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# 拉格朗日方程

## 定义

${\displaystyle {\frac {d}{dt)){\frac {\partial {\mathcal {L))}{\partial {\dot {\mathbf {q} ))))-{\frac {\partial {\mathcal {L))}{\partial \mathbf {q} ))=\mathbf {0} \,\!}$

## 导引

${\displaystyle \mathbf {y} (x)=(y_{1}(x),\ y_{2}(x),\ \ldots ,y_{N}(x))\,\!}$
${\displaystyle {\dot {\mathbf {y} ))(x)=({\dot {y))_{1}(x),\ {\dot {y))_{2}(x),\ \ldots ,\ {\dot {y))_{N}(x))\,\!}$
${\displaystyle f(\mathbf {y} ,\ {\dot {\mathbf {y} )),\ x)=f(y_{1}(x),\ y_{2}(x),\ \ldots ,\ y_{N}(x),\ {\dot {y))_{1}(x),\ {\dot {y))_{2}(x),\ \ldots ,\ {\dot {y))_{N}(x),\ x)\,\!}$

${\displaystyle \mathbf {y} (x)\in (C^{1}[a,\ b])^{N}\,\!}$使泛函${\displaystyle J(\mathbf {y} )=\int _{a}^{b}f(\mathbf {y} ,\ {\dot {\mathbf {y} )),\ x)dx\,\!}$取得局部平稳值，则在区间${\displaystyle (a,\ b)\,\!}$内，欧拉-拉格朗日方程成立：

${\displaystyle {\frac {d}{dx))\left({\frac {\partial }{\partial {\dot {y))_{i))}f(\mathbf {y} ,\ {\dot {\mathbf {y} )),\ x)\right)-{\frac {\partial }{\partial y_{i))}f(\mathbf {y} ,\ {\dot {\mathbf {y} )),\ x)=0\ ,\qquad \qquad \qquad \qquad i=1,\ 2,\ \ldots ,\ N\!}$

• 设定独立变量${\displaystyle x\,\!}$为时间${\displaystyle t\,\!}$
• 设定函数${\displaystyle y_{i}\,\!}$为广义坐标${\displaystyle q_{i}\,\!}$
• 设定泛函${\displaystyle f(\mathbf {y} ,\ {\dot {\mathbf {y} )),\ x)\,\!}$为拉格朗日量${\displaystyle {\mathcal {L))(\mathbf {q} ,\ {\dot {\mathbf {q} )),\ t)\,\!}$

${\displaystyle {\frac {d}{dt)){\frac {\partial {\mathcal {L))}{\partial {\dot {\mathbf {q} ))))-{\frac {\partial {\mathcal {L))}{\partial \mathbf {q} ))=\mathbf {0} \,\!}$
• 为了满足这变换的正确性，广义坐标必须互相独立，所以，这系统必须是完整系统。
• 拉格朗日量是动能减去位势，而位势必须是广义位势。所以，这系统必须是单演系统。

## 半完整系统

${\displaystyle g_{i}(\mathbf {q} ,\ {\dot {\mathbf {q} )))=0\ ,\qquad \qquad \qquad i=1,\ 2,\ 3,\ \dots n\,\!}$

${\displaystyle \sum _{i=1}^{n}\ \lambda _{i}g_{i}=0\,\!}$

${\displaystyle {\frac {d}{dt)){\frac {\partial {\mathcal {L))}{\partial {\dot {\mathbf {q} ))))-{\frac {\partial {\mathcal {L))}{\partial \mathbf {q} ))={\boldsymbol {\mathcal {F))}\,\!}$

${\displaystyle N\,\!}$个广义力运动方程加上${\displaystyle n\,\!}$个约束方程，给出${\displaystyle N+n\,\!}$个方程来解${\displaystyle N\,\!}$个未知广义坐标与${\displaystyle n\,\!}$个拉格朗日乘子。

## 实例

### 自由落体

${\displaystyle {\ddot {x))=g\,\!}$

${\displaystyle T={\frac {1}{2))mv^{2}\,\!}$

${\displaystyle V=-mgx\,\!}$

${\displaystyle {\mathcal {L))=T-V={\frac {1}{2))m{\dot {x))^{2}+mgx\,\!}$

${\displaystyle {\mathcal {L))\,\!}$代入拉格朗日方程，

${\displaystyle 0={\frac {d}{dt)){\frac {\partial {\mathcal {L))}{\partial {\dot {x))))-{\frac {\partial {\mathcal {L))}{\partial x))=m{\frac {d{\dot {x))}{dt))-mg\,\!}$

${\displaystyle {\ddot {x))=g\,\!}$

### 具有质量的移动支撑点的简单摆

${\displaystyle T={\frac {1}{2))M{\dot {X))^{2}+{\frac {1}{2))m\left({\dot {x))^{2}+{\dot {y))^{2}\right)\,\!}$

${\displaystyle V=-mgy\,\!}$

${\displaystyle {\mathcal {L))={\frac {1}{2))M{\dot {X))^{2}+{\frac {1}{2))m\left({\dot {x))^{2}+{\dot {y))^{2}\right)+mgy\,\!}$

${\displaystyle x=X+l\sin \theta \,\!}$
${\displaystyle y=l\cos \theta \,\!}$

${\displaystyle {\mathcal {L))={\frac {1}{2))M{\dot {X))^{2}+{\frac {1}{2))m\left[\left({\dot {X))+l{\dot {\theta ))\cos \theta \right)^{2}+\left(l{\dot {\theta ))\sin \theta \right)^{2}\right]+mgl\cos \theta \,\!}$

${\displaystyle {\frac {d}{dt))\left[(M+m){\dot {X))+ml{\dot {\theta ))\cos \theta \right]=0\,\!}$

${\displaystyle (M+m){\ddot {X))+ml{\ddot {\theta ))\cos \theta -ml{\dot {\theta ))^{2}\sin \theta =0\,\!}$

${\displaystyle p_{X}=(M+m){\dot {X))+ml{\dot {\theta ))\cos \theta =K_{1}\,\!}$

${\displaystyle {\frac {d}{dt))\left[m(l^{2}{\dot {\theta ))+{\dot {X))l\cos \theta )\right]+m({\dot {X))l{\dot {\theta ))+gl)\sin \theta =0\,\!}$

${\displaystyle {\ddot {\theta ))+{\frac {\ddot {X)){l))\cos \theta +{\frac {g}{l))\sin \theta =0\,\!}$

## 参考文献

1. ^ Goldstein, Herbert. Classical Mechanics 3rd. United States of America: Addison Wesley. 1980: pp. 46–47. ISBN 0201657023 （英语）.