拉格朗日方程的三种推导方法之基于汉密顿原理推导

拉格朗日方程是分析力学中的重要方程,其地位相当于牛顿第二定律之于牛顿力学。

哈密顿原理可数学表述为:
δ t 1 t 2 L d t = 0 (1) \delta \int_{{{t}_{1}}}^{{{t}_{2}}}{Ldt=0}\tag{1}
在等时变分情况下,有:
δ q = d d t ( δ q ) (2) \delta \overset{\bullet }{\mathop{q}}\,=\frac{d}{dt}(\delta q)\tag{2}
δ t 1 t 2 L d t = t 1 t 2 ( δ L ) d t = 0 (3) \delta \int_{{{t}_{1}}}^{{{t}_{2}}}{Ldt=\int_{{{t}_{1}}}^{{{t}_{2}}}{(\delta L)dt}}=0\tag{3}
由拉格朗日量定义得,在等时变分情况下有
δ L = L q δ q + L q δ q (4) \delta L=\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\delta \overset{\bullet }{\mathop{q}}\,+\frac{\partial L}{\partial q}\delta q\tag{4}
其中第一项可化为:
L q δ q = L q d d t ( δ q ) = d d t ( L q δ q ) d d t ( L q ) δ q (5) \frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\delta \overset{\bullet }{\mathop{q}}\,=\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\frac{d}{dt}(\delta q)=\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\bullet \delta q)-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q\tag{5}
将(5)代入(4)得:
δ L = d d t ( L q δ q ) d d t ( L q ) δ q + L q δ q (6) \delta L=\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\bullet \delta q)-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q+\frac{\partial L}{\partial q}\delta q\tag{6}
将(6)代入(3)得
( L q δ q ) t 2 t 1 + t 1 t 2 ( d d t ( L q ) δ q + L q δ q ) d t = 0 (7) (\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\bullet \delta q)\left| _{{{t}_{2}}}^{{{t}_{1}}} \right.+\int_{{{t}_{1}}}^{{{t}_{2}}}{(-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q+\frac{\partial L}{\partial q}\delta q})dt=0\tag{7}
t 1 , t 2 {{t}_{1}},{{t}_{2}} δ q = 0 \delta q=0 ,所以(7)变为:
t 1 t 2 ( d d t ( L q ) δ q L q δ q ) d t = 0 (8) \int_{{{t}_{1}}}^{{{t}_{2}}}{(\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q-\frac{\partial L}{\partial q}\delta q})dt=0\tag{8}

t 1 t 2 [ ( d d t ( L q ) + L q ) δ q ] d t = 0 (9) \int_{{{t}_{1}}}^{{{t}_{2}}}{[(-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})+\frac{\partial L}{\partial q}})\delta q]dt=0\tag{9}
q是独立变量,所以得拉格朗日方程:
d d t ( L q ˙ j ) L q j = 0 (10) \frac{d}{dt}\left( \frac{\partial L}{\partial {{{\dot{q}}}_{j}}} \right)-\frac{\partial L}{\partial {{q}_{j}}}=0\tag{10}

发布了26 篇原创文章 · 获赞 14 · 访问量 4907

猜你喜欢

转载自blog.csdn.net/jaysur/article/details/103909348