star2dust

拉格朗日函数

$L(x,\lambda)=f(x,t)+\lambda^T(Ax-b)$

得到

$\begin{aligned} \nabla_z L(z,t)&=\begin{bmatrix} \nabla_x f(x,t)+A^T\lambda\\Ax-b\\ \end{bmatrix}\\ \nabla_{zz}L(z,t)&=\begin{bmatrix} \nabla_{xx}f(x,t)&A^T\\ A&0 \end{bmatrix} \end{aligned}$

控制器

$\dot z=-\nabla_{zz}L^{-1}(z,t)(\alpha \nabla_z L(z,t)+\nabla_{zt}L(z,t))$

注意：$\nabla_{zz}L(z,t)$对称但不正定，满秩可逆。$\nabla_{zz}L^{-1}(z,t)$可由以下公式计算：

$\nabla_{zz}L^{-1}=\begin{bmatrix} H^{-1}(I-A^T(AH^{-1}A^T)^{-1}AH^{-1})&H^{-1}A^T(AH^{-1}A^T)^{-1}\\ (AH^{-1}A^T)^{-1}AH^{-1}&-(AH^{-1}A^T)^{-1} \end{bmatrix}$

以最优资源分配为例，令$A=1_n^T$，则

$\nabla_{zz}L^{-1}=\begin{bmatrix} H^{-1}(I-1_n(1_n^TH^{-1}1_n)^{-1}1_n^TH^{-1})&H^{-1}1_n(1_n^TH^{-1}1_n)^{-1}\\ (1_n^TH^{-1}1_n)^{-1}1_n^TH^{-1}&-(1_n^TH^{-1}1_n)^{-1} \end{bmatrix}$

进一步假设所有智能体有相同的Hessian，即$H=hI_n$，则

$\nabla_{zz}L^{-1}=\begin{bmatrix} (I-\frac{1}{n}1_n1_n^T)/h&1_n/n\\ 1_n^T/n&-h/n \end{bmatrix}$

将控制器写成关于$x,\lambda$的形式

$\begin{aligned} \dot x&=-(I-\frac{1}{n}1_n1_n^T)/h(\alpha\nabla_xf(x,t)+\alpha 1_n\lambda+\nabla_{xt}f(x,t))-\frac{1}{n}1_n1_n^T(x-b)\\ \dot \lambda&=-\frac{1}{n}1_n^T(\alpha\nabla_xf(x,t)+\alpha 1_n\lambda+\nabla_{xt}f(x,t))+\frac{1}{n}h1_n^T(x-b) \end{aligned}$

可以看出

$\begin{aligned} \dot x&=-h^{-1}(\alpha\nabla_xf(x,t)+\alpha 1_n\lambda+\nabla_{xt}f(x,t)+1_n\dot \lambda)\\ \dot \lambda&=-\frac{1}{n}1_n^T(\alpha\nabla_xf(x,t)+\alpha 1_n\lambda+\nabla_{xt}f(x,t))+\frac{1}{n}h1_n^T(x-b) \end{aligned}$