拉格朗日方程的三種推導方法之基於漢密頓原理推導

拉格朗日方程是分析力學中的重要方程，其地位相當於牛頓第二定律之於牛頓力學。

哈密頓原理可數學表述爲：
$\delta \int_{{{t}_{1}}}^{{{t}_{2}}}{Ldt=0}\tag{1}$
在等時變分情況下，有：
$\delta \overset{\bullet }{\mathop{q}}\,=\frac{d}{dt}(\delta q)\tag{2}$
$\delta \int_{{{t}_{1}}}^{{{t}_{2}}}{Ldt=\int_{{{t}_{1}}}^{{{t}_{2}}}{(\delta L)dt}}=0\tag{3}$
由拉格朗日量定義得，在等時變分情況下有
$\delta L=\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\delta \overset{\bullet }{\mathop{q}}\,+\frac{\partial L}{\partial q}\delta q\tag{4}$
其中第一項可化爲：
$\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\delta \overset{\bullet }{\mathop{q}}\,=\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\frac{d}{dt}(\delta q)=\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\bullet \delta q)-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q\tag{5}$
將(5)代入(4)得：
$\delta L=\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\bullet \delta q)-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q+\frac{\partial L}{\partial q}\delta q\tag{6}$
將(6)代入(3)得
$(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,}\bullet \delta q)\left| _{{{t}_{2}}}^{{{t}_{1}}} \right.+\int_{{{t}_{1}}}^{{{t}_{2}}}{(-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q+\frac{\partial L}{\partial q}\delta q})dt=0\tag{7}$
在${{t}_{1}},{{t}_{2}}$處$\delta q=0$，所以(7)變爲：
$\int_{{{t}_{1}}}^{{{t}_{2}}}{(\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})\delta q-\frac{\partial L}{\partial q}\delta q})dt=0\tag{8}$
即
$\int_{{{t}_{1}}}^{{{t}_{2}}}{[(-\frac{d}{dt}(\frac{\partial L}{\partial \overset{\bullet }{\mathop{q}}\,})+\frac{\partial L}{\partial q}})\delta q]dt=0\tag{9}$
q是獨立變量，所以得拉格朗日方程：
$\frac{d}{dt}\left( \frac{\partial L}{\partial {{{\dot{q}}}_{j}}} \right)-\frac{\partial L}{\partial {{q}_{j}}}=0\tag{10}$