# 定常系統

## 應用

$T =\frac{1}{2}m v^2$

$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\sum_i\ \frac{\partial \mathbf{r}}{dq_i}\dot{q}_i+\frac{\partial \mathbf{r}}{dt}$

$T =\frac{1}{2}m\sum_i\ \left(\frac{\partial \mathbf{r}}{\partial q_i}\dot{q}_i+\frac{\partial \mathbf{r}}{\partial t}\right)^2$

$T =T_0+T_1+T_2$

$T_0=\frac{1}{2}m\left(\frac{\partial \mathbf{r}}{\partial t}\right)^2$
$T_1=\sum_i\ m\frac{\partial \mathbf{r}}{\partial t}\cdot \frac{\partial \mathbf{r}}{\partial q_i}\dot{q}_i$
$T_2=\sum_{i,j}\ \frac{1}{2}m\frac{\partial \mathbf{r}}{\partial q_i}\cdot \frac{\partial \mathbf{r}}{\partial q_j}\dot{q}_i\dot{q}_j,\!$

$T_0$$T_1$$T_2$分別為廣義速度$\dot{q}_i$的0次、1次、2次齊次函數。如果這系統是定常系統，位置不顯性地含時間，$\frac{\partial \mathbf{r}}{\partial t}=0$，則只有$T_2$不等於零。所以，$T =T_2$，動能是廣義速度的2次齊次函數。

## 實例1：單擺

$\sqrt{x^2+y^2} - L=0$

## 實例2：受驅擺

$x_t=x_0\cos\omega t$

$\sqrt{(x - x_0\cos\omega t)^2+y^2} - L=0$

## 參考文獻

1. ^ Goldstein, Herbert. Classical Mechanics 3rd. United States of America: Addison Wesley. 1980: pp. 25. ISBN 0201657023 （English）.