优化方法理论合集（6）——多变量问题

多变量优化问题应该是在高阶导问题（看这里优化方法理论合集（2）——高阶导问题）之后更新的，由于笔者失误，特在此补充。所幸不影响之前优化方法的理解与学习。

1. 问题描述

多变量问题描述的是关于质点在$n$维空间中坐标的问题。

给出如下条件：

1. 时间区间
   $$t\in \left[t_0,t_k\right]$$
2. 系统有$2n$个边界条件：
   $$Start=\{\begin{array}{l}{x}_1(t_0)=x_{10}\\ x_2(t_0)=x_{20}\\ ⋮\\ x_n(t_0)=x_{n0}\end{array}$$$$End=\{\begin{array}{l}{x}_1(t_k)=x_{1k}\\ x_2(t_k)=x_{2k}\\ ⋮\\ x_n(t_k)=x_{nk}\end{array}$$
3. 性能指标（拉格朗日形式）：
   $$J={\int }_{t_0}^{t_k}F\left(t,x_1,{\dot{x}}_1,x_2,{\dot{x}}_2,\cdots ,x_n,{\dot{x}}_n\right)dt\to extr.$$

2. 解题步骤

写出解题的必要条件：
$$\{\begin{array}{l}{F}_{x_1}-\frac{d}{dt}{F}_{{\dot{x}}_1}=0\\ F_{x_2}-\frac{d}{dt}{F}_{{\dot{x}}_2}=0\\ ⋮\\ F_{x_n}-\frac{d}{dt}{F}_{{\dot{x}}_n}=0\end{array}\text{\hspace{1em}(1)}$$即对每个$F_i$列出欧拉方程。

与之前所提的不同，极值存在的充分条件不再是勒让德条件，而是盖西条件：
$$H=\left|\begin{array}{cccc}\frac{{\partial }^{2}F}{\partial {\dot{x}}_1{\dot{x}}_1}& \frac{{\partial }^{2}F}{\partial {\dot{x}}_1{\dot{x}}_2}& \cdots & \frac{{\partial }^{2}F}{\partial {\dot{x}}_1{\dot{x}}_n}\\ \frac{{\partial }^{2}F}{\partial {\dot{x}}_2{\dot{x}}_1}& \frac{{\partial }^{2}F}{\partial {\dot{x}}_2{\dot{x}}_2}& \cdots & \frac{{\partial }^{2}F}{\partial {\dot{x}}_2{\dot{x}}_n}\\ ⋮& ⋮& \cdots & ⋮\\ \frac{{\partial }^{2}F}{\partial {\dot{x}}_n{\dot{x}}_1}& \frac{{\partial }^{2}F}{\partial {\dot{x}}_n{\dot{x}}_2}& \cdots & \frac{{\partial }^{2}F}{\partial {\dot{x}}_n{\dot{x}}_n}\end{array}\right|\text{\hspace{1em}(2)}$$同时有
$$\{\begin{array}{l}若H>0，则J\to min\\ 若H<0，则J\to max\end{array}$$

3. 例题

给出如下条件：

1. 时间区间
   $$t\in \left[0,T\right]$$
2. 边界条件
   $$Start=\{\begin{array}{l}{x}_1(0)=x_{10}\\ x_2(0)=x_{20}\end{array}$$$$End=\{\begin{array}{l}{x}_1(T)=x_{1k}\\ x_2(T)=x_{2k}\end{array}$$
3. 性能指标
   $$J={\int }_0^T\left({\dot{x}}_1^2+{\dot{x}}_2^2+2x_1{x}_2\right)dt\to extr.$$
   解：$\square$把$F={\dot{x}}_1^2+{\dot{x}}_2^2+2x_1{x}_2$代入欧拉方程组(1)有
   $$\{\begin{array}{l}{F}_{x_1}-\frac{d}{dt}{F}_{{\dot{x}}_1}=0⟹2x_2-2{\ddot{x}}_1=0\\ F_{x_2}-\frac{d}{dt}{F}_{{\dot{x}}_2}=0⟹2x_1-2{\ddot{x}}_2=0\end{array}$$解出
   $$\begin{gathered} x_2(t)=C_1{e}^t+C_2{e}^{-t}+C_3{e}^{it}+C_4{e}^{-it}, \\ {x}_1(t)={\ddot{x}}_2(t)=C_1{e}^t+C_2{e}^{-t}-C_3{e}^{it}+C_4{e}^{-it} \end{gathered}$$代入初始条件值即可算出$C_i$。

又因为
$$H=\left|\begin{array}{cc}\frac{{\partial }^{2}F}{\partial {\dot{x}}_1\partial {\dot{x}}_1}& \frac{{\partial }^{2}F}{\partial {\dot{x}}_1\partial {\dot{x}}_2}\\ \frac{{\partial }^{2}F}{\partial {\dot{x}}_2\partial {\dot{x}}_1}& \frac{{\partial }^{2}F}{\partial {\dot{x}}_2\partial {\dot{x}}_2}\end{array}\right|=\left|\begin{array}{cc}2& 0\\ 0& 2\end{array}\right|=4>0$$所以$J$有最小值。