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 自动化学报  2017, Vol. 43 Issue (1): 40-59 PDF

Finite Time Formation Control for Multiple Vehicles Based on Pontryagin's Minimum Principle
GENG Zhi-Yong
The State Key Laboratory for Turbulence and Complex Systems, Department of Mechanics and Engineering Science, Peking University, Beijing 100871
Received: 2015-08-27, Accepted: 2016-03-15.
Foundation Item: Supported by National Natural Science Foundation of China (61374033)
Abstract: The paper studies the problem of finite time formation control for multiple vehicles based on Pontryagin's minimum principle. The vehicle is modeled as a fully actuated rigid body with the dynamics evolving on the tangent bundle of Euclidean group. Both the formation maneuver time and the geometric structure of the formation are specified by the formation task. For the required formation, an open loop optimal control law is derived by using Pontryagin's minimum principle. In order to overcome the sensitivity of the open-loop control to the disturbance and increase the robustness of the control law to the initial perturbation, the open loop control law is converted to the closed loop form. This is done by feeding the current state back and initializing the control law at the current time, under the assumption that the mode of communication between the vehicles is all-to-all. For demonstration of the result, some numerical examples of formations for both planar and spacial vehicles are included.
Key words: Finite time formation control     consensus     multiple vehicles     minimum principle

1 预备知识

 $$\begin{array}{c}\mbox{SE}(N)\!=\!\left\{\left[\begin{array}{cc} R & {\pmb p} \\{\pmb 0}_{1× N}&1\\\end{array}\right]\! : R\in\mbox{SO}(N),{\pmb p}\in{\bf R}^N\right\},\\\hfill N=2,3\\\end{array}$$

 $$\mathfrak{se}(N)=\left\{\left[\begin{array}{cc}\hat{\omega}& {\pmb v}\\{\pmb 0}_{1× N}& 0\\\end{array}\right] : \hat{\omega}\in\mathfrak{so}(N),{\pmb v}\in{\bf R}^N\right\}$$

 $$\wedge :{\pmb\eta}=\left[\begin{array}{c}{\pmb\omega}\\{\pmb v}\end{array}\right]\rightarrow\left[\begin{array}{cc}\hat{\omega} & {\pmb v}\\{\pmb 0}_{1× 3}&0\end{array}\right]=\hat{\eta},,{\pmb\omega},{\pmb v}\in{\bf R}^3$$

 \begin{align}\label{eq on inner product}G_q(\hat{x}_q,\hat{y}_q):= G_I(\hat{x}^s,\hat{y}^s):= {\pmb x}^{\rm T}{\pmb y}\end{align} (1)

 $$(B^\flat({\pmb u}))({\pmb v})=B({\pmb v},{\pmb u}),\qquad\forall{\pmb v}\in V$$

 $$B({\pmb v},B^\sharp ({\pmb u}^*))={\pmb u}^*({\pmb v}),\qquad\forall {\pmb v}\in V$$

 \begin{align}\label{eq on dual vector}\hat{x}^*_q =(q^{-1})^*(\hat{x}^b)^* = (\hat{x}^s)^*(q^{-1})^* ,~~~\hat{x}^*=\vee^*({\pmb x})\end{align} (2)
 \begin{align}\label{eq on lineaer functional}\hat{x}^*_q (\hat{y}_q)=(\hat{x}^s)^*(\hat{y}^s)={\pmb x}^{\rm T}{\pmb y}\end{align} (3)

 ${{\hat{x}}^{*}}(\hat{y})={{\left\langle {{\left[ {{{\hat{x}}}^{*}} \right]}^{\text{T}}},\hat{y} \right\rangle }_{{{\mathbf{R}}^{4\times 4}}}}=\text{tr}({{\hat{x}}^{*}}\hat{y})$ (4)
 $\hat{x}_{q}^{*}(\widehat{{{y}_{q}}})={{\left\langle {{\left[ \hat{x}_{q}^{*} \right]}^{\text{T}}},{{{\hat{y}}}_{q}} \right\rangle }_{{{\mathbf{R}}^{4\times 4}}}}=\text{tr}(\hat{x}_{q}^{*}{{{\hat{y}}}_{q}})$ (5)

 $$\begin{array}{c}\begin{array}{rl}[\hat{p}^*_{\xi^s_k}(t_f)]^{pq}=&\displaystyle\sum_\alpha\sum_{i,j}\Gamma_{\alpha\beta}^{ij}\frac{\partial h_{\alpha\beta}^{ij}}{\partial[\hat{\xi}^s_k]^{pq}}(t_f)=\\ &\displaystyle\sum_\alpha\mbox{tr}\displaystyle\left(\Gamma^{\rm T}_{\alpha\beta}\frac{\partial(\hat{\xi}^s_\alpha-\hat{\xi}^s_\beta)}{\partial[\hat{\xi}^s_k]^{pq}}\right)(t_f),\end{array}\\\hfill k=1,\cdots,n,\quad p,q = 1,2,3,4\end{array}$$

 $$\begin{array}{rl}[\hat{p}^*_{\xi^s_k}(t_f)]^{pq}=&\displaystyle\mbox{tr}\displaystyle\left(\Gamma^{\rm T}_{k\beta}\frac{\partial(\hat{\xi}^s_k-\hat{\xi}^s_\beta)}{\partial[\hat{\xi}^s_k]^{pq}}\right)(t_f)=\\ &\mbox{tr}\displaystyle\left(\Gamma^{\rm T}_{k\beta}E_{pq}\right)=\Gamma^{pq}_{k\beta}\end{array}$$

 $$\begin{array}{rl}[\hat{p}^*_{\xi^s_\beta}(t_f)]^{pq}=&\displaystyle\sum_{\alpha\neq\beta}\mbox{tr}\displaystyle\left(\Gamma^{\rm T}_{\alpha\beta}\frac{\partial(\hat{\xi}^s_\alpha-\hat{\xi}^s_\beta)}{\partial[\hat{\xi}^s_\beta]^{pq}}\right)(t_f)=\\ &-\displaystyle\sum_{\alpha\neq\beta}\mbox{tr}\displaystyle\left(\Gamma^{\rm T}_{\alpha\beta} E_{pq}\right)(t_f)=-\displaystyle\sum_{\alpha\neq\beta}\Gamma^{pq}_{\alpha\beta}\end{array}$$

 \begin{align}\label{eq on consensus of velocity}\hat{\eta}^b_{ij}(t_f):=\mbox{Ad}_{\bar{g}_{ij}}\hat{\eta}_j^b(t_f)-\hat{\eta}_i^b(t_f)=0,,i,j= 1,\cdots,n\end{align} (38)

 \begin{align}\label{eq on control decom.}\hat{u}^b_k=\hat{w}^b_k+\mbox{Ad}_{h_k^{-1}}\hat{v}^b_k+[\mbox{Ad}_{h_k^{-1}}\hat{\zeta}^b_k,\hat{\eta}^b_k],,k=1,\cdots,n\end{align} (39)

 \begin{align}\label{eq on velocity decom.}\hat{\xi}^b_k=\hat{\eta}^b_k+\mbox{Ad}_{h_k^{-1}}\hat{\zeta}^b_k,\qquad k=1,\cdots,n\end{align} (40)

 $$J_1(\hat{v}^s_1,\cdots,\hat{v}^s_n)=\int_{t_0}^{t_f}\frac{1}{2}\sum_{k=1}^n G_I(\hat{v}^s_k,\hat{v}^s_k){\rm d}t$$

 $$\hat{v}^b_k=\sum_{j=1}^n\left(\frac{6}{n(t_f-t)^2}\hat{y}_{kj}+\frac{4}{n(t_f-t)}\hat{\zeta}^b_{kj}\right)$$

 $$\begin{array}{rl}&\mbox{Ad}_{h^{-1}_k}\hat{v}^b_k=\\ &\qquad\displaystyle\sum_{j=1}^n\left(\frac{6}{n(t_f-t)^2}\hat{x}_{kj}+\frac{4}{n(t_f-t)}(\hat{\xi}^b_{kj}-\hat{\eta}^b_{kj})\right) \\\end{array}$$

 $$\begin{array}{rl}&\mbox{Ad}_{h^{-1}_k}\hat{\zeta}^b_{kj}(t)=\\ &\qquad(\mbox{Ad}_{\bar{g}_{kj}}\mbox{Ad}_{h^{-1}_j}\hat{\zeta}^b_j-\mbox{Ad}_{h^{-1}_k}\hat{\zeta}^b_k)(t)=\\ &\qquad(\mbox{Ad}_{\bar{g}_{kj}}\hat{\xi}^b_j-\hat{\xi}^b_k-\mbox{Ad}_{\bar{g}_{kj}}\hat{\eta}^b_j+\hat{\eta}^b_k)(t)=\\ &\qquad\hat{\xi}^b_{kj}(t)-\hat{\eta}^b_{kj}(t)\end{array}$$

 \begin{align}\label{eq on dynamic cotroller}\begin{array}{l}\begin{array}{ll}\dot{h}_k = h_k\hat{\eta}^b_k,& h_k(t_0) =I\\\dot{\hat{\eta}}^b_k=\hat{w}^b_k,&\hat{\eta}^b_k(t_0) =\hat{\xi}^{b,0}_k\end{array}\hat{u}^b_k=[\hat{\xi}^b_k,\hat{\eta}^b_k] +\mbox{Ad}_{h^{-1}_k}\hat{v}^b_k + \hat{w}^b_k\end{array},\quad k=1,\cdots,n\end{align} (40)

 \begin{align}\label{eq on dynamic part of sub. sys. 2}\dot{\hat{\eta}}^b_k=\hat{w}^b_k,\qquad k=1,\cdots,n\end{align} (42)

 \begin{align}\label{eq on cost function for consensus}J_2({\pmb w}_1,\cdots,{\pmb w}_n)=\int_{t_0}^{t_f}\frac{1}{2}\sum_{i=1}^n {\pmb w}_i^{\rm T}{\pmb w}_i {\rm d}t\end{align} (43)
4.2 多积分器有限时间最优一致性问题的解

 $$\dot{{\pmb\eta}}={\pmb w},,{\pmb\eta}(t_0) ={\pmb\eta}^0$$

 $[{{\mathbf{1}}_{n-1}}\otimes {{I}_{6}},-{{I}_{n-1}}\otimes {{I}_{6}}]\eta ({{t}_{f}})=0$ (44)

 $$H_{\Sigma_2}=-\frac{1}{2}{\pmb w}^{\rm T}{\pmb w}+{\pmb p}^{\rm T}{\pmb w}$$

 $$\dot{{\pmb\eta}}=\frac{\partial H_{\Sigma_2}}{\partial {\pmb p}}={\pmb w},,\dot{{\pmb p}}=-\frac{\partial H_{\Sigma_2}}{\partial {\pmb\eta}}= 0$$

 $$\frac{\partial H_{\Sigma_2}}{\partial {\pmb w}}=-{\pmb w}+{\pmb p}=0$$

 $${\pmb p}_k=\frac{1}{n(t_f-t_0) }\sum_{i=1}^n({\pmb\eta}_i^0-{\pmb\eta}_k^0) ,\qquad k=1,\cdots,n$$

 $$\begin{array}{l}{\pmb p}_1(t_f)=\displaystyle\sum_{i=1}^n\frac{\partial{\pmb f}_i({\pmb\eta})}{\partial{\pmb\eta}_1}{\pmbλ}_i = {\pmbλ}_1 \\{\pmb p}_2(t_f)=\displaystyle\sum_{i=1}^n\frac{\partial{\pmb f}_i({\pmb\eta})}{\partial{\pmb\eta}_2}{\pmbλ}_i = -{\pmbλ}_1+ {\pmbλ}_2 \\\qquad \qquad\qquad\vdots \\{\pmb p}_{n-1}(t_f)=\displaystyle\sum_{i=1}^n\frac{\partial{\pmb f}_i({\pmb\eta})}{\partial{\pmb\eta}_{n-1}}{\pmbλ}_i = -{\pmbλ}_{n-2}+ {\pmbλ}_{n-1} \\{\pmb p}_n(t_f)=\displaystyle\sum_{i=1}^n\frac{\partial{\pmb f}_i({\pmb\eta})}{\partial{\pmb\eta}_n}{\pmbλ}_i = -{\pmbλ}_{n-1} \\\end{array}$$

 $1_{n}^{\text{T}}p=0$ (45)

 $[{{1}_{n-1}},-{{I}_{n-1}}](\eta ({{t}_{0}})+({{t}_{f}}-{{t}_{0}})p)=0$ (46)

 $p=\frac{Q}{{{t}_{f}}-{{t}_{0}}}\eta ({{t}_{0}})$

 $p=\frac{Q\eta ({{t}_{0}})}{{{t}_{f}}-{{t}_{0}}}=\left[ \begin{matrix} \frac{1}{n({{t}_{f}}-{{t}_{0}})}\sum\limits_{i=1}^{n}{({{\eta }_{i}}({{t}_{0}})-{{\eta }_{1}}({{t}_{0}}))} \\ \frac{1}{n({{t}_{f}}-{{t}_{0}})}\sum\limits_{i=1}^{n}{({{\eta }_{i}}({{t}_{0}})-{{\eta }_{2}}({{t}_{0}}))} \\ \vdots \\ \frac{1}{n({{t}_{f}}-{{t}_{0}})}\sum\limits_{i=1}^{n}{({{\eta }_{i}}({{t}_{0}})-{{\eta }_{n}}({{t}_{0}}))} \\ \end{matrix} \right]$

4.3 协调编队的控制

 \begin{align*}&\mbox{Ad}_{\bar{g}_{1k}}\hat{w}^b_k=\\&\quad\frac{1}{n(t_f-t_0) }\sum_{j=1}^n(\mbox{Ad}_{\bar{g}_{1j}}\hat{\eta}_j^b(t_0) -\mbox{Ad}_{\bar{g}_{1k}}\hat{\eta}_k^b(t_0) )\end{align*}

 $$\begin{array}{c}\hat{w}^b_k=\displaystyle\frac{1}{n(t_f-t_0) }\sum_{j=1}^n(\mbox{Ad}_{\bar{g}_{kj}}\hat{\eta}_j^b(t_0) - \hat{\eta}_k^b(t_0) ),\\\hfill k=1,\cdots,n\\\end{array}$$

 \begin{align}\label{eq on opt. control law for velocity consensus} \hat{w}^b_k=\frac{1}{n(t_f-t)}\sum_{i=1}^n\hat{\eta}_{ki}^b(t),\qquad k=1,\cdots,n\end{align} (47)

 \begin{align}&\dot{h}_k=h_k\hat{\eta}^b_k,h_k(t_0) =I \nonumber\\&\dot{\hat{\eta}}^b_k=\displaystyle\frac{1}{n(t_f-t)}\sum_{i=1}^n\hat{\eta}_{ki}^b(t),\hat{\eta}^b_k(t_0) =\hat{\xi}^{b,0}_k\nonumber\\&\hat{u}^{b,opf}_k=\displaystyle\sum_{i=1}^n\left(\frac{6}{n(t_f-t)^2}\hat{x}_{ki}+\frac{4}{n(t_f-t)}\hat{\xi}^b_{ki}\right.-\nonumber\\&\left.,,,displaystyle\frac{3}{n(t_f-t)}\hat{\eta}^b_{ki}\right)+[\hat{\xi}^b_k,\hat{\eta}^b_k]\end{align} (48)

 $$\hat{u}^b_k=\left\{\begin{array}{ll}\hat{u}^{b,opf}_k,& t_0\leq t\leq t_f\\0,& t> t_f\end{array},\right.\quad k=1,\cdots,n$$

2) 控制器(48) 并不是一个物理系统,它只是产生控制信号当前取值的一个动态解析算法.

3) 不同于基于镇定的控制律,可以在无穷时域上使得控制器的初始状态对系统的影响随时间而趋于消失.对基于有限时间的控制律而言,控制器的初始状态确实对终端时刻的控制结果有影响,但这并不意味着由式(48) 给出的控制器的初始条件是唯一的,对于一般的初始状态,可以考虑相对于给定初始状态的动力学,但这将引起不必要的复杂性.

4) 该控制律只在时间段 $[t_0,t_f]$ 起作用. 系统在时刻$t_f$以后的行为取决于被切换后的控制律.如果在时刻$t_f$以后无控制且无外部扰动,则每一运载体处于由$\dot{\hat{\xi}}^b_i=0,i= 1,\cdots,n$刻画的稳定的相对平衡态. 此外,所有的相对速度$\hat{\xi}^b_j-\mbox{Ad}_{\bar{g}_{ji}}\hat{\xi}^b_i =0$.这就保证了系统保持队形并以恒定的速度运动. 若有外部扰动,为了保持队形可切换到由文献[25]给出的基于镇定的编队控制律.

4.4 协调编队的数值仿真 4.4.1 四个平面运载体反馈最优控制下的协调编队

 图 7 两运载体的质心位置、姿态演化轨迹(左上角为$t = t_f$时的质心位置、姿态演化轨迹 Figure 7 The evolution trajectories of position and attitude of the two vehicles (The figure on the top left are the evolution trajectories of position and attitude at $t = t_f$.)
 图 8 取得一致性过程中两运载体的时间行为(从上到下: 空间坐标系下的位形(质心位置与姿态角);相对于第一个运载体刚体坐标系下的相对位形; 刚体坐标系下的速度;刚体坐标系下的控制. 从左到右: 相应变量的 $x$-轴坐标;$y$-轴坐标; 姿态角$\theta$轴坐标. Figure 8 The time behaviors of the two vehicles during the process of achieving consensus (From the top down: configuration(position and attitude angle) in the space frame; relative configuration with respect to the body frame of first vehicle;velocity in the body frame; control in the body frame. From left to right: the coordinates of the corresponding quantity in$x$-axis,$y$-axis,and attitude angle $\theta$.)

 $$\begin{array}{l}g_{12}=\left[\begin{array}{ccc} 1 & -0.0000 & -40\\ 0.0000 & 1 & -40\\ 0 & 0 & 1\end{array}\right] \0.5cm]g_{13}=\left[\begin{array}{ccc} 1 & 0.0000 & -40\\-0.0000 & 1& 40\\ 0 & 0 & 1\end{array}\right] \0.5cm]g_{14}=\left[\begin{array}{ccc} 1& -0.0000 & -80\\ 0.0000 & 1 & 0\\ 0 & 0 & 1\end{array}\right] \\\end{array}$$

 $$g_{12}=\left[\begin{array}{ccc} 1.0000 & -0.0015 & -39.6954\\ 0.0015 & 1.0000 & -39.9235\\ 0 & 0 & 1.0000\end{array}\right]$$$$g_{13}=\left[\begin{array}{ccc} 1.0000 & 0.0011 & -40.1024\\ -0.0011 & 1.0000& 39.8278\\ 0 & 0 & 1.0000\end{array}\right]$$$$g_{14}=\left[\begin{array}{ccc} 1.0001& -0.0004 & -79.8831\\ 0.0004 & 1.0001 & -0.1594\\ 0 & 0 & 1.0000\end{array}\right]$$

1) 针对不同的编队终端时刻$t_f$为 5,10,$\cdots$,100的情况在同一仿真模型下进行了仿真验证,结果表明在期望的终端时刻达到了编队,精度至少在 $10^{-3}$.随着期望的编队时间$t_f$的增加,控制作用的幅值变小,非零初始速度和外界扰动的作用变得明显,这使得取得编队的轨迹变得复杂.

2) 针对运载体数量为2,3,4,6,8,10及20的情况,对不同的指定编队时间、初始状态、以及外界扰动进行了仿真验证,结果表明该方法对运载体的数量没有限制.

3) 随着外界扰动幅值的加大,为了克服扰动引起的偏差,控制的幅值也随之增大.

4.4.2 四个空间运载体反馈最优控制下的协调编队

 \begin{align} & \begin{array}{*{35}{l}} {{g}_{12}}({{t}_{f}})=\left[ \begin{matrix} 1 & 0.00 & -0.00 & -100 \\ -0.00 & 1 & -0.00 & 0.00 \\ 0.00 & 0.00 & 1 & 0.00 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ \end{array} \\ & \begin{array}{*{35}{l}} {{g}_{13}}({{t}_{f}})=\left[ \begin{matrix} 1 & 0.00 & -0.00 & -50 \\ -0.00 & 1 & 0.00 & 50 \\ 0.00 & -0.00 & 1 & 0.00 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ \end{array} \\ & {{g}_{14}}({{t}_{f}})=\left[ \begin{matrix} 1 & 0.00 & -0.00 & -50 \\ -0.00 & 1 & -0.00 & -50 \\ 0.00 & 0.00 & 1 & 0.00 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ \end{align}
 图 9 系统状态的时间行为 (从上到下:空间坐标系下的欧拉角; 质心位置以及刚体坐标系下的旋转速度和平移速度.从左到右: 相应变量的$x$-轴坐标; $y$-轴坐标;$z$-轴坐标. Figure 9 Time behaviors of system$'$s states (From the top down: Euler angle; position in the space frame; rotation velocity in the body frame; translation velocity in the body frame. From left to right: the coordinates of the corresponding quantity in$x$-axis,$y$-axis,and $z$-axis.
 图 10 系统相对于第一个运载体刚体坐标系下的相对状态 (从上到下: 相对欧拉角;相对位置; 相对旋转速度; 相对平移速度. 从左到右:相应变量的$x$-轴坐标; $y$-轴坐标; $z$-轴坐标. Figure 10 Time behaviors of system$'$s relative states with respect to the body coordinate of the first agent (From top to down: relative Euler angle; relative position; relative rotation velocity; relative translation velocity. From left to right: the coordinates of the corresponding quantity in $x$-axis,$y$-axis,and$z$-axis.)
 图 11 在刚体坐标系下的控制时间行为 (从上到下:广义力矩; 广义力. 从左到右: 相应变量的$x$-轴坐标;$y$-轴坐标; $z$-轴坐标.) Figure 11 Time behaviors of system$'$s control in the body coordinate (From top to down: generalized torque; and generalized force. From left to right: the coordinates of the corresponding quantity in $x$-axis,$y$-axis,and $z$-axis.)
 图 12 空间四运载体质心位置、姿态演化轨迹 Figure 12 The evolution trajectories of position and attitude of the four vehicles

5 结论

 $$\hat{\eta}=\left[\begin{array}{cc}\hat{\omega}& {\pmb v}\\{\pmb 0}_{1× 3}&0\end{array}\right]\neq 0$$

 $$\hat{\xi}=\left[\begin{array}{cc}\hat{\xi}_{\omega}&{\pmb \xi}_v\\{\pmb 0}_{1× 3}&0\\\end{array}\right]\!=\!\left[\begin{array}{cc}\alpha\hat{\omega}& \alpha {\pmb v} + \beta{\pmb\omega}\\{\pmb 0}_{1× 3}&0\\\end{array}\right],\alpha,\beta\in {\bf R}$$

 $$\vee\left([\hat{\eta},\hat{\xi}]\right)= \left[\begin{array}{cc}\hat{\omega}&0\\\hat{v}&\hat{\omega}\end{array}\right]\left[\begin{array}{c}{\pmb\xi}_{\omega}\\{\pmb\xi}_v\end{array}\right]=\left[\begin{array}{c} \hat{\omega}{\pmb\xi}_{\omega}\\\hat{v}{\pmb\xi}_{\omega}+\hat{\omega}{\pmb\xi}_v\end{array}\right]$$

 $$\hat{\xi}_i =\left[\begin{array}{cc} \alpha_i\hat{\omega}&\alpha_i{\pmb v} +\beta_i {\pmb\omega} \\{\pmb 0}_{1× 3}&0\end{array}\right],\qquad i=1,2$$

 $$\begin{array}{rl}&\vee\left([\hat{\xi}_1,\hat{\xi}_2]\right)=\\ &\qquad \left[\begin{array}{cc}\alpha_1\hat{\omega}& 0\\\alpha_1\hat{v} + \beta_1\hat{\omega}&\alpha_1\hat{\omega}\end{array}\right]\left[\begin{array}{c}\alpha_2{\pmb\omega}\\ \alpha_2 {\pmb v} +\beta_2{\pmb\omega} \end{array}\right]=0\\\end{array}$$

 $$\frac{\partial f}{\partial {\pmb x}}=\sum_i{\pmb\varepsilon}_i\frac{\partial f}{\partial x^i}\in V$$

 $\begin{array}{*{35}{l}} \frac{\partial {{{\hat{x}}}^{*}}(\hat{y})}{\partial \hat{x}}=\frac{\partial {{{\hat{y}}}^{*}}(\hat{x})}{\partial \hat{x}}=\sum\limits_{i}{{{{\hat{e}}}_{i}}}\frac{\partial {{x}^{\text{T}}}y}{\partial {{x}^{i}}}=\hat{y} \\ \frac{\partial {{{\hat{x}}}^{*}}(\hat{y})}{\partial {{{\hat{x}}}^{*}}}=\frac{\partial {{{\hat{y}}}^{*}}(\hat{x})}{\partial {{{\hat{x}}}^{*}}}=\sum\limits_{i}{\hat{e}_{i}^{*}}\frac{\partial {{x}^{\text{T}}}y}{\partial {{x}^{i}}}={{{\hat{y}}}^{*}} \\ \end{array}$ (A1)

 $$\frac{\partial \hat{x}_q^*(q\hat{y}^b)}{\partial q^{\rm T}}=\hat{y}^b\hat{x}_q^*$$

 \begin{align} &\frac{\partial \hat{x}_{q}^{*}(q{{{\hat{y}}}^{b}})}{\partial {{q}^{\text{T}}}}= \\ &\sum\limits_{ij}{{{E}_{ji}}}\frac{\partial \hat{x}_{q}^{*}(q{{{\hat{y}}}^{b}})}{\partial {{q}_{ij}}}= \\ &\sum\limits_{ij}{{{E}_{ji}}}{{\left\langle {{[\hat{x}_{q}^{*}]}^{\text{T}}},{{E}_{ij}}{{{\hat{y}}}^{b}} \right\rangle }_{{{\mathbf{R}}_{4\times 4}}}}= \\ &\sum\limits_{ij}{{{E}_{ji}}}{{\left\langle {{E}_{ji}},{{{\hat{y}}}^{b}}\hat{x}_{q}^{*} \right\rangle }_{{{\mathbf{R}}_{4\times 4}}}}={{{\hat{y}}}^{b}}\hat{x}_{q}^{*}. \\ \end{align}

 $\begin{matrix} Z= & X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]+ \\ {} & \frac{1}{12}[Y,[Y,X]]+\cdots \\ \end{matrix}$ (A2)

1) $\exp X \cdot \exp Y=\exp Y \cdot \exp X$;

2) $X \cdot \exp Y=\exp Y \cdot X$.