自从Hopfield^[1]在1984年提出了后人以他名字命名的Hopfield神经网络以来, 这类人工神经网络在很多方面得到了广泛的应用,如组合优化^{[2, 3, 4]}、 图像处理^{[5, 6]}、模式识别^[7]、信号处理^[8]、 通讯技术^[9]等等, 所以在过去的数十年中,作为一个递归神经网络, Hopfield神经网络被持续研究^{[10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]}. 在神经网络的实际应用中,一方面由于两个神经元之间信息传递不免存在时滞, 另一方面由于受到诸如有限的开关速度等硬件的影响,时滞现象也是不可避免的, 所以在神经网络研究中引入时滞得到了广泛的关注^{[15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]}. 随着对神经网络的不断深入的研究,学者们发现由于现实世界中神经细胞的复杂性, 许多现有的神经网络模型很难精确描述神经反应过程的特性, 在神经网络系统中应该包含过去状态的微分信息来进一步描述这样的复杂的神经反应动力系统. 这种新的神经网络模型叫做中立型神经网络模型. 在文献^[20]中,Orman研究了如下中立型时滞Hopfield神经网络模型

$$\begin{equation}\label{eq:sys1} \dot{x}(t)=A_{1}x(t)+A_{2}f(x(t))+A_{3}f(x(t-\tau(t)))+A_{4}\dot{x}(t-\tau(t)),\end{equation}$$

(1.1)

其中各符号的含义相同于以下系统(1.2).

Orman 通过构建李亚普诺夫函数的方法,获得了关于该系统平衡点的存在性、 唯一性和全局渐近稳定性的时滞依赖的充分条件.

本文在系统(1.1)的基础上增加了状态变量为微分时滞的激活函数项, 研究如下中立型时滞神经网络模型的全局渐近稳定性

$$\begin{eqnarray}\label{eq:sys2} \dot{x}(t)&=&A_{1}x(t)+A_{2}f(x(t))+A_{3}f(x(t-\tau_{1}(t)))\nonumber\\ &&+A_{4}\dot{x}(t-\tau_{2}(t))+A_{5}f(\dot{x}(t)) +A_{6}f(\dot{x}(t-\tau_{3}(t))), \end{eqnarray}$$

(1.2)

其中$x(t)=(x_1(t),x_2(t),\cdots,x_n(t))^T$ 表示神经元的状态变量,$f(x)=(f_1(x_1),f_2(x_2),$ $\cdots,$ $f_n(x_n))$ 表示激活函数且有界,$A_1 = {\rm{diag}}\{a_1,a_2,\cdots,a_n\}$ ( $a_i<0,i=1,2,$ $\cdots,a_n$),$A_2=(b_{ij})_{n\times n}$, $A_3=(c_{ij})_{n\times n}$ ,$A_4=(d_{ij})_{n\times n}$,$A_5=(e_{ij})_{n\times n}$ 分别表示对应神经元的连接权重系数矩阵,其中,$a_i$表示第$i$ 个神经元的自反馈强度,$b_{ij}$表示第$j$个神经元的输出$f_j(x_j)$ 对于第$i$个神经元的输入的反馈连接强度,如果第$j$个神经元的输 出使第$i$个神经元激活(或抑制),则$b_{ij}>0$ (或$b_{ij}<0$); $c_{ij}$、$d_{ij}$、$e_{ij}$ 类似; $\tau_1(t)$、$\tau_2(t)$、$\tau_3(t)$ 表示对应项的传输时滞,且满足$0\leq \tau_i(t)\leq \bar{\tau}_i\leq \infty$,$0\leq \dot{\tau}_i(t)\leq \tau^{*}_i\leq 1$, 这里$\dot{\tau}_i(t)$分别表示$\tau_i(t)$一阶导数,$\bar{\tau}_i,\tau^{*}_i$ 为某一正常数,其中$i=1,2,3$.

对于激活函数$f$,我们假设满足以下条件

(H)~ $|f_i(x_i)|\leq M_i$,$f(0)=0$,对于$\forall z_1,z_2 \in {\Bbb R}$,且$z_1 \neq z_2$,有$0\leq \frac{f_i(z_1)-f_i(z_2)}{z_1-z_2}$ $\leq L_i$,其中$M_i$和$L_i$分别表示正常数,$i=1,2,\cdots,a_n$.

我们证明系统(1.2)的零解是全局渐近稳定的.

本节我们将建立关于系统(1.2)的零解的全局渐近稳定性的充分条件,我们先做如下标记

(I)~ $\eta_1=x(t),~ \eta_2=f(x(t)),~ \eta_3=f(x(t-\tau_1(t))),~ \eta_4=\dot{x}(t-\tau_2(t)),~ \eta_5=f(\dot{x}(t)),~ \eta_6=f(\dot{x}(t-\tau_3(t)))$;

(II)~ $L_M=\max \{L_1,L_2,\cdots,L_n\},~ a_M=\max \{a_1,a_2,\cdots,a_n\},~ \lambda=\frac{a_M}{L_M}$;

(III)~ $E$表示单位矩阵,$N$表示自然数集,上标$T$表示矩阵的转置, 标记$\lambda_m(P)$和$\lambda_M(P)$ 分别表示矩阵$P$的最小特征值和最大特征值,${\rm{Diag}}\{\cdots\}$表示对角矩阵.

用标记(I),系统(1.2)可以表示为以下式子

$$\begin{equation}\label{eq:sys3} \dot{x}(t)=A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6. \end{equation}$$

(2.1)

接下来,我们给出以下定理.

定理2.1 如果存在对称正定矩阵$P,Q,R$使得以下对称矩阵$\Omega$ 负定,那么系统$(1.2)$的零解是全局渐近稳定的 $$\begin{eqnarray*} \Omega=\left[\begin{array}{cccccc} \Sigma_{11}~ & A_{2}+\Phi_{12} & ~A_{3}+\Phi_{13} ~& A_{4}+\Phi_{14} & ~ A_{5}+\Phi_{15} ~& A_{6}+\Phi_{16}\\ A_{2}^{T}+\Phi_{12}^{T} ~& \Sigma_{22} & A_{3}+\Phi_{23} & A_{4}+\Phi_{24} & A_{5}+\Phi_{25} & A_{6}+\Phi_{26}\\ A_{3}^{T}+\Phi_{13}^{T} ~& A_{3}^{T}+\Phi_{23}^{T} & \Sigma_{33} & \Phi_{34} & \Phi_{35} & \Phi_{36}\\ A_{4}^{T}+\Phi_{14}^{T} ~& A_{4}^{T}+\Phi_{24}^{T} & \Phi_{34}^{T} & \Sigma_{44} & \Phi_{45} & \Phi_{46}\\ A_{5}^{T}+\Phi_{15}^{T} ~& A_{5}^{T}+\Phi_{25}^{T} & \Phi_{35}^{T} & \Phi_{45}^{T} & \Sigma_{55} & \Phi_{56}\\ A_{6}^{T}+\Phi_{16}^{T} ~& A_{6}^{T}+\Phi_{26}^{T} & \Phi_{36}^{T} & \Phi_{46}^{T} & \Phi_{56}^{T} & \Sigma_{66} \end{array}\right], \end{eqnarray*}$$ 其中 ~$\Phi_{ij}=A_{i}^{T}QA_j ~ (1\leq i\leq ~ j\leq 6,~i,~ j\in N),~ \Sigma_{11}=2A_1+\Phi_{11},~ \Sigma_{22}=2\lambda E+A_{2}+A_{2}^T+P+\Phi_{22},~\\ \Sigma_{33}=-(1-\tau_{1}^{*})P+\Phi_{33},~ \Sigma_{44}=-(1-\tau_{2}^{*})Q+\Phi_{44},~\\ \Sigma_{55}=A_{5}^{T}+R,~ \Sigma_{66}=-(1-\tau_{3}^{*})R+\Phi_{66}.$

证 构造如下李亚普诺夫函数

$$\begin{eqnarray}\label{eq:1} V(t)&=&x^{T}(t)x(t)+2\sum\limits_{i=1}^{n}\int_{0}^{x_{i}(t)}f_{i}(s){\rm d}s \nonumber\\ &&+ \int_{-\tau_{1}(t)}^{0}f^{T}(x(t+s))Pf(x(t+s)){\rm d}s\nonumber\\ &&+\int_{-\tau_{2}(t)}^{0}\dot{x}^{T}(t+s)Q\dot{x}(t+s){\rm d}s\nonumber\\ &&+\int_{-\tau_{3}(t)}^{0}f^T(\dot{x}(t+s))Rf(\dot{x}(t+s)){\rm d}s. \end{eqnarray}$$

(2.2)

由假设(H)可得 $$0\leq\displaystyle{\frac{f_{i}(x_{i}(t))}{x_{i}(t)}}\leq L_{i}\leq L_{M},$$ $$0\leq |f_{i}(x_{i}(t))|\leq L_{i}|x_{i}(t)|\leq L_{M}|x_{i}(t)|,$$ $$0\leq |f_{i}(x_{i}(t))f_{i}(x_{i}(t))|\leq L_{i}|x_{i}(t)f_{i}(x_{i}(t))|\leq L_{M}|x_{i}(t)f_{i}(x_{i}(t))|.$$ 由以上第一个不等式可知,$f_{i}(x_{i}(t))$ 与 $x_{i}(t)$ 是同号的,所以有 $$0\leq f_{i}(x_{i}(t))f_{i}(x_{i}(t))\leq L_{i}x_{i}(t)f_{i}(x_{i}(t))\leq L_{M}x_{i}(t)f_{i}(x_{i}(t)),$$ $$f_{i}(x_{i}(t))x_{i}(t)\geq\frac{1}{L_{M}}f_{i}(x_{i}(t))f_{i}(x_{i}(t)).$$

由${a_i<0,~ a_M=\max \{a_1,a_2,\cdots,a_n\},~ \lambda=\frac{a_M}{L_M}}$,可得 $$\begin{eqnarray*} f^{T}(x(t))Ax(t)&=&\sum\limits_{i=1}^{n}f_{i}(x_{i}(t))a_{i}x_{i}(t)\\ &\leq& \frac{1}{L_{M}}\sum\limits_{i=1}^{n}f_{i}(x_{i}(t))a_{i}f_{i}(x_{i}(t)) \\ &\leq & \frac{a_M}{L_{M}}\sum\limits_{i=1}^{n}f_{i}(x_{i}(t))f_{i}(x_{i}(t))\\ &=& \lambda f^{T}(x(t))f(x(t)). \end{eqnarray*}$$

用标记(I)和(II),以上不等式可表示为$\eta_2^T A_1 \eta_1 \leq \lambda \eta_2^T \eta_2$.

为了便于求导,我们对(2.2)式的第二、三、四项做变量替换$\tau=t+s$,得

$$\begin{eqnarray} \label{eq:2} V(t)&=&x^{T}(t)x(t)+2\sum\limits_{i=1}^{n}\int_{0}^{x_{i}(t)}f_{i}(s){\rm d}s\nonumber\\ &&+ \int_{t-\tau_{1}(t)}^{t}f^{T}(x(\tau))Pf(x(\tau)){\rm d}\tau\nonumber\\ &&+\int_{t-\tau_{2}(t)}^{t}\dot{x}^{T}(\tau)Q\dot{x}(\tau){\rm d}\tau\nonumber\\ &&+\int_{t-\tau_{3}(t)}^{t}f^T(\dot{x}(\tau))Rf(\dot{x}(\tau)){\rm d}\tau. \end{eqnarray}$$

(2.3)

沿系统(1.2)的轨线对式(2.3)两边求导,可得 $$\begin{eqnarray*} \dot{V}(t)& = &2x^T(t)\dot{x}(t)+2f^{T}(x(t))\dot{x}(t)\\ && +f^{T}(x(t))Pf(x(t))-(1-\dot{\tau}_{1}(t))f^{T}(x(t-\tau_{1}(t)))Pf(x(t-\tau_{1}(t)))\\ && +\dot{x}^{T}(t)Q\dot{x}(t)-(1-\dot{\tau}_{2}(t))\dot{x}^{T}(t-\tau_{2}(t))Q\dot{x}(t-\tau_{2}(t))\\ &&+f^{T}(\dot{x}(t))Rf(\dot{x}(t))-(1-\dot{\tau}_{3}(t))f^{T}(\dot{x}(t-\tau_{3}(t)))Rf(\dot{x}(t-\tau_{3}(t))). \end{eqnarray*}$$

于是用标记(I)以及式(2.1)可得 $$\begin{eqnarray*} \dot{V}(t)&= & 2\eta_1^{T}(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)\\ && +2\eta_2^{T}(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)\\ && +\eta_2^{T}P\eta_2-(1-\dot{\tau}_{1}(t))\eta_3^{T}P\eta_3\\ && +(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)^T\\ && Q(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)\\ && -(1-\dot{\tau}_{2}(t))\eta_4^{T}Q\eta_4+\eta_5^{T}R\eta_5-(1-\dot{\tau}_{3}(t))\eta_6^{T}R\eta_6\\ &\leq & 2\eta_1^{T}A_1\eta_1+2\eta_1^{T}A_2\eta_2+2\eta_1^{T}A_3\eta_3+2\eta_1^{T}A_4\eta_4+2\eta_1^{T}A_5\eta_5+2\eta_1^{T}A_6\eta_6\\ && +2\lambda\eta_2^{T}\eta_2+2\eta_2^{T}A_2\eta_2+2\eta_2^{T}A_3\eta_3+2\eta_2^{T}A_4\eta_4+2\eta_2^{T}A_5\eta_5+2\eta_2^{T}A_6\eta_6\\ & &+\eta_2^{T}P\eta_2-(1-\tau_1^*)\eta_3^{T}P\eta_3\\ && +(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)^T\\ && Q(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)\\ & &-(1-\tau_2^*)\eta_4^{T}Q\eta_4+\eta_5^{T}R\eta_5-(1-\tau_{3}^*)\eta_6^{T}R\eta_6 \\ &= & (2\eta_1^{T}A_1\eta_1)+(\eta_1^{T}A_2\eta_2+\eta_2^{T}A_2^T\eta_1)+(\eta_1^{T}A_3\eta_3+\eta_3^{T}A_3^T\eta_1)\\ && +(\eta_1^{T}A_4\eta_4+\eta_4^{T}A_4^T\eta_1)+(\eta_1^{T}A_5\eta_5+\eta_5^{T}A_5^T\eta_1)+(\eta_1^{T}A_6\eta_6+\eta_6^{T}A_6^T\eta_1)\\ && +2\lambda\eta_2^{T}\eta_2+(\eta_2^{T}A_2\eta_2+\eta_2^{T}A_2^T\eta_2)+(\eta_2^{T}A_3\eta_3+\eta_3^{T}A_3^T\eta_2)\\ && +(\eta_2^{T}A_4\eta_4+\eta_4^{T}A_4^T\eta_2)+(\eta_2^{T}A_5\eta_5+\eta_5^{T}A_5^T\eta_2)+(\eta_2^{T}A_6\eta_6+\eta_6^{T}A_6^T\eta_2)\\ && +\eta_2^{T}P\eta_2-(1-\tau_{1}^*)\eta_3^{T}P\eta_3\\ && +(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)^T\\ && Q(A_1\eta_1+A_2\eta_2+A_3\eta_3+A_4\eta_4+A_5\eta_5+A_6\eta_6)\\ && -(1-\tau_2^*)\eta_4^{T}Q\eta_4+\eta_5^{T}R\eta_5-(1-\tau_3^*)\eta_6^{T}R\eta_6\\ &=& \eta_1^{T}(2A_1+A_1^TQA_1)\eta_1+\eta_1^{T}(A_2+A_1^TQA_2)\eta_2+\eta_1^{T}(A_3+A_1^TQA_3)\eta_3\\ && +\eta_1^{T}(A_4+A_1^TQA_4)\eta_4+\eta_1^{T}(A_5+A_1^TQA_5)\eta_5+\eta_1^{T}(A_6+A_1^TQA_6)\eta_6\\ && +\eta_2^{T}(A_2^T+A_2^TQA_1)\eta_1+\eta_2^{T}(2\lambda E+A_2+A_2^T+P+A_2^TQA_2)\eta_2\\ && +\eta_2^{T}(A_3+A_2^TQA_3)\eta_3+\eta_2^{T}(A_4+A_2^TQA_4)\eta_4\\ & &+\eta_2^{T}(A_5+A_2^TQA_5)\eta_5+\eta_2^{T}(A_6+A_2^TQA_6)\eta_6\\ && +\eta_3^{T}(A_3^T+A_3^TQA_1)\eta_1+\eta_3^{T}(A_3^T+A_3^TQA_2)\eta_2+\eta_3^{T}(-(1-\tau_1^*)P\\ &&+ A_3^TQA_3)\eta_3+\eta_3^{T}(A_3^TQA_4)\eta_4+\eta_3^{T}(A_3^TQA_5)\eta_5+\eta_3^{T}(A_3^TQA_6)\eta_6\\ && +\eta_4^{T}(A_4^T+A_4^TQA_1)\eta_1+\eta_4^{T}(A_4^T+A_4^TQA_2)\eta_2+\eta_4^{T}(A_4^TQA_3)\eta_3\\ && +\eta_4^{T}(-(1-\tau_2^*)Q+A_4^TQA_4)\eta_4+\eta_4^{T}(A_4^TQA_5)\eta_5+\eta_4^{T}(A_4^TQA_6)\eta_6\\ & &+\eta_5^{T}(A_5^T+A_5^TQA_1)\eta_1+\eta_5^{T}(A_5^T+A_5^TQA_2)\eta_2+\eta_5^{T}(A_5^TQA_3)\eta_3\\ && +\eta_5^{T}(A_5^TQA_4)\eta_4+\eta_5^{T}(A_5^T+R)\eta_5+\eta_5^{T}(A_5^TQA_6)\eta_6\\ && +\eta_6^{T}(A_6^T+A_6^TQA_1)\eta_1+\eta_6^{T}(A_6^T+A_6^TQA_2)\eta_2+\eta_6^{T}(A_6^TQA_3)\eta_3\\ && +\eta_6^{T}(A_6^TQA_4)\eta_4+\eta_6^{T}(A_6^TQA_5)\eta_5+\eta_6^{T}(-(1-\tau_3^*)R+A_6^TQA_6)\eta_6\\ &=&(\eta_{1}^T ~~\eta_{2}^T ~~\eta_{3}^T ~~ \eta_{4}^T ~~ \eta_{5}^T ~~ \eta_{6}^T )\Omega (\eta_{1}^T ~~\eta_{2}^T ~~\eta_{3}^T ~~ \eta_{4}^T ~~ \eta_{5}^T ~~ \eta_{6}^T )^T, \end{eqnarray*}$$ 其中 $$\Omega=\left[\begin{array}{cccccc} \Sigma_{11}~ & A_{2}+\Phi_{12} & ~A_{3}+\Phi_{13}~ & A_{4}+\Phi_{14} & ~A_{5}+\Phi_{15} ~ & A_{6}+\Phi_{16}\\ A_{2}^{T}+\Phi_{12}^{T} ~& \Sigma_{22} & A_{3}+\Phi_{23} & A_{4}+\Phi_{24} & A_{5}+\Phi_{25} & A_{6}+\Phi_{26}\\ A_{3}^{T}+\Phi_{13}^{T}~ & A_{3}^{T}+\Phi_{23}^{T} & \Sigma_{33} & \Phi_{34} & \Phi_{35} & \Phi_{36}\\ A_{4}^{T}+\Phi_{14}^{T} ~& A_{4}^{T}+\Phi_{24}^{T} & \Phi_{34}^{T} & \Sigma_{44} & \Phi_{45} & \Phi_{46}\\ A_{5}^{T}+\Phi_{15}^{T}~ & A_{5}^{T}+\Phi_{25}^{T} & \Phi_{35}^{T} & \Phi_{45}^{T} & \Sigma_{55} & \Phi_{56}\\ A_{6}^{T}+\Phi_{16}^{T} ~& A_{6}^{T}+\Phi_{26}^{T} & \Phi_{36}^{T} & \Phi_{46}^{T} & \Phi_{56}^{T} & \Sigma_{66} \end{array}\right].$$

根据已知条件,$\Omega$负定,所以$\dot{V}(t)$负定,于是定理2.1成立.

这一节,我们用一个实际例子来验证定理结果的有效性.

证 考虑如下二元神经网络模型

$$\begin{eqnarray} \label{example} \left[\begin{array}{cc} \dot{x}_1(t)\\ \dot{x}_2(t) \end{array}\right] &=& \left[\begin{array}{cccccc} -1 & 0 \\ 0 & -1 \end{array}\right] \left[\begin{array}{cc} x_1(t)\\ x_2(t) \end{array}\right] +\left[\begin{array}{cccccc} 0 & 0 \\ 0.2 & 0 \end{array}\right] \left[\begin{array}{cc} f_1(x_1(t))\\ f_2(x_2(t)) \end{array}\right]\nonumber\\ &&+\left[\begin{array}{cccccc} 0 & 0.3 \\ 0.1 & 0 \end{array}\right] \left[\begin{array}{cc} f_1(x_1(t-0.05\tau(t)))\\ f_2(x_2(t-0.05\tau(t))) \end{array}\right]\nonumber\\ &&+\left[\begin{array}{cccccc} 0.01 & 0 \\ 0.2 & 0.02 \end{array}\right] \left[\begin{array}{cc} \dot{x}_1(t-0.03\tau(t)))\\ \dot{x}_2(t-0.03\tau(t))) \end{array}\right] \nonumber\\ &&+\left[\begin{array}{cccccc} -0.20 & 0.03 \\ 0.02 & -0.10 \end{array}\right] \left[\begin{array}{cccccc} f_1(\dot{x}_1(t))\\ f_2(\dot{x}_2(t)) \end{array}\right] \nonumber\\ &&+\left[\begin{array}{cccccc} 0 & -0.035 \\ 0.025 & 0.100 \end{array}\right] \left[\begin{array}{cc} f_1(\dot{x}_1(t-0.01\tau(t)))\\ f_2(\dot{x}_2(t-0.01\tau(t))) \end{array}\right], \end{eqnarray}$$

(3.1)

其中,${f_i(x_i)=\frac{1-e^{-x_i}}{1+e^{-x_i}}}, i=1,2$,满足激活函数假设条件(H),${\tau(t)=\frac{1}{t^2+1}}$.

在这个系统中,$A_1=\left[\begin{array}{cccccc} -1 & 0 \\ 0 & -1 \end{array}\right]$,$A_2=\left[\begin{array}{cccccc} 0 & 0 \\ 0.2 & 0 \end{array}\right]$, $A_3=\left[\begin{array}{cccccc} 0 & 0.3 \\ 0.1 & 0 \end{array}\right]$, $A_4=\left[\begin{array}{cccccc} 0.01 & 0 \\ 0.2 & 0.02 \end{array}\right]$, $A_5=\left[\begin{array}{cccccc} -0.20 & 0.03 \\ 0.02 & -0.10 \end{array}\right]$, $A_6=\left[\begin{array}{cccccc} 0 & -0.035 \\ 0.025 & 0.100 \end{array}\right]$, $L_M=0.5,a_M=-1,\displaystyle{ \lambda=\frac{a_M}{L_M}=-2},\tau_1^*=0.05,\tau_2^*=0.03,\tau_3^*=0.01.$

这里我们取 $P=\left[\begin{array}{cccccc} 0.49 & 0 \\ 0 & 0.37 \end{array}\right],Q=\left[\begin{array}{cccccc} 0.23 & 0 \\ 0 & 0.56 \end{array}\right],R=\left[\begin{array}{cccccc} 0.0274 & 0 \\ 0 & 0.0212 \end{array}\right]$,根据定理2.1,我们利用Matlab来计算$\Omega$的特征值可得: eig$(\Omega)=(-3.7985 ~ -3.3843 ~ -1.8182 ~ -1.4472$ $-0.5429 ~ -0.4567 ~ -0.2899 ~ -0.1856 ~ -0.1331 ~ -0.0671 ~ -0.0049 ~ -0.0268 )$,从而得知$\Omega$负定, 由定理2.1的结论即可得到系统(3.1)的零解是全局渐近稳定的.

本文从一个新的中立型时滞神经网络的数学模型来研究它的稳定性问题,由定理的结果我们可以看出, 如果时滞为常数或时滞的一阶导数有界,那么系统(1.2)的稳定性将不受影响. 另外,定理的结论中有一个声明条件就是``如果存在对称正定矩阵P、Q、R"满足 负定的这个条件,那么就产生一个新的问题是怎样判断这样的对称正定矩阵是否存在? 这是这个领域研究的一个难题,通常的做法是针对具体问题采用数据测试的方法找到满足 条件的对称正定矩阵. 尽管所得结论带来了一个新的难题, 但是这个难题总是比直接判断系统的原模型是否稳定较容易得多.