自然界中最普遍的问题都是关于非保守非线性问题，而非线性问题可用非标准Lagrange函数的变分问题来解决，因此研究基于非标准Lagrange函数的动力学系统具有重要的理论意义与应用价值。事实上，非标准Lagrange函数的思想可以追溯至1978年Arnold的工作^[1-2]。但是由于缺乏与之对应的哈密顿形式而被忽视，弦理论^[3-5]中被重新考虑。与标准Lagrange函数表示为动能和势能之差不同，非标准Lagrange函数没有关于动能和势能的明显区分，具有指数形式、对数形式等。这种不规则的Lagrange函数凭借其在非线性动力学系统^[6-8]、耗散系统^[9-12]和量子场理论^[13-15]中的重要作用引起了学者们的关注。Musielak^[9-10]研究了耗散系统中获得非标准Lagrange函数的方法及其存在的条件。El-Nabulsi^[15]研究了非线性动力学系统基于两类非标准Lagrange函数的作用量及动力学方程。在宇宙学方面，非标准Lagrange函数也扮演了重要角色，Dimitrijevic^[16]等将非标准Lagrange函数考虑进现代宇宙学模型中，研究了其运动方程。EL-Nabulsi^[17]应用非标准Lagrange函数到Friedmann-Robertson-Walker时空，讨论了它在宇宙学中的影响。关于非标准Lagrange函数应用的研究还有其他一些重要成果^[18-22]，但尚未涉及能量积分及降阶法。

动力学系统的守恒量，表现为深刻的物理规律。分析力学发展后，通过循环积分^[23-26]和广义能量积分^[27-31]可以寻求系统的守恒量从而对系统约化。1904年, Whittaker^[32]利用能量积分降阶了完整保守系统的运动方程，得到了Whittaker方程。之后，Whittaker降阶法引起了学者们的关注。然而，目前为止，对Whittaker方程的研究限于标准Lagrange函数系统。

本文将研究动力学系统基于指数Lagrange函数和Lagrange函数幂函数的广义能量积分及降阶方法。通过变分方法得到系统的Lagrange方程，给出系统广义能量积分存在的条件及形式，建立基于两类非标准Lagrange函数的动力学系统的Whittaker方法。

1 基于指数Lagrange函数的动力学系统的广义能量积分与Whittaker降阶法 1.1 Lagrange方程

假设系统的位形由n个广义坐标q_s(s=1, 2, …, n)确定，则基于指数Lagrange函数的作用量定义为^[15]

$ S = \int_{{t_1}}^{{t_2}} {\exp \left[ {L\left( {t,{q_s},{{\dot q}_s}} \right)} \right]{\rm{d}}t} $

(1)

式中：L=L(t, q_s, ${\dot q_s}$)为经典意义下的Lagrange函数。与基于指数Lagrange函数的作用量式(1) 相应的Hamilton原理可写为

$ \delta S = 0 $

(2)

带有交换关系

$ d\delta {q_s} = \delta {\rm{d}}{q_s}\;\;\;\;\;s = 1,2, \cdots ,n $

(3)

以及边界条件

$ \delta {q_s}\left| {_{t = {t_1}}} \right. = \delta {q_s}\left| {_{t = {t_2}}} \right. = 0\;\;\;\;\;s = 1,2, \cdots ,n $

(4)

因为

$ \begin{array}{*{20}{c}} {\delta \exp \left( L \right) = \sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {q_s}}}\delta {q_s}} + \sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {{\dot q}_s}} = }\\ {\sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {q_s}}}\delta {q_s}} + \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {q_s}} } \right) - }\\ {\sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {q_s}} - \sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{\partial L}}{{\partial t}}\delta {q_s}} } \end{array} $

(5)

将式(5) 代入式(2)，有

$ \begin{array}{*{20}{c}} {\int_{{t_1}}^{{t_2}} {\sum\limits_{s = 1}^n {\left( \begin{array}{l} \exp \left( L \right)\frac{{\partial L}}{{\partial {q_s}}} - \exp \left( L \right)\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) - \\ \;\;\;\;\;\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{\partial L}}{{\partial t}} \end{array} \right)} } \delta {q_s}{\rm{d}}t + }\\ {\sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {q_s}} \left| {_{{t_1}}^{{t_2}}} \right. = 0} \end{array} $

(6)

利用边界条件式(4)，得到

$ \int_{{t_1}}^{{t_2}} {\sum\limits_{s = 1}^n {\left( \begin{array}{l} \exp \left( L \right)\frac{{\partial L}}{{\partial {q_s}}} - \exp \left( L \right)\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) - \\ \;\;\;\;\;\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{\partial L}}{{\partial t}} \end{array} \right)} } \delta {q_s}{\rm{d}}t = 0 $

(7)

对于完整系统，δq_s(s=1, 2, …, n)相互独立，由变分学基本引理^[33]，得到

$ \begin{array}{*{20}{c}} {\exp \left( L \right)\left( {\frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right) = 0}\\ {s = 1,2, \cdots ,n} \end{array} $

(8)

方程式(8) 称为基于指数Lagrange函数的动力学系统的Lagrange方程^[15]。

1.2 广义能量积分

为了得到所论系统的广义能量积分，计算

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\left( {\sum\limits_{s = 1}^n {{{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} - 1} \right)\exp \left( L \right)} \right] = \\ \;\;\;\;\sum\limits_{s = 1}^n {\exp \left( L \right){{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} + \exp \left( L \right){{\ddot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}} + \\ \;\;\;\;\exp \left( L \right){{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \exp \left( L \right)\frac{{{\rm{d}}L}}{{{\rm{d}}t}} = \\ \;\;\;\;\sum\limits_{s = 1}^n {\exp \left( L \right){{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} + \exp \left( L \right){{\ddot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}} + \\ \;\;\;\;\exp \left( L \right){{\dot q}_s}\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \\ \;\;\;\;\exp \left( L \right)\left( {\frac{{{\rm{d}}L}}{{{\rm{d}}t}} + \frac{{\partial L}}{{\partial {q_s}}}{{\dot q}_s} + \frac{{\partial L}}{{\partial {{\dot q}_s}}}{{\ddot q}_s}} \right) = \\ \;\;\;\;\sum\limits_{s = 1}^n {\left[ {\left( {\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}} + \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \frac{{\partial L}}{{\partial {q_s}}}} \right){{\dot q}_s} - \frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right]} \exp \left( L \right) \end{array} $

(9)

由方程式(8，9) 知，如果Lagrange函数不显含时间t，即

$ \frac{{{\rm{d}}L}}{{{\rm{d}}t}} = 0 $

(10)

则沿着系统的动力学轨线有

$ \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\left( {\sum\limits_{s = 1}^n {{{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} - 1} \right)\exp \left( L \right)} \right] = 0 $

(11)

于是系统存在广义能量积分

$ \left( {\sum\limits_{s = 1}^n {{{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} - 1} \right)\exp \left( L \right) = h = {\rm{const}} $

(12)

1.3 Whittaker降阶法

任取一个广义坐标代替t的作用，如取q₁，令

$ {{q'}_r} = \frac{{{\rm{d}}{q_r}}}{{{\rm{d}}{q_1}}}\;\;\;\;\;\;r = 2,3, \cdots ,n $

(13)

则有

$ {{q'}_r} = \frac{{{\rm{d}}{q_r}}}{{{\rm{d}}{q_1}}}\frac{{{\rm{d}}{q_1}}}{{{\rm{d}}t}} = {{q'}_r}{{\dot q}_1}\;\;\;\;r = 2,3, \cdots ,n $

(14)

设

$ \exp {L^ * }\left( {{{\dot q}_1},{{q'}_r},{q_s}} \right) = \exp L\left( {{{\dot q}_1},{{q'}_r}{{\dot q}_1},{q_s}} \right) $

(15)

将式(15) 对${\dot q_1}$, q′_r, q_s分别求偏导数，得

$ \begin{array}{*{20}{c}} {\exp \left( {{L^ * }} \right) = \frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} = \exp \left( L \right)\left( {\frac{{\partial L}}{{\partial {{\dot q}_1}}} + \sum\limits_{r = 2}^n {\frac{{\partial L}}{{\partial {{\dot q}_r}}}\frac{{\partial {{\dot q}_r}}}{{\partial {{\dot q}_1}}}} } \right) = }\\ {\exp \left( L \right)\left( {\frac{{\partial L}}{{\partial {{\dot q}_1}}} + \sum\limits_{r = 2}^n {\frac{{\partial L}}{{\partial {{\dot q}_r}}}\frac{{\partial {{\dot q}_r}}}{{\partial {{\dot q}_1}}}} } \right) = \sum\limits_{s = 1}^n {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{{\dot q}_s}}}{{{{\dot q}_1}}}} } \end{array} $

(16)

$ \begin{array}{*{20}{c}} {\exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {{q'}_r}}} = \exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_r}}}\frac{{\partial {{\dot q}_r}}}{{\partial {{\dot q}_1}}}}\\ {r = 2,3, \cdots ,n} \end{array} $

(17)

$ \begin{array}{*{20}{c}} {\exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {q_s}}} = \exp \left( L \right)\frac{{\partial L}}{{\partial {q_s}}}}\\ {s = 1,2, \cdots ,n} \end{array} $

(18)

利用广义能量积分式(12，14) 解出${\dot q_1}$，记作

$ {{\dot q}_1} = {{\dot q}_1}\left( {{{q'}_r},{q_s}} \right) $

(19)

根据式(15，16)，式(12) 可表示成

$ \exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}}{{\dot q}_1} - \exp \left( {{L^ * }} \right) = h $

(20)

其中${\dot q_1}$由式(19) 确定，将式(20) 分别对q′_r和q_s求偏导数，有

$ \begin{array}{*{20}{c}} {\exp \left( {{L^ * }} \right)\left( {\frac{{\partial {L^ * }}}{{\partial {{q'}_r}}}\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {{q'}_r}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial {{\dot q}_1}\partial {{q'}_r}}}} \right){{\dot q}_1} - }\\ {\exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {{q'}_r}}} = 0} \end{array} $

(21)

$ \begin{array}{*{20}{c}} {\exp \left( {{L^ * }} \right)\left( {\frac{{\partial {L^ * }}}{{\partial {q_s}}}\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {q_s}}} + } \right.}\\ {\left. {\frac{{{\partial ^2}{L^ * }}}{{\partial {{\dot q}_1}\partial {q_s}}}} \right){{\dot q}_1} - \exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {q_s}}} = 0} \end{array} $

(22)

令

$ \exp W\left( {{{q'}_r},{q_s}} \right) = \exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} $

(23)

将式(23) 对q′_r, q_s求偏微分，得

$ \begin{array}{*{20}{c}} {\exp \left( W \right) = \frac{{\partial W}}{{\partial {{q'}_r}}} = }\\ {\exp \left( {{L^ * }} \right)\left( {\frac{{\partial {L^ * }}}{{\partial {{q'}_r}}}\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {{q'}_r}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial {{\dot q}_1}\partial {{q'}_r}}}} \right)} \end{array} $

(24)

$ \begin{array}{*{20}{c}} {\exp \left( W \right) = \frac{{\partial W}}{{\partial {q_s}}} = }\\ {\exp \left( {{L^ * }} \right)\left( {\frac{{\partial {L^ * }}}{{\partial {q_s}}}\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {q_s}}} + \frac{{{\partial ^2}{L^ * }}}{{\partial {{\dot q}_1}\partial {q_s}}}} \right)} \end{array} $

(25)

由式(21，24)，得

$ \exp \left( W \right) = \frac{{\partial W}}{{\partial {{q'}_r}}} = \frac{1}{{{{\dot q}_1}}}\exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {{q'}_r}}} $

(26)

由式(22，25)，得

$ \exp \left( W \right) = \frac{{\partial W}}{{\partial {q_s}}} = \frac{1}{{{{\dot q}_1}}}\exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {q_s}}} $

(27)

再根据式(17，18)，得

$ \begin{array}{*{20}{c}} {\exp \left( W \right) = \frac{{\partial W}}{{\partial {{q'}_r}}} = \exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_r}}}}\\ {r = 2,3, \cdots ,n} \end{array} $

(28)

$ \begin{array}{*{20}{c}} {\exp \left( W \right) = \frac{{\partial W}}{{\partial {q_s}}}\frac{1}{{{{\dot q}_1}}}\exp \left( L \right)\frac{{\partial L}}{{\partial {q_s}}}}\\ {s = 1,2, \cdots ,n} \end{array} $

(29)

方程式(8) 可表示成

$ \exp \left( L \right) = \frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\exp \left( L \right)\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) = 0 $

(30)

将式(28，29) 代入式(30)，有

$ \exp \left( W \right)\frac{{\partial W}}{{\partial {q_r}}}{{\dot q}_1} - \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\exp \left( W \right)\frac{{\partial W}}{{\partial {{q'}_r}}}} \right) = 0 $

或

$ {\exp \left( W \right)\frac{{\partial W}}{{\partial {q_r}}} - \frac{{\rm{d}}}{{{\rm{d}}{q_1}}}\left( {\exp \left( W \right)\frac{{\partial W}}{{\partial {{q'}_r}}}} \right) = 0} $

(31)

展开之后，得到

Lagrange函数的Whittaker方程

$ \begin{array}{*{20}{c}} {\exp \left( W \right)\left( {\frac{{\partial W}}{{\partial {q_r}}} - \frac{{\rm{d}}}{{{\rm{d}}{q_1}}}\frac{{\partial W}}{{\partial {{q'}_r}}} - \frac{{\partial W}}{{\partial {{q'}_r}}}\frac{{\partial W}}{{\partial {q_1}}}} \right) = 0}\\ {r = 2,3, \cdots ,n} \end{array} $

(32)

式中：W为关于q′₂, q′₃, …, q′_n, q₁, q₂, …, q_n的函数，而q₁是相当于时间t的独立变量。Whittaker方程的形式等同于运动方程式(8)，但方程个数变成n-1个，从而达到了降阶的目的。

1.4 算例

设基于指数Lagrange函数的作用量为

$ S = \int_{{t_1}}^{{t_2}} {{e^{ - {q_1}{q_2}}}\left( {\dot q_1^2 + \dot q_2^2} \right){\rm{d}}t} $

(33)

试求广义能量积分，并用其将方程降阶。

本问题中，经典的Lagrange函数为

$ L = \ln \left( {\dot q_1^2 + \dot q_2^2} \right) - {q_1}{q_2} $

(34)

由方程式(8) 得到

$ \begin{array}{l} 2{{\ddot q}_1} - \dot q_1^2{q_2} - 2{{\dot q}_1}{{\dot q}_2}{q_1} + \dot q_2^2{q_2} = 0\\ 2{{\ddot q}_2} - \dot q_2^2{q_1} - 2{{\dot q}_2}{{\dot q}_1}{q_2} + \dot q_1^2{q_1} = 0 \end{array} $

(35)

方程式(35) 为非线性微分方程组。

由式(34) 知

$ \frac{{\partial L}}{{\partial t}} = 0 $

(36)

因此存在能量积分

$ \left( {\dot q_1^2 + \dot q_2^2} \right){e^{ - {q_1}{q_2}}} = h $

(37)

令${\dot q_2} = {\dot q_1}{q'_2}$代入式(37)，有

$ \dot q_1^2\left( {1 + q_2^{'2}} \right){e^{ - {q_1}{q_2}}} = h $

(38)

解出${\dot q_1}$，得

$ {{\dot q}_1} = \sqrt {\frac{{h{e^{{q_1}{q_2}}}}}{{1 + q_2^{'2}}}} $

(39)

根据式(15) 构造expL^*函数，有

$ \begin{array}{*{20}{c}} {\exp {L^ * } = \exp \left( {\ln \left( {\dot q_1^2\left( {1 + q_2^{'2}} \right)} \right) - {q_1}{q_2}} \right) = }\\ {\dot q_1^2\left( {1 + q_2^{'2}} \right){e^{ - {q_1}{q_2}}}} \end{array} $

(40)

构造expW函数

$ \begin{array}{l} \exp W = \exp \left( {{L^ * }} \right)\frac{{\partial {L^ * }}}{{\partial {{\dot q}_1}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;2{{\dot q}_1}\left( {1 + q_2^{'2}} \right){e^{ - {q_1}{q_2}}} \end{array} $

(41)

将式(39) 代入式(41)，整理后得

$ \exp W = 2\sqrt {h{e^{ - {q_1}{q_2}}}\left( {1 + q_2^{'2}} \right)} $

(42)

由方程式(32)，得

$ \exp \left( W \right)\left( {\frac{{\partial W}}{{\partial {q_2}}} - \frac{{\rm{d}}}{{{\rm{d}}{q_1}}}\frac{{\partial W}}{{\partial {{q'}_2}}} - \frac{{\partial W}}{{\partial {{q'}_2}}}\frac{{\partial W}}{{\partial {q_1}}}} \right) = 0 $

(43)

即

$ {q_2}{{q'}_2} - {q_1}q_2^{'2} - {q_1} = 0 $

(44)

2 基于Lagrange函数幂函数的动力学系统的广义能量积分与Whittaker降阶法 2.1 Lagrange方程

假设系统的位形由n个广义坐标q_s(s=1, 2, …, n)确定，则基于Lagrange函数幂函数的作用量定义为^[15]

$ S = \int_{{t_1}}^{{t_2}} {{L^{1 + \gamma }}\left( {t,{q_s},{{\dot q}_s}} \right){\rm{d}}t} $

(45)

式中：L=L(t, q_s, ${\dot q_s}$是经典意义下的Lagrange函数，γ为任意实数。与基于Lagrange函数幂函数的作用量(式(45))相应的Hamilton原理可写为

$ \delta S = 0 $

(46)

带有交换关系

$ d\delta {q_s} = \delta {\rm{d}}{q_s}\;\;\;\;\;s = 1,2, \cdots ,n $

(47)

以及边界条件

$ \begin{array}{*{20}{c}} {\delta {q_s}\left| {_{t = {t_1}}} \right. = \delta {q_s}\left| {_{t = {t_2}}} \right. = 0}\\ {s = 1,2, \cdots ,n} \end{array} $

(48)

因为

$ \begin{array}{*{20}{c}} {\delta {L^{1 + \gamma }} = \left( {1 + \gamma } \right){L^\gamma }\left( {\sum\limits_{s = 1}^n {\frac{{\partial L}}{{\partial {q_s}}}\delta {q_s}} + \sum\limits_{s = 1}^n {\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {{\dot q}_s}} } \right) = }\\ {\left( {1 + \gamma } \right){L^\gamma }\sum\limits_{s = 1}^n {\frac{{\partial L}}{{\partial {q_s}}}\delta {q_s}} + \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\left( {1 + \gamma } \right){L^\gamma }\sum\limits_{s = 1}^n {\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {{\dot q}_s}} } \right] - }\\ {\left( {1 + \gamma } \right){L^\gamma }\sum\limits_{s = 1}^n {\left( {\frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) + \frac{\gamma }{L}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right]\delta {q_s}} } \end{array} $

(49)

将式(49) 代入式(46)，有

$ \begin{array}{*{20}{c}} {\int_{{t_1}}^{{t_2}} {\left( {1 + \gamma } \right){L^\gamma }} \sum\limits_{s = 1}^n {\left[ {\frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) - \frac{\gamma }{L}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right]\delta {q_s}} \cdot }\\ {{\rm{d}}t + \left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {{\dot q}_s}}}\delta {q_s}\left| {_{{t_1}}^{{t_2}}} \right. = 0} \end{array} $

(50)

利用边界条件式(48)，得到

$ \begin{array}{*{20}{c}} {\int_{{t_1}}^{{t_2}} {\left( {1 + \gamma } \right){L^\gamma }} \sum\limits_{s = 1}^n {\left[ {\frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) - \frac{\gamma }{L}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right]} \cdot }\\ {\delta {q_s}{\rm{d}}t = 0} \end{array} $

(51)

对于完整系统，δq_s(s=1, 2, …, n)相互独立，由变分学基本引理^[33]，得到

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){L^\gamma }\left( {\frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \frac{\gamma }{L}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right) = 0}\\ {s = 1,2, \cdots ,n} \end{array} $

(52)

当γ≠-1时，有

$ \begin{array}{*{20}{c}} {\frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \frac{\gamma }{L}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}} = 0}\\ {s = 1,2, \cdots ,n} \end{array} $

(53)

方程式(53) 称为基于Lagrange函数幂函数的动力学系统的Lagrange方程^[15]。

若γ=0，方程式(52) 为经典的Lagrange运动方程，即

$ \frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} = 0\;\;\;\;\;\;s = 1,2, \cdots ,n $

(54)

2.2 广义能量积分

下面计算系统的广义能量积分。因为

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\left( {\sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - L} } \right){L^\gamma }} \right] = \\ \sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){L^\gamma }} \left( {{{\ddot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}} + \frac{\gamma }{L}{{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}} + {{\dot q}_s}\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) - \\ \left( {1 + \gamma } \right){L^\gamma }\frac{{{\rm{d}}L}}{{{\rm{d}}t}} - \\ \sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){L^\gamma }} \left( {{{\ddot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}} + \frac{\gamma }{L}{{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}} + {{\dot q}_s}\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) - \\ \left( {1 + \gamma } \right){L^\gamma }\left( {\frac{{{\rm{d}}L}}{{{\rm{d}}t}} + \frac{{\partial L}}{{\partial {q_s}}}{{\dot q}_s} + \frac{{\partial L}}{{\partial {{\dot q}_s}}}{{\ddot q}_s}} \right) = \\ \sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){L^\gamma }} \cdot \\ \left[ {\left( {\frac{\gamma }{L}\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{\rm{d}}L}}{{{\rm{d}}t}} + \frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial L}}{{\partial {{\dot q}_s}}} - \frac{{\partial L}}{{\partial {q_s}}}} \right){{\dot q}_s} - \frac{{{\rm{d}}L}}{{{\rm{d}}t}}} \right] \end{array} $

(55)

由方程式(53，55) 知，如果Lagrange函数不显含时间t，即

$ \frac{{\partial L}}{{\partial t}} = 0 $

(56)

则沿着系统的动力学轨线有

$ \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\left( {\sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} - L} \right){L^\gamma }} \right] = 0 $

(57)

于是系统存在广义能量积分

$ \left( {\sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){{\dot q}_s}\frac{{\partial L}}{{\partial {{\dot q}_s}}}} - L} \right){L^\gamma } = h = {\rm{const}} $

(58)

2.3 Whittaker降阶法

取广义坐标q₁代替时间t，令

$ {{q'}_r} = \frac{{{\rm{d}}{q_r}}}{{{\rm{d}}{q_1}}}\;\;\;\;\;r = 2,3, \cdots ,n $

(59)

则有

$ {{\dot q}_r} = \frac{{{\rm{d}}{q_r}}}{{{\rm{d}}{q_1}}}\frac{{{\rm{d}}{q_1}}}{{{\rm{d}}t}} = {{q'}_r}{{\dot q}_1}\;\;\;\;\;r = 2,3, \cdots ,n $

(60)

设

$ {{\bar L}^{1 + \gamma }}\left( {{{\dot q}_1},{{q'}_r},{q_s}} \right) = {L^{1 + \gamma }}\left( {{{\dot q}_1},{{q'}_r}{{\dot q}_1},{q_s}} \right) $

(61)

将式(61) 对${\dot q_1}$, q′_r, q_s分别求偏导数，得

$ \begin{array}{l} \left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} = \left( {1 + \gamma } \right){L^\gamma }\left( {\frac{{\partial L}}{{\partial {{\dot q}_1}}} + \sum\limits_{r - 2}^n {\frac{{\partial L}}{{\partial {{\dot q}_r}}}\frac{{\partial {{\dot q}_r}}}{{\partial {{\dot q}_1}}}} } \right) = \\ \left( {1 + \gamma } \right){L^\gamma }\left( {\frac{{\partial L}}{{\partial {{\dot q}_1}}} + \sum\limits_{r - 2}^n {\frac{{\partial L}}{{\partial {{\dot q}_r}}}\frac{{\partial {{\dot q}_r}}}{{\partial {{\dot q}_1}}}} } \right) = \\ \sum\limits_{s = 1}^n {\left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {{\dot q}_s}}}\frac{{{{\dot q}_s}}}{{{{\dot q}_1}}}} \end{array} $

(62)

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{q'}_r}}} = \left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {{\dot q}_r}}}{{\dot q}_1}}\\ {r = 2,3, \cdots ,n} \end{array} $

(63)

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {q_s}}} = \left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {q_s}}}}\\ {s = 1,2, \cdots ,n} \end{array} $

(64)

利用广义能量积分式(58，60) 解出${\dot q_1}$，记作

$ {{\dot q}_1} = {{\dot q}_1}\left( {{{q'}_r},{q_s}} \right) $

(65)

根据式(61，62)，式(58) 可表示成

$ \left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}}{{\dot q}_1} - {{\bar L}^{1 + \gamma }} = h $

(66)

式中${\dot q_1}$由式(65) 确定，将式(66) 分别对q′_r和q_s求偏导数，有

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\bar L}^\gamma }\left( {\frac{\gamma }{{\bar L}}\frac{{\partial \bar L}}{{\partial {{q'}_r}}}\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}\bar L}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_r}}}{{\partial {{q'}_1}}} + } \right.}\\ {\left. {\frac{{{\partial ^2}\bar L}}{{\partial {{\dot q}_1}\partial {{q'}_r}}}} \right){{\dot q}_1} - \left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{q'}_r}}} = 0} \end{array} $

(67)

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\bar L}^\gamma }\left( {\frac{\gamma }{{\bar L}}\frac{{\partial \bar L}}{{\partial {q_s}}}\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}\bar L}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {q_s}}} + } \right.}\\ {\left. {\frac{{{\partial ^2}\bar L}}{{\partial {{\dot q}_1}\partial {q_s}}}} \right){{\dot q}_1} - \left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{q'}_s}}} = 0} \end{array} $

(68)

令

$ {{\tilde L}^{1 + \gamma }}\left( {{{q'}_r},{q_s}} \right) = \left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} $

(69)

将式(69) 对q′_r, q_s求偏微分，得

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {{q'}_r}}} = \left( {1 + \gamma } \right){{\bar L}^\gamma } \cdot }\\ {\left( {\frac{\gamma }{L}\frac{{\partial \bar L}}{{\partial {{q'}_r}}}\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}\bar L}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {{q'}_r}}} + \frac{{{\partial ^2}\bar L}}{{\partial {{\dot q}_1}\partial {{q'}_r}}}} \right)} \end{array} $

(70)

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {q_s}}} = \left( {1 + \gamma } \right){{\bar L}^\gamma } \cdot }\\ {\left( {\frac{\gamma }{L}\frac{{\partial \bar L}}{{\partial {q_s}}}\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} + \frac{{{\partial ^2}\bar L}}{{\partial \dot q_1^2}}\frac{{\partial {{\dot q}_1}}}{{\partial {q_s}}} + \frac{{{\partial ^2}\bar L}}{{\partial {{\dot q}_1}\partial {q_s}}}} \right)} \end{array} $

(71)

由式(67，70)，得

$ \left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {{q'}_r}}} = \frac{1}{{{{\dot q}_1}}}\left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{q'}_r}}} $

(72)

由式(68，71)，得

$ \left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {q_s}}} = \frac{1}{{{{\dot q}_1}}}\left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {q_s}}} $

(73)

再根据式(63，64)，得

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {{q'}_r}}} = \left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {{q'}_r}}}}\\ {r = 2,3, \cdots ,n} \end{array} $

(74)

$ \begin{array}{*{20}{c}} {\left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {q_s}}} = \frac{1}{{{{\dot q}_1}}}\left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {q_s}}}}\\ {s = 1,2, \cdots ,n} \end{array} $

(75)

方程式(52) 可表示成

$ \left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {q_s}}} - \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\left( {1 + \gamma } \right){L^\gamma }\frac{{\partial L}}{{\partial {{\dot q}_s}}}} \right) = 0 $

(76)

将式(74，75) 代入式(76)，有

$ \left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {q_r}}}{{\dot q}_1} - \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {{q'}_r}}}} \right) = 0 $

或

$ \left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {q_r}}} - \frac{{\rm{d}}}{{{\rm{d}}{q_1}}}\left( {\left( {1 + \gamma } \right){{\tilde L}^\gamma }\frac{{\partial \tilde L}}{{\partial {{q'}_r}}}} \right) = 0 $

(77)

整理之后，得到

$ \begin{array}{*{20}{c}} {\frac{{\partial \tilde L}}{{\partial {q_r}}} - \frac{{\rm{d}}}{{{\rm{d}}{q_1}}}\frac{{\partial \tilde L}}{{\partial {{q'}_r}}} - \frac{\gamma }{{\tilde L}}\frac{{\partial \tilde L}}{{\partial {{q'}_r}}}\frac{{{\rm{d}}\tilde L}}{{{\rm{d}}{q_1}}} = 0}\\ {r = 2,3, \cdots ,n} \end{array} $

(78)

方程式(78) 称为基于Lagrange函数幂函数的动力学系统的Whittaker方程。它与方程式(53) 相似，只是时间t的地位被广义坐标q₁取代，但将原n个自由度问题降到了n-1个自由度问题。

2.4 算例

设基于Lagrange函数幂函数的作用量为

$ S = \int_{{t_1}}^{{t_2}} {{{\left[ {{{\dot q}_2}\left( {1 + q_1^2} \right) + {{\dot q}_1} + {q_2}} \right]}^{1 + \gamma }}{\rm{d}}t} $

(79)

试求广义能量积分，并用其将方程降阶。

本问题中，经典的Lagrange函数为

$ L = {{\dot q}_2}\left( {1 + q_1^2} \right) + {{\dot q}_1} + {q_2} $

(80)

若取γ=1，由方程式(53) 得到

$ \begin{array}{*{20}{c}} {2{{\dot q}_2}{q_1}\left( {{{\dot q}_2}\left( {1 + q_1^2} \right) + {q_2}} \right) - }\\ {\left( {{{\ddot q}_2}\left( {1 + q_1^2} \right) + {{\ddot q}_1} + {{\dot q}_2}} \right) = 0}\\ {\left( {1 - 2{{\dot q}_1}{q_1}} \right)\left( {{{\dot q}_2}\left( {1 + q_1^2} \right) + {{\dot q}_1} + {q_2}} \right) - }\\ {\left( {1 + q_1^2} \right)\left( {{{\ddot q}_2}\left( {1 + q_1^2} \right) + 2{{\dot q}_2}{{\dot q}_1}{q_1} + {{\ddot q}_1} + {{\dot q}_2}} \right) = 0} \end{array} $

(81)

运动方程式(81) 为非线性微分方程组。

由式(80) 知

$ \frac{{\partial L}}{{\partial t}} = 0 $

(82)

则存在广义能量积分

$ {\left[ {{{\dot q}_2}\left( {1 + q_1^2} \right) + {{\dot q}_1}} \right]^2} - q_2^2 = h $

(83)

令${\dot q_2} = {\dot q_1}{q'_2}$代入式(83)，有

$ \dot q_1^2{\left[ {{{q'}_2}\left( {1 + q_1^2} \right) + 1} \right]^2} - q_2^2 = h $

(84)

解出${\dot q_1}$，得

$ {{\dot q}_1} = \frac{{\sqrt {h + q_2^2} }}{{{{q'}_2}\left( {1 + q_1^2} \right) + 1}} $

(85)

根据式(61) 构造函数，有

$ \begin{array}{*{20}{c}} {\bar L = {{\dot q}_1}{{q'}_2}\left( {1 + q_1^2} \right) + {{\dot q}_1} + {q_2} = }\\ {{{\dot q}_1}\left[ {{{q'}_2}\left( {1 + q_1^2} \right) + 1} \right] + {q_2}} \end{array} $

(86)

构造${\tilde L^2}$函数

$ \begin{array}{*{20}{c}} {{{\tilde L}^2} = \left( {1 + \gamma } \right){{\bar L}^\gamma }\frac{{\partial \bar L}}{{\partial {{\dot q}_1}}} = }\\ {2\left\{ \begin{array}{l} {{\dot q}_1}\left[ {{{q'}_2}\left( {1 + q_1^2} \right) + } \right.\\ \;\;\;\;\left. 1 \right] + {q_2} \end{array} \right\}\left[ {{{q'}_2}\left( {1 + q_1^2} \right) + 1} \right]} \end{array} $

(87)

将式(85) 代入式(87)，整理后得

$ {{\tilde L}^2} = 2\left( {\sqrt {h + q_2^2} + {q_2}} \right)\left[ {{{q'}_2}\left( {1 + q_1^2} \right) + 1} \right] $

(88)

由方程式(78)，得

$ \frac{{\partial \bar L}}{{\partial {q_2}}} - \frac{{\rm{d}}}{{{\rm{d}}{q_1}}}\frac{{\partial \tilde L}}{{\partial {{q'}_2}}} - \frac{1}{{\tilde L}}\frac{{\partial \tilde L}}{{\partial {{q'}_2}}}\frac{{{\rm{d}}\tilde L}}{{{\rm{d}}{q_1}}} = 0 $

(89)

即

$ \begin{array}{*{20}{c}} {\left[ {{{q'}_2}\left( {1 + q_1^2} \right) + 1} \right]\left( {1 + \frac{{2{q_2}}}{{\sqrt {h + q_2^2} }}} \right) - }\\ {2{q_1}\left( {\sqrt {h + q_2^2} + {q_2}} \right) = 0} \end{array} $

(90)

3 结束语

利用非标准Lagrange函数可以解决非线性问题等，算例中通过Whittaker降阶法分别将方程式(35，81) 降阶为方程式(44，90)，显示了基于非标准Lagrange函数的Whittaker降阶法在非线性动力学中的优势。文章研究了基于两类非标准Lagrange函数的动力学系统的广义能量积分与降阶问题，利用系统的Lagrange方程得到了广义能量积分存在的约束条件，并将著名的Whittaker降阶法推广到基于非标准Lagrange函数的动力学系统，得到了基于非标准Lagrange函数的Whittaker方程。本文的方法具有普遍意义，可进一步推广到其他非标准Lagrange函数系统。同时，也可研究基于非标准Lagrange函数的循环积分及Routh降阶法。