{\gamma K_2(\gamma )}\left[ (6K_3(\gamma )+\gamma K_2(\gamma ))u^0u_iu_j+\delta _{ij}K_3(\gamma )u^0\right] , \end{aligned}$$
and
$$\begin{aligned} T^{ijk}[{\mathbf {M}}]&= {\bar{\Lambda }}^i_{\alpha }{\bar{\Lambda }}^j_{\beta }{\bar{\Lambda }}^k_{\gamma }{\bar{T}}^{\alpha \beta \gamma }\\&=\sum _{\alpha =0}{\tilde{r}}v_{i}\Big ({\tilde{r}}^2v_{j}v_{k}\frac{n_0[3K_3(\gamma )+\gamma K_2(\gamma )]}{\gamma K_2(\gamma )}\\&\quad +\sum _{l=1}^3\Big (\frac{{\tilde{r}}-1}{|v|^2}v_jv_l+\delta _{jl}\Big )\Big (\frac{{\tilde{r}}-1}{|v|^2}v_kv_l+\delta _{kl}\Big ) \frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\Big )\\&\quad +\sum _{\alpha =1}^3\Big (\frac{{\tilde{r}}-1}{|v|^2}v_iv_{\alpha }+\delta _{i\alpha }\Big ) \Big [{\tilde{r}}v_{j}\Big (\frac{{\tilde{r}}-1}{|v|^2}v_kv_{\alpha }+\delta _{k\alpha }\Big )\\&\quad +{\tilde{r}}v_{k}\Big (\frac{{\tilde{r}}-1}{|v|^2}v_jv_{\alpha }+\delta _{j\alpha }\Big )\Big ]\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}\\&= \frac{n_0}{\gamma K_2(\gamma )}\{[3K_3(\gamma )+\gamma K_2(\gamma )]u_iu_ju_k+u_i(u_ju_k+\delta _{jk})K_3(\gamma )\}\\&\quad +\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}[u_j(u_iu_k+\delta _{ik})+u_k(u_iu_j+\delta _{ij})]\\&=\frac{n_0}{\gamma K_2(\gamma )}[6K_3(\gamma )+\gamma K_2(\gamma )]u_iu_ju_k \\&\quad +\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}(u_i\delta _{jk}+u_j\delta _{ik}+u_k\delta _{ij}). \end{aligned}$$
\(\square \)
With the preparation, now we construct the coefficients \((F_n, E_n, B_n), 1\le n\le 2k-1\) in a conductive way, and estimate their regularities.
Theorem 8.1
For any \(n\in [0,2k-2]\), assume that \((F_i, E_i, B_i)\) have been constructed for all \(0\le i\le n\). Then the microscopic part \({\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\) can be written as:
$$\begin{aligned} \begin{aligned} {\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )&=L^{-1}\Big [-\frac{1}{\sqrt{{\mathbf {M}}}}\Big (\partial _tF_n +{\hat{p}}\cdot \nabla _xF_n-\sum _{\begin{array}{c} i+j=n+1\\ i,j\ge 1 \end{array}}Q(F_i,F_j)\\&-\sum _{\begin{array}{c} i+j=n\\ i,j\ge 0 \end{array}}\Big (E_i+{\hat{p}} \times B_i \Big )\cdot \nabla _pF_j\Big )\Big ]. \end{aligned} \end{aligned}$$
And \(a_{n+1}(t,x), b_{n+1}(t,x), c_{n+1}(t,x)\), \(E_{n+1}(t,x), B_{n+1}(t,x)\) satisfy the following system:
$$\begin{aligned}&\partial _t\Big (n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})+[e_0(u^0)^2+P_0|u|^2]c_{n+1}\Big )\nonumber \\&\qquad +\nabla _x\cdot \Big (n_0u a_{n+1}+(e_0+P_0)u (u\cdot b_{n+1})+P_0 b_{n+1}+(e_0+P_0)u^0 u c_{n+1}\Big )\nonumber \\&\qquad +\nabla _x\cdot \int _{{\mathbb {R}}^3} \frac{p}{p^0}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp=0, \end{aligned}$$
(8.13)
$$\begin{aligned}&\partial _t\Big ((e_0+P_0)u^0u_j a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma ))u^0 u_j (u\cdot b_{n+1})\nonumber \\&\qquad +K_3(\gamma )u^0 b_{n+1,j}]+\frac{n_0}{\gamma K_2(\gamma )}[(5K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +K_3(\gamma )|u|^2]u_jc_{n+1}\Big )\nonumber \\&\qquad +\nabla _x\cdot \Big ((e_0+P_0)u_j u a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u_j u [(u\cdot b_{n+1})+u^0c_{n+1}]\Big )\nonumber \\&\qquad +\partial _{x_j}(P_0a_{n+1})+\nabla _x\cdot \Big [\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}(ub_{n+1,j}+u_jb_{n+1})\Big ]\nonumber \\&\qquad + \partial _{x_j}\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}[(u\cdot b_{n+1})+u^0c_{n+1}]\Big )\nonumber \\&\qquad +E_{0,j}\Big (n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})+[e_0(u^0)^2+P_0|u|^2]c_{n+1}\Big )\nonumber \\&\qquad +\Big [\Big (n_0u a_{n+1}+(e_0+P_0)u (u\cdot b_{n+1})+P_0 b_{n+1}+(e_0+P_0)u^0 u c_{n+1}\Big )\times B_0\Big ]_j\nonumber \\&\qquad +n_0u^0E_{n+1,j}+\Big (n_0u\times B_{n+1}\Big )_j\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}E_{k,j}\Big (n_0u^0a_{l}+(e_0+P_0)u^0(u\cdot b_{l})+P_0b_{l}+[e_0(u^0)^2+P_0|u|^2]c_{l}\Big )\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [\Big (n_0u a_{l}+(e_0+P_0)u (u\cdot b_{l})+(e_0+P_0)u^0 u c_{l}\Big )\times B_k\Big ]_j\nonumber \\&\qquad +\nabla _x\cdot \int _{{\mathbb {R}}^3} \frac{p_jp}{p^0}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp+\Big [\int _{{\mathbb {R}}^3} {\hat{p}}\times B_0\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\Big ]_j\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [\int _{{\mathbb {R}}^3} {\hat{p}}\times B_k\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{l}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\Big ]_j=0, \end{aligned}$$
(8.14)
for \(j=1, 2, 3\) with \(b_{n+1}=(b_{n+1,1}, b_{n+1,2}, b_{n+1,3})\), \(E_{n+1}=(E_{n+1,1}, E_{n+1,2}, E_{n+1,3})\),
$$\begin{aligned}&\partial _t\Big ([e_0(u^0)^2+P_0|u|^2] a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}[(5K_3(\gamma )+\gamma K_2(\gamma )) (u^0)^2+K_3(\gamma )|u|^2]\nonumber \\&\qquad \times (u\cdot b_{n+1})+\frac{n_0}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +3K_3(\gamma )|u|^2]u^0c_{n+1}\Big )\nonumber \\&\qquad +\nabla _x\cdot \Big ((e_0+P_0)u^0 u a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u^0 u(u\cdot b_{n+1})\nonumber \\&\qquad +\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0b_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}[(5K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2\nonumber \\&\qquad +K_3(\gamma )|u|^2]uc_{n+1} \Big )+n_0u\cdot E_{n+1}+ n_0u\cdot E_0 a_{n+1}\nonumber \\&\qquad +(e_0+P_0)(u\cdot b_{n+1})(u\cdot E_0)+P_0E_0\cdot b_{n+1}\nonumber \\&\qquad +(e_0+P_0)u^0(u\cdot E_0)c_{n+1}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_0\nonumber \\&\qquad + \sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [n_0u\cdot E_k a_{l}+(e_0+P_0)(u\cdot b_{l})(u\cdot E_k)+P_0E_k\cdot b_{l}\nonumber \\&\qquad +(e_0+P_0)u^0(u\cdot E_k)c_{l}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{l}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_k\Big ]=0, \end{aligned}$$
(8.15)
$$\begin{aligned}&\partial _tE_{n+1}-\nabla _x \times B_{n+1} \nonumber \\&\quad =n_0u a_{n+1}+(e_0+P_0)u (u\cdot b_{n+1})\nonumber \\&\qquad +(e_0+P_0)u^0 u c_{n+1}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp, \nonumber \\&\qquad \partial _t B_{n+1}+ \nabla _x \times E_{n+1}=0,\nonumber \\&\qquad \nabla _x\cdot E_{n+1}=-\Big (n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})\nonumber \\&\qquad +[e_0(u^0)^2+P_0|u|^2]c_{n+1}\Big ), \nonumber \\&\nabla _x\cdot B_{n+1}=0. \end{aligned}$$
(8.16)
Furthermore, assume \(a_{n+1}(0,x), b_{n+1}(0,x), c_{n+1}(0,x), E_{n+1}(0,x)\), \(B_{n+1}(0,x)\in H^N, N\ge 0\) be given initial data to the system consisted of equations (8.13), (8.14), (8.15) and (8.16). Then the linear system is well-posed in \(C^0([0,\infty );H^N)\). Moreover, it holds that
$$\begin{aligned}&|F_{n+1}|\lesssim (1+t)^{n}{\mathbf {M}}^{1_-},\qquad |\nabla _pF_{n+1}|\lesssim (1+t)^{n}{\mathbf {M}}^{1_-},\nonumber \\&\quad |\nabla _xF_{n+1}|\lesssim (1+t)^{n}{\mathbf {M}}^{1_-},\qquad |\nabla _p^2F_{n+1}|\lesssim (1+t)^{n}{\mathbf {M}}^{1_-},\nonumber \\&\quad |\nabla _x\nabla _pF_{n+1}|\lesssim (1+t)^{n}{\mathbf {M}}^{1_-},\nonumber \\&\quad |E_{n+1}|+|B_{n+1}|+|\nabla _xE_{n+1}|+|\nabla _xE_{n+1}|\lesssim (1+t)^{n}. \end{aligned}$$
(8.17)
Proof
From the equation of \(F_n\) in (1.10), the microscopic part \({\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\) can be written as
$$\begin{aligned} \begin{aligned} {\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )&=L^{-1}\Big [-\frac{1}{\sqrt{{\mathbf {M}}}}\Big (\partial _tF_n+{\hat{p}}\cdot \nabla _xF_n-\sum _{\begin{array}{c} i+j=n+1\\ i,j\ge 1 \end{array}}Q(F_i,F_j)\\&\qquad -\sum _{\begin{array}{c} i+j=n\\ i,j\ge 0 \end{array}}\Big (E_i+{\hat{p}} \times B_i \Big )\cdot \nabla _pF_j\Big )\Big ]. \end{aligned} \end{aligned}$$
We first prove the equation (8.13). Note that, from (1.14),
$$\begin{aligned} \int _{{\mathbb {R}}^3} F_{n+1} dp&=\int _{{\mathbb {R}}^3} [a_{n+1}+b_{n+1}\cdot p+c_{n+1} p_0]{\mathbf {M}} dp\nonumber \\&=n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})+[e_0(u^0)^2+P_0|u|^2]c_{n+1},\nonumber \\ \int _{{\mathbb {R}}^3} {\hat{p}}_jF_{n+1}\,dp&=n_0u_j a_{n+1}+(e_0+P_0)u_j (u\cdot b_{n+1})+P_0b_{n+1,j}\nonumber \\&+(e_0+P_0)u^0 u_j c_{n+1}+\int _{{\mathbb {R}}^3} {\hat{p}}_j\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp, \end{aligned}$$
(8.18)
for \(j=1, 2, 3,\) and
$$\begin{aligned} \int _{{\mathbb {R}}^3} \Big (E_{n+1}+{\hat{p}} \times B_{n+1} \Big )\cdot \nabla _pF_0dp=\int _{{\mathbb {R}}^3}\Big (E_0+{\hat{p}} \times B_0 \Big )\cdot \nabla _pF_{n+1} dp=0. \end{aligned}$$
Then, we integrate the equation of \(F_{n+1}\) in (1.10) w.r.t. p to get (8.13).
Now we prove (8.14). From (1.14) and Lemma 8.3, it holds that
$$\begin{aligned}&\int _{{\mathbb {R}}^3} p_j F_{n+1} dp\\&\quad =\int _{{\mathbb {R}}^3} p_j [a_{n+1}+b_{n+1}\cdot p+c_{n+1} p_0]{\mathbf {M}} dp\\&\quad =(e_0+P_0)u^0u_j a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma ))u^0 u_j (u\cdot b_{n+1})\\&\qquad +K_3(\gamma )u^0 b_{n+1,j}]+\frac{n_0}{\gamma K_2(\gamma )}[(5K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +K_3(\gamma )|u|^2]u_jc_{n+1},\\&\int _{{\mathbb {R}}^3} \frac{p_jp}{p^0} F_{n+1} dp\\&\quad =\int _{{\mathbb {R}}^3} \frac{p_jp}{p^0} [a_{n+1}+b_{n+1}\cdot p+c_{n+1} p_0]{\mathbf {M}} dp+\int _{{\mathbb {R}}^3} \frac{p_jp}{p^0}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\\&\quad =(e_0+P_0)u_j u a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u_j u [(u\cdot b_{n+1})+u^0c_{n+1}]\nonumber \\&\qquad +e_j P_0a_{n+1}+\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}(ub_{n+1,j}+u_jb_{n+1})\nonumber \\&\qquad + e_j\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}[(u\cdot b_{n+1})+u^0c_{n+1}]+\int _{{\mathbb {R}}^3} \frac{p_jp}{p^0}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp, \end{aligned}$$
where \(e_j, (j=1,2,3),\) are the unit base vectors in \({\mathbb {R}}^3\), and
$$\begin{aligned}&-\int _{{\mathbb {R}}^3} p_j\Big (E_{n+1}+{\hat{p}} \times B_{n+1} \Big )\cdot \nabla _pF_0dp\\&\quad =\int _{{\mathbb {R}}^3}E_{n+1,j}F_{0}dp+\int _{{\mathbb {R}}^3}\Big ({\hat{p}} \times B_{n+1} \Big )_jF_{0} dp \\&\quad =n_0u^0E_{n+1,j}+\Big (n_0u\times B_{n+1}\Big )_j,\\&-\int _{{\mathbb {R}}^3}p_j\Big (E_0+{\hat{p}} \times B_0 \Big )\cdot \nabla _pF_{n+1} dp\\&\quad =\int _{{\mathbb {R}}^3}E_{0,j}F_{n+1}dp+\int _{{\mathbb {R}}^3}\Big ({\hat{p}} \times B_0 \Big )_jF_{n+1} dp\\&\quad =E_{0,j}\Big (n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})+[e_0(u^0)^2+P_0|u|^2]c_{n+1}\Big )\nonumber \\&\qquad +\Big [\Big (n_0u a_{n+1}+(e_0+P_0)u (u\cdot b_{n+1})+P_0b_{n+1}+(e_0+P_0)u^0 u c_{n+1}\Big )\times B_0\Big ]_j\nonumber \\&\qquad +\int _{{\mathbb {R}}^3}\Big ({\hat{p}} \times B_0 \Big )_j\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big ) dp . \end{aligned}$$
Then, we multiply the equation of \(F_{n+1}\) in (1.10) by \(p_j\) and integrate the resulting equation w.r.t. p to get (8.14).
Next, we show that (8.15) holds. By (1.14) and Lemma 8.3, one has
$$\begin{aligned}&\int _{{\mathbb {R}}^3} p_0 F_{n+1} dp\\&\quad =\int _{{\mathbb {R}}^3} p_0 [a_{n+1}+b_{n+1}\cdot p+c_{n+1} p_0]{\mathbf {M}} dp\\&\quad =[e_0(u^0)^2+P_0|u|^2] a_{n+1}+\frac{n_0}{\gamma K_2(\gamma )}[(5K_3(\gamma )+\gamma K_2(\gamma )) (u^0)^2+K_3(\gamma )|u|^2]\\&\qquad \times (u\cdot b_{n+1})+\frac{n_0}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +3K_3(\gamma )|u|^2]u^0c_{n+1}, \end{aligned}$$
and
$$\begin{aligned}&-\int _{{\mathbb {R}}^3} p^0\Big (E_{n+1}+{\hat{p}} \times B_{n+1} \Big )\cdot \nabla _pF_0dp\\&\quad =\int _{{\mathbb {R}}^3}E_{n+1}\cdot {\hat{p}}F_{0}dp=n_0u\cdot E_{n+1},\\&-\int _{{\mathbb {R}}^3} p^0\Big (E_{0}+{\hat{p}} \times B_0 \Big )\cdot \nabla _pF_{n+1}dp\\&\quad =\int _{{\mathbb {R}}^3} E_{0}\cdot {\hat{p}}F_{n+1}dp\\&\quad = n_0u\cdot E_0 a_{n+1}+(e_0+P_0)(u\cdot b_{n+1})(u\cdot E_0)+P_0E_0\cdot b_{n+1}\\&\qquad +(e_0+P_0)u^0(u\cdot E_0)c_{n+1}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_0. \end{aligned}$$
We integrate the equation of \(F_n\) in (1.10) with \(p^0\) over \({\mathbb {R}}^3_p\) to obtain (8.15). Finally, it is straightforward to obtain the Maxwell system (8.16) of \(E_{n+1}, B_{n+1}\) from (8.18).
Now we prove the well-posedness of the system (8.13), (8.14), (8.15) and (8.16). By conditions (1.16) and the equation \( \partial _t(n_0 u^0) + \nabla _x\cdot (n_0 u) =0\) in (1.17), we simplify equations (8.13), (8.14), (8.15) as follows:
$$\begin{aligned}&n_0u^0\partial _t\Big (a_{n+1}+h(u\cdot b_{n+1})+hu^0c_{n+1}\Big )-\partial _t(P_0c_{n+1})\nonumber \\&\qquad +n_0u\cdot \nabla _x\Big ( a_{n+1}+h (u\cdot b_{n+1})+hu^0 c_{n+1}\Big )+\nabla _x\cdot (P_0b_{n+1})\nonumber \\&\qquad +\nabla _x\cdot \int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp=0, \end{aligned}$$
(8.19)
$$\begin{aligned}&n_0u^0\partial _t\Big (hu_j a_{n+1}+\frac{1}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma )) u_j (u\cdot b_{n+1}+u^0 c_{n+1})\nonumber \\&\qquad +K_3(\gamma ) b_{n+1,j}]\Big )-\partial _t\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_jc_{n+1}\Big )\nonumber \\&\qquad +n_0u\cdot \nabla _x\Big (hu_j a_{n+1}+\frac{1}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma ))u_j[(u\cdot b_{n+1})+u^0c_{n+1}]\nonumber \\&\qquad +K_3(\gamma ) b_{n+1,j}]\Big )+\partial _{x_j}(P_0a_{n+1})+\nabla _x\cdot \Big [\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_jb_{n+1}\Big ]\nonumber \\&\qquad + \partial _{x_j}\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}[(u\cdot b_{n+1})+u^0c_{n+1}]\Big )\nonumber \\&\qquad +E_{0,j}\Big (n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})+[e_0(u^0)^2+P_0|u|^2]c_{n+1}\Big )\nonumber \\&\qquad +\Big [\Big (n_0u a_{n+1}+(e_0+P_0)u (u\cdot b_{n+1})+P_0b_{n+1}+(e_0+P_0)u^0 u c_{n+1}\Big )\times B_0\Big ]_j\nonumber \\&\qquad +n_0u^0E_{n+1,j}+\Big (n_0u^0\times B_{n+1}\Big )_j\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}E_{k,j}\Big (n_0u^0a_{l}+(e_0+P_0)u^0(u\cdot b_{l})+[e_0(u^0)^2+P_0|u|^2]c_{l}\Big )\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [\Big (n_0u a_{l}+(e_0+P_0)u (u\cdot b_{l})+P_0b_l+(e_0+P_0)u^0 u c_{l}\Big )\times B_k\Big ]_j\nonumber \\&\qquad +\nabla _x\cdot \int _{{\mathbb {R}}^3} \frac{p_jp}{p^0}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp+\Big [\int _{{\mathbb {R}}^3} {\hat{p}}\times B_0\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\Big ]_j\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [\int _{{\mathbb {R}}^3} {\hat{p}}\times B_k\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{l}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\Big ]_j=0, \end{aligned}$$
(8.20)
and
$$\begin{aligned}&n_0u^0\partial _t\Big (hu^0 a_{n+1}+\frac{1}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma )) u^0](u\cdot b_{n+1})\nonumber \\&\qquad +\frac{1}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +3K_3(\gamma )|u|^2]c_{n+1}\Big )\nonumber \\&\qquad -\partial _t(P_0a_{n+1})-\partial _t\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}(u\cdot b_{n+1})\Big )\nonumber \\&\qquad +n_0u\cdot \nabla _x\Big (hu^0 a_{n+1}+\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u^0(u\cdot b_{n+1})\nonumber \\&\qquad +\frac{1}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2+3K_3(\gamma )|u|^2]c_{n+1} \Big )\nonumber \\&\qquad +\nabla _x\cdot \Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0b_{n+1}\Big )+\nabla _x\cdot \Big (\frac{2n_0K_3(\gamma )}{\gamma K_2(\gamma )}uc_{n+1}\Big )\nonumber \\&\qquad +n_0u\cdot E_{n+1}+ n_0u\cdot E_0 a_{n+1}+(e_0+P_0)(u\cdot b_{n+1})(u\cdot E_0)\nonumber \\&\qquad +P_0E_0\cdot b_{n+1}+(e_0+P_0)u^0(u\cdot E_0)c_{n+1}\nonumber \\&\qquad +\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_0\nonumber \\&\qquad + \sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [n_0u\cdot E_k a_{l}+(e_0+P_0)(u\cdot b_{l})(u\cdot E_k)+P_0E_k\cdot b_{l}\nonumber \\&\qquad +(e_0+P_0)u^0(u\cdot E_k)c_{l}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{l}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_k\Big ]=0. \end{aligned}$$
(8.21)
Equations (8.19), (8.20), (8.21) can be further written as:
$$\begin{aligned}&n_0u^0\Big (\partial _ta_{n+1}+h(u\cdot \partial _t b_{n+1})+hu^0\partial _tc_{n+1}\Big )-P_0\partial _tc_{n+1}\nonumber \\&\qquad +n_0u^0\Big (\partial _t\Big (hu\Big )\cdot b_{n+1})+\partial _t(hu^0)c_{n+1}\Big )-(\partial _tP_0)c_{n+1}\nonumber \\&\qquad +n_0u\cdot \Big ( \nabla _xa_{n+1}+h \nabla _xb_{n+1}\cdot u+hu^0 \nabla _x c_{n+1}\Big )+P_0\nabla _x\cdot b_{n+1}\nonumber \\&\qquad +n_0u\cdot \Big ( \nabla _x\Big (h u\Big )\cdot b_{n+1})+\nabla _x\Big (h u^0\Big ) c_{n+1}\Big ) +b_{n+1}\cdot \nabla _xP_0\nonumber \\&\qquad +\nabla _x\cdot \int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp=0, \end{aligned}$$
(8.22)
$$\begin{aligned}&n_0u^0\Big (hu_j \partial _ta_{n+1}+\frac{1}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma )) u_j (u\cdot \partial _tb_{n+1}+u^0 \partial _tc_{n+1})\nonumber \\&\qquad +K_3(\gamma ) \partial _tb_{n+1,j}]\Big )-\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_j\partial _tc_{n+1}-\partial _t\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_j\Big )c_{n+1}\nonumber \\&\qquad +n_0u^0\Big [\partial _t\Big (hu_j\Big ) a_{n+1}+\partial _t\Big (\frac{m^2(6K_3(\gamma )+\gamma K_2(\gamma ))}{\gamma K_2(\gamma )} u_j u\Big )\cdot b_{n+1}\nonumber \\&\qquad +\partial _t\Big (\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma )) u_ju^0\Big ) c_{n+1}+\partial _t\Big (\frac{K_3(\gamma )}{\gamma K_2(\gamma )}\Big )b_{n+1,j}\Big ]\nonumber \\&\qquad +n_0u\cdot \Big (hu_j \nabla _xa_{n+1}+\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u_j[( \nabla _xb_{n+1}\cdot u)+u^0\nabla _xc_{n+1}]\nonumber \\&\qquad +K_3(\gamma ) \nabla _xb_{n+1,j}]\Big )+P_0\partial _{x_j}a_{n+1}+\partial _{x_j}P_0a_{n+1}\nonumber \\&\qquad +n_0u\cdot \Big [\nabla _x\Big (hu_j\Big ) a_{n+1}+\nabla _x\Big (\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u_ju\Big )\cdot b_{n+1}\nonumber \\&\qquad +\nabla _x\Big (\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u_ju^0\Big )c_{n+1}\nonumber \\&\qquad +\nabla _x\Big (\frac{K_3(\gamma )}{\gamma K_2(\gamma )}\Big ) b_{n+1,j}\Big ]+\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_j\nabla _x\cdot b_{n+1}\nonumber \\&\qquad + \frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}[(u\cdot \partial _{x_j}b_{n+1})+u^0\partial _{x_j}c_{n+1}]\nonumber \\&\qquad +\nabla _x\cdot \Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u\Big )b_{n+1,j}+\nabla _x\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u_j\Big )\cdot b_{n+1}\nonumber \\&\qquad + \partial _{x_j}\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u\Big )\cdot b_{n+1}+\partial _{x_j}\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0\Big )c_{n+1}\nonumber \\&\qquad +E_{0,j}\Big (n_0u^0a_{n+1}+(e_0+P_0)u^0(u\cdot b_{n+1})+[e_0(u^0)^2+P_0|u|^2]c_{n+1}\Big )\nonumber \\&\qquad +\Big [\Big (n_0u a_{n+1}+(e_0+P_0)u (u\cdot b_{n+1})+(e_0+P_0)u^0 u c_{n+1}\Big )\times B_0\Big ]_j\nonumber \\&\qquad +n_0u^0E_{n+1,j}+\Big (n_0u^0\times B_{n+1}\Big )_j\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}E_{k,j}\Big (n_0u^0a_{l}+(e_0+P_0)u^0(u\cdot b_{l})+[e_0(u^0)^2+P_0|u|^2]c_{l}\Big )\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [\Big (n_0u a_{l}+(e_0+P_0)u (u\cdot b_{l})+(e_0+P_0)u^0 u c_{l}\Big )\times B_k\Big ]_j\nonumber \\&\qquad +\nabla _x\cdot \int _{{\mathbb {R}}^3} \frac{p_jp}{p^0}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\nonumber \\&\qquad +\Big [\int _{{\mathbb {R}}^3} {\hat{p}}\times B_0\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\Big ]_j\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [\int _{{\mathbb {R}}^3} {\hat{p}}\times B_k\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{l}}{\sqrt{{\mathbf {M}}}}\Big )\,dp\Big ]_j=0, \end{aligned}$$
(8.23)
and
$$\begin{aligned}&n_0u^0\Big (hu^0 \partial _ta_{n+1}+\frac{1}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma )) u^0](u\cdot \partial _tb_{n+1})\nonumber \\&\qquad +\frac{1}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +3K_3(\gamma )|u|^2]\partial _tc_{n+1}\Big )\nonumber \\&\qquad + n_0u^0\Big [\partial _t(hu^0)a_{n+1}+\partial _t\Big (\frac{1}{\gamma K_2(\gamma )}[(6K_3(\gamma )+\gamma K_2(\gamma )) u^0]u\Big )\cdot b_{n+1}\nonumber \\&\qquad +\partial _t\Big (\frac{1}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2 +3K_3(\gamma )|u|^2]\Big )c_{n+1}\Big ]\nonumber \\&\qquad -P_0\partial _ta_{n+1}-\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}(u\cdot \partial _tb_{n+1})\nonumber \\&\qquad -\partial _tP_0a_{n+1}-\partial _t\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u\Big )\cdot b_{n+1}\nonumber \\&\qquad +n_0u\cdot \Big (hu^0 \nabla _x a_{n+1}+\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u^0( \nabla _xb_{n+1}\cdot u)\nonumber \\&\qquad +\frac{1}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2+3K_3(\gamma )|u|^2]\nabla _xc_{n+1} \Big )\nonumber \\&\qquad +n_0u\cdot \Big [\nabla _x\Big (hu^0\Big ) a_{n+1}+\nabla _x\Big (\frac{1}{\gamma K_2(\gamma )}(6K_3(\gamma )+\gamma K_2(\gamma ))u^0u\Big )\cdot b_{n+1}\nonumber \\&\qquad +\nabla _x\Big (\frac{1}{\gamma K_2(\gamma )}[(3K_3(\gamma )+\gamma K_2(\gamma ))(u^0)^2+3K_3(\gamma )|u|^2]\Big )c_{n+1} \Big ]\nonumber \\&\qquad +\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0\Big )\nabla _x\cdot b_{n+1}+\Big (\frac{2n_0K_3(\gamma )}{\gamma K_2(\gamma )}u\Big )\cdot \nabla _x c_{n+1}\nonumber \\&\qquad +\nabla _x\Big (\frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )}u^0\Big )\cdot b_{n+1}+\nabla _x\cdot \Big (\frac{2n_0K_3(\gamma )}{\gamma K_2(\gamma )}u\Big )c_{n+1}+n_0u\cdot E_{n+1}\nonumber \\&\qquad + n_0u\cdot E_0 a_{n+1}+(e_0+P_0)(u\cdot b_{n+1})(u\cdot E_0)+P_0E_0\cdot b_{n+1}\nonumber \\&\qquad +(e_0+P_0)u^0(u\cdot E_0)c_{n+1}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{n+1}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_0\nonumber \\&\qquad +\sum _{\begin{array}{c} k+l=n+1\\ k,l\ge 1 \end{array}}\Big [n_0u\cdot E_k a_{l}+(e_0+P_0)(u\cdot b_{l})(u\cdot E_k)+P_0E_k\cdot b_{l}\nonumber \\&\qquad +(e_0+P_0)u^0(u\cdot E_k)c_{l}+\int _{{\mathbb {R}}^3} {\hat{p}}\sqrt{{\mathbf {M}}}{\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{l}}{\sqrt{{\mathbf {M}}}}\Big )\,dp \cdot E_k\Big ]=0. \end{aligned}$$
(8.24)
Now we can write the equations (8.22), (8.3), (8.24) as a linear symmetric hyperbolic system:
$$\begin{aligned} \mathbf{A}_0\partial _t U+\sum _{i=1}^3\mathbf{A}_i\partial _i U +\mathbf{B}_1 U+\mathbf{B}_2{\bar{U}}=S, \end{aligned}$$
(8.25)
where U and \({\bar{U}}\) are
$$\begin{aligned} U=\left( \begin{array}{c} \displaystyle a_{n+1} \\ \displaystyle b_{n+1}\\ \displaystyle c_{n+1} \end{array} \right) ,\qquad {\bar{U}}=\left( \begin{array}{c} \displaystyle E_{n+1} \\ \displaystyle B_{n+1} \end{array} \right) . \end{aligned}$$
For simplicity, denote
$$\begin{aligned} h_1=h_1(t,x)\equiv \frac{n_0}{\gamma K_2(\gamma )} (6K_3(\gamma )+\gamma K_2(\gamma )), \quad h_2=h_2(t,x)\equiv \frac{n_0K_3(\gamma )}{\gamma K_2(\gamma )} . \end{aligned}$$
\(5\times 5\) Matrixes \(\mathbf{A}_0\), \(\mathbf{A}_i, (i=1,2,3)\) in (8.25) are
$$\begin{aligned} \mathbf{A}_0=\left( \begin{array}{ccc} \displaystyle n_0u^0 &{} n_0u^0hu^t &{} [e_0(u^0)^2+P_0|u|^2]\\ \displaystyle n_0u^0h u &{} (h_1u\otimes u+h_2\mathbf{I})u^0 &{} (h_1(u^0)^2-h_2)u\\ \displaystyle [e_0(u^0)^2+P_0|u|^2] &{} (h_1(u^0)^2-h_2 )u^t &{} (h_1(u^0)^2-3h_2 )u^0 \end{array} \right) , \end{aligned}$$
and
$$\begin{aligned} \mathbf{A}_i=\left( \begin{array}{ccc} \displaystyle n_0u_i &{} n_0hu_iu^t+P_0 e_i^t &{} n_0hu^0u_i\\ \displaystyle n_0hu_iu+P_0 e_i &{} h_1u_iu\otimes u+h_2(u_i \mathbf{I}+\tilde{\mathbf{A}}_i) &{} (h_1u_iu+h_2 e_i)u^0\\ \displaystyle n_0hu^0u_i &{} (h_1u_iu^t+h_2 e_i^t)u^0 &{} (h_1(u^0)^2-h_2 )u_i \end{array} \right) , \end{aligned}$$
where \((\cdot )^t\) denotes the transpose of a vector in \({{\mathbb {R}}^3}\), \(\mathbf{I}\) is the \(3\times 3\) identity matrix and \(\tilde{\mathbf{A}}_i\) is a \(3\times 3\) matrix with its components
$$\begin{aligned} (\tilde{\mathbf{A}}_i)_{jk}=\delta _{ij}u_k+\delta _{ik}u_j,\qquad 1\le j,k\le 3. \end{aligned}$$
The components of matrixes \(\mathbf{B}_1\) and \(\mathbf{B}_2\), and the remainder terms S can be written explicitly as functions of \(F_i, E_i, B_i,~ (0\le i\le n\)) and their first order derivatives. For the matrix \(\mathbf{A}_0\), its determinant is
$$\begin{aligned}&n_0^5\left| \begin{array}{ccc} \displaystyle u_0 &{} u^0u^th &{} h(u^0)^2-\frac{P_0}{n_0}\\ \displaystyle \mathbf{0} &{} u^0[h_3u\otimes u+h_4\mathbf{I}] &{} (u^0)^2h_3u\\ \displaystyle 0 &{} (u^0)^2h_3u^t &{}(u^0)^3h_3-u^0h_4+\frac{1}{\gamma ^2u^0} \end{array} \right| \\&=n_0^5(u^0)^6\left| \begin{array}{cc} \displaystyle h_3u\otimes u+h_4\mathbf{I} &{} h_3u\\ \displaystyle h_3u^0u^t &{} u^0h_3-\frac{1}{u^0}h_4+\frac{1}{\gamma ^2(u^0)^3} \end{array} \right| \\&=n_0^5(u^0)^6\left| \begin{array}{cc} \displaystyle h_4\mathbf{I} &{} h_3u\\ \displaystyle \frac{u^t}{ u^0}\Big [h_4-\frac{1}{\gamma ^2(u^0)^2}\Big ]&{} u^0h_3-\frac{1}{u^0}h_4+\frac{1}{\gamma ^2(u^0)^3} \end{array} \right| \\&=n_0^5(u^0)^6\left| \begin{array}{cc} \displaystyle h_4\mathbf{I} &{} h_3u\\ \displaystyle \mathbf{0}^t&{} \Big [\frac{-|u|^2}{ u^0h_4}\Big (h_4-\frac{1}{\gamma ^2(u^0)^2}\Big )+u^0\Big ]h_3-\frac{1}{u^0}h_4+\frac{1}{\gamma ^2(u^0)^3} \end{array} \right| \\&=n_0^5(u^0)^6h_4^3\Big \{\Big [\frac{-|u|^2}{ u^0h_4}\Big (h_4-\frac{1}{\gamma ^2(u^0)^2}\Big )+u^0\Big ]h_3-\frac{1}{u^0}h_4+\frac{c^4}{\gamma ^2(u^0)^3}\Big \}, \end{aligned}$$
where \(\mathbf{0}\) denotes the column vector \((0,0,0)^t\), \(h_3\) and \(h_4\) are functions with the following forms:
$$\begin{aligned} h_3=-\Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2-\frac{2}{\gamma }\frac{K_1(\gamma )}{K_2(\gamma )}+1+\frac{8}{\gamma ^2},\quad h_4=\frac{K_1(\gamma )}{\gamma K_2(\gamma )}+\frac{4}{\gamma ^2}. \end{aligned}$$
Noting
$$\begin{aligned} -\Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2-\frac{3}{\gamma }\frac{K_1(\gamma )}{K_2(\gamma )} +1+\frac{3}{\gamma ^2}>0 \end{aligned}$$
by Proposition 6.3 in Appendix 3 [31], we can further obtain
$$\begin{aligned} |\mathbf{A}_0|&>n_0^5(u^0)^6h_4^3\frac{1}{u^0}(h_3-h_4)\\&=n_0^5(u^0)^5h_4^3\Big [-\Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2-\frac{3}{\gamma }\frac{K_1(\gamma )}{K_2(\gamma )} +1+\frac{4}{\gamma ^2}\Big ]\\&>\frac{n_0^5(u^0)^5h_4^3}{\gamma ^2}. \end{aligned}$$
On the other hand, the system (8.16) can also be written as a linear symmetric hyperbolic system of \((E_{n+1}, B_{n+1})\):
$$\begin{aligned} \partial _t {\bar{U}}+\sum _{i=1}^3{ {\bar{\mathbf{A}}}}_i\partial _i {\bar{U}} +\mathbf{B}U=0, \end{aligned}$$
(8.26)
where \(\mathbf{B}\) is a \(6\times 5\) matrix whose components are functions of \(n_0, u\). Denote \(\mathbf{O}\) as a \(3\times 3\) matrix with all components be 0, and define matrixes:
$$\begin{aligned} {\bar{\mathbf{A}}}_{11}=\left( \begin{array}{ccc} \displaystyle 0 &{} 0 &{} 0\\ \displaystyle 0 &{} 0 &{} 1\\ \displaystyle 0 &{} -1 &{} 0 \end{array} \right) , \quad {\bar{\mathbf{A}}}_{12}=\left( \begin{array}{ccc} \displaystyle 0 &{} 0 &{} 0\\ \displaystyle 0 &{} 0 &{} -1\\ \displaystyle 0 &{} 1 &{} 0 \end{array} \right) ,\quad {\bar{\mathbf{A}}}_{21}=\left( \begin{array}{ccc} \displaystyle 0 &{} 0 &{} -1\\ \displaystyle 0 &{} 0 &{} 0\\ \displaystyle 1 &{} 0 &{} 0 \end{array} \right) , \\ {\bar{\mathbf{A}}_{22}}=\left( \begin{array}{ccc} \displaystyle 0 &{} 0 &{} 1\\ \displaystyle 0 &{} 0 &{} 0\\ \displaystyle -1 &{} 0 &{} 0 \end{array} \right) ,\quad {\bar{\mathbf{A}}}_{31}=\left( \begin{array}{ccc} \displaystyle 0 &{} 1 &{} 0\\ \displaystyle -1 &{} 0 &{} 0\\ \displaystyle 0 &{} 0 &{} 0 \end{array} \right) ,\quad {\bar{\mathbf{A}}}_{32}=\left( \begin{array}{ccc} \displaystyle 0 &{} -1 &{} 0\\ \displaystyle 1 &{} 0 &{} 0\\ \displaystyle 0 &{} 0 &{} 0 \end{array} \right) . \end{aligned}$$
Then \({ {\bar{\mathbf{A}}}}_i\) in (8.26) can be expressed as
$$\begin{aligned} { {\bar{\mathbf{A}}}}_i=\left( \begin{array}{cc} \displaystyle \mathbf{O} &{} { {\bar{\mathbf{A}}}}_{i1}\\ \displaystyle { {\bar{\mathbf{A}}}}_{i2} &{} \mathbf{O} \end{array}\right) . \end{aligned}$$
Combining (8.25) and (8.26), we can obtain the wellposedness of \((a_{n+1}, b_{n+1}, c_{n+1}, E_{n+1}\), \(B_{n+1}) \in C^0([0,\infty );H^N) \) with initial data \(a_{n+1}(0,x), b_{n+1}(0,x), c_{n+1}(0,x), E_{n+1}(0,x), B_{n+1}(0,x)\) \(\in H^N, N\ge 0\) by the standard theorem of linear symmetric system [3] (Chapter 4.2). Moreover, for \(n=0\), one has
$$\begin{aligned} \Vert \partial _t\mathbf{A}_0\Vert _{\infty }+\sum _{i=1}^{3}\Vert \nabla _x\mathbf{A}_i\Vert _{\infty }+\sum _{i=1}^{2}\Vert \mathbf{B}_i\Vert _{\infty }+\Vert S\Vert _{\infty }\lesssim (1+t)^{-\beta _0}. \end{aligned}$$
Therefore, standard energy estimates on systems (8.25) and (8.26) yield
$$\begin{aligned}&\frac{d}{dt}\left( \Vert (a_1(t),b_1(t),c_1(t))\Vert _{H^N}^2+\Vert [E_1(t),B_1(t)]\Vert _{H^N}^2\right) \\&\quad \lesssim (1+t)^{-\beta _0}[\left( \Vert (a_1(t),b_1(t),c_1(t))\Vert _{H^N}^2+\Vert [E_1(t),B_1(t)]\Vert _{H^N}^2\right) \\&\qquad +\left( \Vert (a_1(t),b_1(t),c_1(t))\Vert _{H^N}+\Vert [E_1(t),B_1(t)]\Vert _{H^N}\right) ]. \end{aligned}$$
Namely,
$$\begin{aligned} \Vert (a_1(t),b_1(t),c_1(t))\Vert _{H^N}+\Vert [E_1(t),B_1(t)]\Vert _{H^N}\lesssim 1 \end{aligned}$$
(8.27)
by Grönwall’s inequality. Let \(Lf_1=\nu f_1+K(f_1)=f_2=w_0(p) \mathbf{M}\), where \(w_0(p)\) is any polynomial of p. Note that, by (1.2), (1.21) and (8.1),
$$\begin{aligned} |K(f_1)|\lesssim \int _{{\mathbb {R}}^3}\frac{s^{\frac{3}{2}}}{gp^0q^0}U_1(p,q)\exp \{-[1+O(1)|u|]\frac{\sqrt{s}|p-q|}{2T_0g} \}|f_1(q)| dq, \end{aligned}$$
and \(f_1=\frac{f_2}{\nu }-\frac{K(f_1)}{\nu }\). Then, we can verify that \(|f_1|\lesssim {\mathbf {M}}^{1_-}\). Therefore, it holds that
$$\begin{aligned} {\{\mathbf{I}-\mathbf{P}\}}\Big (\frac{F_{1}}{\sqrt{{\mathbf {M}}}}\Big )\lesssim (1+|p|){\mathbf {M}}^{1_-}. \end{aligned}$$
It is straightforward to obtain (8.17) by the structure of \(F_1\) and (8.27). Assuming (8.17) holds for \(1\le i\le n\), the case \(i=n+1\) for (8.17) holds again by the structure of the equation for \(F_{n+1}\) in (1.10), the induction assumption and similar analysis as the case \(i=1\) ( \(n=0\)). \(\quad \square \)
