这是自己的一份笔记，主要是和格子Boltzmann方法的DnQb离散速度模型相关的。还请读者确定自己已经知道常见的DnQb模型长什么样再来读这篇文章，这样会好理解一些。
一般来说，密度分布函数的Maxwell-Boltzmann平衡态分布为：
f e q = ρ ( 2 π R T ) D / 2 exp [ − ( ξ − u ) 2 2 R T ] , (1-1) f^{eq} = \frac{\rho}{ (2 \pi R T)^{D/2} } \exp \left[ - \frac{(\boldsymbol{\xi} - \boldsymbol{u})^2}{2 R T} \right],
\tag{1-1}
f e q = ( 2 π R T ) D / 2 ρ exp [ − 2 R T ( ξ − u ) 2 ] , ( 1 - 1 )
其中 ξ \boldsymbol{\xi} ξ R R R D D D ρ \rho ρ u \boldsymbol{u} u c α \boldsymbol{c}_\alpha c α W α W_\alpha W α c s 2 = R T c_s^2 = RT c s 2 = R T
f α e q = ρ W α × [ 1 + ξ α ⋅ u c s 2 + ( ξ α ⋅ u ) 2 2 c s 4 − ∣ u ∣ 2 2 c s 2 ] . (1-2) f^{eq}_{\alpha} = \rho W_{\alpha} \times
\left[
1 + \frac{\boldsymbol{\xi}_{\alpha} \cdot \boldsymbol{u}}{c_s^2} +
\frac{(\boldsymbol{\xi}_{\alpha} \cdot \boldsymbol{u})^2}{2 c_s^4} -
\frac{|\boldsymbol{u}|^2}{2 c_s^2}
\right].
\tag{1-2}
f α e q = ρ W α × [ 1 + c s 2 ξ α ⋅ u + 2 c s 4 ( ξ α ⋅ u ) 2 − 2 c s 2 ∣ u ∣ 2 ] . ( 1 - 2 )
那么这两者是怎么联系起来的呢？参考文献 [1] 讲的就是这个事情。最近自己总算是略懂一点了，这里把笔记写下来，作为备忘录。
当马赫数远小于1时，我们可以对式(1)的平衡态做如下离散：
f e q ( ξ ) = ρ ( 2 π R T ) D / 2 exp ( − ∣ ξ ∣ 2 2 R T ) × [ 1 + ξ ⋅ u R T + ( ξ ⋅ u ) 2 2 ( R T ) 2 − ∣ u ∣ 2 2 R T ] (1-3) \begin{aligned}
f^{eq} (\boldsymbol{\xi}) &= \frac{\rho}{ (2 \pi R T)^{D/2} } \exp \left( - \frac{|\boldsymbol{\xi}|^2}{2 R T} \right)\\
& \quad \times
\left[
1 + \frac{\boldsymbol{\xi} \cdot \boldsymbol{u}}{R T} +
\frac{(\boldsymbol{\xi} \cdot \boldsymbol{u})^2}{2 (R T)^2} -
\frac{|\boldsymbol{u}|^2}{2 R T}
\right]
\end{aligned}
\tag{1-3}
f e q ( ξ ) = ( 2 π R T ) D / 2 ρ exp ( − 2 R T ∣ ξ ∣ 2 ) × [ 1 + R T ξ ⋅ u + 2 ( R T ) 2 ( ξ ⋅ u ) 2 − 2 R T ∣ u ∣ 2 ] ( 1 - 3 )
式(1-3)中，我们对 u \boldsymbol{u} u
∫ − ∞ + ∞ ψ ( ξ ) f e q ( ξ ) d ξ \int_{-\infty}^{+\infty} \psi(\boldsymbol{\xi}) f^{eq} (\boldsymbol{\xi}) \mathrm{d}\boldsymbol{\xi}
∫ − ∞ + ∞ ψ ( ξ ) f e q ( ξ ) d ξ
用离散的方式近似计算，其中 ψ ( ξ ) \psi(\boldsymbol{\xi}) ψ ( ξ ) ξ \boldsymbol{\xi} ξ
由于这一节只讨论一维问题，分子速度 ξ \xi ξ u u u D = 1 D=1 D = 1
f e q ( ξ ) = ρ ( 2 π R T ) 1 / 2 exp ( − ξ 2 2 R T ) × [ 1 + ξ u R T + ( ξ u ) 2 2 ( R T ) 2 − u 2 2 R T ] . (1-4) \begin{aligned}
f^{eq} (\xi) &= \frac{\rho}{ (2 \pi R T)^{1/2} } \exp \left( - \frac{\xi^2}{2 R T} \right)\\
& \quad \times
\left[
1 + \frac{\xi u}{R T} +
\frac{(\xi u)^2}{2 (R T)^2} -
\frac{u^2}{2 R T}
\right].
\end{aligned}
\tag{1-4}
f e q ( ξ ) = ( 2 π R T ) 1 / 2 ρ exp ( − 2 R T ξ 2 ) × [ 1 + R T ξ u + 2 ( R T ) 2 ( ξ u ) 2 − 2 R T u 2 ] . ( 1 - 4 )
式(1-4)的中括号内的四项可以用积分的加法拆成四个积分，并且每一项都可以看成如下形式：
Constant ⋅ ∫ − ∞ + ∞ exp ( − ξ 2 2 R T ) ψ ( ξ ) d ξ , \text{Constant} \cdot \int_{-\infty}^{+\infty} \exp \left( - \frac{\xi^2}{2 R T} \right) \psi(\xi) \mathrm{d}\xi,
Constant ⋅ ∫ − ∞ + ∞ exp ( − 2 R T ξ 2 ) ψ ( ξ ) d ξ ,
其中 ψ ( ξ ) \psi(\xi) ψ ( ξ ) ξ \xi ξ ψ m ( ξ ) = ξ m \psi_{m}(\xi) = \xi^{m} ψ m ( ξ ) = ξ m m m m
ρ ( 2 π R T ) 1 / 2 ⋅ ∫ − ∞ + ∞ exp ( − ξ 2 2 R T ) ψ m ( ξ ) d ξ . (1-5) \frac{\rho}{ (2 \pi R T)^{1/2} } \cdot \int_{-\infty}^{+\infty} \exp \left( - \frac{\xi^2}{2 R T} \right) \psi_{m}(\xi) \mathrm{d}\xi.
\tag{1-5}
( 2 π R T ) 1 / 2 ρ ⋅ ∫ − ∞ + ∞ exp ( − 2 R T ξ 2 ) ψ m ( ξ ) d ξ . ( 1 - 5 )
为简化这个积分的计算，这里需要引入变量代换
ζ = ξ / 2 R T , \zeta = \xi / \sqrt{2 R T},
ζ = ξ / 2 R T ,
则
d ξ = 2 R T d ζ . \mathrm{d}\xi = \sqrt{2 R T} \mathrm{d}\zeta.
d ξ = 2 R T d ζ .
代换后，有：
ρ ( 2 π R T ) 1 / 2 ⋅ ∫ − ∞ + ∞ exp ( − ζ 2 ) ψ m ( 2 R T ζ ) 2 R T d ζ = ρ ( 2 π R T ) 1 / 2 ⋅ ( 2 R T ) ( m + 1 ) / 2 ⋅ I m = ρ π ⋅ ( 2 R T ) m / 2 ⋅ I m \begin{aligned}
&
\frac{\rho}{ (2 \pi R T)^{1/2} } \cdot \int_{-\infty}^{+\infty} \exp \left( - \zeta^2 \right) \psi_{m}(\sqrt{2 R T} \zeta) \sqrt{2 R T} \mathrm{d}\boldsymbol{\zeta}
\\=& \frac{\rho}{ (2 \pi R T)^{1/2} } \cdot (2 R T)^{(m+1)/2} \cdot I_m
\\=& \frac{\rho}{\sqrt{\pi}} \cdot (2 R T)^{m/2} \cdot I_m
\end{aligned}
= = ( 2 π R T ) 1 / 2 ρ ⋅ ∫ − ∞ + ∞ exp ( − ζ 2 ) ψ m ( 2 R T ζ ) 2 R T d ζ ( 2 π R T ) 1 / 2 ρ ⋅ ( 2 R T ) ( m + 1 ) / 2 ⋅ I m π ρ ⋅ ( 2 R T ) m / 2 ⋅ I m
其中 I m = ∫ − ∞ + ∞ exp ( − ζ 2 ) ζ m d ζ \displaystyle I_m = \int_{-\infty}^{+\infty} \exp \left( - \zeta^2 \right) \zeta^m \mathrm{d}\boldsymbol{\zeta} I m = ∫ − ∞ + ∞ exp ( − ζ 2 ) ζ m d ζ m m m
所以，下面的积分会被展开为
I = ∫ − ∞ + ∞ ψ m ( ξ ) f e q ( ξ ) d ξ = ρ π [ ( 1 − u 2 2 R T ) ( 2 R T ) m / 2 I m + u R T ( 2 R T ) ( m + 1 ) / 2 I m + 1 + u 2 2 ( R T ) 2 ( 2 R T ) ( m + 2 ) / 2 I m + 2 ] . (1-6) \begin{aligned}
I=& \int_{-\infty}^{+\infty} \psi_m(\xi) f^{eq} (\xi) \mathrm{d}\xi
\\=& \frac{\rho}{\sqrt{\pi}} \left[
\left(1 - \frac{u^2}{2 R T}\right) (2 R T)^{m/2} I_m +
\frac{u}{RT} (2 R T)^{(m+1)/2} I_{m+1} \right.
\\&\qquad \left. +
\frac{u^2}{2(RT)^2} (2 R T)^{(m+2)/2} I_{m+2}
\right].
\end{aligned}
\tag{1-6}
I = = ∫ − ∞ + ∞ ψ m ( ξ ) f e q ( ξ ) d ξ π ρ [ ( 1 − 2 R T u 2 ) ( 2 R T ) m / 2 I m + R T u ( 2 R T ) ( m + 1 ) / 2 I m + 1 + 2 ( R T ) 2 u 2 ( 2 R T ) ( m + 2 ) / 2 I m + 2 ] . ( 1 - 6 )
假设我们想用一个三点格式的Gauss-Hermite数值积分 进行近似，则：
∫ − ∞ + ∞ exp ( − ζ 2 ) ϕ ( ζ ) d ζ = ∑ i ∈ [ − 1 , 0 , 1 ] w i ϕ ( ζ i ) . \int_{-\infty}^{+\infty} \exp(- \zeta^2) \phi(\zeta) \mathrm{d} \zeta = \sum_{i \in [-1, 0, 1]} w_{i} \phi(\zeta_{i}).
∫ − ∞ + ∞ exp ( − ζ 2 ) ϕ ( ζ ) d ζ = i ∈ [ − 1 , 0 , 1 ] ∑ w i ϕ ( ζ i ) .
其中：
ζ 0 = 0 , ζ 1 = 3 / 2 , ζ − 1 = − 3 / 2 , w 0 = 2 π / 3 , w 1 = π / 6 , w − 1 = π / 6. \begin{aligned}
\zeta_0 = 0 ,\quad& \zeta_{1} = \sqrt{3/2} ,\quad& \zeta_{-1} = -\sqrt{3/2} ,\\
w_0 = 2\sqrt{\pi} / 3 ,\quad& w_{1} = \sqrt{\pi} / 6 ,\quad& w_{-1} = \sqrt{\pi} / 6.
\end{aligned}
ζ 0 = 0 , w 0 = 2 π / 3 , ζ 1 = 3 / 2 , w 1 = π / 6 , ζ − 1 = − 3 / 2 , w − 1 = π / 6 .
如果用变换关系代换回 ξ \xi ξ ξ 0 = 0 \xi_0 = 0 ξ 0 = 0 ξ − 1 = − 3 R T \xi_{-1} = -\sqrt{3RT} ξ − 1 = − 3 R T ξ 1 = 3 R T \xi_{1} = \sqrt{3RT} ξ 1 = 3 R T
所以，积分 I m I_m I m
I m = ∫ − ∞ + ∞ exp ( − ζ 2 ) ζ m d ζ ≈ ∑ i ∈ [ − 1 , 0 , 1 ] w i ζ i m = ( 2 R T ) − m / 2 ∑ i ∈ [ − 1 , 0 , 1 ] w i ξ i m \begin{aligned}
I_m &= \int_{-\infty}^{+\infty} \exp \left( - \zeta^2 \right) \zeta^m \mathrm{d}\boldsymbol{\zeta}
\\&\approx \sum_{i \in [-1, 0, 1]} w_{i} \zeta_{i}^m
\\&= (2RT)^{-m/2} \sum_{i \in [-1, 0, 1]} w_{i} \xi_{i}^m
\end{aligned}
I m = ∫ − ∞ + ∞ exp ( − ζ 2 ) ζ m d ζ ≈ i ∈ [ − 1 , 0 , 1 ] ∑ w i ζ i m = ( 2 R T ) − m / 2 i ∈ [ − 1 , 0 , 1 ] ∑ w i ξ i m
也就是说，
( 2 R T ) m / 2 I m ≈ ∑ i ∈ [ − 1 , 0 , 1 ] w i ξ i m (2 R T)^{m/2} I_m \approx \sum_{i \in [-1, 0, 1]} w_{i} \xi_{i}^m
( 2 R T ) m / 2 I m ≈ i ∈ [ − 1 , 0 , 1 ] ∑ w i ξ i m
综上所述，式(1-6)就会被改写为：
I = ∫ − ∞ + ∞ ψ m ( ξ ) f e q ( ξ ) d ξ ≈ ∑ i ∈ [ − 1 , 0 , 1 ] ξ i m ⋅ { ρ w i π ⋅ [ 1 + ξ i u R T + ( ξ i u ) 2 2 ( R T ) 2 − u 2 2 R T ] } . \begin{aligned}
I=& \int_{-\infty}^{+\infty} \psi_m(\xi) f^{eq} (\xi) \mathrm{d}\xi
\\ \approx& \sum_{i \in [-1, 0, 1]} \xi_{i}^{m} \cdot \left\{
\rho \frac{w_{i}}{\sqrt{\pi}} \cdot
\left[
1 + \frac{\xi_{i} u}{R T} + \frac{(\xi_{i} u)^2}{2 (R T)^2} - \frac{u^2}{2 R T}
\right]
\right\}.
\end{aligned}
I = ≈ ∫ − ∞ + ∞ ψ m ( ξ ) f e q ( ξ ) d ξ i ∈ [ − 1 , 0 , 1 ] ∑ ξ i m ⋅ { ρ π w i ⋅ [ 1 + R T ξ i u + 2 ( R T ) 2 ( ξ i u ) 2 − 2 R T u 2 ] } .
到这一步，我们就可以定义 ξ i \xi_{i} ξ i f i e q f^{eq}_{i} f i e q
f i e q = ρ W i ⋅ [ 1 + ξ i u c s 2 + ( ξ i u ) 2 2 c s 4 − u 2 2 c s 2 ] . (1-7) f^{eq}_{i} = \rho W_{i} \cdot
\left[
1 + \frac{\xi_{i} u}{c_s^2} + \frac{(\xi_{i} u)^2}{2 c_s^4} - \frac{u^2}{2 c_s^2}
\right].
\tag{1-7}
f i e q = ρ W i ⋅ [ 1 + c s 2 ξ i u + 2 c s 4 ( ξ i u ) 2 − 2 c s 2 u 2 ] . ( 1 - 7 )
其中的权重 W i = w i / π W_{i} = w_i / \sqrt{\pi} W i = w i / π c = 3 R T c = \sqrt{3RT} c = 3 R T c s = R T = c / 3 c_s = \sqrt{RT} = c / \sqrt{3} c s = R T = c / 3
至此，我们就整理出了 D1Q3 模型：
i {i} i 离散速度 ξ i \xi_{i} ξ i
权重系数 W i W_i W i
0
0
2/3
-c
-c
1/6
c
c
1/6
注意：在多数LBM实践中，我们通常将 c c c c = 1 c=1 c = 1
由于各个坐标轴之间是正交的，这种做法同样可以拓展到高维空间。
打个比方，对于三维问题， ξ = [ ξ x , ξ y , ξ z ] T \boldsymbol{\xi} = [\xi_x, \xi_y, \xi_z]^{\mathrm{T}} ξ = [ ξ x , ξ y , ξ z ] T d ξ = d ξ x d ξ y d ξ z \mathrm{d}\boldsymbol{\xi} = \mathrm{d}\xi_x \mathrm{d}\xi_y \mathrm{d}\xi_z d ξ = d ξ x d ξ y d ξ z
对于二维问题，记 ψ m , n ( ξ ) = ξ x m ξ y n \psi_{m,n}(\boldsymbol{\xi})=\xi_x^{m} \xi_y^{n} ψ m , n ( ξ ) = ξ x m ξ y n
I = ∫ ψ m , n ( ξ ) f ( s t ) d ξ = ρ π ( 2 R T ) m + n { ( 1 − u 2 2 R T ) I m I n + 1 2 R T 2 ( u x I m + 1 I n + u y I m I n + 1 ) + 1 R T ( u x 2 I m + 2 I n + 2 u x u y I m + 1 I n + 1 + u y 2 I m I n + 2 ) } (1-8) \begin{aligned}
I =& \int\psi_{m , n} (\boldsymbol{\xi}) f^{( \mathrm{s t} )} \mathrm{d}\boldsymbol{\xi}
\\=& \frac{\rho}{\pi} ( \sqrt{2 R T} )^{m+n} \Biggl\{\left( 1-\frac{u^{2}} {2 R T} \right) I_{m} I_{n}
\\&+ \frac{1} {\sqrt{2 R T}} 2 ( u_{x} I_{m+1} I_{n}+u_{y} I_{m} I_{n+1} )
\\&+ \frac{1} {R T} (u_{x}^{2} I_{m+2} I_{n}+2 u_{x} u_{y} I_{m+1} I_{n+1}+u_{y}^{2} I_{m} I_{n+2}) \Biggr\}
\end{aligned}
\tag{1-8}
I = = ∫ ψ m , n ( ξ ) f ( s t ) d ξ π ρ ( 2 R T ) m + n { ( 1 − 2 R T u 2 ) I m I n + 2 R T 1 2 ( u x I m + 1 I n + u y I m I n + 1 ) + R T 1 ( u x 2 I m + 2 I n + 2 u x u y I m + 1 I n + 1 + u y 2 I m I n + 2 ) } ( 1 - 8 )
式(1-8)同样可以整理成：
I = ∑ i , j ∈ [ − 1 , 0 , 1 ] ψ ( ξ i , j ) ⋅ { ρ w i w j π [ 1 + ξ i , j ⋅ u c s 2 + ( ξ i , j ⋅ u ) 2 2 c s 4 − ∥ u ∥ 2 2 c s 2 ] } (1-9) I = \sum_{i,j\in[-1,0,1]} \psi(\boldsymbol{\xi}_{i,j}) \cdot \left\{ \rho \frac{w_i w_j}{\pi}
\left[
1 + \frac{\boldsymbol{\xi}_{i,j} \cdot \boldsymbol{u}}{c_s^2} + \frac{(\boldsymbol{\xi}_{i,j} \cdot \boldsymbol{u})^2}{2 c_s^4} - \frac{\|\boldsymbol{u}\|^2}{2 c_s^2}
\right]
\right\}
\tag{1-9}
I = i , j ∈ [ − 1 , 0 , 1 ] ∑ ψ ( ξ i , j ) ⋅ { ρ π w i w j [ 1 + c s 2 ξ i , j ⋅ u + 2 c s 4 ( ξ i , j ⋅ u ) 2 − 2 c s 2 ∥ u ∥ 2 ] } ( 1 - 9 )
其中 ξ i , j = 2 R T ( ζ i , ζ j ) \boldsymbol{\xi}_{i,j}=\sqrt{2RT} (\zeta_i,\zeta_j) ξ i , j = 2 R T ( ζ i , ζ j ) W i , j = w i w j / π W_{i,j} = w_i w_j / \pi W i , j = w i w j / π
类似的，对于三维问题，则可以整理出
I = ∑ i , j , k ∈ [ − 1 , 0 , 1 ] ψ ( ξ i , j , k ) ⋅ { ρ w i w j w k π 3 / 2 [ 1 + ξ i , j , k ⋅ u c s 2 + ( ξ i , j , k ⋅ u ) 2 2 c s 4 − ∥ u ∥ 2 2 c s 2 ] } (1-10) I = \sum_{i,j,k\in[-1,0,1]} \psi(\boldsymbol{\xi}_{i,j,k}) \cdot \left\{ \rho \frac{w_i w_j w_k}{\pi^{3/2}}
\left[
1 + \frac{\boldsymbol{\xi}_{i,j,k} \cdot \boldsymbol{u}}{c_s^2} + \frac{(\boldsymbol{\xi}_{i,j,k} \cdot \boldsymbol{u})^2}{2 c_s^4} - \frac{\|\boldsymbol{u}\|^2}{2 c_s^2}
\right]
\right\}
\tag{1-10}
I = i , j , k ∈ [ − 1 , 0 , 1 ] ∑ ψ ( ξ i , j , k ) ⋅ { ρ π 3 / 2 w i w j w k [ 1 + c s 2 ξ i , j , k ⋅ u + 2 c s 4 ( ξ i , j , k ⋅ u ) 2 − 2 c s 2 ∥ u ∥ 2 ] } ( 1 - 1 0 )
其中 ξ i , j , k = 2 R T ( ζ i , ζ j , ζ k ) \boldsymbol{\xi}_{i,j,k}=\sqrt{2RT} (\zeta_i,\zeta_j,\zeta_k) ξ i , j , k = 2 R T ( ζ i , ζ j , ζ k ) W i , j , k = w i w j w k / π 3 / 2 W_{i,j,k} = w_i w_j w_k / \pi^{3/2} W i , j , k = w i w j w k / π 3 / 2
上面的内容可以说从数学上给出了明确的证明。然而实际使用中，其实还是用待定系数法来敲定 DnQb 模型更多一些。下面的内容里，为了保持和何雅玲老师的书标号一致，这里将第 α \alpha α e α \boldsymbol{e}_{\alpha} e α ω α \omega_{\alpha} ω α
一般来说，d 维空间下的格子速度的 n 阶张量为：
E i 1 i 2 . . . i n ( n ) = ∑ α ( e α ) i 1 ( e α ) i 2 . . . ( e α ) i n (2-1) E_{i_1 i_2 ... i_n}^{(n)} = \sum_{\alpha} (\boldsymbol{e}_{\alpha})_{i_1} (\boldsymbol{e}_{\alpha})_{i_2} ... (\boldsymbol{e}_{\alpha})_{i_n}
\tag{2-1}
E i 1 i 2 . . . i n ( n ) = α ∑ ( e α ) i 1 ( e α ) i 2 . . . ( e α ) i n ( 2 - 1 )
假如你选取的离散速度集（共有 M M M
E i 1 i 2 . . . i 2 n + 1 ( 2 n + 1 ) = 0 , E^{(2n+1)}_{i_1 i_2 ... i_{2n+1}} = 0,
E i 1 i 2 . . . i 2 n + 1 ( 2 n + 1 ) = 0 ,
和
E i 1 i 2 . . . i 2 n ( 2 n ) = M d ( d + 2 ) . . . ( d + 2 n − 2 ) Δ i 1 i 2 . . . i 2 n ( 2 n ) E^{(2n)}_{i_1 i_2 ... i_{2n}} = \frac{M}{d (d+2) ... (d+2n-2)} \Delta^{(2n)}_{i_1 i_2 ... i_{2n}}
E i 1 i 2 . . . i 2 n ( 2 n ) = d ( d + 2 ) . . . ( d + 2 n − 2 ) M Δ i 1 i 2 . . . i 2 n ( 2 n )
下面我们首先从张量算子 Δ ( 2 n ) \Delta^{(2n)} Δ ( 2 n )
张量算子 Δ ( 2 n ) \Delta^{(2n)} Δ ( 2 n ) 2 n 2n 2 n
Δ i j ( 2 ) = δ i j Δ i j k l ( 4 ) = δ i j δ k l + δ i k δ j l + δ i l δ j k Δ i j k l p q ( 6 ) = δ i j Δ i j k l ( 4 ) + δ i k Δ j l p q ( 4 ) + δ i l Δ j k p q ( 4 ) + δ i p Δ j k l q ( 4 ) + δ i q Δ j k l p ( 4 ) \begin{aligned}
\Delta^{(2)}_{ij}&=\delta_{ij} \\
\Delta^{(4)}_{ijkl}&=\delta_{ij}\delta_{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk} \\
\Delta^{(6)}_{ijklpq}&=\delta_{ij}\Delta^{(4)}_{ijkl} + \delta_{ik}\Delta^{(4)}_{jlpq} + \delta_{il}\Delta^{(4)}_{jkpq} + \delta_{ip}\Delta^{(4)}_{jklq} + \delta_{iq}\Delta^{(4)}_{jklp}
\end{aligned}
Δ i j ( 2 ) Δ i j k l ( 4 ) Δ i j k l p q ( 6 ) = δ i j = δ i j δ k l + δ i k δ j l + δ i l δ j k = δ i j Δ i j k l ( 4 ) + δ i k Δ j l p q ( 4 ) + δ i l Δ j k p q ( 4 ) + δ i p Δ j k l q ( 4 ) + δ i q Δ j k l p ( 4 )
并存在如下递归关系：
Δ i 1 i 2 … i 2 n ( 2 n ) = ∑ j = 2 2 n δ i 1 , i j ⋅ Δ i 2 … i j − 1 i j + 1 … i 2 n ( 2 n − 2 ) \Delta^{(2n)}_{i_1 i_2 \dots i_{2n}} = \sum_{j=2}^{2n} \delta_{i_1, i_{j}} \cdot \Delta^{(2n-2)}_{i_2 \dots i_{j-1} i_{j+1} \dots i_{2n}}
Δ i 1 i 2 … i 2 n ( 2 n ) = j = 2 ∑ 2 n δ i 1 , i j ⋅ Δ i 2 … i j − 1 i j + 1 … i 2 n ( 2 n − 2 )
其中 δ i j \delta_{ij} δ i j
2 n 2n 2 n n n n (2n−1)!! 项。计算思路为 ：第一个元素有 (2n−1) 个选择配对；剩下 (2n−2) 个元素之中的第一个有 (2n−3) 个选择；后续依此类推。有的书会从排列组合视角出发，写成 ( 2 n ) ! 2 n n ! = 总积 偶数部分的积 \frac{(2n)!}{2^n n!} = \frac{\text{总积}}{\text{偶数部分的积}} 2 n n ! ( 2 n ) ! = 偶数部分的积 总积 计算思路为 ：2 n 2n 2 n 2 n 2n 2 n n n n n ! n! n !
为了简洁表示 Δ ( 2 n ) \Delta^{(2n)} Δ ( 2 n ) ( i 1 , i 2 , … , i k ) (i_1,i_2,\dots,i_k) ( i 1 , i 2 , … , i k ) i 1 ≥ i 2 ≥ ⋯ ≥ i k i_1 \geq i_2 \geq \dots \geq i_k i 1 ≥ i 2 ≥ ⋯ ≥ i k Δ i j k l ( 4 ) \Delta^{(4)}_{ijkl} Δ i j k l ( 4 ) i , j , k , l i,j,k,l i , j , k , l
1111（全1，非递增）→ 值3
2111（1个2、3个1，非递增）→ 值0
2211（2个2、2个1，非递增）→ 值1
2221（3个2、1个1，非递增）→ 值0
2222（全2，非递增）→ 值3
即 Δ ( 4 ) = [ 3 , 0 , 1 , 0 , 3 ] \Delta^{(4)}=[3,0,1,0,3] Δ ( 4 ) = [ 3 , 0 , 1 , 0 , 3 ]
在后续的描述中，为了避免写一大堆下标，我们采用如下简记
E ( n ) = E i 1 i 2 … i n ( n ) , Δ ( 2 n ) = Δ i 1 i 2 … i 2 n ( 2 n ) , δ ( 2 n ) = δ i 1 i 2 … i 2 n ( 2 n ) , E^{(n)} = E^{(n)}_{i_1 i_2 \dots i_{n}}, \quad
\Delta^{(2n)} = \Delta^{(2n)}_{i_1 i_2 \dots i_{2n}}, \quad
\delta^{(2n)} = \delta^{(2n)}_{i_1 i_2 \dots i_{2n}},
E ( n ) = E i 1 i 2 … i n ( n ) , Δ ( 2 n ) = Δ i 1 i 2 … i 2 n ( 2 n ) , δ ( 2 n ) = δ i 1 i 2 … i 2 n ( 2 n ) ,
这里的 Δ ( 2 n ) \Delta^{(2n)} Δ ( 2 n ) { i 1 , i 2 , … i 2 n } \{i_1, i_2, \dots i_{2n}\} { i 1 , i 2 , … i 2 n } Δ i 1 i 2 … i 2 n ( 2 n ) \Delta^{(2n)}_{i_1 i_2 \dots i_{2n}} Δ i 1 i 2 … i 2 n ( 2 n ) δ ( 2 n ) \delta^{(2n)} δ ( 2 n )
从 E ( 2 n ) E^{(2n)} E ( 2 n ) e α \boldsymbol{e}_{\alpha} e α
1 M ∑ α ( e α ⋅ v ) 2 n = ( 2 n − 1 ) ! ! d ( d + 2 ) . . . ( d + 2 n − 2 ) ⏟ Q 2 n ∥ v ∥ 2 n \frac{1}{M} \sum_{\alpha} (\boldsymbol{e}_{\alpha} \cdot \boldsymbol{v})^{2n} =
\underbrace{\frac{(2n-1)!!}{d (d+2) ... (d+2n-2)}}_{Q_{2n}} \|\boldsymbol{v}\|^{2n}
M 1 α ∑ ( e α ⋅ v ) 2 n = Q 2 n d ( d + 2 ) . . . ( d + 2 n − 2 ) ( 2 n − 1 ) ! ! ∥ v ∥ 2 n
对于二维情况 (d = 2 d=2 d = 2 Q 2 = 1 / 2 Q_2 = 1/2 Q 2 = 1 / 2 Q 4 = 3 / 8 Q_4 = 3/8 Q 4 = 3 / 8 Q 6 = 5 / 16 Q_6 = 5/16 Q 6 = 5 / 1 6 Q 8 = 35 / 128 Q_8 = 35/128 Q 8 = 3 5 / 1 2 8 d = 3 d=3 d = 3 Q 2 n = 1 / ( 2 n + 1 ) Q_{2n} = 1 / (2n+1) Q 2 n = 1 / ( 2 n + 1 )
同理，拓展到奇数阶，则有：
1 M ∑ α ( e α ⋅ v ) 2 n ( e α ⋅ v ) = Q 2 n ∥ v ∥ 2 n v \frac{1}{M} \sum_{\alpha} (\boldsymbol{e}_{\alpha} \cdot \boldsymbol{v})^{2n} (\boldsymbol{e}_{\alpha} \cdot \boldsymbol{v}) =
Q_{2n} \|\boldsymbol{v}\|^{2n} \boldsymbol{v}
M 1 α ∑ ( e α ⋅ v ) 2 n ( e α ⋅ v ) = Q 2 n ∥ v ∥ 2 n v
因此，对于一个所有 e α \boldsymbol{e}_{\alpha} e α Q 2 n Q_{2n} Q 2 n
假如 e α {\boldsymbol{e}_{\alpha}} e α
E i 1 i 2 . . . i n ( n ) = ∑ α ω ( ∥ e α ∥ ) ⋅ ( e α ) i 1 ( e α ) i 2 . . . ( e α ) i n E_{i_1 i_2 ... i_n}^{(n)} = \sum_{\alpha}
\omega(\|\boldsymbol{e}_{\alpha}\|) \cdot (\boldsymbol{e}_{\alpha})_{i_1} (\boldsymbol{e}_{\alpha})_{i_2} ... (\boldsymbol{e}_{\alpha})_{i_n}
E i 1 i 2 . . . i n ( n ) = α ∑ ω ( ∥ e α ∥ ) ⋅ ( e α ) i 1 ( e α ) i 2 . . . ( e α ) i n
其中 ω ( ∥ e α ∥ ) \omega(\|\boldsymbol{e}_{\alpha}\|) ω ( ∥ e α ∥ ) ∥ e α ∥ \|\boldsymbol{e}_{\alpha}\| ∥ e α ∥
Qian等的 DnQb 模型采用如下的平衡态分布：
f α e q ( ρ , u ; e α , ω α ) = ω α ρ ⋅ [ 1 + e α ⋅ u c s 2 + ( e α ⋅ u ) 2 2 c s 4 − ∥ u ∥ 2 2 c s 2 ] . f^{eq}_{\alpha} (\rho, \boldsymbol{u}; \boldsymbol{e}_{\alpha}, \omega_{\alpha}) = \omega_{\alpha} \rho \cdot
\left[
1 + \frac{\boldsymbol{e}_{\alpha} \cdot \boldsymbol{u}}{c_s^2} + \frac{(\boldsymbol{e}_{\alpha} \cdot \boldsymbol{u})^2}{2 c_s^4} - \frac{\|\boldsymbol{u}\|^2}{2 c_s^2}
\right].
f α e q ( ρ , u ; e α , ω α ) = ω α ρ ⋅ [ 1 + c s 2 e α ⋅ u + 2 c s 4 ( e α ⋅ u ) 2 − 2 c s 2 ∥ u ∥ 2 ] .
其中 ρ \rho ρ u \boldsymbol{u} u c s 2 = R T c_s^2 = RT c s 2 = R T
∑ α f α e q = ρ , ∑ α f α e q e α = ρ u , ∑ α f α e q e α i e α j = ρ u i u j + p δ i j . (2-2) \begin{aligned}
\sum_{\alpha} f^{eq}_{\alpha} &= \rho ,\\
\sum_{\alpha} f^{eq}_{\alpha} \boldsymbol{e}_{\alpha} &= \rho \boldsymbol{u} ,\\
\sum_{\alpha} f^{eq}_{\alpha} e_{\alpha i} e_{\alpha j} &= \rho u_i u_j + p \delta_{ij}.
\end{aligned}
\tag{2-2}
α ∑ f α e q α ∑ f α e q e α α ∑ f α e q e α i e α j = ρ , = ρ u , = ρ u i u j + p δ i j . ( 2 - 2 )
在待定系数法的框架下，就是把已经设计好的离散速度集 { e α } \{\boldsymbol{e}_{\alpha}\} { e α } ω α \omega_{\alpha} ω α
在这种视角下，我们要怎么确定就是要用 D2Q9、D3Q15、D3Q9 这些形式呢？
首先，正四边形最多能保证三阶格子张量是各向同性的，而到了四阶就变成了 E i j k l ( 4 ) ∣ M = 4 = 2 δ i j k l E_{ijkl}^{(4)}|_{M=4} = 2 \delta_{ijkl} E i j k l ( 4 ) ∣ M = 4 = 2 δ i j k l { e α } \{\boldsymbol{e}_{\alpha}\} { e α } M = 8 > 4 M=8>4 M = 8 > 4 E ( 4 ) E^{(4)} E ( 4 ) M M M
同样的道理，在三维空间中若是使用立方晶格，此时速度方向为立方体顶点（M = 8 M=8 M = 8 E ( 4 ) E^{(4)} E ( 4 ) E ( 4 ) E^{(4)} E ( 4 )
首先，我们把 D2Q9 分成三组
组别
速度配置
权重
1
( 0 , 0 ) (0,0) ( 0 , 0 ) ω 0 \omega_0 ω 0
2
cyc:( ± c , 0 ) (\pm c,0) ( ± c , 0 )
ω 1 = ω 2 = ω 3 = ω 4 \omega_1=\omega_2=\omega_3=\omega_4 ω 1 = ω 2 = ω 3 = ω 4
3
( ± c , ± c ) (\pm c,\pm c) ( ± c , ± c ) ω 5 = ω 6 = ω 7 = ω 8 \omega_5=\omega_6=\omega_7=\omega_8 ω 5 = ω 6 = ω 7 = ω 8
对于组别 2，根据格子张量的表格，我们有
∑ α ∈ [ 1 − 4 ] e α i = 0 , ∑ α ∈ [ 1 − 4 ] e α i e α j = 2 δ i j ⋅ c 2 ∑ α ∈ [ 1 − 4 ] e α i e α j e α k = 0 , ∑ α ∈ [ 1 − 4 ] e α i e α j e α k e α l = 2 δ i j k l ⋅ c 4 \begin{aligned}
\sum_{\alpha\in[1-4]} e_{\alpha i} &= 0 ,& \sum_{\alpha\in[1-4]} e_{\alpha i} e_{\alpha j} &= 2\delta_{ij} \cdot c^2\\
\sum_{\alpha\in[1-4]} e_{\alpha i} e_{\alpha j} e_{\alpha k} &= 0 ,& \sum_{\alpha\in[1-4]} e_{\alpha i} e_{\alpha j} e_{\alpha k} e_{\alpha l} &= 2\delta_{ijkl} \cdot c^4\\
\end{aligned}
α ∈ [ 1 − 4 ] ∑ e α i α ∈ [ 1 − 4 ] ∑ e α i e α j e α k = 0 , = 0 , α ∈ [ 1 − 4 ] ∑ e α i e α j α ∈ [ 1 − 4 ] ∑ e α i e α j e α k e α l = 2 δ i j ⋅ c 2 = 2 δ i j k l ⋅ c 4
对于组别 3，同理可得
∑ α ∈ [ 5 − 8 ] e α i = 0 , ∑ α ∈ [ 5 − 8 ] e α i e α j = 2 δ i j ⋅ ( 2 c ) 2 ∑ α ∈ [ 5 − 8 ] e α i e α j e α k = 0 , ∑ α ∈ [ 5 − 8 ] e α i e α j e α k e α l = ( 4 Δ i j k l ( 4 ) − 8 δ i j k l ( 4 ) ) ⋅ ( 2 c ) 4 \begin{aligned}
\sum_{\alpha\in[5-8]} e_{\alpha i} &= 0 ,& \sum_{\alpha\in[5-8]} e_{\alpha i} e_{\alpha j} &= 2\delta_{ij} \cdot (\sqrt{2} c)^2\\
\sum_{\alpha\in[5-8]} e_{\alpha i} e_{\alpha j} e_{\alpha k} &= 0 ,& \sum_{\alpha\in[5-8]} e_{\alpha i} e_{\alpha j} e_{\alpha k} e_{\alpha l} &= (4 \Delta^{(4)}_{ijkl} - 8 \delta^{(4)}_{ijkl}) \cdot (\sqrt{2} c)^4\\
\end{aligned}
α ∈ [ 5 − 8 ] ∑ e α i α ∈ [ 5 − 8 ] ∑ e α i e α j e α k = 0 , = 0 , α ∈ [ 5 − 8 ] ∑ e α i e α j α ∈ [ 5 − 8 ] ∑ e α i e α j e α k e α l = 2 δ i j ⋅ ( 2 c ) 2 = ( 4 Δ i j k l ( 4 ) − 8 δ i j k l ( 4 ) ) ⋅ ( 2 c ) 4
代入式(2-2)，依次得：
ρ ( ω 0 + 4 ω 1 + 4 ω 5 ) + ρ ∥ u ∥ 2 [ ω 1 ( c 2 c s 4 − 2 c s 2 ) + ω 5 ( 2 c 2 c s 4 − 2 c s 2 ) − ω 0 2 c s 2 ] = ρ (2-3) \rho (\omega_0 + 4 \omega_1 + 4 \omega_5) +
\rho \|\boldsymbol{u}\|^2 \left[ \omega_1 \left(\frac{c^2}{c_s^4} - \frac{2}{c_s^2}\right) \right.\\
\left. + \omega_5 \left(\frac{2 c^2}{c_s^4} - \frac{2}{c_s^2}\right) - \frac{\omega_0}{2 c_s^2} \right] = \rho
\tag{2-3}
ρ ( ω 0 + 4 ω 1 + 4 ω 5 ) + ρ ∥ u ∥ 2 [ ω 1 ( c s 4 c 2 − c s 2 2 ) + ω 5 ( c s 4 2 c 2 − c s 2 2 ) − 2 c s 2 ω 0 ] = ρ ( 2 - 3 )
ρ u ( ω 1 2 c 2 c s 2 + ω 5 4 c 2 c s 2 ) = ρ u (2-4) \rho \boldsymbol{u} \left(
\omega_1 \frac{2 c^2}{c_s^2} + \omega_5 \frac{4 c^2}{c_s^2}
\right) = \rho \boldsymbol{u}
\tag{2-4}
ρ u ( ω 1 c s 2 2 c 2 + ω 5 c s 2 4 c 2 ) = ρ u ( 2 - 4 )
ρ ω 1 ∑ α = 1 4 [ e α i e α j ( 1 − u 2 2 c s 2 ) + e α i e α j e α k e α l u k u l 2 c s 4 ] + ρ ω 5 ∑ α = 5 8 [ e α i e α j ( 1 − u 2 2 c s 2 ) + e α i e α j e α k e α l u k u l 2 c s 4 ] = ρ u i u j + p δ i j (2-5) \begin{aligned}
\rho \omega_1 \sum_{\alpha=1}^{4} \left[
e_{\alpha i} e_{\alpha j} \left(1 - \frac{u^2}{2 c_s^2}\right) +
\frac{e_{\alpha i} e_{\alpha j} e_{\alpha k} e_{\alpha l} u_k u_l}{2 c_s^4}
\right] &+ \\
\rho \omega_5 \sum_{\alpha=5}^{8} \left[
e_{\alpha i} e_{\alpha j} \left(1 - \frac{u^2}{2 c_s^2}\right) +
\frac{e_{\alpha i} e_{\alpha j} e_{\alpha k} e_{\alpha l} u_k u_l}{2 c_s^4}
\right] &= \rho u_i u_j + p \delta_{ij}
\end{aligned}
\tag{2-5}
ρ ω 1 α = 1 ∑ 4 [ e α i e α j ( 1 − 2 c s 2 u 2 ) + 2 c s 4 e α i e α j e α k e α l u k u l ] ρ ω 5 α = 5 ∑ 8 [ e α i e α j ( 1 − 2 c s 2 u 2 ) + 2 c s 4 e α i e α j e α k e α l u k u l ] + = ρ u i u j + p δ i j ( 2 - 5 )
由式(2-3)，得：
ω 0 + 4 ω 1 + 4 ω 5 = 1 ω 1 ( c 2 c s 4 − 2 c s 2 ) + ω 2 ( 2 c 2 c s 4 − 2 c s 2 ) − ω 0 2 c s 2 = 0 \omega_0 + 4 \omega_1 + 4 \omega_5 = 1 \\
\omega_1 \left(\frac{c^2}{c_s^4} - \frac{2}{c_s^2}\right) + \omega_2 \left(\frac{2 c^2}{c_s^4} - \frac{2}{c_s^2}\right) - \frac{\omega_0}{2 c_s^2} = 0
ω 0 + 4 ω 1 + 4 ω 5 = 1 ω 1 ( c s 4 c 2 − c s 2 2 ) + ω 2 ( c s 4 2 c 2 − c s 2 2 ) − 2 c s 2 ω 0 = 0
由式(2-4)，得：
( ω 1 + 2 ω 5 ) 2 c 2 c s 2 = 1 (\omega_1 + 2 \omega_5) \frac{2 c^2}{c_s^2} =1
( ω 1 + 2 ω 5 ) c s 2 2 c 2 = 1
式(2-5)需要用到如下化简：
Δ i j k l ( 4 ) u k u l = δ i j ( δ k l u k u l ) + ( δ i k δ j l u k u l ) + ( δ i l δ j k u k u l ) = δ i j ∥ u ∥ 2 + 2 u i u j \begin{aligned}
\Delta^{(4)}_{ijkl} u_{k} u_{l}
&= \delta_{ij} (\delta_{kl} u_{k} u_{l}) + (\delta_{ik}\delta_{jl}u_{k} u_{l}) + (\delta_{il}\delta_{jk} u_{k} u_{l}) \\
&=\delta_{ij} \|\boldsymbol{u}\|^2 + 2 u_i u_j
\end{aligned}
Δ i j k l ( 4 ) u k u l = δ i j ( δ k l u k u l ) + ( δ i k δ j l u k u l ) + ( δ i l δ j k u k u l ) = δ i j ∥ u ∥ 2 + 2 u i u j
其中：
第一项仅当 k = l k=l k = l δ k l u k u l = ∑ k = l u k u l = ∑ k u k u k = ∥ u ∥ 2 \delta_{kl} u_k u_l = \sum_{k=l} u_k u_l = \sum_{k} u_k u_k = \|\boldsymbol{u}\|^2 δ k l u k u l = ∑ k = l u k u l = ∑ k u k u k = ∥ u ∥ 2
第二项 δ i k \delta_{ik} δ i k k = i k=i k = i δ j l \delta_{jl} δ j l l = j l=j l = j k , l k,l k , l k = i k=i k = i l = j l=j l = j u i u j u_i u_j u i u j
第三项 δ i l \delta_{il} δ i l l = i l=i l = i δ j k \delta_{jk} δ j k k = j k=j k = j l = i l=i l = i k = j k=j k = j u i u j u_i u_j u i u j
由于式(2-5)的左侧可被简化为
( ω 1 − 4 ω 5 ) c 4 c s 4 ρ δ i j k l u k u l + ρ c 2 δ i j [ ( 1 − ∥ u ∥ 2 2 c s 2 ) ( 2 ω 1 + 4 ω 5 ) + 2 ω 5 ∥ u ∥ 2 c 2 c s 4 ] + ρ u i u j w 5 4 c 2 c s 4 (2-6) \begin{aligned}
(\omega_1 - 4 \omega_5) \frac{c^4}{c_s^4} \rho \delta_{ijkl} u_k u_l &+
\rho c^2 \delta_{ij} \left[ (1 - \frac{\|\boldsymbol{u}\|^2}{2 c_s^2}) (2\omega_1 + 4\omega_5) \right. \\
&\left. + 2 \omega_5 \frac{\|\boldsymbol{u}\|^2 c^2}{c_s^4} \right] +
\rho u_i u_j w_5 \frac{4 c^2}{c_s^4}
\tag{2-6}
\end{aligned}
( ω 1 − 4 ω 5 ) c s 4 c 4 ρ δ i j k l u k u l + ρ c 2 δ i j [ ( 1 − 2 c s 2 ∥ u ∥ 2 ) ( 2 ω 1 + 4 ω 5 ) + 2 ω 5 c s 4 ∥ u ∥ 2 c 2 ] + ρ u i u j w 5 c s 4 4 c 2 ( 2 - 6 )
所以
w 5 4 c 2 c s 4 = 1 ω 1 − 4 ω 5 = 0 ( 1 − ∥ u ∥ 2 2 c s 2 ) ( 2 ω 1 + 4 ω 5 ) + 2 ω 5 ∥ u ∥ 2 c 2 c s 4 = p ρ c 2 \begin{aligned}
w_5 \frac{4 c^2}{c_s^4} &= 1 \\
\omega_1 - 4 \omega_5 &= 0 \\
\left(1 - \frac{\|\boldsymbol{u}\|^2}{2 c_s^2}\right) (2\omega_1 + 4\omega_5) + 2 \omega_5 \frac{\|\boldsymbol{u}\|^2 c^2}{c_s^4} &= \frac{p}{\rho c^2}
\end{aligned}
w 5 c s 4 4 c 2 ω 1 − 4 ω 5 ( 1 − 2 c s 2 ∥ u ∥ 2 ) ( 2 ω 1 + 4 ω 5 ) + 2 ω 5 c s 4 ∥ u ∥ 2 c 2 = 1 = 0 = ρ c 2 p
联立这些方程就可以解得： ω 0 = 4 / 9 \omega_0 = 4/9 ω 0 = 4 / 9 ω 1 = 1 / 9 \omega_1 = 1/9 ω 1 = 1 / 9 ω 5 = 1 / 36 \omega_5 = 1/36 ω 5 = 1 / 3 6 c s = c / 3 c_s = c / \sqrt{3} c s = c / 3
（1）对于二维正M边形网格，e α = e ( cos ( 2 π α M ) , sin ( 2 π α M ) ) \boldsymbol{e}_{\alpha} = e \left( \cos(\frac{2\pi\alpha}{M}), \sin(\frac{2\pi\alpha}{M})\right) e α = e ( cos ( M 2 π α ) , sin ( M 2 π α ) ) n n n M M M
各向同性张量 E ( n ) E^{(n)} E ( n )
正多边形边数 M M M
E ( 2 ) E^{(2)} E ( 2 ) M > 2 M \gt 2 M > 2
E ( 3 ) E^{(3)} E ( 3 ) M ≥ 2 , M ≠ 3 M \ge 2, M \neq 3 M ≥ 2 , M = 3
E ( 4 ) E^{(4)} E ( 4 ) M > 2 , M ≠ 4 M \gt 2, M \neq 4 M > 2 , M = 4
E ( 5 ) E^{(5)} E ( 5 ) M ≥ 2 , M ≠ [ 3 , 5 ] M \ge 2, M \neq [3,5] M ≥ 2 , M = [ 3 , 5 ]
E ( 6 ) E^{(6)} E ( 6 ) M > 4 , M ≠ 6 M \gt 4, M \neq 6 M > 4 , M = 6
E ( 7 ) E^{(7)} E ( 7 ) M ≥ 2 , M ≠ [ 3 , 5 , 7 ] M \ge 2, M \neq [3,5,7] M ≥ 2 , M = [ 3 , 5 , 7 ]
（2）文献[2]给出了三维情况下 E ( n ) E^{(n)} E ( n ) ϕ = ( 1 + 5 ) / 2 \phi=(1+\sqrt{5})/2 ϕ = ( 1 + 5 ) / 2
几何形状
速度配置
M
E ( 2 ) E^{(2)} E ( 2 ) E ( 3 ) E^{(3)} E ( 3 ) E ( 4 ) E^{(4)} E ( 4 ) E ( 5 ) E^{(5)} E ( 5 ) E ( 6 ) E^{(6)} E ( 6 )
四面体
( 1 , 1 , 1 ) (1,1,1) ( 1 , 1 , 1 ) ( 1 , − 1 , − 1 ) (1,-1,-1) ( 1 , − 1 , − 1 ) 4
Y
N
N
N
N
立方体
cyc:( ± 1 , ± 1 , ± 1 ) (\pm 1,\pm 1,\pm 1) ( ± 1 , ± 1 , ± 1 )
8
Y
Y
N
Y
N
八面体
cyc:( ± 1 , 0 , 0 ) (\pm 1,0,0) ( ± 1 , 0 , 0 )
6
Y
Y
N
Y
N
十二面体
( ± 1 , ± 1 , ± 1 ) (\pm 1,\pm 1,\pm 1) ( ± 1 , ± 1 , ± 1 ) ( 0 , ± ϕ − 1 , ± ϕ ) (0, \pm \phi-1, \pm \phi) ( 0 , ± ϕ − 1 , ± ϕ ) 20
Y
Y
Y
Y
N
二十面体
cyc: ( 0 , ± ϕ , ± 1 ) (0, \pm \phi, \pm 1) ( 0 , ± ϕ , ± 1 )
12
Y
Y
Y
Y
N
（3）两种基本正方形格子模型的偶数阶张量
速度配置
M
E ( 2 ) E^{(2)} E ( 2 ) E ( 4 ) E^{(4)} E ( 4 ) E ( 6 ) E^{(6)} E ( 6 )
cyc:( ± 1 , 0 ) (\pm 1,0) ( ± 1 , 0 )
4
2 δ ( 2 ) 2 \delta^{(2)} 2 δ ( 2 ) 2 δ ( 4 ) 2 \delta^{(4)} 2 δ ( 4 ) 2 δ ( 6 ) 2 \delta^{(6)} 2 δ ( 6 )
( ± 1 , ± 1 ) (\pm 1,\pm 1) ( ± 1 , ± 1 ) 8
4 δ ( 2 ) 4 \delta^{(2)} 4 δ ( 2 ) 4 Δ ( 4 ) − 8 δ ( 4 ) 4 \Delta^{(4)} - 8 \delta^{(4)} 4 Δ ( 4 ) − 8 δ ( 4 ) 4 3 Δ ( 6 ) − 16 δ ( 6 ) \frac{4}{3} \Delta^{(6)} - 16 \delta^{(6)} 3 4 Δ ( 6 ) − 1 6 δ ( 6 )
（4）几种最对称的三维Bravais晶格的偶数阶张量
形状
速度配置
M
E ( 2 ) E^{(2)} E ( 2 ) E ( 4 ) E^{(4)} E ( 4 ) E ( 6 ) E^{(6)} E ( 6 )
Primitive cubic
cyc:( ± 1 , 0 , 0 ) (\pm 1,0,0) ( ± 1 , 0 , 0 )
6
2 δ ( 2 ) 2 \delta^{(2)} 2 δ ( 2 ) 2 δ ( 4 ) 2 \delta^{(4)} 2 δ ( 4 ) 2 δ ( 6 ) 2 \delta^{(6)} 2 δ ( 6 )
Body-centered cubic
( ± 1 , ± 1 , ± 1 ) (\pm 1,\pm 1,\pm 1) ( ± 1 , ± 1 , ± 1 ) 8
8 δ ( 2 ) 8 \delta^{(2)} 8 δ ( 2 ) 8 ( Δ ( 4 ) − 2 δ ( 4 ) ) 8 (\Delta^{(4)} - 2\delta^{(4)}) 8 ( Δ ( 4 ) − 2 δ ( 4 ) ) 8 ( Δ ( 6 ) − 2 Δ ( 4 , 2 ) + 16 δ ( 6 ) ) 8 (\Delta^{(6)} - 2 \Delta^{(4,2)} + 16\delta^{(6)}) 8 ( Δ ( 6 ) − 2 Δ ( 4 , 2 ) + 1 6 δ ( 6 ) )
Face-centered cubic
cyc:( ± 1 , ± 1 , 0 ) (\pm 1,\pm 1,0) ( ± 1 , ± 1 , 0 )
12
8 δ ( 2 ) 8 \delta^{(2)} 8 δ ( 2 ) 4 ( Δ ( 4 ) − δ ( 4 ) ) 4 (\Delta^{(4)} - \delta^{(4)}) 4 ( Δ ( 4 ) − δ ( 4 ) ) 4 ( Δ ( 4 , 2 ) + 13 δ ( 6 ) ) 4 (\Delta^{(4,2)} + 13\delta^{(6)}) 4 ( Δ ( 4 , 2 ) + 1 3 δ ( 6 ) )
在文献[2]的原话中：
Δ ( n , m ) \Delta^{(n,m)} Δ ( n , m ) n n n m m m
具体到 Δ ( 4 , 2 ) = Δ i j k l m n ( 4 , 2 ) \Delta^{(4,2)} = \Delta^{(4,2)}_{ijklmn} Δ ( 4 , 2 ) = Δ i j k l m n ( 4 , 2 ) C 6 2 = 15 \mathrm{C}_{6}^{2} = 15 C 6 2 = 1 5 δ i j Δ k l m n ( 4 ) \delta_{ij}\Delta^{(4)}_{klmn} δ i j Δ k l m n ( 4 ) δ i k Δ j l m n ( 4 ) \delta_{ik}\Delta^{(4)}_{jlmn} δ i k Δ j l m n ( 4 )
[1] HE X, LUO L S. Theory of the lattice Boltzmann method: From the Boltzmann equation to the lattice Boltzmann equation [J]. Physical Review E , 1997, 56(6): 6811-6817. DOI:10.1103/PhysRevE.56.6811 .Cellular automaton fluids 1: Basic theory [J]. Journal of Statistical Physics , 1986, 45(3-4): 471-526. DOI:10.1007/BF01021083 .格子Boltzmann方法的理论及应用 [M]. 1 版. 北京: 高等教育出版社, 2023.