## 物理代写|分析力学代写Analytical Mechanics代考|Lagrange Multipliers

As a consequence of (2.68), the equation
$$\int_{t_1}^{t_2} d t \sum_{l=1}^p \lambda_l\left(\sum_{k=1}^n a_{l k} \delta q_k\right)=\int_{t_1}^{t_2} d t \sum_{k=1}^n\left(\sum_{l=1}^p \lambda_l a_{l k}\right) \delta q_k=0$$
is obviously valid for all values of the Lagrange multipliers $\lambda_1(t), \ldots, \lambda_p(t)$. Adding (2.69) to (2.67) there results
$$\int_{t_1}^{t_2} d t \sum_{k=1}^n\left{\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}\right} \delta q_k=0 .$$
Since the $\delta q$ s are not independent, nothing can he said about the coefficient of each $\delta q_k$ in this last equation. With an adequate numbering of the variables, we can take the first $n-p$ variations $\delta q_1, \ldots, \delta q_{n-p}$ as mutually independent, the last $p$ variations being determined in terms of the first ones by solving the $p$ equations (2.68). On the other hand, we have $p$ Lagrange multipliers at our disposal and they can be chosen such that the coefficients of the last $p$ variations in (2.70) vanish – that is,
$$\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}=0, \quad k=n-p+1, \ldots, n .$$
With the $\lambda .$ s determined by Eqs. (2.71), Eq. (2.70) reduces to
$$\int_{t_1}^{t_2} d t \sum_{k=1}^{n-p}\left{\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}\right} \delta q_k=0,$$
which involves only the independent variations $\delta q_1, \ldots, \delta q_{n-p}$, implying
$$\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}=0, \quad k=1, \ldots, n-p .$$

## 物理代写|分析力学代写Analytical Mechanics代考|Constraint Forces

It remains to investigate the physical meaning of the Lagrange multipliers, which are also determined in the process of solving the equations of motion. Let us imagine the constraints removed and generalised forces $Q_k^{\prime}$ acting without causing any change to the motion of the system. This would only occur if $Q_k^{\prime}$ were the constraint forces because only then the system would obey the restrictions (2.75). But, taking into account that $L$ includes only the applied forces, the additional forces $Q_k^{\prime}$ would appear in the equations of motion in the form $(1.145)$ – that is,
$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)-\frac{\partial L}{\partial q_k}=Q_k^{\prime} .$$ In order for the motions to be identical, the equations of motion (2.74) and (2.76) must be identical. In other words, $$Q_k^{\prime}=\sum{l=1}^p \lambda_l a_{l k}$$
is the $k$ th component of the generalised constraint force. Thus, in addition to the motion of the system, in the present formulation the constraint forces appear as part of the answer. It must not be forgotten that the components of the generalised constraint force and the constraint forces proper are related by ( $1.75)$.

The conclusion of the above analysis is that, suitably modified, Lagrange’s equations are valid even in the presence of an important class of non-holonomic constraints. Over and above the motion of the mechanical system, the constraint forces are determined as an important subproduct of the Lagrange multiplier formalism. ${ }^{11}$

Exercise 2.4.1 Prove that the total virtual work of the generalised constraint forces (2.77) is zero. This shows that even in the presence of non-holonomic constraints Hamilton’s principle can be in harmony with d’Alembert’s principle.

## 物理代写|分析力学代写分析力学代考|拉格朗日乘子

$$\int_{t_1}^{t_2} d t \sum_{l=1}^p \lambda_l\left(\sum_{k=1}^n a_{l k} \delta q_k\right)=\int_{t_1}^{t_2} d t \sum_{k=1}^n\left(\sum_{l=1}^p \lambda_l a_{l k}\right) \delta q_k=0$$

$$\int_{t_1}^{t_2} d t \sum_{k=1}^n\left{\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}\right} \delta q_k=0 .$$

$$\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}=0, \quad k=n-p+1, \ldots, n .$$
， $\lambda .$ s由等式确定。(2.71)，式(2.70)简化为
$$\int_{t_1}^{t_2} d t \sum_{k=1}^{n-p}\left{\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}\right} \delta q_k=0,$$
，其中只涉及独立变量$\delta q_1, \ldots, \delta q_{n-p}$，意味着
$$\frac{\partial L}{\partial q_k}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)+\sum{l=1}^p \lambda_l a_{l k}=0, \quad k=1, \ldots, n-p .$$

## 物理代写|分析力学代写分析力学代考|约束力

$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}k}\right)-\frac{\partial L}{\partial q_k}=Q_k^{\prime} .$$为了使运动相同，运动方程(2.74)和(2.76)必须相同。换句话说，$$Q_k^{\prime}=\sum{l=1}^p \lambda_l a_{l k}$$

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