Robust Quasistatic Finite Elements and Flesh Simulation

最新推荐文章于 2018-06-25 20:14:48 发布

seamanj

最新推荐文章于 2018-06-25 20:14:48 发布

阅读量942

点赞数

分类专栏： paper

本文链接：https://blog.youkuaiyun.com/seamanj/article/details/50230187

版权

paper 专栏收录该内容

106 篇文章

订阅专栏

本文探讨了刚体力学中的准静态有限元方法，包括应变能量、全局刚度矩阵的对称性原理，以及如何确保正定性的数学推导。通过详细的数学公式和推导过程，阐述了在特定条件下全局刚度矩阵保持对称和正定的原因。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

文章来源:
Robust Quasistatic Finite Elements and Flesh Simulation

3.Quasistatic Formulation

{
公式 $\vec{x_t}=\vec{v}$ 以及 $\vec{v_t}=M^{-1}\vec{f}(t,\vec{x},\vec{t})$ 中下标 $t$ 表示对 $t$ 求导
}

4.Strain Energy

That is, the global stiffness matrix $-\partial{\vec{f}}/\partial{\vec{x}}$ is always symmetric, as a result of the hyperelastic energy having continuous second derivatives with respect to the spatial configuration.

{
[引用于https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Schwarz.27_theorem]
Schwarz’ theorem
In mathematical analysis, Schwarz’ theorem (or Clairaut’s theorem) named after Alexis Clairaut and Hermann Schwarz, states that if
$f \colon \mathbb{R}^n \to \mathbb{R}$
has continuous second partial derivatives at any given point in $\mathbb{R}^n$ , say, $(a_1, \dots, a_n)$ , then $\forall i, j \in \{ 1,2,\ldots, n\}$ ,

\partial 2 f \partial x i \partial x j (a 1, \dots, a n) = \partial 2 f \partial x j \partial x i (a 1, \dots, a n),

$\frac{\partial^2 f}{\partial x_i\, \partial x_j}(a_1, \dots, a_n) = \frac{\partial^2 f}{\partial x_j\, \partial x_i}(a_1, \dots, a_n),$
The partial derivations of this function are commutative at that point.
}
{
所以对于

Ψ $\Psi$ 的黑塞

H $H$ 矩阵中的元素来说

Hij=∂2Ψ∂xi∂xj=∂2Ψ∂xj∂xi=Hji $H_{ij}=\frac{\partial^2 \Psi}{\partial x_i\, \partial x_j}=\frac{\partial^2 \Psi}{\partial x_j\, \partial x_i}=H_{ji}$ , 所以这里说the global stiffness matrix是对称的.
}

Furthermore, a steady state corresponds to a local minimum of the hyperelastic energy indicating that the energy Hessian, $\partial^2 \Psi / \partial \vec{x}^2$ , (or equivalently the global stiffness matrix)is positive definite in the vicinity of an isolated steady state.
{
[引用于Hessian Matrix]
也就是在临界点时,黑塞矩阵正定则对应函数值的最小点
}
{
Critical points
If the gradient (the vector of the partial derivatives) of a function f is zero at some point x, then f has a critical point (or stationary point) at x.
}

7.Diagonalization

公式(4)的推导

δ P = \partial U P ( U T F V ) V T \partial ( U T F V ) ∣ ∣ ∣ F : δ (U T F V) = U {\partial P ( F ) \partial F ∣ ∣ ∣ U T F V : U T δ F V} V T

$\begin{align} \delta P &= \frac{\partial{UP(U^TFV)V^T}}{\partial(U^TFV)}\bigg|_F:\delta(U^TFV) \\ &= U\left\{ \frac{\partial{P(F)}}{\partial F}\bigg|_{U^TFV}:U^T\delta FV \right\}V^T \end{align}$

{
这里要证明两点:
(1) $\delta(U^TFV)=U^T\delta FV$
(2) $\frac{\partial{UP(U^TFV)V^T}}{\partial(U^TFV)}\bigg|_F:\delta(U^TFV) = U\left\{ \frac{\partial{P(F)}}{\partial F}\bigg|_{U^TFV}:U^T\delta FV \right\}V^T$
证明:
(1)

δ (U T F V) = δ (U i j e^j \otimes e^i F m n e^m \otimes e^n V p q e^p \otimes e^q) = U i j e^j \otimes e^i δ F m n e^m \otimes e^n V p q e^p \otimes e^q = U T δ F m n V

$\begin{align} \delta(U^TFV) &=\delta(U_{ij}\hat{e}_j\otimes\hat{e}_iF_{mn}\hat{e}_m\otimes\hat{e}_nV_{pq}\hat{e}_p\otimes\hat{e}_q)\\&=U_{ij}\hat{e}_j\otimes\hat{e}_i\delta F_{mn}\hat{e}_m\otimes\hat{e}_nV_{pq}\hat{e}_p\otimes\hat{e}_q\\&=U^T\delta F_{mn}V \end{align}$
(2)

∂UP(UTFV)VT∂(UTFV)∣∣∣F:δ(UTFV)=∂UP(F)VT∂(F)∣∣∣UTFV:UTδ(F)V=∂(Uije^i⊗e^jPmne^m⊗e^nVpqe^q⊗e^p)∂(Fxye^x⊗e^y):ϵabe^a⊗e^b=Uij∂Pmn∂FxyVpq(e^i⊗e^j)⋅(e^m⊗e^n)⋅(e^q⊗e^p)⊗(e^x⊗e^y):ϵabe^a⊗e^b=Uij(e^i⊗e^j){∂Pmn∂Fxy(e^m⊗e^n)⋅(e^x⊗e^y:ϵabe^a⊗e^b)}Vpq(e^q⊗e^p)=Uij(e^i⊗e^j){∂Pmn∂Fxye^m⊗e^n⊗e^x⊗e^y:ϵabe^a⊗e^b}Vpq(e^q⊗e^p)=U{∂P(F)∂F∣∣∣UTFV:UTδFV}VT

$\begin{align} &\frac{\partial{UP(U^TFV)V^T}} {\partial(U^TFV)}\bigg|_F:\delta(U^TFV) \\&= \frac{\partial{UP(F)V^T}}{\partial(F)}\bigg|_{U^TFV}:U^T\delta(F)V \\&=\frac{\partial{(U_{ij}\hat{e}_i\otimes\hat{e}_jP_{mn}\hat{e}_m\otimes\hat{e}_nV_{pq}\hat{e}_q\otimes\hat{e}_p)}}{\partial(F_{xy}\hat{e}_x\otimes\hat{e}_y)}:\epsilon_{ab}\hat{e}_a\otimes\hat{e}_b \\&=U_{ij}\frac{\partial{P_{mn}}}{\partial F_{xy}}V_{pq}(\hat{e}_i\otimes\hat{e}_j)\cdot(\hat{e}_m\otimes\hat{e}_n)\cdot(\hat{e}_q\otimes\hat{e}_p)\otimes(\hat{e}_x\otimes\hat{e}_y):\epsilon_{ab}\hat{e}_a\otimes\hat{e}_b\\&=U_{ij}(\hat{e}_i\otimes\hat{e}_j)\left\{ \frac{\partial{P_{mn}}}{\partial F_{xy}}(\hat{e}_m\otimes\hat{e}_n)\cdot(\hat{e}_x\otimes\hat{e}_y:\epsilon_{ab}\hat{e}_a\otimes\hat{e}_b)\right\} V_{pq}(\hat{e}_q\otimes\hat{e}_p)\\&=U_{ij}(\hat{e}_i\otimes\hat{e}_j)\left\{ \frac{\partial{P_{mn}}}{\partial F_{xy}}\hat{e}_m\otimes\hat{e}_n\otimes\hat{e}_x\otimes\hat{e}_y:\epsilon_{ab}\hat{e}_a\otimes\hat{e}_b\right\} V_{pq}(\hat{e}_q\otimes\hat{e}_p)\\&=U\left\{ \frac{\partial{P(F)}}{\partial F}\bigg|_{U^TFV}:U^T\delta FV \right\}V^T \end{align}$

注意这里运用到的一些公式:
引用来源:tensors.pdf p31,p83

这里写图片描述

}

8. Enforcing Positive Definiteness

{
我们先看下P与F的关系
引用于The classical FEM method and discretization methodology: Course Notes (Eftychios Sifakis) p23
这里写图片描述

将F写成 $F =U\hat FV^T$ ，则 $P=U(\Psi_I\cdot2\hat F+\Psi_{II}\cdot4\hat F^3+\Psi_{III}\cdot2III\cdot\hat F^{-1})V^T$
写成索引形式:
$P_{ij}=f(\sigma_1,\sigma_2,\sigma_3)_kU_{ik}V_{jk}$
将F写成索引形式:
$F_{ij}=\sigma_kU_{ik}V_{jk}$
那么
$\frac{\partial P_{ij}}{\partial F_{ij}}\bigg|_\hat{F}=\frac{\partial P_{ij}}{\partial F_{ij}}\bigg|_{\sigma_1,\sigma_2,\sigma_3}$
注意这里自变量为 $\hat F$ ,即把变量限制为 $\sigma_1,\sigma_2,\sigma_3$ ,当然此时 $U_{ik},V_{jk}$ 此时就变成了 $\delta_{ik},\delta_{jk}$
另外约定: $\frac{\partial c}{\partial a},\frac{\partial a}{\partial c}$ 为0, $\frac{\partial c_1}{\partial c_2}$ 是无意义的,其中 $a$ 为变量， $c$ , $c_1$ , $c_2$ 为变量
所以通过这种方式可以确定9*9矩阵里面的前三行和前三列，交叉部分的偏导可以把变量限制为 ${\sigma_1,\sigma_2,\sigma_3}$ ,其余部分为0,那么求 $A$ 的时候我们会用这种形式来求. $B_{12}$ , $B_{13}$ , $B_{23}$ 我们用另外一种方式来求.

先来求 $B$
另外注意一点 ${\sigma_1,\sigma_2,\sigma_3}$ 与F的非对角线元素关于变量 ${\sigma_1,\sigma_2,\sigma_3}$ 相互独立的
即
$\frac{\partial f(\sigma_1,\sigma_2,\sigma_3)}{\partial F_{ij}}\bigg|_\hat F=\frac{\partial f(\sigma_1,\sigma_2,\sigma_3)}{\partial \sigma_kU_{ik}V_{jk}}\bigg|_\hat F=\frac{\partial f(\sigma_1,\sigma_2,\sigma_3)}{\partial \sigma_k \delta_{ij}}\bigg|_{(\sigma_1,\sigma_2,\sigma_3)}=\frac{\partial f(\sigma_1,\sigma_2,\sigma_3)}{\partial 0}\bigg|_{(\sigma_1,\sigma_2,\sigma_3)}=0$
注意由于是非对角线元素所以， $i\neq j$ ,即 $\delta_{ij}=0$
下面开始求B,注意下面中的 $i\neq j$ , $p\neq q$
$\frac{\partial F_{ij}}{\partial F_{pq}}=\delta_{ip}\delta_{jq}$

\partial ( F F T F ) i j \partial F p q = \partial F i k F l k F l j \partial F p q = \partial F i k \partial F p q F l k F l j + F i k \partial F l k \partial F p q F l j + F i k F l k \partial F l j \partial F p q = δ i p δ k q F l k F l j + δ l p δ k q F i k F l j + δ l p δ j q F i k F l k = δ i p δ k q F l k F l j + F i q F p j + δ l p δ j q F i k F l k

$\frac{\partial (FF^TF)_{ij}}{\partial F_{pq}}=\frac{\partial F_{ik}F_{lk}F_{lj}}{\partial F_{pq}}=\frac{\partial F_{ik}}{\partial F_{pq}}F_{lk}F_{lj}+F_{ik}\frac{\partial F_{lk}}{\partial F_{pq}}F_{lj}+F_{ik}F_{lk}\frac{\partial F_{lj}}{\partial F_{pq}}\\=\delta_{ip}\delta_{kq}F_{lk}F_{lj}+\delta_{lp}\delta_{kq}F_{ik}F_{lj}+\delta_{lp}\delta_{jq}F_{ik}F_{lk}=\delta_{ip}\delta_{kq}F_{lk}F_{lj}+F_{iq}F_{pj}+\delta_{lp}\delta_{jq}F_{ik}F_{lk}$
这里强调下

δ $\delta$ 的化简方式通常是往常量去化,比如说:

δliδkjFikFlj $\delta_{li}\delta_{kj}F_{ik}F_{lj}$ 这里

i,j $i,j$ 为常量,

l,k $l,k$ 为变量 ,所以说应该化

FijFij $F_{ij}F_{ij}$

下面引用来源:tensors.pdf p50
$A^{-1}=\frac{adj(A)}{|A|}=\frac{cof(A)^T}{|A|}$
PDF上其实有点错误,修正因为
$C_{1i}=\epsilon_{ijk}A_{2j}A_{3k},C_{2i}=\epsilon_{ijk}A_{3j}A_{1k},C_{3i}=\epsilon_{ijk}A_{1j}A_{2k}$
这里C代表A的余子阵cofacter

所以

F - T i j = c o f ( F ) i j | F | = ϵ j m n F ( i \oplus 1 ) m F ( i \oplus 2 ) n | F |

$F^{-T}_{ij}=\frac{cof(F)_{ij}}{|F|}=\frac{\epsilon_{jmn}F_{(i\oplus1)m}F_{(i\oplus2)n}}{|F|}$
注意这里\oplus记成模3循环加法,我也不知道怎么定义=.=,自创的,即

2⊕2=1 $2\oplus2=1$

\partial F - T i j \partial F p q = \partial ϵ j m n F ( i \oplus 1 ) m F ( i \oplus 2 ) n \partial | F | F p q

$\frac{\partial F^{-T}_{ij}}{\partial F_{pq}}=\frac{\partial \epsilon_{jmn}F_{(i\oplus1)m}F_{(i\oplus2)n}}{\partial |F|F_{pq}}$

现在小结一下:

P 1 = \partial F i j \partial F p q = δ i p δ j q

$P1=\frac{\partial F_{ij}}{\partial F_{pq}}=\delta_{ip}\delta_{jq}$

P 2 = \partial ( F F T F ) i j \partial F p q = δ i p δ k q F l k F l j + F i q F p j + δ l p δ j q F i k F l k

$P2=\frac{\partial (FF^TF)_{ij}}{\partial F_{pq}}=\delta_{ip}\delta_{kq}F_{lk}F_{lj}+F_{iq}F_{pj}+\delta_{lp}\delta_{jq}F_{ik}F_{lk}$

P 3 = \partial F - T i j \partial F p q = \partial ϵ j m n F ( i \oplus 1 ) m F ( i \oplus 2 ) n \partial | F | F p q

$P3=\frac{\partial F^{-T}_{ij}}{\partial F_{pq}}=\frac{\partial \epsilon_{jmn}F_{(i\oplus1)m}F_{(i\oplus2)n}}{\partial |F|F_{pq}}$

这里写图片描述
这样就把 $B$ 中的 $\alpha,\beta$ 求出来了

下面来求 $A$
$P_{ii}=U_{ik}V_{ik}(\Psi_I\cdot2\sigma_k+\Psi_{II}\cdot4\sigma_k^3+\Psi_{III}\cdot2III\cdot\frac{1}{\sigma_k})$
$F_{jj}=U_{jk}V_{jk}\sigma_k$
如果 $i=j$

\partial P i i \partial F i i ∣ ∣ ∣ F^= \partial ( Ψ I \cdot 2 σ i + Ψ I I \cdot 4 σ 3 i + Ψ I I I \cdot 2 I I I \cdot 1 σ i ) \partial σ i = \partial ( [ 2 σ i , 4 σ 3 i , 2 I I I σ i ] ⎡ ⎣ ⎢ Ψ I Ψ I I Ψ I I I ⎤ ⎦ ⎥ ) \partial σ i = \partial ( [ 2 σ i , 4 σ 3 i , 2 I I I σ i ] ) \partial σ i \cdot ⎡ ⎣ ⎢ Ψ I Ψ I I Ψ I I I ⎤ ⎦ ⎥ + [2 σ i, 4 σ 3 i, 2 I I I σ i] \cdot \partial ⎡ ⎣ ⎢ Ψ I Ψ I I Ψ I I I ⎤ ⎦ ⎥ \partial σ i = α i i + β i i + 4 I I I Ψ I I I σ 2 i + [2 σ i, 4 σ 3 i, 2 I I I σ i] \cdot ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ Ψ I, I \cdot \partial I \partial σ i Ψ I, I I \cdot \partial I I \partial σ i Ψ I, I I I \cdot \partial I I I \partial σ i Ψ I I, I \cdot \partial I \partial σ i Ψ I I, I I \cdot \partial I I \partial σ i Ψ I, I I I \cdot \partial I I I \partial σ i Ψ I I I, I \cdot \partial I \partial σ i Ψ I I I, I I \cdot \partial I I \partial σ i Ψ I I I, I I I \cdot \partial I I I \partial σ i ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = α i i + β i i + 4 I I I Ψ I I I σ 2 i + [2 σ i, 4 σ 3 i, 2 I I I σ i] \cdot ⎡ ⎣ ⎢ Ψ I, I Ψ I, I I Ψ I, I I I Ψ I I, I Ψ I I, I I Ψ I, I I I Ψ I I I, I Ψ I I I, I I Ψ I I I, I I I ⎤ ⎦ ⎥ \cdot ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ \partial I \partial σ i \partial I I \partial σ i \partial I I I \partial σ i ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

$\begin{align} \frac{\partial P_{ii}}{\partial F_{ii}}\bigg|_\hat F&=\frac{\partial (\Psi_I\cdot2\sigma_i+\Psi_{II}\cdot4\sigma_i^3+\Psi_{III}\cdot2III\cdot\frac{1}{\sigma_i})}{\partial \sigma_i}\\&=\frac{\partial ([2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\begin{bmatrix} \Psi_I \\ \Psi_{II}\\ \Psi_{III}\\ \end{bmatrix})}{\partial \sigma_i}\\&=\frac{\partial ([2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}])}{\partial \sigma_i}\cdot\begin{bmatrix} \Psi_I \\ \Psi_{II}\\ \Psi_{III}\\ \end{bmatrix}+[2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\cdot\frac{\partial \begin{bmatrix} \Psi_I \\ \Psi_{II}\\ \Psi_{III}\\ \end{bmatrix}}{\partial \sigma_i}\\&=\alpha_{ii}+\beta_{ii}+\frac{4III\Psi_{III}}{\sigma_i^2}+[2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\cdot\begin{bmatrix} \Psi_{I,I}\cdot\frac{\partial I}{\partial \sigma_i } \quad \Psi_{I,II}\cdot\frac{\partial II}{\partial \sigma_i } \quad \Psi_{I,III}\cdot\frac{\partial III}{\partial \sigma_i }\\ \Psi_{II,I}\cdot\frac{\partial I}{\partial \sigma_i } \quad \Psi_{II,II}\cdot\frac{\partial II}{\partial \sigma_i } \quad \Psi_{I,III}\cdot\frac{\partial III}{\partial \sigma_i }\\ \Psi_{III,I}\cdot\frac{\partial I}{\partial \sigma_i } \quad \Psi_{III,II}\cdot\frac{\partial II}{\partial \sigma_i } \quad \Psi_{III,III}\cdot\frac{\partial III}{\partial \sigma_i }\\ \end{bmatrix}\\&=\alpha_{ii}+\beta_{ii}+\frac{4III\Psi_{III}}{\sigma_i^2}+[2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\cdot\begin{bmatrix} \Psi_{I,I} \quad \Psi_{I,II} \quad \Psi_{I,III}\\ \Psi_{II,I} \quad \Psi_{II,II} \quad \Psi_{I,III}\\ \Psi_{III,I} \quad \Psi_{III,II}\quad \Psi_{III,III}\\ \end{bmatrix}\cdot\begin{bmatrix} \frac{\partial I}{\partial \sigma_i } \\ \frac{\partial II}{\partial \sigma_i }\\ \frac{\partial III}{\partial \sigma_i }\\ \end{bmatrix} \end{align}$
引用于The classical FEM method and discretization methodology: Course Notes (Eftychios Sifakis) p23
这里写图片描述

\partial P i i \partial F i i ∣ ∣ ∣ F^= α i i + β i i + 4 I I I Ψ I I I σ 2 i + [2 σ i, 4 σ 3 i, 2 I I I σ i] \cdot ⎡ ⎣ ⎢ Ψ I, I Ψ I, I I Ψ I, I I I Ψ I I, I Ψ I I, I I Ψ I, I I I Ψ I I I, I Ψ I I I, I I Ψ I I I, I I I ⎤ ⎦ ⎥ \cdot ⎡ ⎣ ⎢ ⎢ ⎢ 2 σ i 4 σ 3 i 2 I I I σ i ⎤ ⎦ ⎥ ⎥ ⎥ = α i i + β i i + γ i i

$\begin{align} \frac{\partial P_{ii}}{\partial F_{ii}}\bigg|_\hat F&=\alpha_{ii}+\beta_{ii}+\frac{4III\Psi_{III}}{\sigma_i^2}+[2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\cdot\begin{bmatrix} \Psi_{I,I} \quad \Psi_{I,II} \quad \Psi_{I,III}\\ \Psi_{II,I} \quad \Psi_{II,II} \quad \Psi_{I,III}\\ \Psi_{III,I} \quad \Psi_{III,II}\quad \Psi_{III,III}\\ \end{bmatrix}\cdot\begin{bmatrix} 2\sigma_i \\ 4\sigma_i^3\\ \frac{2III}{\sigma_i}\\ \end{bmatrix}&=\alpha_{ii}+\beta_{ii}+\gamma_{ii} \end{align}$

当 $i \neq j$

\partial P i i \partial F j j ∣ ∣ ∣ F^= \partial ( [ 2 σ i , 4 σ 3 i , 2 I I I σ i ] ⎡ ⎣ ⎢ Ψ I Ψ I I Ψ I I I ⎤ ⎦ ⎥ ) \partial σ j = \partial ( [ 2 σ i , 4 σ 3 i , 2 I I I σ i ] ) \partial σ j \cdot ⎡ ⎣ ⎢ Ψ I Ψ I I Ψ I I I ⎤ ⎦ ⎥ + [2 σ i, 4 σ 3 i, 2 I I I σ i] \cdot \partial ⎡ ⎣ ⎢ Ψ I Ψ I I Ψ I I I ⎤ ⎦ ⎥ \partial σ j = 4 I I I Ψ I I I σ i σ j + [2 σ i, 4 σ 3 i, 2 I I I σ i] \cdot ⎡ ⎣ ⎢ Ψ I, I Ψ I, I I Ψ I, I I I Ψ I I, I Ψ I I, I I Ψ I, I I I Ψ I I I, I Ψ I I I, I I Ψ I I I, I I I ⎤ ⎦ ⎥ \cdot ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 2 σ j 4 σ 3 j 2 I I I σ j ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = γ i j

$\begin{align} \frac{\partial P_{ii}}{\partial F_{jj}}\bigg|_\hat F&=\frac{\partial ([2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\begin{bmatrix} \Psi_I \\ \Psi_{II}\\ \Psi_{III}\\ \end{bmatrix})}{\partial \sigma_j}\\&=\frac{\partial ([2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}])}{\partial \sigma_j}\cdot\begin{bmatrix} \Psi_I \\ \Psi_{II}\\ \Psi_{III}\\ \end{bmatrix}+[2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\cdot\frac{\partial \begin{bmatrix} \Psi_I \\ \Psi_{II}\\ \Psi_{III}\\ \end{bmatrix}}{\partial \sigma_j}\\&=\frac{4III\Psi_{III}}{\sigma_i\sigma_j}+[2\sigma_i,4\sigma_i^3,\frac{2III}{\sigma_i}]\cdot\begin{bmatrix} \Psi_{I,I} \quad \Psi_{I,II} \quad \Psi_{I,III}\\ \Psi_{II,I} \quad \Psi_{II,II} \quad \Psi_{I,III}\\ \Psi_{III,I} \quad \Psi_{III,II}\quad \Psi_{III,III}\\ \end{bmatrix}\cdot\begin{bmatrix} 2\sigma_j \\ 4\sigma_j^3\\ \frac{2III}{\sigma_j}\\ \end{bmatrix}&=\gamma_{ij} \end{align}$
}