半个世纪以来, 经过无数科学工作者的艰苦研究和探索, 人们对混沌运动的特点、规律及其在各个学科领域的表现已经有了深刻的理解. 1963年, Lorenz在一个三维自治系统首次发现了混沌吸引子^[1]; 1999年, Chen和Ueta也发现了一种和Lorenz系统族相似但不同的混沌吸引子^[2]; 根据文献[3]对混沌系统的代数结构划分, Lorenz系统满足${a_{12}}{a_{21}} > 0$, Chen系统满足${a_{12}}{a_{21}} < 0$, 自然的问题是是否存在过渡系统满${a_{12}}{a_{21}} = 0$; 2002年Lü和Chen发现了过渡系统Lü系统^[4]; 2008年Yang和Chen^[5]给出了混沌系统代数结构的分类条件, 发现Lorenz系统满足${a_{11}}{a_{22}} > 0$, Chen系统, Lü系统满足${a_{11}}{a_{22}} < 0$, 一个自然的问题是是否有过渡混沌系统满足${a_{11}}{a_{22}} = 0$.事实上, 满足${a_{11}}{a_{22}} = 0$的混沌系统在2005年, 2008年, 2010年等分别被Gh. Tigan, Yang和作者构造^[5-8].人们构造混沌系统, 其主要目的是为了研究混沌的形成机理, 探究混沌系统间的内在联系, 以期更为深入的探索混沌的奥秘.然而目前混沌基础理论研究方面多集中在混沌吸引子的发现, 通向混沌的道路, 混沌的动力学行为等^[9-12], 但从微分方程定性理论层面来看, 对系统的周期解的探究, 轨道结构研究并不多见.又由于混沌运动具有初值敏感性, 混沌控制成了混沌应用的关键环节.到目前为止, 国内外的科学工作者基于不同的策略提出了大量的混沌控制方法^[13-15].为了研究振子状态及周期解, 国内外研究者提出基于摄动思想的控制方法^[16-17], 通过扰动使系统达到某一状态.文献[18-20]运用Melnikov方法对Duffing振子, 上田振子, 布鲁塞尔振子, Lorenz系统, 扩散Lorenz系统进行了分析, 并通过参数扰动进行了控制.又由于混沌系统在工程领域的重要性, 因此有目的的构造简单混沌系统来产生或强化混沌现象, 并对混沌系统进行结构分析, 周期轨道探测等已经成为一个关键性的课题.本文在文献[7, 21]的基础上继续对此类T混沌系统

$ \begin{equation} \frac{{dx}}{{dt}} = a(y - x), \frac{{dy}}{{dt}} = cx - axz, \frac{{dz}}{{dt}} = - bz + xy \end{equation} $

(1.1)

($a$, $b$, $c$为实数, $a \ne 0$)进行了参数扰动控制, 并通过广义Hamilton系统扰动理论, 运用Melnikov方法对周期轨道进行了分析, 解析地给出系统周期轨道, 并进行了数值仿真验证.

令$x = \tilde x$, $y = \frac{1}{{\sqrt a }}\tilde y$, $z = \frac{1}{a}(c - \tilde z)$, 则系统(1.1) 可变为

$ \begin{equation} \frac{{d\tilde x}}{{dt}} = - a\tilde x + \sqrt a \tilde y, \frac{{d\tilde y}}{{dt}} = \sqrt a \tilde x\tilde z, \frac{{d\tilde z}}{{dt}} = b(c - \tilde z) - \sqrt a \tilde x\tilde y. \end{equation} $

(2.1)

为了便于控制系统(2.1), 运用周期参数扰动控制方法, 将参数$c$变为${c_0} + {c_1}\sin \omega t$, 则有

$ \begin{equation} \frac{{d\tilde x}}{{dt}} = - a\tilde x + \sqrt a \tilde y, \frac{{d\tilde y}}{{dt}} = \sqrt a \tilde x\tilde z, \frac{{d\tilde z}}{{dt}} = b({c_0} + {c_1}\sin \omega t - \tilde z) - \sqrt a \tilde x\tilde y. \end{equation} $

(2.2)

显然当$c_1=0$时, 系统(2.2) 即为系统(2.1).为了便于分析系统(2.1), 令$\tilde x = \frac{{\hat x}}{\varepsilon }$, $\tilde y = \frac{{\hat y}}{{{\varepsilon ^2}}}$, $\tilde z = \frac{{\hat z}}{{{\varepsilon ^2}}}$, $t = \varepsilon \tau $, $\omega = \frac{{{\omega _1}}}{\varepsilon }$, $\varepsilon = \frac{1}{{\sqrt {b{c_0}} }}$, 有

$ \begin{equation} \frac{{d\hat x}}{{d\tau }} = \sqrt a \hat y - \varepsilon a\hat x, \frac{{d\hat y}}{{d\tau }} = \sqrt a \hat x\hat z, \frac{{d\hat z}}{{d\tau }} = - \sqrt a \hat x\hat y + \varepsilon (1 + \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}\tau - b\hat z). \end{equation} $

(2.3)

记$\hat x = x$, $\hat y = y$, $\hat z = z$, $\tau = t$, 且导数用符号“$\cdot$”, 则系统(2.3) 可变为

$ \begin{equation} \dot x = \sqrt a y - \varepsilon ax, \dot y = \sqrt a xz, \dot z = - \sqrt a xy + \varepsilon (1 + \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}t - bz). \end{equation} $

(2.4)

显然当$\varepsilon = 0$, 系统(2.4) 为广义Hamilton系统

$ \left( {\begin{array}{*{20}{c}} {\dot x}\\ {\dot y}\\ {\dot z} \end{array}} \right) = \left( \begin{array}{l} \;\;\;\;0\;\;\;\;\;\; - \sqrt a z\;\;\;\sqrt a y\\ \;\;\sqrt a z\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;0\\ - \sqrt a y\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;0 \end{array} \right)\left( {\begin{array}{*{20}{c}} x\\ 0\\ 1 \end{array}} \right) = J \cdot \nabla H, $

(2.5)

其中Hamilton函数为$H(x, y, z) = \frac{1}{2}{x^2} + z = A, $ Casimir函数为

$ C(x, y, z) = a({y^2} + {z^2}) = a{B^2}, $

且在平衡点$(0, 0, B)$处存在“8形”同宿轨道, 在同宿轨道“内外”分别被振荡周期轨道和旋转周期轨道充满, 在三维空间中, 这些周期轨道在柱面${y^2} + {z^2} = {B^2}$和$\frac{1}{2}{x^2} + z = A$上.

为了应用广义Hamilton系统中次谐波Melnikov向量函数理论^[22], 取

$ \left\{ \begin{array}{l} x = x, \\ y = - (B + \rho )\sin \theta, \\ z = - (B + \rho )\cos \theta \end{array} \right. $

且$B > 0$, $\left| {\frac{\rho }{B}} \right| \ll 1$, 则系统(2.4) 可变化为慢变系统^[23]

$ \begin{equation} \left\{ \begin{array}{l} \dot \theta = \sqrt a x + \varepsilon (\frac{1}{{B + \rho }}(\sin \theta + \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}t\sin \theta ) + b\cos \theta \sin \theta ), \\ \dot x = - \sqrt a (B + \rho )\sin \theta - \varepsilon ax, \\ \dot \rho = - \varepsilon (\cos \theta + \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}t\cos \theta + b(B + \rho ){\cos ^2}\theta ). \end{array} \right. \end{equation} $

(2.6)

当$\varepsilon = 0$时, 系统(2.6) 退化为$\theta - x$平面上的单摆系统, 其Hamiltonian函数

$ \tilde H(\theta, x, \rho ) = \sqrt a (\frac{1}{2}{x^2} - (B + \rho )\cos \theta ) = \sqrt a A. $

容易得知, 当$ - (B + \rho ) < A < (B + \rho )$, 有振荡周期轨道$\{ \Gamma _o^\kappa \} $

$ \begin{equation} \left\{ \begin{array}{l} {\theta _o}(t, \kappa ) = 2\arcsin \left( {\kappa {\rm{sn}}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)} \right), \\ {x_o}(t, \kappa ) = 2\kappa \sqrt {B + \rho } {\rm{cn}}\left( {\sqrt {a(B + \rho )} t, \kappa } \right), \end{array} \right. \end{equation} $

(2.7)

其中${\kappa ^2} = \frac{{A + (B + \rho )}}{{2(B + \rho )}}$, 周期${T_o}(\kappa ) = \frac{{4K(\kappa )}}{{\sqrt {a(B + \rho )} }}$, ${\rm{sn}}(u, \kappa )$和${\rm{cn}}(u, \kappa )$表示含模数$\kappa$的Jacobian椭圆函数, $K(\kappa )$表示第一类完全椭圆积分.当$A > (B + \rho )$, 存在旋转周期轨道$\{ {\Gamma _ \pm }\} $

$ \begin{equation} \left\{ \begin{array}{l} {\theta _{r \pm }}(t, {\kappa _1}) = \pm 2\arcsin \left( {{\rm{sn}}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)} \right), \\ {x_{r \pm }}(t, {\kappa _1}) = \pm \frac{{2\sqrt {(B + \rho )} }}{{{\kappa _1}}}{\rm{dn}}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right), \end{array} \right. \end{equation} $

(2.8)

其中$\kappa _1^2 = \frac{{2(B + \rho )}}{{A + (B + \rho )}}$, 周期${T_r}({\kappa _1}) = \frac{{2{\kappa _1}K({\kappa _1})}}{{\sqrt {a(B + \rho )} }}$, ${\rm{dn}}(u, \kappa )$表示含模数$\kappa$的Jacobian椭圆函数.

为了对慢变系统(2.6) 的周期轨道进行分析, 定义满足$mT = n{T_p}$的未扰周期轨道的次谐波Melnikov向量函数两分量为

$ M_1^p = \int_0^{mT} {\left\{ \begin{array}{l} - a\sqrt a x_p^2 + \sqrt a {\sin ^2}{\theta _p} + b\sqrt a (B + \rho )\cos {\theta _p}{\sin ^2}{\theta _p}\\ + \sqrt a \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}{t_0}{\sin ^2}{\theta _p}\cos {\omega _1}t \end{array} \right\}dt}, $

(3.1)

$ M_3^p = \int_0^{mT} {[\cos {\theta _p} + b(B + \rho ){{\cos }^2}{\theta _p} + \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}{t_0}\cos {\theta _p}\cos {\omega _1}t]dt}, $

(3.2)

其中$p = o$为振荡周期轨道, $p = r$为旋转周期轨道.

振荡周期轨道$\{ \Gamma _o^\kappa \} $的周期${T_o}(\kappa ) = \frac{{4K(\kappa )}}{{\sqrt {a(B + \rho )} }} = \frac{m}{n}\frac{{2\pi }}{{{\omega _1}}} = \frac{m}{n}T$, 取$p = o, n = 1$, 运用(3.1), (3.2) 式有

$ \begin{equation} M_1^o = - a\sqrt a {U_1} + \sqrt a {U_2} + b\sqrt a (B + \rho ){U_3} + \sqrt a \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}{t_0}{U_4}, \end{equation} $

(3.3)

其中

$ {U_1} = \int_0^{mT} {x_o^2dt} \nonumber\\ \;\;\;\;\;\;\;\;= \int_0^{mT} {\left[{4{\kappa ^2}(B + \rho ){\rm{c}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)} \right]dt} \nonumber\\ \;\;\;\;\;\;\;\;= 16\sqrt {\frac{{B + \rho }}{a}} \left[{E(\kappa )-(1-{\kappa ^2})K(\kappa )} \right]. $

$E(\kappa )$表示第二类完全椭圆积分

$ {U_2} = \int_0^{mT} {{{\sin }^2}{\theta _o}dt} \nonumber \\ = \int_0^{mT} {\left\{ {4{\kappa ^2}{\rm{s}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)\left[{1-{\kappa ^2}{\rm{s}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)} \right]} \right\}dt} \nonumber \\ = \frac{{16}}{{3\sqrt {a(B + \rho )} }}\left[{(1-{\kappa ^2})K(\kappa )-(1-2{\kappa ^2})E(\kappa )} \right], \nonumber \\ {U_3} = \int_0^{mT} {{{\sin }^2}{\theta _o}\cos {\theta _o}dt} \nonumber \\ = \int_0^{mT} {\left\{ {4{\kappa ^2}{\rm{s}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)\left[{1-{\kappa ^2}{\rm{s}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)} \right]\frac{{2{\kappa ^2}(B + \rho ){\rm{c}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right) - A}}{{B + \rho }}} \right\}dt} \nonumber \\ = \frac{{16}}{{15\sqrt a {{(\sqrt {B + \rho } )}^3}}}\left[\begin{array}{l} \left[{5({\kappa ^2}-1)A-2({\kappa ^4}-3{\kappa ^2} + 2)(B + \rho )} \right]K(\kappa )\\ + \left[{5(2{\kappa ^2}-1)A-4({\kappa ^4}-{\kappa ^2} + 1)(B + \rho )} \right]E(\kappa ) \end{array} \right], \nonumber \\ {U_4} = \int_0^{mT} {{{\sin }^2}{\theta _o}\cos {\omega _1}tdt} \nonumber \\ = \int_0^{mT} {\left\{ {4{\kappa ^2}{\rm{s}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)\left[{1-{\kappa ^2}{\rm{s}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)} \right]\cos {\omega _1}t} \right\}dt} \nonumber \\ \mathop = \limits^{u = \sqrt {a(B + \rho )} t} \frac{{4{\kappa ^2}}}{{\sqrt {a(B + \rho )} }}\int_0^{4mK(\kappa )} {\left\{ {\left[{{\rm{s}}{{\rm{n}}^2}\left( {u, \kappa } \right)-{\kappa ^2}{\rm{s}}{{\rm{n}}^4}\left( {u, \kappa } \right)} \right]\cos \left( {\frac{{m\pi }}{{2nK(\kappa )}}u} \right)} \right\}du}. \nonumber $

可以看到, 当$\frac{m}{2} = j$时, 则当$m = 2, j = 1$, 有

$ U_4^2 = - \frac{{8{\pi ^2}}}{{3K(\kappa )\sqrt {a(B + \rho )} }}\left[{{{\left( {\frac{\pi }{{K(\kappa )}}} \right)}^2} + 13-2{\kappa ^2}} \right]{\rm{csch}}\frac{{\pi K'(\kappa )}}{{K(\kappa )}}, $

当$m = 4, j = 2$, 有

$ U_4^4 = - \frac{{16{\pi ^2}}}{{3K(\kappa )\sqrt {a(B + \rho )} }}\left[{2{{\left( {\frac{\pi }{{K(\kappa )}}} \right)}^2} + 11-{\kappa ^2}} \right]{\rm{csch}}\frac{{2\pi K'(\kappa )}}{{K(\kappa )}}, $

所以当

$ \begin{equation} \left| {\frac{{ - a{U_1} + {U_2} + b(B + \rho ){U_3}}}{{\frac{{{c_1}}}{{{c_0}}}{U_4}}}} \right| < 1 \end{equation} $

(3.4)

成立时, $M_1^o = 0$有解.同样, 可以计算

$ \begin{equation} M_3^o = {U_5} + b(B + \rho ){U_6} + \frac{{{c_1}}}{{{c_0}}}\sin ({\omega _1}{t_0}){U_7}, \end{equation} $

(3.5)

其中

$ {U_5} = \int_0^{mT} {\cos {\theta _o}dt} = \int_0^{mT} {\frac{{2{\kappa ^2}(B + \rho ){\rm{c}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right) - A}}{{B + \rho }}dt} \nonumber \\ = \frac{4}{{\sqrt a {{(\sqrt {B + \rho } )}^3}}}\left[{\left[ {2({\kappa ^2}-1)(B + \rho )-A} \right]K(\kappa ) + 2(B + \rho )E(\kappa )} \right], \nonumber \\ {U_6} = \int_0^{mT} {{{\cos }^2}{\theta _o}dt} = \int_0^{mT} {\frac{{{{\left[{2{\kappa ^2}(B + \rho ){\rm{c}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right)-A} \right]}^2}}}{{{{(B + \rho )}^2}}}dt} \nonumber \\ = \frac{4}{{3\sqrt a {{(\sqrt {B + \rho } )}^5}}}\left[\begin{array}{l} \left[{4(1-{\kappa ^2})(2-3{\kappa ^2}){{(B + \rho )}^2} + 12A(1-{\kappa ^2})(B + \rho ) + 3{A^2}} \right]K(\kappa )\\ - \left[{8(1-2{\kappa ^2}){{(B + \rho )}^2} + 12A(B + \rho )} \right]E(\kappa ) \end{array} \right], \nonumber \\ {U_7} = \int_0^{mT} {\cos {\theta _o}\cos {\omega _1}tdt} = \int_0^{mT} {\frac{{2{\kappa ^2}(B + \rho ){\rm{c}}{{\rm{n}}^2}\left( {\sqrt {a(B + \rho )} t, \kappa } \right) - A}}{{B + \rho }}\cos {\omega _1}tdt} \nonumber \\ = \frac{{2{\kappa ^2}}}{{\sqrt {a(B + \rho )} }}\int_0^{4K(\kappa )} {\frac{{E(\kappa ) - {{\kappa '}^2}K(\kappa )}}{{{\kappa ^2}K(\kappa )}}\cos \left( {\frac{{m\pi }}{{2K(\kappa )}}u} \right)du} \nonumber \\ ~~~~+ \frac{{2{\kappa ^2}}}{{\sqrt {a(B + \rho )} }}\int_0^{4K(\kappa )} {\frac{\pi }{{{\kappa ^2}{K^2}(\kappa )}}\sum\limits_{j = 1}^\infty {\left[{j \cdot {\rm{csch}}\left( {j \cdot \frac{{\pi K'(\kappa )}}{{K(\kappa )}}} \right) \cdot \cos \left( {j \cdot \frac{\pi }{{K(\kappa )}}u} \right)} \right]\cos \left( {\frac{{m\pi }}{{2K(\kappa )}}u} \right)du} }, \nonumber $

其中$\kappa ' = \sqrt {1 - {\kappa ^2}} $, 且取$\frac{m}{2} = j$, 则当$m = 2, j = 1$, 有

$ U_7^2 = \frac{{4\pi }}{{K(\kappa )\sqrt {a(B + \rho )} }}{\rm{csch}}\frac{{\pi K'(\kappa )}}{{K(\kappa )}}, $

当$m = 4, j = 2$, 有

$ U_7^4 = \frac{{8\pi }}{{K(\kappa )\sqrt {a(B + \rho )} }}{\rm{csch}}\frac{{2\pi K'(\kappa )}}{{K(\kappa )}}, $

所以当

$ \begin{equation} \left| {\frac{{{U_5} + b(B + \rho ){U_6}}}{{\frac{{{c_1}}}{{{c_0}}}{U_7}}}} \right| < 1 \end{equation} $

(3.6)

成立时, $M_3^o = 0$有解.由以上运算, 可得

定理 3.1  如果系统(2.2) 的各参数同时满足(3.4), (3.6) 式, 则相应的子谐波Melnikov函数为零, 其振荡周期轨道的周期为$\frac{{2m\pi }}{{{\omega _1}}}$.

旋转周期轨道$\{ {\Gamma _ \pm }\} $的周期${T_r}({\kappa _1}) = \frac{{2{\kappa _1}K({\kappa _1})}}{{\sqrt {a(B + \rho )} }} = \frac{m}{n}\frac{{2\pi }}{{{\omega _1}}} = \frac{m}{n}T$, 取$p = r, ~n = 1$, 运用(3.1), (3.2) 式有

$ \begin{equation} M_1^r = - a\sqrt a {G_1} + \sqrt a {G_2} + b\sqrt a (B + \rho ){G_3} + \sqrt a \frac{{{c_1}}}{{{c_0}}}\sin {\omega _1}{t_0}{G_4}, \end{equation} $

(3.7)

其中

$ {G_1} \;=\; \int_0^{mT} {x_r^2dt} = \int_0^{mT} {\left[{\frac{{4(B + \rho )}}{{\kappa _1^2}}{\rm{d}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)} \right]dt} = \frac{8}{{{\kappa _1}}}\sqrt {\frac{{B + \rho }}{a}} E({\kappa _1}), \nonumber \\ {G_2}\; =\; \int_0^{mT} {{{\sin }^2}{\theta _r}dt} = \int_0^{mT} {\left\{ {4{\rm{s}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)\left[{1-{\rm{s}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)} \right]} \right\}dt} \nonumber \\ \;\;\;\;\;\;\;\;\;\;\;\;= \frac{{16}}{{3\kappa _1^3\sqrt {a(B + \rho )} }}\left[{2(\kappa _1^2-1)K({\kappa _1})-(\kappa _1^2-2)E({\kappa _1})} \right], \nonumber \\ {G_3} = \int_0^{mT} {{{\sin }^2}{\theta _r}\cos {\theta _r}dt} \nonumber \\ = \int_0^{mT} {\left\{ {4{\rm{s}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)\left[{1-{\rm{s}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)} \right]\frac{{\frac{2}{{\kappa _1^2}}(B + \rho ){\rm{d}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right) - A}}{{B + \rho }}} \right\}dt} \nonumber \\ = \frac{8}{{15\kappa _1^5\sqrt a {{(\sqrt {B + \rho } )}^3}}}\left[\begin{array}{l} \left[{10\kappa _1^2(1-\kappa _1^2)A-2(\kappa _1^4-3\kappa _1^2 + 2)(B + \rho )} \right]K({\kappa _1})\\ + \left[{4(\kappa _1^4-\kappa _1^2 + 1)(B + \rho )-5\kappa _1^2(2-\kappa _1^2)A} \right]E({\kappa _1}) \end{array} \right], \nonumber \\ {G_4} = \int_0^{mT} {{{\sin }^2}{\theta _r}\cos {\omega _1}tdt} \nonumber \\ = \int_0^{mT} {\left\{ {4{\rm{s}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)\left[{1-{\rm{s}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)} \right]\cos {\omega _1}t} \right\}dt}, \nonumber $

取$m = j$, 则当$m = 3, j = 3$, 有

$ G_4^3 = \frac{{4{\pi ^2}}}{{3\kappa _1^3K({\kappa _1})\sqrt {a(B + \rho )} }}\left[{9{{\left( {\frac{\pi }{{K({\kappa _1})}}} \right)}^2} + 22-11\kappa _1^2} \right]{\rm{csch}}\frac{{3\pi K'({\kappa _1})}}{{K({\kappa _1})}}, $

当$m = 4, j = 4$, 有

$ G_4^4 = \frac{{8{\pi ^2}}}{{3\kappa _1^3K({\kappa _1})\sqrt {a(B + \rho )} }}\left[{8{{\left( {\frac{\pi }{{K({\kappa _1})}}} \right)}^2} + 14-7\kappa _1^2} \right]{\rm{csch}}\frac{{4\pi K'({\kappa _1})}}{{K({\kappa _1})}}, $

所以当

$ \begin{equation} \left| {\frac{{ - a{G_1} + {G_2} + b(B + \rho ){G_3}}}{{\frac{{{c_1}}}{{{c_0}}}{G_4}}}} \right| < 1 \end{equation} $

(3.8)

成立时, $M_1^r = 0$有解.同样, 可以计算

$ \begin{equation} M_3^r = {G_5} + b(B + \rho ){G_6} + \frac{{{c_1}}}{{{c_0}}}\sin ({\omega _1}{t_0}){G_7}, \end{equation} $

(3.9)

其中

$ {G_5} = \int_0^{mT} {\cos {\theta _r}dt} = \int_0^{mT} {\frac{{\frac{2}{{\kappa _1^2}}(B + \rho ){\rm{d}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right) - A}}{{B + \rho }}dt} \nonumber \\ = \frac{2}{{{\kappa _1}\sqrt a {{(\sqrt {B + \rho } )}^3}}}\left[{2(B + \rho )E({\kappa _1})-A\kappa _1^2K({\kappa _1})} \right], \nonumber \\ {G_6} = \int_0^{mT} {{{\cos }^2}{\theta _r}dt} = \int_0^{mT} {\frac{{{{\left[{\frac{2}{{\kappa _1^2}}(B + \rho ){\rm{d}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right)-A} \right]}^2}}}{{{{(B + \rho )}^2}}}dt} \nonumber \\ = \frac{2}{{3\kappa _1^3\sqrt a {{(\sqrt {B + \rho } )}^5}}}\left[\begin{array}{l} \left[{4(\kappa _1^2-1){{(B + \rho )}^2} + 3{A^2}\kappa _1^4} \right]K({\kappa _1})\\ + \left[{8(2-\kappa _1^2){{(B + \rho )}^2} + 12\kappa _1^2(B + \rho )} \right]E({\kappa _1}) \end{array} \right], \nonumber \\ {G_7} = \int_0^{mT} {\cos {\theta _r}\cos {\omega _1}tdt} = \int_0^{mT} {\frac{{\frac{2}{{\kappa _1^2}}(B + \rho ){\rm{d}}{{\rm{n}}^2}\left( {\frac{{\sqrt {a(B + \rho )} }}{{{\kappa _1}}}t, {\kappa _1}} \right) - A}}{{B + \rho }}\cos {\omega _1}tdt}, \nonumber \\ $

取$m = j$, 则当$m = 3, j = 3$时, 有

$ G_7^3 = \frac{{4{\pi ^2}}}{{{\kappa _1}K({\kappa _1})\sqrt {a(B + \rho )} }}{\rm{csch}}\frac{{3\pi K'({\kappa _1})}}{{K({\kappa _1})}}, $

当$m = 4, j = 4$时, 有

$ G_7^4 = \frac{{8{\pi ^2}}}{{{\kappa _1}K({\kappa _1})\sqrt {a(B + \rho )} }}{\rm{csch}}\frac{{4\pi K'({\kappa _1})}}{{K({\kappa _1})}}, $

所以当

$ \begin{equation} \left| {\frac{{{G_5} + b(B + \rho ){G_6}}}{{\frac{{{c_1}}}{{{c_0}}}{G_7}}}} \right| < 1 \end{equation} $

(3.10)

成立时, $M_3^r = 0$有解.由以上运算, 可得

定理 3.2  如果系统(2.2) 的各参数同时满足(3.8), (3.10) 式, 则相应的子谐波Melnikov函数为零, 其旋转周期轨道的周期为$\frac{{2m\pi }}{{{\omega _1}}}$.

为了论证上述理论, 我们给出数值仿真.取$a = 2$和$a = 1.05$, 图 1给出未扰系统${(2.4)_{\varepsilon = 0}}$在三维空间及在相应平面上的投影“8形”同宿轨道内外的振荡周期轨道和旋转周期轨道.取$B = 1$, 图 2分别给出了$A=1$和$A=3$时抛物柱面$\frac{1}{2}{x^2} + z = A$与圆柱面${y^2} + {z^2} = 1$的交线, 即振荡周期轨道和旋转周期轨道.

当系统(2.4) 变为慢变系统(2.6) 时, 相应的空间柱面上的同宿轨道退化为平面$\theta - x$平面内的异宿轨道, 在保持上述参数的不变的情形下, 图 3给出了未扰系统${(2.6)_{\varepsilon = 0}}$在退化$\theta - x$平面内的异宿轨道, 及振荡周期轨道(解析式(2.7) 和数值方法)和旋转周期轨道(解析式(2.8) 和数值方法). 图 4作出了系统(2.6) 的三维空间振荡周期轨道和旋转周期轨道及在$\theta - x$平面内的投影, 从投影图形可以看到, 当$\varepsilon $在零的附近取值, 系统(2.6) 在定理3.1和定理3.2所给参数条件下, 在三维空间异宿轨道的内外存在振荡周期轨道和旋转周期轨道.

本文通过符号与数值运算, 运用次谐波Melnikov函数分析和计算了周期参数扰动的T混沌系统的振荡周期轨道和旋转周期轨道, 并通过数值仿真进行了实验, 验证了理论分析的正确性.根据数值仿真, 可以看到周期参数扰动策略不仅可以探究系统的周期轨道解析式, 同时可以对原混沌系统进行控制.当然还可以继续分析该系统的全局结构, 如分叉等动力学行为, 限于篇幅, 此处讨论从略.