Euler-Lagrange方程的严格推导
Euler-Lagrange方程的严格推导

Euler-Lagrange方程的严格推导

状态变量x(t):[t_0,t_f] \rightarrow \mathbb{R}^n连续可微,在给定初值时刻t_0状态为x(t_0)=x_0,在给定的终端时刻t_f状态为x(t_f)=x_f。\\函数g取值与\mathbb{R},二阶连续可微。则状态变量x最小化性能指标
J(x)=\int_{t_0}^{t_f}{g(x(t),\dot{x}(t),t)\mathsf{d}t}
的必要条件是对任意时刻t\in[t_0,t_f],
\frac{\partial g}{\partial x}(x(t), \dot{x}(t), t)-\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}(t), t)\right]=0

证:

\begin{aligned}
\Delta J(x, \delta x) &=J(x+\delta x)-J(x) \\
&=\int_{t_{0}}^{t_{f}} g(x(t)+\delta x(t), \dot{x}(t)+\dot{x}(t), t) \mathrm{d} t-\int_{t_{0}}^{t_{f}} g(x(t), \dot{x}(t), t) \mathrm{d} t \\
&=\int_{t_{0}}^{t_{f}}\{g(x(t)+\delta x(t), \dot{x}(t)+\dot{x}(t), t)-g(x(t), \dot{x}(t), t)\} \mathrm{d} t
\end{aligned}

使用Taylor公式化简,有:

\begin{aligned}
\Delta J(x, \delta x)=& \int_{t_{0}}^{t_{1}}\left\{\frac{\partial g}{\partial x}(x(t), \dot{x}(t), t) \cdot \delta x(t)+\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}(t), t) \cdot \delta \dot{x}(t)\right\} \mathrm{d} t \\
&+o\left(\|\delta x\|_{1}\right)
\end{aligned}

使用分部积分法,有:

\begin{aligned}
& \Delta J(x, \delta x) \\
=& \int_{t_{0}}^{t_{f}}\left\{\frac{\partial g}{\partial x}(x(t), \dot{x}(t), t)-\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}, t)\right]\right\} \cdot \delta x(t) \mathrm{d} t \\
&+\left.\left[\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}(t), t) \cdot \delta x(t)\right]\right|_{t_{0}} ^{t_{f}}+o\left(\|\delta x\|_{1}\right) \\
=& \int_{t_{0}}^{t_{f}}\left\{\frac{\partial g}{\partial x}(x(t), \dot{x}(t), t)-\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}, t)\right]\right\} \cdot \delta x(t) \mathrm{d} t+o\left(\|\delta x\|_{1}\right)
\end{aligned}
上式的积分项是\delta x的线性泛函,后项为其高阶无穷小量项,所以泛函变分为:
\delta J(x, \delta x)=\int_{t_{0}}^{t_{f}}\left\{\frac{\partial g}{\partial x}(x(t), \dot{x}(t), t)-\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}, t)\right]\right\} \cdot \delta x(t) \mathrm{d} t
令\delta J(x,\delta x)=0

有:

\frac{\partial g}{\partial x}(x(t), \dot{x}(t), t)-\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{\partial g}{\partial \dot{x}}(x(t), \dot{x}(t), t)\right]=0

QED.

一条评论

发表回复

您的电子邮箱地址不会被公开。

此站点使用Akismet来减少垃圾评论。了解我们如何处理您的评论数据