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拉格朗日方程

一、第二类拉格朗日方程#

一般地,质点系各个质点的矢量坐标可以表示为广义坐标的函数r=ri(q1,q2,,qk;t)\vec{r}=\vec{r}_i(q_1,q_2,\cdots,q_k;t)。质点速度相应地通过广义速度表示为:

r˙i=l=1kriqlq˙l+r˙it\dot{\vec{r}}_i=\sum_{l=1}^k\frac{\partial {\vec{r}}_i}{\partial q_l}\dot{q}_l+\frac{\partial \dot{\vec{r}}_i}{\partial t}

上式两边对广义速度q˙j\dot q_j求偏导数,得到消点规则:

riqj=r˙q˙\frac{\partial {\vec{r}}_i}{\partial q_j}=\frac{\partial \dot{\vec{r}}}{\partial\dot{q}}

质点速度关于广义坐标的导数为

r˙iqj=l=1k2riqlqjq˙l+2ritqj\frac{\partial \dot{\vec{r}}_i}{\partial q_j} = \sum_{l=1}^k \frac{\partial^2 \vec{r}_i}{\partial q_l \partial q_j} \dot{q}_l + \frac{\partial^2 \vec{r}_i}{\partial t \partial q_j}

而质点矢量坐标关于广义坐标的导数仍为广义坐标的函数,

riqj=riqj(q1,q2,,qk;t)\frac{\partial \vec{r}_i}{\partial q_j} = \frac{\partial \vec{r}_i}{\partial q_j}(q_1, q_2, \cdots, q_k; t)

再将它关于时间求全导数,可得

ddt(riqj)=l=1kql(riqj)q˙l+t(riqj)=l=1k2riqjqlq˙l+2riqjt\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \vec{r}_i}{\partial q_j}\right) = \sum_{l=1}^k \frac{\partial}{\partial q_l} \left(\frac{\partial \vec{r}_i}{\partial q_j}\right) \dot{q}_l + \frac{\partial}{\partial t} \left(\frac{\partial \vec{r}_i}{\partial q_j}\right) = \sum_{l=1}^k \frac{\partial^2 \vec{r}_i}{\partial q_j \partial q_l} \dot{q}_l + \frac{\partial^2 \vec{r}_i}{\partial q_j \partial t}

这里,假定矢量坐标 ri\vec{r}_i 一阶与二阶连续可微,即得另一个恒等关系式,即交换规则:

ddt(riqj)=r˙iqj\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \vec{r}_i}{\partial q_j}\right)=\frac{\partial \dot{\vec{r}}_i}{\partial q_j}

于是可将广义惯性力表示为:

Qgj=i=1nmidr˙idtriqj=i=1nmiddt(r˙iriqj)+i=1nmir˙iddt(riqj)Q_{gj} = -\sum_{i=1}^n m_i \frac{\mathrm{d} \dot{\vec{r}}_i}{\mathrm{d} t} \cdot \frac{\partial \vec{r}_i}{\partial q_j} = -\sum_{i=1}^n m_i \frac{\mathrm{d}}{\mathrm{d} t} \left( \dot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} \right) + \sum_{i=1}^n m_i \dot{\vec{r}}_i \cdot \frac{\mathrm{d}}{\mathrm{d} t} \left( \frac{\partial \vec{r}_i}{\partial q_j} \right)

Qgj=ddt(i=1nmir˙iriqj)+i=1nmir˙ir˙iqj(2.10)Q_{gj} = -\frac{\mathrm{d}}{\mathrm{d} t} \left( \sum_{i=1}^n m_i \dot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} \right) + \sum_{i=1}^n m_i \dot{\vec{r}}_i \cdot \frac{\partial \dot{\vec{r}}_i}{\partial q_j} \tag{2.10}

注意到:

r˙ir˙iq˙j=q˙j(12r˙ir˙i)=q˙j(12vi2)\dot{\vec{r}}_i \cdot \frac{\partial \dot{\vec{r}}_i}{\partial \dot{q}_j} = \frac{\partial}{\partial \dot{q}_j} \left( \frac{1}{2} \dot{\vec{r}}_i \cdot \dot{\vec{r}}_i \right) = \frac{\partial}{\partial \dot{q}_j} \left( \frac{1}{2} v_i^2 \right)r˙ir˙iqj=qj(12r˙ir˙i)=qj(12vi2)\dot{\vec{r}}_i \cdot \frac{\partial \dot{\vec{r}}_i}{\partial q_j} = \frac{\partial}{\partial q_j} \left( \frac{1}{2} \dot{\vec{r}}_i \cdot \dot{\vec{r}}_i \right) = \frac{\partial}{\partial q_j} \left( \frac{1}{2} v_i^2 \right)

那么广义惯性力可以表示为:

Qgj=ddt[i=1nmiq˙j(12vi2)]+i=1nmiqj(12vi2)=ddt[q˙j(i=1n12mivi2)]+qj(i=1n12mivi2)=ddt(Tq˙j)+Tqj\begin{aligned} Q_{gj} &= -\frac{\mathrm{d}}{\mathrm{d}t} \left[ \sum_{i=1}^n m_i \frac{\partial}{\partial \dot{q}_j} \left( \frac{1}{2} v_i^2 \right) \right] + \sum_{i=1}^n m_i \frac{\partial}{\partial q_j} \left( \frac{1}{2} v_i^2 \right) \\&= -\frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{\partial}{\partial \dot{q}_j} \left( \sum_{i=1}^n \frac{1}{2} m_i v_i^2 \right) \right] + \frac{\partial}{\partial q_j} \left( \sum_{i=1}^n \frac{1}{2} m_i v_i^2 \right) \\&= -\frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial T}{\partial \dot{q}_j} \right) + \frac{\partial T}{\partial q_j} \end{aligned}

其中,T=i=1n12mivi2T = \sum_{i=1}^n \frac{1}{2} m_i v_i^2 是系统的动能。因此,广义惯性力可以表示成动能的导数表达,将广义惯性力代入到广义坐标系中的动力学普遍方程中,得到:

ddt(Tq˙j)Tqj=Qjj=1,2,,k\frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} = Q_j \quad j = 1, 2, \cdots, k

这就是(第二类)拉格朗日方程

二、广义速度表示的动能#

利用速度表达式,可以将质点系的动能表示为:

T=l=1n12mlvl2=l=1n12mlr˙lr˙l=l=1n12ml(i=1krlqiq˙i+rlt)(j=1krlqjq˙j+rlt)=12i=1kj=1kaijq˙iq˙j+i=1kbiq˙i+c=T2+T1+T0\begin{aligned} T &= \sum_{l=1}^n \frac{1}{2} m_l v_l^2 = \sum_{l=1}^n \frac{1}{2} m_l \dot{\mathbf{r}}_l \cdot \dot{\mathbf{r}}_l \\&= \sum_{l=1}^n \frac{1}{2} m_l \left( \sum_{i=1}^k \frac{\partial \mathbf{r}_l}{\partial q_i} \dot{q}_i + \frac{\partial \mathbf{r}_l}{\partial t} \right) \cdot \left( \sum_{j=1}^k \frac{\partial \mathbf{r}_l}{\partial q_j} \dot{q}_j + \frac{\partial \mathbf{r}_l}{\partial t} \right) \\&= \frac{1}{2} \sum_{i=1}^k \sum_{j=1}^k a_{ij} \dot{q}_i \dot{q}_j + \sum_{i=1}^k b_i \dot{q}_i + c \\&= T_2 + T_1 + T_0 \end{aligned}

其中动能关于广义速度的二次项、一次项与零次项部分分别为:

  • 二次项:
T2=12i=1kj=1kaijq˙iq˙j,aij=l=1nmlrlqirlqjT_2 = \frac{1}{2} \sum_{i=1}^k \sum_{j=1}^k a_{ij} \dot{q}_i \dot{q}_j, \qquad a_{ij} = \sum_{l=1}^n m_l \frac{\partial \mathbf{r}_l}{\partial q_i} \cdot \frac{\partial \mathbf{r}_l}{\partial q_j}
  • 一次项:
T1=i=1kbiq˙i,bi=l=1nmlrlqirltT_1 = \sum_{i=1}^k b_i \dot{q}_i, \qquad b_i = \sum_{l=1}^n m_l \frac{\partial \mathbf{r}_l}{\partial q_i} \cdot \frac{\partial \mathbf{r}_l}{\partial t}
  • 零次项:
T0=c=12l=1nmlrltrltT_0 = c = \frac{1}{2} \sum_{l=1}^n m_l \frac{\partial \mathbf{r}_l}{\partial t} \cdot \frac{\partial \mathbf{r}_l}{\partial t}

这里的aij,bi,ca_{ij},b_i,c都是广义坐标及时间的函数,aij=ajia_{ij}=a_{ji}具有对称性。

三、保守系统的拉格朗日方程#

保守系统受到的主动力都是有势力,而有势力相应地广义力可以表示为势能的导数的负数,所以拉格朗日方程成为:

ddt(Tq˙j)Tqj+Vqj=0(j=1,2,,k)\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial\dot{q}_j}\right)-\frac{\partial T}{\partial q_j}+\frac{\partial V}{\partial q_j}=0\quad (j=1,2,\cdots,k)

势能仅仅取决于质点系的位形,只是广义坐标的函数,从而:

Vq˙j=0\frac{\partial V}{\partial \dot{q}_j}=0

引入拉格朗日函数(或者说动势):

L=TVL=T-V

拉格朗日函数L=L(q˙1,q˙2,,q˙k;q1,q2,,qk)L=L(\dot{q}_1,\dot{q}_2,\cdots,\dot{q}_k;q_1,q_2,\cdots,q_k),则拉格朗日方程可以表示为:

ddt(Lq˙j)Lqj=0(j=1,2,,k)\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{q}_{j}}\right)-\frac{\partial L}{\partial q_{j}}=0\quad (j=1,2,\cdots,k)

上面括号中的内容叫做广义动,后一项为拉格朗日力。此时,只需计算系统的动能与势能,无需计算广义力。利用动能表达式,可以将拉格朗日函数表示为:

L=T2+T1+T0VL=T_2+T_1+T_0-V

四、非保守系统的拉格朗日方程#

对于非保守系统,主动力可以分为有势力和非有势力两类:

ddt(Tq˙j)Tqj=Qj=Vqj+Q~j\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial\dot{q}_j}\right)-\frac{\partial T}{\partial q_j}=Q_j=-\frac{\partial\mathrm{V}}{\partial q_j}+\tilde{Q}_j

系统的拉格朗日方程为:

ddt(Lq˙j)Lqj=Q~j(j=1,2,...,k)\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{q}_j}\right)-\frac{\partial L}{\partial q_j}=\tilde{Q}_j\quad (j=1,2,...,k)

其中Q~\tilde{Q}是非有势力的广义力。

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拉格朗日方程
https://www.laoguantx.cn/posts/lagrangianequations/
作者
老官童鞋gogo
发布于
2025-12-19
许可协议
CC BY-NC-SA 4.0

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