Noether定理首次从变分学背景下揭示了对称性与守恒量之间的相互关系[1],即Hamilton作用量在关于广义坐标和时间的变换群的无限小变换下的不变性意味着沿着系统的动力学真实运动轨道存在一个守恒量.Noether定理阐释了牛顿力学的所有守恒量,如:时间的均匀性导致质点系的能量守恒;空间的均匀性导致质点系的动量守恒;空间各向同性导致质点系的动量矩守恒[2].Djukić等[3]将结果拓展到完整非保守系统并考虑无限小变换的生成元包含广义速度的情形;李子平[4],Bahar等[5]给出线性非完整约束系统的Noether定理;刘端[6]进一步将结果拓展到一般非完整非保守力学系统;梅凤翔[7-8]建立了Birkhoff系统和广义Birkhoff系统的Noether理论;文献[9-10]研究了分数阶Birkhoff系统的Noether对称性与守恒量.关于Noether定理及其应用的研究已取得重要进展[11-23].但是,上述Noether定理都是基于经典形式的变分原理,即:变分问题的作用量是由积分泛函定义的,它们不能应用于作用量由微分方程定义的一般情形.
1930年,Herglotz[24]在研究接触变换及其与Hamilton系统和Poisson括号的联系时提出了一类广义变分原理,其作用量是由微分方程来定义的.Herglotz广义变分原理不仅能够描述所有采用经典变分原理能够描述的物理过程,而且能够应用于解决经典变分原理不能适用的问题,如:非保守动力学过程或耗散系统的变分描述[25-26].2002年,Georgieva等[27]研究了Herglotz广义变分原理和Noether定理.但是迄今为止基于Herglotz广义变分原理的Noether对称性的研究尚不多见且研究限于位形空间中的Lagrange系统[28-30].由于力学系统的相空间具有自然辛结构,在数学描述上比Lagrange力学要容易[31].对于某些系统,在位形空间中其对称性并不明显地表现出来,但在相空间中却具有一定的对称性质,用正则形式的Noether定理则可导出相应的守恒量[2].本文将研究相空间中非保守系统的Herglotz广义变分原理,并基于非等时变分与等时变分之间的关系导出相空间中Hamilton-Herglotz作用量的变分公式,建立在此情形下Noether对称性的判据,提出并证明相空间中Herglotz变分问题的Noether定理及其逆定理.文末以著名的Emden方程和平方阻尼振子系统为例说明结果的应用.
1 相空间中Herglotz广义变分原理与Hamilton正则方程根据位形空间中Herglotz广义变分原理[29],相空间中非保守系统的Herglotz变分问题可定义为:
确定函数
$ \dot {z}(t) = p_s (t)\dot {q}_s (t) - H\left( {t, q_s (t), p_s (t), z(t)} \right) $ | (1) |
定义的泛函
$ \left. {q_s (t)} \right|_{t = t_1 } = q_{s1} \, , \ \ \left. {q_s \left( t \right)} \right|_{t = t_2 } = q_{s2} \, \ \ \left( {s = 1, 2, \cdots, n} \right) $ | (2) |
和初始条件
$ \left. {z(t)} \right|_{t = t_1 } = z_1 $ | (3) |
下,在
由式(1)确定的泛函z可称为Hamilton-Herglotz作用量,上述变分问题可称为相空间中非保守系统的Herglotz广义变分原理.
对方程(1)进行等时变分运算,有
$ \delta \dot {z} = \dot {q}_s \delta p_s + p_s \delta \dot {q}_s - \dfrac{\partial H}{\partial q_s }\delta q_s -\\ \qquad \dfrac{\partial H}{\partial p_s }\delta p_s - \dfrac{\partial H}{\partial z}\delta z $ | (4) |
这里及文中采用Einstein求和约定,即同一项中两个相同的活动指标表示对其求和.利用交换关系
$ \frac{{{\rm{d}}\delta z}}{{{\rm{d}}t}} = A - \dfrac{\partial H}{\partial z}\delta z $ | (5) |
其中
$ A = \dot {q}_s \delta p_s + p_s \delta \dot {q}_s - \dfrac{\partial H}{\partial q_s }\delta q_s - \dfrac{\partial H}{\partial p_s }\delta p_s $ | (6) |
方程(5)是关于
$ \delta z(t)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right) - \delta z\left( {t_1 } \right) = \\ \qquad \int_{t_1 }^t {A\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}} t } } \right){\rm{d}}t} $ | (7) |
由初始条件(3),且考虑到
$ \delta z\left( {t_1 } \right) = \delta z\left( {t_2 } \right) = 0 $ | (8) |
方程(7)在所有
$ \int_{t_1 }^{t_2 } {A\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right){\rm{d}}t} = 0 $ | (9) |
将式(6)代入方程(9),对
$ \begin{array}{l} \int_{{t_1}}^{{t_2}} {\exp \left( {\int_{{t_1}}^t {\frac{{\partial H}}{{\partial z}}{\rm{d}}t} } \right)} \left[ {\left( { - {{\dot p}_s} - \frac{{\partial H}}{{\partial {q_s}}} - {p_s}\frac{{\partial H}}{{\partial z}}} \right)\delta {q_s} + } \right.\\ \;\;\;\;\;\;\;\left. {\left( {{{\dot q}_s} - \frac{{\partial H}}{{\partial {p_s}}}} \right)\delta {p_s}} \right]{\rm{d}}t = 0 \end{array} $ | (10) |
由
$ \left. \begin{array}{l} \exp \left( {\int_{{t_1}}^t {\frac{{\partial H}}{{\partial z}}{\rm{d}}t} } \right)\left( {{{\dot q}_s} - \frac{{\partial H}}{{\partial {p_s}}}} \right) = 0\\ \exp \left( {\int_{{t_1}}^t {\frac{{\partial H}}{{\partial z}}{\rm{d}}t} } \right)\left( { - {{\dot p}_s} - \frac{{\partial H}}{{\partial {q_s}}} - {p_s}\frac{{\partial H}}{{\partial z}}} \right) = 0\\ \;\;\;\;\;\;\left( {s = 1,2, \cdots ,n} \right) \end{array} \right\} $ | (11) |
或
$ \dot {q}_s = \dfrac{\partial H}{\partial p_s }\, , \ \ \dot {p}_s = - \dfrac{\partial H}{\partial q_s } - p_s \dfrac{\partial H}{\partial z} \ \ \left( {s = 1, 2, \cdots, n} \right) $ | (12) |
方程(12)可称为相空间中非保守系统基于Herglotz广义变分原理的Hamilton正则方程.
如果Hamilton函数不显含z,即
$ z = \int_{t_1 }^{t_2 } {\left[{p_s (t)\dot {q}_s \left( t \right)-H\left( {t, q_s (t), p_s (t)} \right)} \right]{\rm{d}}t} $ | (13) |
而方程(12)成为
$ \dot {q}_s = \dfrac{\partial H}{\partial p_s } \, , \ \ \dot {p}_s = - \dfrac{\partial H}{\partial q_s } \ \ \ \left( {s = 1, 2, \cdots, n} \right) $ | (14) |
式(13)是相空间中Hamilton作用量,方程(14)为经典Hamilton正则方程.
方程(12)中第二组方程的项
引进时间t,广义坐标
$ \left. \begin{array}{l} \bar t = t + \Delta t\\ {{\bar q}_s}\left( {\bar t{\mkern 1mu} } \right) = {q_s}(t) + \Delta {q_s}\;\;\left( {s = 1,2, \cdots ,n} \right)\\ {{\bar p}_s}\left( {\bar t{\mkern 1mu} } \right) = {p_s}(t) + \Delta {p_s} \end{array} \right\} $ | (15) |
或其展开式
$ \left. \begin{array}{l} \bar t = t + {\varepsilon _\sigma }{\tau ^\sigma }\left( {t,{q_k},{p_k},z} \right)\\ {{\bar q}_s}\left( {\bar t{\mkern 1mu} } \right) = {q_s}(t) + {\varepsilon _\sigma }\xi _s^\sigma \left( {t,{q_k},{p_k},z} \right)\\ {{\bar p}_s}\left( {\bar t{\mkern 1mu} } \right) = {p_s}(t) + {\varepsilon _\sigma }\eta _s^\sigma \left( {t,{q_k},{p_k},z} \right)\\ \;\;\;\;\;\;\left( {s,k = 1,2, \cdots ,n} \right) \end{array} \right\} $ | (16) |
其中
$ \bar {z}\left( \bar {t} \, \right) = z(t) + \Delta z\left( t \right) $ | (17) |
作用量z的非等时变分
$ \Delta F = \delta F + \dot {F}\Delta t $ | (18) |
由式(18),并注意到交换关系
$ \frac{{{\rm{d}}\delta F}}{{{\rm{d}}t}} = \delta \frac{{{\rm{d}}F}}{{{\rm{d}}t}} = \delta \dot F $ | (19) |
易得
$ \Delta \dot F = \frac{{{\rm{d}}\Delta F}}{{{\rm{d}}t}} - \dot F\frac{{{\rm{d}}\Delta t}}{{{\rm{d}}t}} $ | (20) |
由方程(1),得到
$ \Delta \dot {z} = \dot {q}_s \Delta p_s + p_s \Delta \dot {q}_s - \dfrac{\partial H}{\partial t}\Delta t - \\ \qquad \dfrac{\partial H}{\partial q_s }\Delta q_s - \dfrac{\partial H}{\partial p_s }\Delta p_s - \dfrac{\partial H}{\partial z}\Delta z $ | (21) |
利用式(20),并考虑到方程(1),式(21)成为
$ \dfrac{{\rm{d}} \Delta z }{{\rm{d}}t} = \dot {q}_s \Delta p_s + p_s \dfrac{{\rm{d}} \Delta q_s }{{\rm{d}}t} - H\dfrac{{\rm{d}} \Delta t }{{\rm{d}}t} - \dfrac{\partial H}{\partial t}\Delta t - \\ \qquad \dfrac{\partial H}{\partial q_s }\Delta q_s - \dfrac{\partial H}{\partial p_s }\Delta p_s - \dfrac{\partial H}{\partial z}\Delta z $ | (22) |
解方程(22),可得
$ \Delta z(t)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right) - \Delta z\left( {t_1 } \right) = \\ \qquad \int_{t_1 }^t {\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)} \left( {\dot {q}_s \Delta p_s + p_s \dfrac{{\rm{d}} \Delta q_s }{{\rm{d}}t} - H\dfrac{{\rm{d}} \Delta t }{{\rm{d}}t} - }\right.\\ \qquad \left.{ \dfrac{\partial H}{\partial t}\Delta t - \dfrac{\partial H}{\partial q_s }\Delta q_s - \dfrac{\partial H}{\partial p_s }\Delta p_s } \right){\rm{d}}t $ | (23) |
显然
$ \Delta z(t)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right) = \\ \quad \int_{t_1 }^t \left\{ {\dfrac{{\rm{d}} }{{\rm{d}}t}\left[{\left( {p_s \Delta q_s-H\Delta t} \right)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)} \right]} \right. +\\ \quad \exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)\left[{\left( {- \dot {p}_s- \dfrac{\partial H}{\partial q_s }- p_s \dfrac{\partial H}{\partial z}} \right)\left( {\Delta q_s - \dot {q}_s \Delta t} \right) + }\right.\\ \quad \left.{\left.{\left( {\dot {q}_s -\dfrac{\partial H}{\partial p_s }} \right)\left( {\Delta p_s -\dot {p}_s \Delta t} \right)} \right]} \right\} {\rm{d}}t $ | (24) |
由于
$ \Delta t = \varepsilon _\sigma \tau ^\sigma\, , \ \ \Delta q_s = \varepsilon _\sigma \xi _s^\sigma \\ \Delta p_s = \varepsilon _\sigma \eta _s^\sigma \quad \left( {s = 1, 2, \cdots, n} \right) $ | (25) |
将式(25)代入式(23)和式(24),得到
$ \Delta z(t)\exp \left( {\int_{t_1 }^t \dfrac{\partial H}{\partial z}{\rm{d}}t } \right) =\\ \qquad \int_{t_1 }^t \left[{\exp \left( {\int_{t_1 }^t \dfrac{\partial H}{\partial z}{\rm{d}}t } \right)\left( {\dot {q}_s \eta _s^\sigma + p_s \dot {\xi }_s^\sigma- H\dot {\tau }^\sigma- }\right.}\right.\\ \qquad \left.{\left.{\dfrac{\partial H}{\partial t}\tau ^\sigma - \dfrac{\partial H}{\partial q_s }\xi _s^\sigma -\dfrac{\partial H}{\partial p_s }\eta _s^\sigma } \right)} \right] \varepsilon _\sigma {\rm{d}}t $ | (26) |
以及
$ \Delta z(t)\exp \left( {\int_{t_1 }^t \dfrac{\partial H}{\partial z}{\rm{d}}t } \right)= \\ \quad \int_{t_1 }^t \left\{ {\dfrac{{\rm{d}} }{{\rm{d}}t}\left[{\left( {p_s \xi _s^\sigma-H\tau ^\sigma } \right)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)} \right]} \right. +\\ \quad \left.{ \exp \left( {\int_{t_1 }^t \dfrac{\partial H}{\partial z}{\rm{d}}t } \right)\left[{\left( {- \dot {p}_s- \dfrac{\partial H}{\partial q_s }- p_s \dfrac{\partial H}{\partial z}} \right)\left( {\xi _s^\sigma - \dot {q}_s \tau ^\sigma } \right) +}\right.}\right. \\ \quad \left.{\left.{ \left( {\dot {q}_s -\dfrac{\partial H}{\partial p_s }} \right)\left( {\eta _s^\sigma - \dot {p}_s \tau ^\sigma } \right)} \right]} \right\} \varepsilon _\sigma {\rm{d}}t $ | (27) |
式(26)和式(27)是相空间中Hamilton-Herglotz作用量变分的两个基本公式.
3 相空间中基于Herglotz变分问题的Noether定理下面根据Noether对称性的概念[12],给出相空间中非保守系统基于Herglotz变分问题的Noether对称变换的定义和判据.
定义1 对于相空间中非保守系统的Herglotz变分问题,如果Hamilton -Herglotz作用量是无限小群变换的不变量,即:对每一个无限小变换,始终成立
$ \Delta z\left( {t_2 } \right) = 0 $ | (28) |
则称无限小群变换为系统(12)的Noether对称变换.
由定义1和式(26),可以得到如下判据:
判据1 对于无限小群变换(16),如果无限小生成元
$ \dot {q}_s \eta _s^\sigma + p_s \dot {\xi }_s^\sigma - H\dot {\tau }^\sigma -\dfrac{\partial H}{\partial t}\tau ^\sigma - \\ \qquad \dfrac{\partial H}{\partial q_s }\xi _s^\sigma - \dfrac{\partial H}{\partial p_s }\eta _s^\sigma = 0 \quad \left( {\sigma = 1, 2, \cdots, r} \right) $ | (29) |
则变换为相空间中非保守系统基于Herglotz变分问题的Noether对称变换.
下面建立相空间中非保守系统基于Herglotz变分问题的Noether定理,有:
定理1 对于相空间中非保守系统的Herglotz变分问题,如果无限小群变换(16)是系统(12)的Noether对称变换,则该系统存在r个线性独立的守恒量,形如
$ \left\{ \begin{array}{l} I_N^\sigma = \left( {{p_s}\xi _s^\sigma - H{\tau ^\sigma }} \right)\exp \left( {\int_{{t_1}}^t {\frac{{\partial H}}{{\partial z}}{\rm{d}}t} } \right) = {c^\sigma }\\ \left( {\sigma = 1,2, \cdots ,r} \right) \end{array} \right. $ | (30) |
证明 因为无限小群变换(16)是系统(12)的Noether对称变换,由定义1,有
$ \Delta z\left( {t_2 } \right) = 0 $ |
将上式代入式(27),得
$ \int_{t_1 }^{t_2 } {\left\{ {\dfrac{{\rm{d}} }{{\rm{d}}t}\left[{\left( {p_s \xi _s^\sigma-H\tau ^\sigma } \right)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)} \right]} \right.} + \\ \quad \exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right) \cdot \left[{\left( {- \dot {p}_s- \dfrac{\partial H}{\partial q_s }- p_s \dfrac{\partial H}{\partial z}} \right)\left( {\xi _s^\sigma - \dot {q}_s \tau ^\sigma } \right) +}\right.\\ \quad \left.{ \left. { \left( {\dot {q}_s -\dfrac{\partial H}{\partial p_s }} \right)\left( {\eta _s^\sigma -\dot {p}_s \tau ^\sigma } \right)} \right]} \right\} \varepsilon _\sigma {\rm{d}}t = 0 $ | (31) |
将方程(12)代入式(31),并考虑到
$ \left. \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\left( {{p_s}\xi _s^\sigma - H{\tau ^\sigma }} \right)\exp \left( {\int_{{t_1}}^t {\frac{{\partial H}}{{\partial z}}{\rm{d}}t} } \right)} \right] = 0\\ \left( {\sigma = 1,2, \cdots ,r} \right) \end{array} \right\} $ | (32) |
积分之,便得式(30).证毕.
定理1可称为相空间中非保守系统基于Herglotz变分问题的Noether定理.利用该定理可以通过Noether对称性找到非保守动力学系统或耗散系统的守恒量.如果Hamilton函数不显含z,则定理1成为
推论1 对于经典Hamilton系统(14),如果无限小群变换是系统的Noether对称变换,则系统存在r个线性独立的守恒量,形如
$ I_N^\sigma = p_s \xi _s^\sigma - H\tau ^\sigma = c^\sigma \quad \left( {\sigma = 1, 2, \cdots, r} \right) $ | (33) |
推论1是经典Hamilton系统的Noether定理[12].
4 相空间中基于Herglotz变分问题的Noether逆定理假设系统(12)有r个线性独立的守恒量
$ I^\sigma = I^\sigma \left( {t, q_s, p_s } \right) = \mbox{const} \ \ \ \left( {\sigma = 1, 2, \cdots, r} \right) $ | (34) |
将式(34)对时间t求导,得到
$ \dfrac{{\rm{d}} I^\sigma }{{\rm{d}}t} = \dfrac{\partial I^\sigma }{\partial t} + \dfrac{\partial I^\sigma }{\partial q_s }\dot {q}_s + \dfrac{\partial I^\sigma }{\partial p_s }\dot {p}_s = 0 $ | (35) |
将Hamilton正则方程(11)的第二组方程乘以
$ \exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)\left( { - \dot {p}_s - \dfrac{\partial H}{\partial q_s } - p_s \dfrac{\partial H}{\partial z}} \right)\left( {\xi _s^\sigma - \dfrac{\partial H}{\partial p_s }\tau ^\sigma } \right) = 0 $ | (36) |
比较式(35)和式(36)中含
$ \exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right)\left( {\xi _s^\sigma - \dfrac{\partial H}{\partial p_s }\tau ^\sigma } \right) = \dfrac{\partial I^\sigma }{\partial p_s } $ |
即
$ \xi _s^\sigma = \dfrac{\partial H}{\partial p_s }\tau ^\sigma + \dfrac{\partial I^\sigma }{\partial p_s }\exp \left( { - \int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right) \\ \qquad \left( {s = 1, 2, \cdots, n\, ; \ \sigma = 1, 2, \cdots, r} \right) $ | (37) |
再令已知积分(34)等于守恒量(30),即
$ \left( {p_s \xi _s^\sigma - H\tau ^\sigma } \right)\exp \left( {\int_{t_1 }^t {\dfrac{\partial H}{\partial z}{\rm{d}}t } } \right) = I^\sigma \\ \qquad \left( {\sigma = 1, 2, \cdots, r} \right) $ | (38) |
方程(37)和(38)是关于
定理2 如果已知系统(12)的r个线性独立的守恒量(34),则由式(37)和式(38)确定的无限小群变换是系统的Noether对称变换.
定理2可称为相空间中Herglotz变分问题的Noether逆定理.利用该定理可由已知积分找到相应的Noether对称性.
如果Hamilton函数不显含z,则式(37)和式(38)成为
$ \xi _s^\sigma = \dfrac{\partial H}{\partial p_s }\tau ^\sigma + \dfrac{\partial I^\sigma }{\partial p_s }\, \ \ \left( {s = 1, 2, \cdots, n\, ; \ \sigma = 1, 2, \cdots, r} \right) $ | (39) |
$ p_s \xi _s^\sigma - H\tau ^\sigma = I^\sigma \, \ \ \left( {\sigma = 1, 2, \cdots, r} \right) $ | (40) |
于是定理2成为
推论2 如果已知经典Hamilton系统(14)的r个线性独立的守恒量(34),则由式(39)和式(40)确定的无限小群变换是系统的Noether对称变换.
推论2是经典Hamilton系统的Noether逆定理[12].
5 算例例1 考虑著名的Emden方程[12]
$ \ddot {q} + \dfrac{2}{t}\dot {q} + q^5 = 0 $ | (41) |
它是一个单自由度完整非保守系统,试用本文方法研究其对称性与守恒量.
方程(41)可化为Herglotz变分问题来研究,其Lagrange函数为
$ L = \dfrac{1}{2}\dot {q}^2 - \dfrac{1}{6}q^6 - \dfrac{2}{t}z $ | (42) |
取广义动量和Hamilton函数为
$ p = \dfrac{\partial L}{\partial \dot {q}} = \dot {q}\, , \ \ H = p\dot {q} - L = \dfrac{1}{2}p^2 + \dfrac{1}{6}q^6 + \dfrac{2}{t}z $ | (43) |
其中作用量z满足微分方程
$ \dot {z} = p\dot {q} - \left( {\dfrac{1}{2}p^2 + \dfrac{1}{6}q^6 + \dfrac{2}{t}z} \right) $ | (44) |
方程(12)给出
$ \dot {q} = p\, , \ \ \dot {p} = - q^5 - p\dfrac{2}{t} $ | (45) |
方程(45)是方程(41)在Herglotz变分问题下的Hamilton正则形式.判据方程(29)给出
$ \dot {q}\eta + p\dot {\xi } - \left( {\dfrac{1}{2}p^2 + \dfrac{1}{6}q^6 + \dfrac{2}{t}z} \right)\dot {\tau } +\\ \qquad \dfrac{2}{t^2}z\tau - q^5\xi - p\eta = 0 $ | (46) |
方程(46)有解
$ \tau = - t - \dfrac{12zt}{3p^2t - q^6t - 12z} \\ \xi = \dfrac{1}{2}q - \dfrac{12pzt}{3p^2t - q^6t - 12z} \\ \eta = 1 $ | (47) |
生成元(47)相应于Emden方程(41)在相空间中基于Herglotz变分问题的Noether对称性,由定理1,得到
$ I_N = \dfrac{t^2} t_1^2 \left( {\dfrac{1}{2}pq + \dfrac{1}{2}tp^2 + \dfrac{1}{6}tq^6} \right) = \mbox{const}. $ | (48) |
式(48)是与生成元(47)相应的Noether守恒量.
其次,利用Noether逆定理由已知积分求相应的Noether对称变换.假设系统有积分(48),式(37)和式(38)分别给出
$ \xi = p\tau + \dfrac{1}{2}q + pt $ | (49) |
$ p\xi - \left( {\dfrac{1}{2}p^2 + \dfrac{1}{6}q^6 + \dfrac{2}{t}z} \right)\tau = \dfrac{1}{2}pq + \dfrac{1}{2}tp^2 + \dfrac{1}{6}tq^6 $ | (50) |
联立求解方程(49)和(50),得到
$ \tau = - t - \dfrac{12zt}{3p^2t - q^6t - 12z} \\ \xi = \dfrac{1}{2}q - \dfrac{12pzt}{3p^2t - q^6t - 12z} $ | (51) |
由定理2,生成元(51)相应于系统的Noether对称变换.
例2 研究平方阻尼振子,其运动微分方程为[12]
$ \ddot {q} + \gamma \dot {q}^2 + q = 0 $ | (52) |
其中
$ L = \dfrac{1}{2}\dot {q}^2 - \dfrac{q}{2\gamma } + \dfrac{1}{4\gamma ^2} - 2\gamma \dot {q}z $ | (53) |
广义动量和Hamilton函数为
$ p = \dfrac{\partial L}{\partial \dot {q}} = \dot {q} - 2\gamma z \\ H = p\dot {q} - L = \dfrac{1}{2}\left( {p + 2\gamma z} \right)^2 + \dfrac{q}{2\gamma } -\dfrac{1}{4\gamma ^2} $ | (54) |
微分方程(1)给出
$ \dot {z} = \dfrac{1}{2}\left( {p + 2\gamma z} \right)^2 - 2\gamma z\left( {p + 2\gamma z} \right) - \dfrac{q}{2\gamma } + \dfrac{1}{4\gamma ^2} $ | (55) |
Hamilton正则方程(12)给出
$ \left. \begin{array}{l} \dot q = p + 2\gamma z\\ \dot p = - 2p\gamma \left( {p + 2\gamma z} \right) - \frac{1}{{2\gamma }} \end{array} \right\} $ | (56) |
判据方程(29)给出
$ \dot {q}\eta + p\dot {\xi } - \left[{\dfrac{1}{2}\left( {p + 2\gamma z} \right)^2 + \dfrac{q}{2\gamma }-\dfrac{1}{4\gamma ^2}} \right]\dot {\tau } - \\ \qquad \dfrac{1}{2\gamma }\xi - \left( {p + 2\gamma z} \right)\eta = 0 $ | (57) |
方程(57)有解
$ \tau = - 1\, , \ \ \xi = 0\, , \ \ \eta = t $ | (58) |
生成元(58)相应于平方阻尼振子系统(52)在相空间中基于Herglotz变分问题的Noether对称性,由定理1,得到
$ I_N = \dfrac{{\rm e}^{2\gamma q}}{{\rm e}^{2\gamma q_1 }}\left[{\dfrac{1}{2}\left( {p + 2\gamma z} \right)^2 + \dfrac{q}{2\gamma }-\dfrac{1}{4\gamma ^2}} \right] = \mbox{const} $ | (59) |
其中
其次,研究Noether逆定理的应用.假设系统有积分(59),由式(37)和式(38)得到
$ \left. \begin{array}{l} \xi = \left( {p + 2\gamma z} \right)\tau + p + 2\gamma z\\ p\xi - H\tau = \frac{1}{2}{\left( {p + 2\gamma z} \right)^2} + \frac{q}{{2\gamma }} - \frac{1}{{4{\gamma ^2}}} \end{array} \right\} $ | (60) |
由此解得
$ \tau = - 1\, , \ \ \ \xi = 0 $ | (61) |
Herglotz广义变分原理为研究非保守或耗散系统动力学提供了一个有效途径,文章研究了相空间中非保守系统的Herglotz广义变分原理及其Noether对称性与守恒量.文章的主要贡献在于:提出了相空间中非保守系统的Herglotz广义变分原理,该原理给出了相空间中非保守系统的一个变分描述,建立了系统的Hamilton正则方程(12);导出了Hamilton-Herglotz作用量的非等时变分式(26)和式(27);建立了相空间中非保守系统基于Herglotz变分问题的Noether定理及其逆定理,即定理1和2.显然,通常的相空间中的Noether定理仅适用于作用量由积分泛函定义的情形,而不能用于作用量由微分方程定义的情形.但当Hamilton函数不显含z时,后者退化为前者.因此,定理包含了经典Hamilton系统的Noether定理和逆定理.文章的方法和结果可进一步推广到非完整系统,Birkhoff系统等.
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