针对网络节点动力学是二维的情况, 考虑如下含随机噪声的复杂动力网络:

$ \left\{ \begin{array}{l} {{\dot x}_i} = {f_1}\left( {t,{x_i},{y_i}} \right) + \sum\limits_{j = 1}^N {{c_{ij}}{h_j}\left( {{x_j}} \right)} + {\sigma _1}\left( {{x_i},t} \right)\dot \omega ,\\ {{\dot y}_i} = {f_2}\left( {t,{x_i},{y_i}} \right) + {\sigma _2}\left( {{y_i},t} \right)\dot \omega , \end{array} \right. $

(1.1)

其中$1 \leqslant i, j \leqslant N$, 网络仅通过分量${x_i}$互相耦合, 并假设只有这个分变量可测, ${h_j}\left( {{x_j}} \right)$是内部耦合函数, ${c_{ij}}$是外部耦合强度, $C = {\left( {{c_{ij}}} \right)_{N \times N}}$是网络的外联耦合矩阵, 其定义为若节点$j$对节点$i$有影响($j \ne i$), 则${c_{ij}} \ne 0$; 否则, ${c_{ij}} =0$.另外, $C$满足${c_{ii}} = -\sum\limits_{j = 1, j \ne i} {{c_{ij}}} $, ${\sigma _1}\left( {{x_i}, t} \right)$, ${\sigma _2}\left( {{y_i}, t} \right)$: ${R_ + } \times {R^n} \to {R^{n \times m}}$是与节点状态相关的噪声强度函数, $\omega $为定义在一个完备概率空间$\left( \Omega, f, P \right)$上, 并含有一个自然代数流${{\{{{f}_{t}}\}}_{t\ge 0}}$的$m$维布朗运动.

为了识别(1.1) 的拓扑结构, 将(1.1) 看作驱动网络, 构造相应的响应网络如下:

$ \left\{ \begin{array}{*{35}{l}} {{{\dot{\hat{x}}}}_{i}}={{f}_{1}}\left( t,{{{\hat{x}}}_{i}},{{{\hat{y}}}_{i}} \right)+\sum\limits_{j=1}^{N}{{{{\hat{c}}}_{ij}}{{h}_{j}}\left( {{{\hat{x}}}_{j}} \right)}+{{u}_{i}}+{{\sigma }_{1}}\left( {{{\hat{x}}}_{i}},t \right)\dot{\omega }, \\ {{{\dot{\hat{y}}}}_{i}}={{f}_{2}}\left( t,{{{\hat{x}}}_{i}},{{{\hat{y}}}_{i}} \right)+{{\sigma }_{2}}\left( {{{\hat{y}}}_{i}},t \right)\dot{\omega }, \\ \end{array} \right. $

(1.2)

其中${u_i}$是控制项, $\hat C = {\left( {{{\hat c}_{ij}}} \right)_{N \times N}}$是对原网络结构$C$的估计.

设${\tilde x_i} = {\hat x_i} - {x_i}$, ${\tilde y_i} = {\hat y_i} - {y_i}$, ${\tilde c_{ij}} = {\hat c_{ij}} - {c_{ij}}$, ${\sigma _1}\left( {{e_1}, t} \right)\dot \omega = {\sigma _1}\left( {{{\hat x}_i}, t} \right)\dot \omega - {\sigma _1}\left( {{x_i}, t} \right)\dot \omega $, ${\sigma _2}\left( {{e_2}, t} \right)\dot \omega = {\sigma _2}\left( {{{\hat y}_i}, t} \right)\dot \omega - {\sigma _2}\left( {{y_i}, t} \right)\dot \omega $, 则可得如下误差系统

$ \left\{ \begin{align} &{{{\dot{\tilde{x}}}}_{i}}={{f}_{1}}\left( t,{{{\hat{x}}}_{i}},{{{\hat{y}}}_{i}} \right)-{{f}_{1}}\left( t,{{x}_{i}},{{y}_{i}} \right)+\sum\limits_{j=1}^{N}{{{{\hat{c}}}_{ij}}{{h}_{j}}\left( {{{\hat{x}}}_{j}} \right)}-\sum\limits_{j=1}^{N}{{{c}_{ij}}{{h}_{j}}\left( {{x}_{j}} \right)}+{{u}_{i}}+{{\sigma }_{1}}\left( {{e}_{1}},t \right)\dot{\omega }, \\ &{{{\dot{\tilde{y}}}}_{i}}={{f}_{2}}\left( t,{{{\hat{x}}}_{i}},{{{\hat{y}}}_{i}} \right)-{{f}_{2}}\left( t,{{x}_{i}},{{y}_{i}} \right)+{{\sigma }_{2}}\left( {{e}_{2}},t \right)\dot{\omega }, \\ \end{align} \right. $

(1.3)

其中$1 \leqslant i \leqslant N$.

假设 1.1  假设${h_j}\left( x \right)\left( {1 \leqslant j \leqslant {l_2}} \right)$是线性无关的, 满足${m_1} \leqslant \left| {{{h'}_j}\left( x \right)} \right| \leqslant {m_2}$, 其中${m_1}$和${m_2}$是正常数.

假设 1.2  设噪声强度函数${\sigma _i}\left( {{e_i}, t} \right)$有界, 满足$ \frac{1}{2}{\hbox{tr}}\left( {{\sigma _i}^T{\sigma _i}} \right) \leqslant \sum\limits_{i = 1}^N {{p_i}{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde y}_i}^2}, $其中${p_i}$为非负实数.

假设 1.3  存在一个正的$q>0$使得$ \left| {f\left( {t, x, z} \right) - f\left( {t, y, w} \right)} \right| \leqslant q\left( {\left| {x - y} \right| + \left| {z - w} \right|} \right), $其中$x, y, z, w \in {\mathbb{R}^n}$.

考虑如下$n$维随机微分方程^[1]:

$ \begin{equation} dx\left( t \right) = f\left( {t, x\left( t \right)} \right)dt + g\left( {t, x\left( t \right)} \right)dW\left( t \right), \end{equation} $

(1.4)

其中$f\left( {t, x} \right):\left[{0, + \infty } \right] \times {R^n} \to {R^n}$$g\left( {t, x} \right):\left[{0, + \infty } \right] \times {R^n} \to {R^{n \times m}}.$

定义 1.1  如果开集$G \subset {R^n}$满足$P\left\{ {x\left( {t, {x_0}} \right) \in G, \forall t \geqslant 0} \right\} = 1, {x_0} \in G$, 则称$G$为关于方程(1.4) 的解的不变集, 也就是说, 所有从$G$出发的轨道永远在$G$内.

这里, 设${C^{1, 2}}\left( {{R_ + } \times G;{R_ + }} \right)$表示所有定义在${R_ + } \times G$上的非负函数集合$V\left( {t, x} \right)$, 其中$V\left( {t, x} \right)$关于$x$二阶连续可微, 关于$t$一阶连续可微, $\bar G$表示G的闭包.

引理 1.1  如果存在函数$V \in {C^{1, 2}}\left( {{R_ + } \times G;{R_ + }} \right), \gamma \in {L^1}\left( {{R_ + };{R_ + }} \right)$以及连续函数$w:\bar G \to {R_ + }$使得$LV\left( {t, x} \right) \leqslant \gamma \left( t \right) - w\left( x \right)$, $\left( {t, x} \right) \in {R_ + } \times G, $如果$G$有界; 或者

$ \mathop {\mathop {\lim }\limits_{x \in G} }\limits_{\left| x \right| \to \infty } \mathop {\inf }\limits_{0 \leqslant t < \infty } V\left( {t, x} \right) = \infty $

且$G$有界, 则$\forall {x_0} \in G$, $\mathop {\lim }\limits_{t \to \infty } V\left( {t, x\left( {t;{x_0}} \right)} \right)$存在而且几乎处处有界.进一步地, 有

$ \mathop {\lim }\limits_{t \to \infty } w\left( {x\left( {t;{x_0}} \right)} \right) = 0\quad {\hbox{a.s.}}. $

也就是说, 从$G$内出发的方程(1.4) 的解最终会以概率$1$渐进稳定到$D_w^G: = \left\{ {x \in \bar G:w\left( x \right) = 0} \right\}$.

引理 1.2  (Schur complement)下列线性矩阵不等式

$ \left( {\begin{array}{*{20}{c}} {A\left( x \right)}&{B\left( x \right)} \\ {B{{\left( x \right)}^T}}&{C\left( x \right)} \end{array}} \right) < 0 $

等价于下列条件:

1) $A\left( x \right) < 0$, $C\left( x \right) - B{\left( x \right)^T}A{\left( x \right)^{ - 1}}B\left( x \right) < 0$;

2) $C\left( x \right) < 0$, $A\left( x \right) - B\left( x \right)A{\left( x \right)^{ - 1}}B{\left( x \right)^T} < 0$, 其中$A{\left( x \right)^T} = A\left( x \right)$, $C{\left( x \right)^T} = C\left( x \right).$

定理 1.1  当假设1.1--1.3成立时, 采用下列控制率

$ \begin{equation} \dot{\hat{c}}_{ij} = - {\delta _i}{\tilde x_i}^T{h_j}\left( {{{\hat x}_j}} \right), 1 \leqslant i \leqslant {l_1}, 1 \leqslant j \leqslant {l_2}, \end{equation} $

(1.5)

$ {u_i} = - {d_i}{\tilde x_i}, $

(1.6)

$ \begin{cases} {{\dot d}_i} = {k_i}{{\tilde x}_i}^2, 1 \leqslant i \leqslant {l_1}, \hfill \\ {d_i} = 0, {\hbox{其它}}, \end{cases} $

(1.7)

则$\mathop {\lim }\limits_{t \to \infty } {\hat c_{ij}} = {c_{ij}}$.

证 令${\tilde c_{ij}} \triangleq {\hat c_{ij}} - {c_{ij}}$, $1 \leqslant i, j \leqslant N$, ${h_i}\left( {{{\hat x}_i}} \right) - {h_i}\left( {{x_i}} \right) = {h'_i}\left( {{\xi _i}} \right)\left( {{{\hat x}_i} - {x_i}} \right), $构造如下Lyapunov函数

$ V = \frac{1}{2}\sum\limits_{i = 1}^N {\left( {{{\tilde x}_i}^T{{\tilde x}_i} + {{\tilde y}_i}^T{{\tilde y}_i}} \right)} + \frac{1}{2}\sum\limits_{i = 1}^N {\sum\limits_{j = 1}^N {\frac{{{{\tilde c}_{ij}}^2}}{{{\delta _i}}}} } + \frac{1}{2}\sum\limits_{i = 1}^{{l_1}} {\frac{{{{\left( {{d_i} - d} \right)}^2}}}{{{k_i}}}}, \\ \;\;\;\;\;\;\;\;V \in {C^{1, 2}}\left( {{R_ + } \times G;{R_ + }} \right), G = {R^{nN + N + {N^2}}}, \\ LV = \sum\limits_{i = 1}^N {\left( {{{\tilde x}_i}^T\dot{\tilde{x}}_i + {{\tilde y}_i}^T\dot{\tilde{y}}_i} \right)} + \sum\limits_{i = 1}^N {\sum\limits_{j = 1}^N {\frac{{{{\tilde c}_{ij}} \cdot \dot{\tilde{c}}_{ij}}}{{{\delta _i}}}} } + \sum\limits_{i = 1}^{{l_1}} {\frac{{\left( {{d_i} - d} \right) \cdot {{\dot d}_i}}}{{{k_i}}}} + \frac{1}{2}tr\left( {{\sigma _i}^T{\sigma _i}} \right) \\ \;\;\;\;\;\;\;\;\leqslant \sum\limits_{i = 1}^N {\left[{q{{\tilde x}_i}^2 + q\left( {\left| {{{\tilde x}_i}^T} \right|\left| {{{\tilde y}_i}} \right| + \left| {{{\tilde y}_i}^T} \right|\left| {{{\tilde x}_i}} \right|} \right) + q{{\tilde y}_i}^2} \right]}\\ \;\;\;\;\;\;\;\;+\sum\limits_{i = 1}^N {\sum\limits_{j = 1}^N {{c_{ij}}{{\tilde x}_i}^T\left( {{h_j}\left( {{{\hat x}_j}} \right) - {h_j}\left( {{x_j}} \right)} \right)} } - \sum\limits_{i = 1}^{{l_1}} {d{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde y}_i}^2} \\ \;\;\;\;\;\;\;\;= \sum\limits_{i = 1}^N {\left[{q{{\tilde x}_i}^2 + q\left( {\left| {{{\tilde x}_i}^T} \right|\left| {{{\tilde y}_i}} \right| + \left| {{{\tilde y}_i}^T} \right|\left| {{{\tilde x}_i}} \right|} \right) + q{{\tilde y}_i}^2} \right]} + \sum\limits_{i = 1}^N {{c_{ii}}{{\tilde x}_i}^T{{h'}_i}\left( {{\xi _i}} \right){{\tilde x}_i}} \\ \;\;\;\;\;\;\;\;+ \sum\limits_{i = 1}^N {\sum\limits_{j = 1, j \ne i}^N {{c_{ij}}{{\tilde x}_i}^T{{h'}_i}\left( {{\xi _i}} \right){{\tilde x}_j}} } - \sum\limits_{i = 1}^{{l_1}} {d{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde y}_i}^2} \\ \;\;\;\;\;\;\;\;\leqslant \sum\limits_{i = 1}^N {\left[{q{{\tilde x}_i}^2 + q\left( {\left| {{{\tilde x}_i}^T} \right|\left| {{{\tilde y}_i}} \right| + \left| {{{\tilde y}_i}^T} \right|\left| {{{\tilde x}_i}} \right|} \right) + q{{\tilde y}_i}^2} \right]} + \sum\limits_{i = 1}^N {{c_{ii}}{m_1}{{\tilde x}_i}^2} \\ \;\;\;\;\;\;\;\;+ \sum\limits_{i = 1}^N {\sum\limits_{j = 1, j \ne i}^N {{c_{ij}}{m_2}\left| {{{\tilde x}_i}^T} \right|\left| {{{\tilde x}_j}} \right|} } - \sum\limits_{i = 1}^{{l_1}} {d{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde x}_i}^2} + \sum\limits_{i = 1}^N {{p_i}{{\tilde y}_i}^2} \\ \;\;\;\;\;\;\;\;= \left( {{{\tilde X}^T}, {{\tilde Y}^T}} \right)\left( {\begin{array}{*{20}{c}}{{m_2}\bar C - D + \left( {q + p} \right){I_{NN}}}{q{I_{NN}}} \\ {q{I_{NN}}}{\left( {p + q} \right){I_{NN}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\tilde X} \\ {\tilde Y} \end{array}} \right) \\ \;\;\;\;\;\;\;\;\triangleq \left( {{{\tilde X}^T}, {{\tilde Y}^T}} \right)R\left( {\begin{array}{*{20}{c}} {\tilde X} \\ {\tilde Y} \end{array}} \right), $

其中

$ \tilde{X}={{\left( \left| {{{\tilde{x}}}_{1}} \right|,\cdots ,\left| {{{\tilde{x}}}_{N}} \right| \right)}^{T}},\tilde{Y}={{\left( \left| {{{\tilde{y}}}_{1}} \right|,\cdots ,\left| {{{\tilde{y}}}_{N}} \right| \right)}^{T}},p=\mathop {\max }\limits_{1≤i≤N}\,\left\{ {{p}_{i}} \right\}, \\R=\left( \begin{matrix} {{R}_{1}}\ \ \ \ \ \ \ \ \ \ \ \ q{{I}_{NN}} \\ q{{I}_{NN}}\ \ \ \ \ \left( p+p \right){{I}_{NN}} \\ \end{matrix} \right), $

${R_1} = {m_2}\bar C - D + \left( {q + p} \right){I_{NN}}$, ${I_{NN}}$是$N \times N$的单位矩阵, $D$是对角矩阵, 其中第$i $ ($1 \leqslant i \leqslant {l_1}$)个元素为$d, $其余元素为0, 定义${C^S} = \left( {{C^T} + C} \right)/2$, $\bar C$是将矩阵${C^S}$的对角元素${c_{ii}}$变成$\left( {{m_1}/{m_2}} \right){c_{ii}}$后所形成的矩阵.取一个合适的正常数d可以使${R_1} < 0$, 并且$\left( {p + q} \right){I_{NN}} - {q^2} \cdot {R_1}^{ - 1} < 0$, 由引理1.2可知, $R < 0$.令$\lambda < 0$是矩阵$R$的最大特征值, 则$LV \leqslant \lambda {\tilde x^T}\tilde x \triangleq w\left( {\tilde x} \right)$.

又因为$\mathop {\mathop {\lim }\limits_{x \in G} }\limits_{\left| x \right| \to \infty } \mathop {\inf }\limits_{0 \leqslant t < \infty } V = \infty $且${\sigma _i}$有界, 由引理1.1可知对于$G$内任意初值, $\mathop {\lim }\limits_{t \to \infty } V$存在而且几乎处处有界, 且$\mathop {\lim }\limits_{t \to \infty } w\left( {\tilde x} \right) = 0$.进一步地, 从$G$出发的关于方程(1.3), (1.5)-(1.6) 的解最终会以概率1渐进稳定到

$ D_w^G: = \left\{ {x \in \bar G:w\left( {\tilde x} \right) = 0} \right\} = \left\{ {x \in \bar G:\tilde x = 0} \right\} = \left\{ {\tilde x = 0;{{\tilde c}_{ij}} = 0;{d_i} = {\hbox{cons}}\tan ts} \right\}, $

其中$i, j = 1, 2, \cdots, N$, 也就是说, 采用控制率(1.5)-(1.7) 响应网络(1.2) 与网络(1.1) 同步, 且$\forall i, j \in \left\{ {1, 2, \cdots, N} \right\}$, $\mathop {\lim }\limits_{t \to \infty } {\hat c_{ij}} = {c_{ij}}$.

本节中, 我们将用FHN方程作为模型来验证上述方法的有效性. FHN方程表述为

$ \left\{ \begin{align} &\dot{V}=V-\frac{1}{3}{{V}^{3}}-W+\text{ }{{I}_{ex}}, \\ &\dot{W}=\varepsilon \left( V+a-bW \right), \\ \end{align} \right. $

其中$V$代表膜电位, $W$是恢复变量, ${I_{_{ex}}}$是外部刺激电流, $ a, $ $b, $ $\varepsilon $是正数, 通常情况下, 在$0 < \varepsilon \ll 1$时, $V$和$W$分别作为快变量和慢变量, 参数$b$是显著影响系统动态特性的其中一个决定性条件.

在这里, 考虑含噪声的FHN系统

$ \left\{ \begin{align} &{{{\dot{V}}}_{i}}=V-\frac{1}{3}{{V}^{3}}-W+\text{ }{{I}_{ex}}-\sum\limits_{j=1,j\ne i}^{N}{{{g}_{ij}}\left( {{h}_{i}}\left( {{V}_{i}} \right)-{{h}_{j}}\left( {{V}_{j}} \right) \right)}+{{\sigma }_{1}}\left( {{V}_{i}},t \right)\dot{\omega }, \\ &\ \ \ \ \triangleq {{f}_{1}}\left( {{V}_{i}},{{W}_{i}} \right)+\sum\limits_{j=1}^{N}{{{g}_{ij}}{{h}_{j}}\left( {{V}_{j}} \right)}+{{\sigma }_{1}}\left( {{V}_{i}},t \right)\dot{\omega }, \\ &\ \ \ \ {{{\dot{W}}}_{i}}=\varepsilon \left( V+a-bW \right)+{{\sigma }_{2}}\left( {{W}_{i}},t \right)\dot{\omega }\triangleq {{f}_{2}}\left( {{V}_{i}},{{W}_{i}} \right)+{{\sigma }_{2}}\left( {{W}_{i}},t \right)\dot{\omega }. \\ \end{align} \right. $

令$\varepsilon =0.08, a=0.7, b=0.8, {I_{ex}} = \left( {0.1 + 0.1i} \right)\cos \left( {t/50} \right), $ ${k_i}=20, $ ${\delta _i}=30, $易证FHN方程满足3个假设条件.

当耦合为线性时, 以含有20个节点的环状网络为例来进行仿真, 为了识别

$ {c_{ij}}\left( {1 \leqslant i \leqslant 4, 1\leqslant j \leqslant 2} \right), $

其中${c_{11}} = - 2, {c_{21}} = 1, {c_{31}} = 0, {c_{41}} = 0, {c_{12}} = 1, {c_{22}} = - 2, {c_{32}} = 1, $${c_{42}} = 0$未知, 图 1给出了当耦合强度取1, 噪声取1, $ p=1$时, 识别${\hat c_{ij}}$的演化结果, 由图可以看出, 识别结果与原网络耦合矩阵$C$的结果相同, 由此可以看出理论的正确性.

由图 2, 3可以看出, 在其它值不变的情况下, $p$值增大时, 适用的耦合强度也随之增大, 但是当$p$太小时(图 4), 不能正确识别出结果, 这是因为网络内部达到了近似同步.

由仿真可以得到, 网络结构可识别时, 噪声强度越大, 适用的耦合强度的范围越大.

令${h_j}\left( x \right) = \exp ( - {x^2}/2)$, 易证满足假设1.1.在其它条件相同的情况下, 在2邻接网络上进行加边, 产生小世界网络模型, 令$N=20, $为了识别${c_{ij}}\left( {1 \leqslant i \leqslant 4, 1 \leqslant j \leqslant 2} \right)$, 其中${c_{11}} = 0, {c_{21}} = 1, {c_{31}} = 1, {c_{41}} = 0, {c_{12}} = 1, $${c_{22}} = 0, {c_{32}} = 1, {c_{42}} = 1$未知, 噪声、耦合强度和$p$都为1. 图 5给出了${\hat c_{ij}}$的识别结果, 由此可见, 当耦合取为非线性时, 上述理论依然正确.

同样的, 当耦合为非线性时, 也会有2.1节中的结论, $p$越大, 耦合强度也越大. 图 6, 7给出了将$p$取不同值时, 网络结构识别的结果, 由图可以看出, 都可以正确识别.

由上述图形可以得出与2.1节相同的规律, 也即噪声强度的范围将影响网络结构正确识别时的耦合强度范围, 噪声强度越大, 可识别的耦合强度也越大.

本文针对节点动力学含随机噪声的复杂网络, 基于随机微分方程理论, 用牵制控制方法来进行拓扑识别, 设计自适应反馈控制器和识别率, 反演网络结构.不同于节点全部控制的识别方法, 牵制控制方法是通过控制部分节点来识别网络中未知的连接, 数值仿真验证了该方法的有效性.数值仿真部分主要分为两个部分:耦合为线性的和非线性的.通过仿真结果表明, 耦合不管为线性的或者非线性的, 都能达到很好的识别效果.噪声强度越大时, 耦合强度的适用范围也越大, 当噪声强度太小时, 反而不能达到很好的识别效果, 这是因为此时网络内部达到了近似同步.