控制系统的混沌同步问题近年来备受关注^[1], 随着分数阶微积分的发展越来越多的学者开始研究关于分数阶混沌系统的控制与同步问题^[2-3], 文献[4]研究了一类不确定分数阶混沌系统的自适应滑模混沌同步问题, 能够使驱动系统与响应系统达到同步; 文献[5]基于主动滑模控制方法实现了分数阶混沌系统的同步控制; 文献[6]分别采用线性反馈和主动控制法研究了两个不同Sprott混沌系统的控制与同步问题; 文献[7]研究了一类简单二次非线性Sprott混沌系统的分析与控制, 得到了平衡点的稳定性与Hopf分岔; 文献[8]研究了一类不确定混沌系统的自适应滑模终端控制问题.本文研究了分数阶二次非线性Sprott混沌系统的滑模同步控制及滑模终端控制问题, 得到了分数阶Sprott系统取得滑模混沌同步的充分条件.

定义1^[9] Caputo分数阶导数定义为

$ {}{_c}D{_{t_0, t}^{\alpha}}= D {_{t_0, t}^{-(n-\alpha)}} \displaystyle\frac {d^n}{dt^n}x(t)=\displaystyle\frac 1 {\Gamma(n-\alpha) } \int ^t_{t_0} (t-\tau )^{n-\alpha-1}x^{(n)}(\tau)d\tau, n-1<\alpha<n\in Z^+. $

二次非线性Sprott混沌系统^[8]

$ \begin{equation} \label{eq:1} \left\{ \begin{aligned} \dot {x_1}&= -2x_2, \\ \dot{x_2}&=x_1+{x_3}^2, \\ \dot{x_3}&=c+x_2-bx_3, \end{aligned} \right. \end{equation} $

(2.1)

其中$x_1, x_2, x_3\in R^3$为系统的状态变量, 当$b=2, c=1$时出现混沌吸引子, 设计对应的分数阶系统为主系统

$ \begin{equation} \label{eq:2} \left\{ \begin{aligned} D_t^\alpha {x_1}&= -2x_2, \\ D_t^\alpha {x_2}&=x_1+{x_3}^2, \\ D_t^\alpha {x_3}&=c+x_2-bx_3. \end{aligned} \right. \end{equation} $

(2.2)

当$\alpha=0.95, b=2.1, c=1.2$时系统呈现混沌态, 对应的从系统设计为

$ \begin{equation} \label{eq:3} \left\{ \begin{aligned} D_t^\alpha {y_1}&= -2y_2 +u_1, \\ D_t^\alpha {y_2}&=y_1+{y_3}^2+u_2, \\ D_t^\alpha {y_3}&=c+y_2-by_3+u_3. \end{aligned} \right. \end{equation} $

(2.3)

定义系统误差

$ e_1=y_1-x_1, e_2=y_2-x_2, e_3=y_3-x_3. $

上述两式相减得到误差系统为

$ \begin{equation} \label{eq:4} \left\{ \begin{aligned} D_t^\alpha {e_1}&= -2e_2 +u_1, \\ D_t^\alpha {e_2}&=e_1+{y_3}^2-{x_3}^2+u_2, \\ D_t^\alpha {e_3}&=e_2-be_3+u_3. \end{aligned} \right. \end{equation} $

(2.4)

引理1^[10] (Barbalat引理)若函数$f(t)$在$[0, +\infty)$上一致连续, 并且广义积分$\displaystyle\int ^{+\infty}_0 f(t)dt$存在, 则有$\lim\limits_{t\rightarrow \infty}f(t) = 0.$

定理1  选取滑模面$s(t)=D_t^{\alpha-1} (e_1+e_2+e_3)$, 控制器

$ u_1(t)=2e_2 -e_1 , u_2(t)={x_3}^2-{y_3}^2-e_1-e_2 , u_3(t)=e_1+be_3-\eta {\rm sgn}(s(t)), \eta >0, $

则分数阶系统(2.2), (2.3)是滑模混沌同步的.

证  当状态轨迹位于滑模面上时, $s(t)=0, \dot s(t)=0.$

由$s(t)=D_t^{\alpha-1} (e_1+e_2+e_3)=0\Rightarrow D_t^{1-\alpha}D_t^{\alpha-1} (e_1+e_2+e_3)=0\Rightarrow(e_1+e_2+e_3)=0.$由$ D_t^\alpha {e_1}= -2e_2 +u_1, u_1(t)=2e_2 -e_1 , $得到$D_t^\alpha {e_1}= -e_1, $从而根据分数阶微积分理论得$e_1(t)\rightarrow 0.$同理, 由$D_t^\alpha {e_2}=e_1+{y_3}^2-{x_3}^2+u_2, u_2(t)={x_3}^2-{y_3}^2-e_1-e_2 \Rightarrow D_t^\alpha {e_2}= -e_2, $所以得$e_2(t)\rightarrow 0.$又$D_t^\alpha {e_3}=e_2-be_3+u_3, u_3(t)=e_1+be_3-\eta {\rm sgn}(s(t)), $由滑模面上$s(t)=0, $所以$\eta {\rm sgn}(s(t))=0, $又由于$e_1+e_2+e_3=0\Rightarrow D_t^\alpha {e_3}= -e_3\Rightarrow e_3(t)\rightarrow 0.$

当状态轨迹不位于滑模面上时, 选取Lyapunov函数$V(t)=\displaystyle\frac 1 2 s^2(t)\Rightarrow \dot{V}(t) =s(t)\dot s(t), $

$ \begin{aligned} s(t)&=D_t^{\alpha-1} (e_1+e_2+e_3) \Rightarrow \dot s(t)=D_t^{\alpha} (e_1+e_2+e_3), \\ \dot{V}(t)&=s(t)\dot s(t)=s(t)[D_t^{\alpha} e_1+D_t^{\alpha}e_2+D_t^{\alpha}e_3] \\ &= s(t)[- e_1-e_2+e_1+e_2-\eta {\rm sgn}(s(t))]=-\eta |s(t)|<0. \end{aligned} $

由于$\dot{V}(t)\leq -\eta |s(t)|, $积分可以得到

$ \displaystyle\int ^t_0 |s(\tau)| d\tau \leq \displaystyle\frac { \displaystyle\int ^t_0 \dot{V}(\tau) d\tau} {-\eta} \leq \displaystyle\frac {V(0)-V(\infty)}{-\eta}\leq V(0)<\infty, $

所以$s(t)$是可积的且有界, 根据引理1 (Barbalat引理)可知$s(t)\rightarrow 0\Rightarrow e_i(t)\rightarrow 0.$

由以上分析可知, 误差系统将收敛于零.

以系统(2.2)为驱动系统, 如下系统为响应系统

$ \begin{equation} \left\{ \begin{aligned} D_t^\alpha {y_1}&= -2y_2 +\triangle f_1(y)+d_1(t)+u_1(t), \\ D_t^\alpha {y_2}&=y_1+{y_3}^2+\triangle f_2(y)+d_2(t)+u_2(t), \\ D_t^\alpha {y_3}&=c+y_2-by_3+\triangle f_3(y)+d_3(t)+u_3(t). \end{aligned} \right. \end{equation} $

(3.1)

假设1  设不确定项$\triangle f_i(y)$和外部扰动$d_i(t)$有界, 即存在$m_i, n_i>0$使得

$ |\triangle f_i(y)|<m_i, |d_i(t)|<n_i. $

假设2   $~m_i, n_i(i=1, 2, 3)$未知.

定义系统误差$e_1=y_1-x_1, e_2=y_2-x_2, e_3=y_3-x_3$, 很容易得到误差方程

$ \begin{equation} \label{eq:5} \left\{ \begin{aligned} D_t^\alpha {e_1}&= -2e_2 +\triangle f_1(y)+d_1(t)+u_1(t), \\ D_t^\alpha {e_2}&=e_1+{y_3}^2-{x_3}^2+\triangle f_2(y)+d_2(t)+u_2(t), \\ D_t^\alpha {e_3}&=e_2-be_3+\triangle f_3(y)+d_3(t)+u_3(t). \end{aligned} \right. \end{equation} $

(3.2)

引理2^[11] 假设存在连续正定函数$V(t)$满足微分不等式

$ \dot{V}(t)\leq-pV^\eta(t) , \forall t \geq t_0, V(t_0)\geq 0, $

式中$p>0, 0<\eta<1$是两个正常数, 则对于任意给定的$t_0, $$V(t)$满足如下不等式

$ V^{1-\eta}(t)\leq V^{1-\eta}(t_0)-p(1-\eta)(t-t_0), t_0<t<T, $

并且$V(t)\equiv 0, t\geq 0, $其中$T=t_0+\displaystyle\frac {V^{1-\eta}(t_0)}{p(1-\eta)}.$

引理3^[12] 设有实数$a_1, a_2, \cdots, a_n, 0<q<2$, 则有下列不等式成立

$ |a_1|^q+|a_2|^q+\cdots+|a_n|^q\geq({a_1}^2+{a_2}^2+\cdots+{a_n}^2)^{q/2}. $

针对误差系统(3.2)设计非奇异终端滑模面

$ \begin{equation}s_i(t)=D_t^{\alpha-1} e_i(t)+\lambda_i \displaystyle\int ^t_0|D_t^{\alpha-1} e_i(\tau)|^r {\rm sgn}(D_t^{\alpha-1} e_i(\tau))d\tau, \lambda_i>0, 0<\alpha<1.\end{equation} $

(3.3)

定理2  误差系统(3.2)在非奇异滑模面(3.3)上, 系统的轨迹在有限时间$t_s$内到达平衡点, 其中

$ \begin{equation} t_s\leq \displaystyle\frac{\left(\sum \limits^3_{i=1}(D_t^{\alpha-1} e_i(0))^2 \right)^{(1-r)/2}}{(1-r)\mu}, \mu=\min \left\{\lambda_1, \lambda_2, \lambda_3\right\}. \end{equation} $

(3.4)

证  误差系统满足滑模面方程$s_i(t)=0, \dot s_i(t)=0, $于是有

$ D_t^\alpha e_i(t)=-\lambda_i|D_t^{\alpha-1} e_i(t)|^r {\rm sgn}(D_t^{\alpha-1} e_i(t)). $

选取Lyapunov函数$V(t)=\displaystyle\frac 1 2\sum \limits^3_{i=1}(D_t^{\alpha-1} e_i(t))^2, $则

$ \dot{V} =\sum \limits^3_{i=1}D_t^{\alpha-1} e_i\cdot D_t^{\alpha} e_i=-\sum \limits^3_{i=1}\lambda_i|D_t^{\alpha-1} e_i(t)|^{r+1}\leq -\mu-\sum \limits^3_{i=1}|D_t^{\alpha-1} e_i(t)|^{r+1}, $

由引理3得

$ \dot{V} \leq -2^{\frac {1+r} 2}\left(\displaystyle\frac 1 2\sum \limits^3_{i=1}(D_t^{\alpha-1} e_i(t))^2\right)^{\frac {1+r} 2}=-2^{\frac {1+r} 2}\mu V^{\frac {1+r}2}, $

又由引理2易得误差轨迹会在有限时间$t_s$内达到平衡点且

$ t_s\leq \displaystyle\frac{\left(\sum \limits^3_{i=1}(D_t^{\alpha-1} e_i(0))^2 \right)^{(1-r)/2}}{(1-r)\mu}, \mu=\min \left\{\lambda_1, \lambda_2, \lambda_3\right\}. $

定理3  在控制器(3.5)和自适应律(3.6)的作用下, 误差系统(3.2)的状态轨迹能达到滑模面.

$ \begin{equation}\begin{array}{ll} u_1=2e_2 -\lambda_1|D_t^{\alpha-1} e_1|^r {\rm sgn}(D_t^{\alpha-1} e_1)-(\hat m_1+\hat n_1){\rm sgn}(s_1)-k_1{\rm sgn}(s_1) , \\ u_2=-e_1+{x_3}^2-{y_3}^2-\lambda_2|D_t^{\alpha-1} e_2|^r {\rm sgn}(D_t^{\alpha-1} e_2)-(\hat m_2+\hat n_2){\rm sgn}(s_2)-k_2{\rm sgn}(s_2) , \\ u_3=-e_2+be_3-\lambda_3|D_t^{\alpha-1} e_3|^r {\rm sgn}(D_t^{\alpha-1} e_3)-(\hat m_3+\hat n_3){\rm sgn}(s_3)-k_3{\rm sgn}(s_3), \end{array}\end{equation} $

(3.5)

控制器选取趋近律控制, $k_i>0$为增益系数, 表示趋近速度, 式中$\hat m_i, \hat n_i$分别为$m_i, n_i$的估计值, 设计如下自适应律

$ \begin{equation} \left\{ \begin{array}{ll} \dot{ \hat m}_i=|s_i|, \hat m_i(0)=\hat m_{i0}, \\ \dot{ \hat n}_i=|s_i|, \hat n_i(0)=\hat n_{i0}, \end{array} \right. i=1, 2, 3. \end{equation} $

(3.6)

证  选择Lyapunov函数$V(t)=\displaystyle\frac 1 2\sum \limits^3_{i=1}\left({s_i}^2+(\hat m_i- m_i)^2+(\hat n_i-n_i)^2\right ), $求导得

$ \begin{aligned} \dot{V}=&\sum \limits^3_{i=1}\left\{s_i[D_t^{\alpha} e_i(t)+\lambda_i|D_t^{\alpha-1} e_i(t)|^r {\rm sgn}(D_t^{\alpha-1} e_i(t))]+[(\hat m_i- m_i)|s_i|+(\hat n_i-n_i)|s_i|]\right\}\\ =&s_1[-2e_2 +\triangle f_1(y)+d_1(t)+u_1(t)+\lambda_1|D_t^{\alpha-1} e_1|^r {\rm sgn}(D_t^{\alpha-1} e_1)]\\ &+s_2[e_1+{y_3}^2-{x_3}^2+\triangle f_2(y)+d_2(t)+u_2(t)+\lambda_2|D_t^{\alpha-1} e_2|^r {\rm sgn}(D_t^{\alpha-1} e_2)] \\ &+s_3[e_2-be_3+\triangle f_3(y)+d_3(t)+u_3(t)+\lambda_3|D_t^{\alpha-1} e_3|^r {\rm sgn}(D_t^{\alpha-1} e_3)]\\ &+\sum \limits^3_{i=1}[(\hat m_i- m_i+\hat n_i-n_i)|s_i|]. \end{aligned} $

由于$s_i\cdot {\rm sgn}(s_i)=|s_i|, $再根据假设条件1, 2, 很容易得到

$ \begin{aligned}\dot{V} \leq& |s_1|(m_1+n_1)-|s_1|[(\hat m_1+\hat n_1)+k_1]+|s_2|(m_2+n_2)-|s_2|[(\hat m_2+\hat n_2)+k_2]\\ &+|s_3|(m_3+n_3)-|s_3|[(\hat m_3+\hat n_3)+k_3] +\sum \limits^3_{i=1}[(\hat m_i- m_i)|s_i|+(\hat n_i-n_i)|s_i|]\\ =&\sum \limits^3_{i=1}[(m_i+n_i)-(\hat m_i+\hat n_i)-k_i]|s_i|+\sum \limits^3_{i=1}[(\hat m_i-m_i)|s_i|+(\hat n_i-n_i)|s_i|]\\ =&\sum \limits^3_{i=1}[(m_i-\hat m_i)+(n_i-\hat n_i)-k_i]|s_i|+\sum \limits^3_{i=1}[(\hat m_i-m_i)|s_i|+(\hat n_i-n_i)|s_i|]\\ =&-\sum \limits^3_{i=1}k_i|s_i|<0, \end{aligned} $

由

$ \dot{V} \leq-\sum \limits^3_{i=1}k_i|s_i|\leq-k\sum \limits^3_{i=1}|s_i|<0, $

其中$k=\min \left\{\lambda_1, \lambda_2, \lambda_3\right\}.$不难得到

$ \displaystyle\int ^t_0 \sum \limits^3_{i=1}|s_i(\tau)| d\tau \leq \displaystyle\frac {-1} k \displaystyle\int ^t_0 \dot{V}(\tau) d\tau \leq \displaystyle\frac {V(0)-V(\infty)}{k}\leq \displaystyle\frac {V(0)} k<\infty, $

所以$s_i(t)$是可积的且有界, 根据引理1 (Barbalat引理)可知$s(t)\rightarrow 0\Rightarrow e_i(t)\rightarrow 0.$

为了说明方法的正确性, 利用四阶龙格-库塔法对系统进行仿真研究.

定理1中, 系统参数选取$\alpha=0.95, b=2.1, c=1.2$, 选取滑模面$s(t)=D_t^{\alpha-1} (e_1+e_2+e_3), $控制器

$ u_1(t)=2e_2-e_1, u_2(t)={x_3}^2-{y_3}^2-e_1-e_2, u_3(t)=e_1+be_3-\eta {\rm sgn}(s(t)), $

驱动系统与响应系统的初始值分别设置为

$ (x_1, x_2, x_3)=(7.3, 6.4, 9.2), (y_1, y_2, y_3)=(8.5, 5.7, 3.6), \eta=3, $

其系统的误差曲线如图 1所示.图 2, 3分别对不加和加上控制器两种情况进行仿真, 误差系统(2.3)中的不确定项分别为

$ \triangle f_1(y)=\cos(2\pi y_2 ), \triangle f_2(y)=0.5\cos(2\pi y_3 ), \triangle f_3(y)=0.3\cos(2\pi y_2 ), $

外部扰动取$d_1(t)=0.2\cos t, d_2(t)=0.6\sin t, d_3(t)=\cos 3t, $驱动系统与响应系统的初始值分别设置为$(x_1, x_2, x_3)=(7.3, 6.4, 9.2), (y_1, y_2, y_3)=(8.5, 5.7, 3.6)$无控制器和有控制器系统状态的两个仿真结果分别如图 2, 3所示, 如果滑模面参数取$\lambda_1=3, \lambda_2=4, \lambda_3=7, r=0.6, $控制器中的参数选取为$\mu=3, k_1=9, k_2=8, k_3=5, (\hat m_1, \hat m_2, \hat m_3)=(0.3, 0.5, 1), (\hat n_1, \hat n_2, \hat n_3)=(0.8, 0.6, 0.3), $此时的系统误差曲线和仿真结果如图 4所示, 图 4看出系统的误差很快趋近于零.

基于稳定性理论研究了分数阶二次非线性Sprott系统的滑模混沌同步控制及滑模终端同步控制问题, 并给出了严格的证明, 数值仿真表明了方法的有效性, 文中分数阶滑模面的设计可以用来解决一类分数阶混沌系统的滑模终端同步控制问题.