在近几十年，神经网络在信号处理、联想记忆、模式识别、定点计算等科学领域有着广泛的用途^[1-2].在神经网络的应用方面，由于其信号传输和有限运算放大器切换速度的影响，可能导致神经网络失去稳定或者振动.因此，时滞神经网络的稳定性得到了广泛研究，并取得了重大成果^[3-7].众所周知，耗散性理论在动态系统、非线性控制等领域的稳定性分析有着重要的地位^[8-9].作为耗散性特例的无源性，不仅仅是系统稳定性的体现，还涉及系统的输入输出存储函数，在分析电子电路、机械系统、非线性系统有着重要的作用^[10-11].同时，时滞系统的无源性问题引起了大家的广泛关注，以及时滞神经网络的无源性分析也取得了重大发展.最初时滞独立神经网络的无源性得到了研究，并取得了成果^[12-13].相比时滞依赖神经网络，时滞独立神经网络的无源性结果具有更大的保守性，探讨时滞依赖神经网络的无源已成为当代主流，且得到的神经网络无源性的结果保守性也越来越小^[14-16].

传统的李雅普诺夫泛函构造方法要求李雅普诺夫泛函中的每个对称矩阵是正定的.文献[17]构造了不需要所有的对阵矩阵是正定的李雅普诺夫泛函，证明了时滞神经网络的无源性.受文献[17]的启发，本文构造一个不需要所有二次项对称矩阵全都正定的李雅普诺夫泛函，通过引入Jensen不等式、Schur补引理及自由权矩阵等方法，证明了李雅普诺夫泛函正定性与李雅普诺夫泛函导数的负定性.通过实验结果的对比，本文得到了比文献[17]保守性更小且满足神经网络无源的结果.

1 问题描述和预备知识

考虑时滞和参数不确定的神经网络系统：

$ \left\{ {\begin{array}{*{20}{l}} {\boldsymbol{\dot x}(t) = - \boldsymbol{A}(t)\boldsymbol{x}(t) + \boldsymbol{W}(t)\boldsymbol{f}(\boldsymbol{x}(t))}+ \\ {\;\;\;\;\;\;\;\;\; {\boldsymbol{W}_1}(t)\boldsymbol{f}(\boldsymbol{x}(t - \tau (t))) + \boldsymbol{u}(t), {\rm{ }}}\\ {\boldsymbol{y}(t) = \boldsymbol{f}(\boldsymbol{x}(t)), }\\ {\boldsymbol{x}(t) = \boldsymbol{\phi} (t), t \in [\bar \tau, 0].} \end{array}} \right. $

(1)

其中$\boldsymbol{x}(t) = {[{x_1}(t){\rm{ }}{x_2}(t) \ldots {x_n}(t)]^{\rm{T}}} \in {R^n}$表示状态向量，$\boldsymbol{f}(t) = {[{f_1}(t){\rm{ }}{f_2}(t) \ldots {f_n}(t)]^{\rm{T}}} \in {R^n}$表示激活函数$\boldsymbol{u}(t) = {[{u_1}(t){\rm{ }}{{\rm{u}}_2}(t) \ldots {u_n}(t)]^{\rm{T}}}$，$\boldsymbol{y}(t) = {[{y_1}(t){\rm{ }}{{\rm{y}}_2}(t) \ldots {y_n}(t)]^{\rm{T}}}$分别表示神经网络的输入和输出向量，连续函数$\boldsymbol{\phi} (t) \in {R^n}$是初始条件.时变时滞τ(t)是时变连续函数，满足

$ 0 < \tau (t) \leqslant \bar \tau, \dot \tau (t) \leqslant \mu, $

(2)

其中$\bar \tau $与μ是正的常量.不确定矩阵A(t)，W(t)，W₁(t)满足：

$ \left\{ \begin{array}{l} \boldsymbol{A}(t) = \boldsymbol{A} + \Delta \boldsymbol{A}(t), \\ \boldsymbol{W}(t) = \boldsymbol{W} + \Delta \boldsymbol{W}(t), \\ {\boldsymbol{W}_1}(t) = {\boldsymbol{W}_1} + \Delta {\boldsymbol{W}_1}(t), \end{array} \right. $

(3)

其中$\boldsymbol{A} = {\rm{diag}}\{ {a_1}, {a_2}, \ldots, {a_n}\} $是正定的对角矩阵，$\boldsymbol{W} = ({w_{ij}}) \in {R^{n \times n}}$和${\boldsymbol{W}_1} = (w_{ij}^1) \in {R^{n \times n}}$是已知的连接权值矩阵.ΔA(t)、ΔW(t)、ΔW₁(t)是未知矩阵，表示时变参数的不确定性，具有以下形式：

$ \left\{ \begin{array}{l} \Delta \boldsymbol{A}(t) = {\boldsymbol{H}_1}{\boldsymbol{F}_1}(t){\boldsymbol{E}_1}{\rm{, }}\\ \Delta \boldsymbol{W}(t) = {\boldsymbol{H}_2}{\boldsymbol{F}_2}(t){\boldsymbol{E}_3}, \\ \Delta {\boldsymbol{W}_1}(t) = {\boldsymbol{H}_3}{\boldsymbol{F}_3}(t){\boldsymbol{E}_3}{\rm{, }} \end{array} \right. $

(4)

其中H_i和E_i，i=1, 2, 3，是已知常数实矩阵，${\boldsymbol{F}_i}( \cdot ):\boldsymbol{R} \to {\boldsymbol{R}^{k \times l}}, i = 1, 2, 3$，是一个未知时变矩阵函数，满足

$ {\boldsymbol{F}_i}{(t){\rm{^T}}}{\boldsymbol{F}_i}(t) \leqslant \boldsymbol{I}, i = 1, 2, 3.{\rm{ }} $

(5)

假定激活函数满足以下假设.

假设1 (Wang et al.^[18]).函数f_i(·)连续有界，且满足：

$ F_i^ - \leqslant \frac{{{f_i}({a_1}) - {f_i}({a_2})}}{{{a_1} - {a_2}}} \leqslant F_i^ +, i = 1, 2, \cdots, n, $

(6)

其中${f_i}(0) = 0, {a_1}, {a_2} \in R, {a_1} \ne {a_2}, F_i^ - $和$F_i^ + $是已知实标量.

注1：上述假设的激活函数首先在文献[18]提出，该假设比文献[15, 17]更具一般性，因为$F_i^ - $和$F_i^ + $可能为正数，零或者负数，即假设1的激活函数可能是非单调，非可微的，无界的.因此，本文的无源条件比文献[15, 17]更弱保守性.

为推导本文的结论，首先给出以下定义和引理.

定义1 (Li and Liao^[12]).在零初始条件下，对神经网络(1)如果存在一标量γ$ \geqslant $0在所有t_f$ \geqslant $0都满足：

$ 2\int_0^{{t_f}} {\boldsymbol{y}{{(s)}^{\rm{T}}}\boldsymbol{u}(s){\rm{d}}s \geqslant } - \gamma \int_0^{{t_f}} {\boldsymbol{u}{{(s)}^{\rm{T}}}\boldsymbol{u}(s){\rm{d}}s}, $

(7)

则神经网络(1)是无源的.

引理1 (Gu^[19]).对任意对称正定矩阵${\boldsymbol{M}_1} \in {R^{n \times n}}$且M₁ > 0，标量b > a，向量函数$\boldsymbol{\omega }:[a, b] \to {R^n}$，经积分定义得到不等式：

$ \begin{split}{l} (b - a)\int_a^b {\boldsymbol{\omega }{{(s)}^{\rm{T}}}{\boldsymbol{M}_1}} \boldsymbol{\omega }(s){\rm{d}}s \geqslant \\ {\left[{\int_a^b {\boldsymbol{\omega }(s){\rm{d}}s} } \right]^{\rm{T}}}{\boldsymbol{M}_1}\left[{\int_a^b {\boldsymbol{\omega }(s){\rm{d}}s} } \right]. \end{split} $

(8)

引理2 (Park and Ko^[20]).对任意矩阵$\left[{\begin{array}{*{20}{c}} {{\boldsymbol{M}_2}} & \boldsymbol{S}\\ * & {{\boldsymbol{M}_2}} \end{array}} \right] \geqslant 0$，标量$\bar \tau$ > 0，τ(t) > 0满0 < τ(t)$ \leqslant $$\bar \tau$，向量函数$\boldsymbol{\dot x}:\left[{-\tau, 0} \right] \to $Rⁿ，即有如下积分的定义：

$ - (b - a)\int_a^b {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}{\boldsymbol{M}_2}} \boldsymbol{\dot x}(s){\rm{d}}s \leqslant \boldsymbol{\varsigma }{(t)^{\rm{T}}}\boldsymbol{{{\varPi }}\varsigma }(t), $

(9)

其中$\boldsymbol{\varsigma }(t) = {\left[{\begin{array}{*{20}{c}}{\boldsymbol{x}{{(t)}^{\rm{T}}}} &\;\;\; {\boldsymbol{x}{{(t-\tau (t))}^{\rm{T}}}} &\;\;\; {\boldsymbol{x}{{(t-\bar \tau )}^T}}\end{array}} \right]^{\rm{T}}}$，$\boldsymbol{\varPi } = \left[{\begin{array}{*{20}{c}} {-{\boldsymbol{M}_2}} & {{\boldsymbol{M}_2}-\boldsymbol{S}} & \boldsymbol{S}\\ * & {-2{\boldsymbol{M}_2} + \boldsymbol{S} + {\boldsymbol{S}^T}} & {{\boldsymbol{M}_2} - \boldsymbol{S}}\\ * & * & { - {\boldsymbol{M}_2}} \end{array}} \right].$

引理3 (Boyd et al.^[21] Schur complement).给定常数矩阵Ω₁、Ω₂、Ω₃满足Ω₁=Ω₁T和Ω₂ > 0，仅当$\left[{\begin{array}{*{20}{c}} {{\boldsymbol{\varOmega }_1}} & {{\boldsymbol{\varOmega }_3}^{\rm T}}\\ {{\boldsymbol{\varOmega }_3}} & {-{\boldsymbol{\varOmega }_2}} \end{array}} \right] < 0$或$\; \left[{\begin{array}{*{20}{c}} {-{\boldsymbol{\varOmega }_2}} & {{\boldsymbol{\varOmega }_3}}\\ {{\boldsymbol{\varOmega }_3}^T} & {{\boldsymbol{\varOmega }_1}} \end{array}} \right] < 0$时，可得

$ {\boldsymbol{\varOmega }_1} + {\boldsymbol{\varOmega }_3}{\boldsymbol{\varOmega }_2}^{ - 1}{\boldsymbol{\varOmega }_3} < 0. $

(10)

引理4 (Petersen and Hollot^[22]).对适当维实矩阵H、E和F(t)，F(t)满足F(t)^TF(t)$ \leqslant $I对任意标量ε < 0则有：

$ \boldsymbol{HF}(t)\boldsymbol{E} + {(\boldsymbol{HF}(t)\boldsymbol{E})^{\rm{T}}} \leqslant {\varepsilon ^{ - 1}}\boldsymbol{H}{\boldsymbol{H}^{\rm{T}}} + \varepsilon {\boldsymbol{E}^{\rm{T}}}\boldsymbol{E}. $

(11)

首先考虑标称神经网络系统如下：

$ \left\{ {\begin{array}{*{20}{c}} {\boldsymbol{\dot x}(t) = - \boldsymbol{Ax}(t) + \boldsymbol{Wf}(\boldsymbol{x}(t)){\rm{ }}} +\\ {{\boldsymbol{W}_1}\boldsymbol{f}(\boldsymbol{x}(t - \tau (t))) + \boldsymbol{u}(t), }\\ \!{\boldsymbol{y}(t) = \boldsymbol{f}(\boldsymbol{x}(t)), {\rm{ }}}\\ {\boldsymbol{x}(t) = \boldsymbol{\phi}(t), t \in [\bar \tau, 0].{\rm{ }}} \end{array}} \right. $

(12)

为方便，本文有以下表示：

$ \begin{array}{c} {\boldsymbol{K}_1} = {\rm{diag}}\{ F_1^ +, F_2^ +, \cdots, F_n^ + \}, \\ {\boldsymbol{K}_2} = {\rm{diag}}\{ F_1^ -, F_2^ -, \cdots, F_n^ - \}, \\ {\boldsymbol{F}_1} = {\rm{diag}}\{ F_1^ - F_1^ +, F_2^ - F_2^ +, \cdots, F_n^ - F_n^ + \}, \\ {\boldsymbol{F}_2} = {\rm{diag}}\{ \frac{{F_1^ - + F_1^ + }}{2}, \frac{{F_2^ - + F_2^ + }}{2}, \cdots, \frac{{F_n^ - + F_n^ + }}{2}\} . \end{array} $

2 主要结果

定理1对标称神经网络(12)，由假设1，对给定标量$\bar \tau $ > 0和μ，对于任意时滞τ(t)满足0 < τ(t)$ \leqslant $$\bar \tau $和$\dot \tau (t) \leqslant \mu $，如果存在对称矩阵P > 0，Z > 0，Ψ₁ > 0，Ψ₂ > 0，Ξ < 0，$\mathbb{Z} = \left[{\begin{array}{*{20}{c}}\boldsymbol{Z} & \boldsymbol{S}\\* & \boldsymbol{Z}\end{array}} \right] > 0$，${\boldsymbol{Q}_1} = {\boldsymbol{Q}_1}^{\rm{T}}, {\boldsymbol{Q}_2}, {\boldsymbol{Q}_3} = {\boldsymbol{Q}_3}^{\rm{T}}, $${\boldsymbol{R}_1} = {\boldsymbol{R}_1}^{\rm{T}}, {\boldsymbol{R}_2}, {\boldsymbol{R}_3} = {\boldsymbol{R}_3}^{\rm{T}}$，对角矩阵L_i > 0，其中i=1，2，$cdots$，4，${\boldsymbol{D}_1} = {\rm{diag}}\!\left\{ {{d_{11}}, } \right.\left. {{d_{12}}, \cdots, {d_{1n\!}}} \right\} > 0$，${\boldsymbol{D}_2} = {\rm{diag}}\left\{ {{d_{21}}, {d_{22}}, } \right.$$\left. \cdots, {{d_{2n}}} \right\} > 0$，标量γ $ \geqslant $0，则神经网络(12)是无源的.

$ {\boldsymbol{\psi }_1} = \left[{\begin{array}{*{20}{c}} {{\boldsymbol{Q}_1} + {\boldsymbol{F}_1}{\boldsymbol{L}_1}} & {{\boldsymbol{Q}_2}-{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ * & {{\boldsymbol{Q}_3} + {\boldsymbol{L}_1}} \end{array}} \right] > 0, $

(13)

$ {\boldsymbol{\psi }_2} = \left[{\begin{array}{*{20}{c}} {{{\bar \tau }^{-1}}\boldsymbol{P} + \boldsymbol{Z}} & {-\boldsymbol{Z}} & 0\\ * & {{\boldsymbol{R}_1} + \boldsymbol{Z} + {\boldsymbol{F}_1}{\boldsymbol{L}_2}} & {{\boldsymbol{R}_2}-{\boldsymbol{F}_2}{\boldsymbol{L}_2}}\\ * & * & {{\boldsymbol{R}_3} + {\boldsymbol{L}_2}} \end{array}} \right] \geqslant 0, $

(14)

$ {\boldsymbol{\varXi}} \! = \! \left[{\begin{array}{*{20}{c}} {{\boldsymbol{\psi }_{\rm{1}}}}&{{\boldsymbol{\psi }_{\rm{2}}}}&{{\boldsymbol{\psi }_{\rm{3}}}}&{{\boldsymbol{\psi }_{\rm{4}}}}&{S}&{\rm{0}}&{{\boldsymbol{\psi }_{\rm{5}}}}&{{-}\bar \tau \boldsymbol{AZ}} \\ {\rm{*}}&{{\boldsymbol{\psi }_{\rm{6}}}}&{{\boldsymbol{\psi }_{\rm{7}}}}&{{-}{\boldsymbol{F}_{\rm{2}}}{\boldsymbol{L}_{\rm{4}}}}&{\rm{0}}&{\rm{0}}&{{\boldsymbol{\psi }_{\rm{8}}}}&{\bar \tau {\boldsymbol{W}^{\rm{T}}}\boldsymbol{Z}} \\ {\rm{*}}&{\rm{*}}&{{\boldsymbol{\psi }_{\rm{9}}}}&{{\boldsymbol{\psi }_{{\rm{10}}}}}&{\rm{0}}&{\rm{0}}&{\rm{0}}&{\bar \tau {\boldsymbol{W}_{\rm{1}}}^{\rm{T}}\boldsymbol{Z}} \\ {\rm{*}}&{\rm{*}}&{\rm{*}}&{{\boldsymbol{\psi }_{{\rm{11}}}}}&{\boldsymbol{Z}{-}\boldsymbol{S}}&{\rm{0}}&{\rm{0}}&{\rm{0}} \\ {\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{{ - }{\boldsymbol{R}_{\rm{1}}}{ - }\boldsymbol{Z}}&{{ - }{\boldsymbol{R}_{\rm{2}}}}&{\rm{0}}&{\rm{0}} \\ {\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{{ - }{\boldsymbol{R}_{\rm{3}}}}&{\rm{0}}&{\rm{0}} \\ {\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{{ - }\gamma {I}}&{\bar \tau \boldsymbol{Z}} \\ {\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{\rm{*}}&{{ - }\boldsymbol{Z}} \end{array}} \right] \! \leqslant \! 0, $

(15)

其中

$ \begin{split} \begin{array}{l} {\boldsymbol{\varPsi }_1} = - \boldsymbol{A}{(\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1})^{\rm{T}}} - (\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - \\ \;\;\;\;\;\;\;\; {\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{A} + {\boldsymbol{Q}_1} + {\boldsymbol{R}_1} - \boldsymbol{Z} - {\boldsymbol{F}_1}{\boldsymbol{L}_1} - {\boldsymbol{F}_1}{\boldsymbol{L}_4}, \\ {\boldsymbol{\varPsi }_2} = (\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{W} + {\boldsymbol{Q}_2} - \boldsymbol{A}({\boldsymbol{D}_1} - {\boldsymbol{D}_2}) + \\ \;\;\;\;\;\;\;\; {\boldsymbol{R}_2} + {\boldsymbol{F}_2}{\boldsymbol{L}_1} + {\boldsymbol{F}_2}{\boldsymbol{L}_4}, \\ {\boldsymbol{\varPsi }_3} = (\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1}){\boldsymbol{W}_1} - {\boldsymbol{F}_2}{\boldsymbol{L}_4}, \\ {\boldsymbol{\varPsi }_4} = {\boldsymbol{F}_1}{\boldsymbol{L}_4} + \boldsymbol{Z} - \boldsymbol{S, }\\ {\boldsymbol{\varPsi }_5} = \boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1}, \\ {\boldsymbol{\varPsi }_6} = ({\boldsymbol{D}_1}\! - \!{\boldsymbol{D}_2})\boldsymbol{W} + {\boldsymbol{W}^{\rm{T}}}({\boldsymbol{D}_1} \!- \!{\boldsymbol{D}_2}) + {\boldsymbol{Q}_3} + {\boldsymbol{R}_3} - {\boldsymbol{L}_1} - {\boldsymbol{L}_4}, \\ {\boldsymbol{\varPsi }_7} = ({\boldsymbol{D}_1} - {\boldsymbol{D}_2}){\boldsymbol{W}_1} + {\boldsymbol{L}_4}, \\ {\boldsymbol{\varPsi }_8} = ({\boldsymbol{D}_1} - {\boldsymbol{D}_2}) - \boldsymbol{I, }\\ {\boldsymbol{\varPsi }_9} = - (1 - \mu ){\boldsymbol{Q}_3} - {\boldsymbol{L}_3} - {\boldsymbol{L}_4}, \\ {\boldsymbol{\varPsi }_{10}} = - (1 - \mu ){\boldsymbol{Q}_2}^{\rm{T}} + {\boldsymbol{F}_2}{\boldsymbol{L}_4} + {\boldsymbol{F}_2}{\boldsymbol{L}_3}, \\ {\boldsymbol{\varPsi }_{11}} = - (1 - \mu ){\boldsymbol{Q}_1} - 2\boldsymbol{Z} + \boldsymbol{S} + {\boldsymbol{S}^{\rm{T}}} - {\boldsymbol{F}_1}{\boldsymbol{L}_3} - {\boldsymbol{F}_1}{\boldsymbol{L}_4}. \end{array} \end{split} $

证明构造以下李雅普诺夫-克拉索夫斯基泛函：

$ \boldsymbol{V}(\boldsymbol{x}(t)) = {\boldsymbol{V}_1}(\boldsymbol{x}(t)) + {\boldsymbol{V}_2}(\boldsymbol{x}(t)) + {\boldsymbol{V}_3}(\boldsymbol{x}(t)), $

(16)

其中

$ \begin{array}{l} {\boldsymbol{V}_1}(\boldsymbol{x}(t)) = \boldsymbol{x}{(t)^{\rm{T}}}\boldsymbol{Px}(t) + 2\sum\limits_{i = 1}^n {{d_{1i}}\int_0^{{x_i}(t)} {({f_i}(s)} }- \\ \;\;\;\;\;\;\;\;\;\;\;\;\;F_i^ - s){\rm{d}}s + 2\sum\limits_{i = 1}^n {{d_{2i}}\int_0^{{x_i}(t)} {(F_i^ + s - {f_i}(s))} } {\rm{d}}s, \\ {\boldsymbol{V}_2}(\boldsymbol{x}(t)) = \int_{t - \tau (t)}^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}} \boldsymbol{Q\eta }(s){\rm{d}}s + \int_{t - \bar \tau }^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}} \boldsymbol{R\eta }(s){\rm{d}}s, \\ {\boldsymbol{V}_3}(\boldsymbol{x}(t)) = \bar \tau \int\limits_{ - \bar \tau }^0 {\int\limits_{t + \theta }^t {\boldsymbol{\dot x}{{(s)}^T}} } \boldsymbol{Z\dot x}(s){\rm{d}}s{\rm{d}}\theta, \\ \boldsymbol{\eta }(s)\! =\! \left[{\begin{array}{*{20}{c}} {\boldsymbol{x}(s)}\\ {\boldsymbol{f}(\boldsymbol{x}(s))} \end{array}} \right], \boldsymbol{Q} \!=\! \left[{\begin{array}{*{20}{c}} {{\boldsymbol{Q}_1}}\\ {\boldsymbol{Q}_2^{\rm{T}}} \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{Q}_2}}\\ {{\boldsymbol{Q}_3}} \end{array}} \right], \boldsymbol{R} \!=\! \left[{\begin{array}{*{20}{c}} \!{{\boldsymbol{R}_1}}\!\\ \!{\boldsymbol{R}_2^T}\! \end{array}} \right.\left. {\begin{array}{*{20}{c}} \!{{\boldsymbol{R}_2}}\!\\ \!{{\boldsymbol{R}_3}}\! \end{array}} \right]. \end{array} $

证明泛函V(x(t))正定.对任意θ < 0，由引理1得

$ \begin{split} \bar \tau \int\limits_{ - \bar \tau }^0 {\int\limits_{t + \theta }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}} } \boldsymbol{Z\dot x}(s){\rm{d}}s{\rm{d}}\theta \geqslant \\ \bar \tau \int\limits_{ - \bar \tau }^0 {\left\{ { - \frac{1}{\theta }{{\left[{\int\limits_{t + \theta }^t {\boldsymbol{\dot x}(s){\rm{d}}s} } \right]}^{\rm{T}}}\boldsymbol{Z}\left[{\int\limits_{t + \theta }^t {\boldsymbol{\dot x}(s){\rm{d}}s} } \right]} \right\}} {\rm{d}}\theta, \end{split} $

(17)

对任意标量θ满足-$\bar \tau$$ \leqslant $θ < 0，即$ - \displaystyle{\frac{{\bar \tau }}{\theta }} \geqslant 1 > 0$，得

$ \begin{split} \bar \tau \int\limits_{ - \bar \tau }^0 {\int\limits_{t + \theta }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}} } \boldsymbol{Z\dot x}(s){\rm{d}}s{\rm{d}}\theta \geqslant \\ \bar \tau \int\limits_{ - \bar \tau }^0 {\left\{ { - \frac{1}{\theta }{{\left[{\int\limits_{t + \theta }^t {\boldsymbol{\dot x}(s){\rm{d}}s} } \right]}^{\rm{T}}}\boldsymbol{Z}\left[{\int\limits_{t + \theta }^t {\boldsymbol{\dot x}(s){\rm{d}}s} } \right]} \right\}} {\rm{d}}\theta \geqslant \\ \int\limits_{ - \bar \tau }^0 {\left\{ {{{\left[{\int\limits_{t + \theta }^t {\boldsymbol{\dot x}(s){\rm{d}}s} } \right]}^{\rm{T}}}\boldsymbol{Z}\left[{\int\limits_{t + \theta }^t {\boldsymbol{\dot x}(s){\rm{d}}s} } \right]} \right\}} {\rm{d}}\theta = \\ {\int\limits_{t - \bar \tau }^t {\left[{\begin{array}{*{20}{c}} {\boldsymbol{x}(t)}\\ {\boldsymbol{x}(s)} \end{array}} \right]} ^{\rm{T}}}\boldsymbol{\varGamma }\left[{\begin{array}{*{20}{c}} {\boldsymbol{x}(t)}\\ {\boldsymbol{x}(s)} \end{array}} \right]{\rm{d}}s, \end{split} $

(18)

其中$\boldsymbol{\varGamma } = \left[{\begin{array}{*{20}{c}} \boldsymbol{Z} & {-\boldsymbol{Z}}\\ {-\boldsymbol{Z}} & \boldsymbol{Z} \end{array}} \right]$.

由假设1，存在任意${l_{1i}} \geqslant 0, {l_{2i}} \geqslant 0, i = 1, 2, \cdots, n, $满足不等式

$ \left[{{f_i}({x_i}(t))-F_i^-{x_i}(t)} \right]{l_{1i}}\left[{F_i^ + {x_i}(t)-{f_i}({x_i}(t))} \right] \geqslant 0, $

(19)

$ \left[{{f_i}({x_i}(t))-F_i^-{x_i}(t)} \right]{l_{2i}}\left[{F_i^ + {x_i}(t)-{f_i}({x_i}(t))} \right] \geqslant 0, $

(20)

可得$\boldsymbol{\eta }{(t)^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_i}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_i}}\\ {-{\boldsymbol{L}_i}} \end{array}} \right]\boldsymbol{\eta }(t) \geqslant 0, i = 1, 2$.

其中${\boldsymbol{L}_1} = {\rm{diag}}\{ {l_{11}}, {l_{12}}, \cdots, {l_{1n}}\} $，${L_2} = {\rm{diag}}\left\{ {{l_{21}}, {l_{22}}, \cdots, } \right.$${\left. {{l_{2n}}} \right\}}$.综合式(19)与(20)得

$ \begin{split}{l} \int\limits_{t - \tau (t)}^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_1}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ {-{\boldsymbol{L}_1}} \end{array}} \right]\boldsymbol{\eta }(s)} {\rm{d}}s + \\ \int\limits_{t - \bar \tau }^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_2}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_2}}\\ {-{\boldsymbol{L}_2}} \end{array}} \right]\boldsymbol{\eta }(s)} {\rm{d}}s \geqslant 0. \end{split} $

(21)

假设1确保

$ \begin{split}{l} 2\sum\limits_{i = 1}^n {{d_{1i}}\int_0^{{x_i}(t)} {({f_i}(s) - F_i^ - s)} } {\rm{d}}s + \\ 2\sum\limits_{i = 1}^n {{d_{2i}}\int_0^{{x_i}(t)} {(F_i^ + s - {f_i}(s))} } {\rm{d}}s \geqslant 0, \end{split} $

(22)

由不等式(17)~(22)得

$ \begin{array}{l} \boldsymbol{V}(\boldsymbol{x}(t)) \geqslant \boldsymbol{x}{(t)^{\rm{T}}}\boldsymbol{Px}(t) + \int_{t - \tau (t)}^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}} \boldsymbol{Q\eta }(s){\rm{d}}s + \\ \int_{t - \bar \tau }^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}} \boldsymbol{R\eta }(s){\rm{d}}s + \bar \tau \int\limits_{ - \bar \tau }^0 {\int\limits_{t + \theta }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}} } \boldsymbol{Z\dot x}(s){\rm{d}}s{\rm{d}}\theta - \\ \int\limits_{t - \tau (t)}^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_1}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ {-{\boldsymbol{L}_1}} \end{array}} \right]\boldsymbol{\eta }(s)} {\rm{d}}s - \\ \int\limits_{t - \bar \tau }^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_2}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_2}}\\ {-{\boldsymbol{L}_2}} \end{array}} \right]\boldsymbol{\eta }(s)} {\rm{d}}s \geqslant \\ \int_{t - \tau (t)}^t {\boldsymbol{\eta }{{(s)}^{\rm{T}}}} {\boldsymbol{\psi }_1}\boldsymbol{\eta }(s){\rm{d}}s +\int_{t - \bar \tau }^t {\boldsymbol{\tilde \eta }{{(t, s)}^{\rm{T}}}} {\boldsymbol{\psi }_2}\boldsymbol{\tilde \eta }(t, s){\rm{d}}s, \end{array} $

(23)

其中

$ \begin{array}{l} \boldsymbol{\tilde \eta }(t, s) = {\left[{\begin{array}{*{20}{c}} {\boldsymbol{x}{{(t)}^{\rm{T}}}} & {\boldsymbol{x}{{(s)}^{\rm{T}}}} & {\boldsymbol{f}{{(\boldsymbol{x}(s))}^{\rm{T}}}} \end{array}} \right]^{\rm{T}}}, \\ {\boldsymbol{\psi }_1} = \left[{\begin{array}{*{20}{c}} {{\boldsymbol{Q}_1} + {\boldsymbol{F}_1}{\boldsymbol{L}_1}} & {{\boldsymbol{Q}_2}-{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ * & {{\boldsymbol{Q}_3} + {\boldsymbol{L}_1}} \end{array}} \right], \\ {\boldsymbol{\psi }_2} = \left[{\begin{array}{*{20}{c}} {{{\bar \tau }^{-1}}\boldsymbol{P} + \boldsymbol{Z}} & {-\boldsymbol{Z}} & 0\\ * & {{\boldsymbol{R}_1} + \boldsymbol{Z} + {\boldsymbol{F}_1}{\boldsymbol{L}_2}} & {{\boldsymbol{R}_2}-{\boldsymbol{F}_2}{\boldsymbol{L}_2}}\\ * & * & {{\boldsymbol{R}_3} + {\boldsymbol{L}_2}} \end{array}} \right]. \end{array} $

Ψ₁ > 0、Ψ₂ > 0即可得V(x(t))正定，证毕.

证明$\boldsymbol{\dot V}(\boldsymbol{x}(t))$负定.由式(16)可得，

$ \begin{split} & \boldsymbol{\dot V}(\boldsymbol{x}(t)) - \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - 2\boldsymbol{y}{(t)^{\rm{T}}}\boldsymbol{u}(t) \leqslant 2\boldsymbol{x}{(t)^{\rm{T}}}(\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - \\ & \quad {\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{\dot x}(t) + 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\boldsymbol{\dot x}(t) + \boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{Q\eta }(t) - \\ & \quad \quad \left( {1 - \dot \tau (t)} \right)\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)) + \boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{R\eta }(t) - \\ & \quad \quad \quad \boldsymbol{\eta }{(t - \bar \tau )^{\rm{T}}}\boldsymbol{R\eta }(t - \bar \tau ) + {{\bar \tau }^2}\boldsymbol{\dot x}{(t)^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(t) - \\ & \quad \quad \quad \quad \bar \tau \int\limits_{t - \bar \tau }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(s){\rm{d}}s} - \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - \\ & \quad \quad \quad \quad \quad 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}\boldsymbol{u}(t). \end{split} $

(24)

由${\boldsymbol{\psi }_1} = \left[{\begin{array}{*{20}{c}} {{\boldsymbol{Q}_1} + {\boldsymbol{F}_1}{\boldsymbol{L}_1}} & {{\boldsymbol{Q}_2}-{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ * & {{\boldsymbol{Q}_3} + {\boldsymbol{L}_1}} \end{array}} \right] > 0, $得$\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}$${\boldsymbol{\psi }_1}\boldsymbol{\eta }(t - \tau (t)) > 0.$

由$\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_1}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ {-{\boldsymbol{L}_1}} \end{array}} \right]\boldsymbol{\eta }(t - \tau (t)) \geqslant 0$，可得$\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)) > 0$.

由$\dot \tau (t) \leqslant \mu $，可得

$ \begin{array}{l} - \left( {1 - \mu } \right)\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)) \geqslant \\ - \left( {1 - \dot \tau (t)} \right)\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)). \end{array} $

因此

$ \begin{split} & \boldsymbol{\dot V}(\boldsymbol{x}(t)) - \gamma \!\boldsymbol{u}{(t)\!^{\rm{T}}}\!\boldsymbol{u}(t) - 2\boldsymbol{y}{(t)^{\rm{T}}}\boldsymbol{u}(t) \leqslant 2\boldsymbol{x}{(t)^{\rm{T}}}\!(\boldsymbol{P} + \\ & \quad{\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{\dot x}(t) + 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\boldsymbol{\dot x}(t) + \\ & \quad \quad\boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{Q\eta }(t) - \left( {1 - \mu } \right)\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)) + \\ & \quad \quad \quad\boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{R\eta }(t) - \boldsymbol{\eta }{(t - \bar \tau )^{\rm{T}}}\boldsymbol{R\eta }(t - \bar \tau ) + \\ & \quad \quad \quad {{\bar \tau }^2}\boldsymbol{\dot x}{(t)^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(t) - \bar \tau \int\limits_{t - \bar \tau }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(s){\rm{d}}s} - \\ & \quad \quad \quad \quad \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}\boldsymbol{u}(t). \end{split} $

(25)

由假设1，存在任意${l_{3i}} \geqslant 0, {l_{4i}} \geqslant 0, i = 1, 2, \cdots, n, $满足不等式$\left[{{f_i}({x_i}(t-\tau (t)))-F_i^-{x_i}(t - \tau (t))} \right] \times $ $ {l_{3i}} \left[{F_i^ + {x_i}(t-\tau (t))} \right.\left. {-{f_i}({x_i}(t-\tau (t)))} \right] \geqslant 0, $可得

$ \boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_3}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_3}}\\ {-{\boldsymbol{L}_3}} \end{array}} \right]\boldsymbol{\eta }(t - \tau (t)) \geqslant 0, $

(26)

$\left[{{f_i}({x_i}(t)) \;-\; {f_i}({x_i}(t \;-\; \tau (t)))-F_i^ - ({x_i}(t)} \right. - {x_i}(t\left. { - \tau (t)))} \right] \times $$ {l_{4i}} [F_i^ + {x_i}(t) \!-\! {x_i}(t-\tau (t)))\!-{f_i}({x_i}(t))\left. { \!-\! {f_i}({x_i}(t \!-\! \tau (t)))} \right]0, $可得

$ \begin{array}{l} {\left[{\begin{array}{*{20}{c}} {\boldsymbol{\eta }(t)}\\ {\boldsymbol{\eta }(t-\tau (t))} \end{array}} \right]^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_4}}\! & \!{{\boldsymbol{F}_2}{\boldsymbol{L}_4}} & {{\boldsymbol{F}_1}{\boldsymbol{L}_4}} & {-{\boldsymbol{F}_2}{\boldsymbol{L}_4}}\\ * & {-{\boldsymbol{L}_4}} & { - {\boldsymbol{F}_2}{\boldsymbol{L}_4}} & {{\boldsymbol{L}_4}}\\ * & * \!& { - {\boldsymbol{F}_1}{\boldsymbol{L}_4}} & {{\boldsymbol{F}_2}{\boldsymbol{L}_4}}\\ * & * & * & { - {\boldsymbol{L}_4}} \end{array}} \right] \times \\ \left[{\begin{array}{*{20}{c}} {\boldsymbol{\eta }(t)}\\ {\boldsymbol{\eta }(t-\tau (t))} \end{array}} \right] \geqslant 0, \end{array} $

(27)

其中${\boldsymbol{L}_3}\!=\!{\rm{diag}}\{ {l_{31}}, {l_{32}}, \! \cdots \!, {l_{3n}}\} $，${\boldsymbol{L}_4}\!=\!{\rm{diag}}\{ {l_{41}}, {l_{42}}, \! \cdots \!, $${l_{4n}}\} $.

由引理2可得：

$ \begin{aligned} & \boldsymbol{\dot V}(\boldsymbol{x}(t)) - \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - 2\boldsymbol{y}(t)\boldsymbol{u}(t) \leqslant 2\boldsymbol{x}{(t)^{\rm{T}}}(\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - \\ & {\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{\dot x}(t) + 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\boldsymbol{\dot x}(t) + \boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{Q\eta }(t) - \\ & \left( {1 - \mu } \right)\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)) + \boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{R\eta }(t) - \\ & \boldsymbol{\eta }{(t - \bar \tau )^{\rm{T}}}\boldsymbol{R\eta }(t - \bar \tau ) + {{\bar \tau }^2}\boldsymbol{\dot x}{(t)^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(t) - \\ & \bar \tau \int\limits_{t - \bar \tau }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(s){\rm{d}}s} - \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - \\ & 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}\boldsymbol{u}(t) + \boldsymbol{\eta }{(t)^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_1}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_1}}\\ {-{\boldsymbol{L}_1}} \end{array}} \right]\boldsymbol{\eta }(t) + \\ & \boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_3}}\\ * \end{array}} \right.\left. {\begin{array}{*{20}{c}} {{\boldsymbol{F}_2}{\boldsymbol{L}_3}}\\ {-{\boldsymbol{L}_3}} \end{array}} \right]\boldsymbol{\eta }(t - \tau (t)) \!+\! {\left[{\begin{array}{*{20}{c}} {\boldsymbol{\eta }(t)}\\ {\boldsymbol{\eta }(t-\tau (t))} \end{array}} \right]^{\rm{T}}} \times \\ & \left[{\begin{array}{*{20}{c}} {-{\boldsymbol{F}_1}{\boldsymbol{L}_4}} & {{\boldsymbol{F}_2}{\boldsymbol{L}_4}} & {{\boldsymbol{F}_1}{\boldsymbol{L}_4}} & {-{\boldsymbol{F}_2}{\boldsymbol{L}_4}}\\ * & {-{\boldsymbol{L}_4}} & { - {\boldsymbol{F}_2}{\boldsymbol{L}_4}} & {{\boldsymbol{L}_4}}\\ * & * & { - {\boldsymbol{F}_1}{\boldsymbol{L}_4}} & {{\boldsymbol{F}_2}{\boldsymbol{L}_4}}\\ * & * & * & { - {\boldsymbol{L}_4}} \end{array}} \right]\left[{\begin{array}{*{20}{c}} {\boldsymbol{\eta }(t)}\\ {\boldsymbol{\eta }(t-\tau (t))} \end{array}} \right] + \\ & {\left[{\begin{array}{*{20}{c}} {\boldsymbol{x}(t)}\\ {\boldsymbol{x}(t-\tau (t))}\\ {\boldsymbol{x}(t-\bar \tau )} \end{array}} \right]^{\rm{T}}}\left[{\begin{array}{*{20}{c}} {-\boldsymbol{Z}} & {\boldsymbol{Z}-\boldsymbol{S}} & \boldsymbol{S}\\ * & {-2\boldsymbol{Z} + \boldsymbol{S} + {\boldsymbol{S}^T}} & {\boldsymbol{Z} - \boldsymbol{S}}\\ * & * & { - \boldsymbol{Z}} \end{array}} \right] \times \\ & \left[{\begin{array}{*{20}{c}} {\boldsymbol{x}(t)}\\ {\boldsymbol{x}(t-\tau (t))}\\ {\boldsymbol{x}(t-\bar \tau )} \end{array}} \right] = \boldsymbol{\xi }{(t)^{\rm{T}}}(\boldsymbol{\Lambda } + \boldsymbol{\Upsilon} )\boldsymbol{\xi }(t). \end{aligned} $

(28)

其中

$ \begin{array}{l} {\boldsymbol{\zeta }}{\rm{(}}t{\rm{)}}\left[{\begin{array}{*{20}{c}} {x{{{\rm{(}}t{\rm{)}}}^{\rm{T}}}}&{f{{{\rm{(}}x{\rm{(}}t{\rm{))}}}^{\rm{T}}}}&{f{{{\rm{(}}x{\rm{(}}t-\tau {\rm{(}}t{\rm{)))}}}^{\rm{T}}}} \end{array}} \right.\\ {\left. {\begin{array}{*{20}{c}} {{\boldsymbol{x}}{{{\rm{(}}t-\tau {\rm{(}}t{\rm{))}}}^{\rm{T}}}}&{x{{{\rm{(}}t-\bar \tau {\rm{)}}}^{\rm{T}}}}&{f{{{\rm{(}}x{\rm{(}}t - \bar \tau {\rm{))}}}^{\rm{T}}}}&{{\boldsymbol{u}}{{{\rm{(}}t{\rm{)}}}^{\rm{T}}}} \end{array}} \right]^{\rm{T}}, } \end{array} $

设$\boldsymbol{J}({t_f}) = \int\limits_0^{{t_f}} {\left[{-\gamma \boldsymbol{u}{{(t)}^{\rm{T}}}\boldsymbol{u}(t)-2\boldsymbol{f}{{(\boldsymbol{x}(t))}^{\rm{T}}}\boldsymbol{u}(t)} \right]} {\rm{d}}t$，其中t_f$ \leqslant $0，在零初始条件下，满足

$ \begin{array}{l} \boldsymbol{V}(\boldsymbol{x}({t_f})) \leqslant \int\limits_0^{{t_f}} {\left[{\boldsymbol{\dot V}(\boldsymbol{x}(t))-\gamma \boldsymbol{u}{{(t)}^{\rm{T}}}\boldsymbol{u}(t)-2\boldsymbol{f}{{(\boldsymbol{x}(t))}^{\rm{T}}}\boldsymbol{u}(t)} \right]} {\rm{d}}t, \end{array} $

由引理3知，Ξ < 0可满足(Λ+Υ) < 0，由J(t_f)$ \leqslant $0可得神经网络(12)是无源的.

定理2对参数不确定神经网络(1)，在假设1情况下，对给定标量$\bar \tau$ > 0和μ，对于任意时滞τ(t)满足0 < τ(t)$ \leqslant $$\bar \tau$和$\dot \tau (t) \leqslant \mu $，如果存在对称矩阵，P > 0，Z > 0，Ψ₁ > 0，Ψ₂ > 0，$\mathbb{Z} = \left[{\begin{array}{*{20}{c}} \boldsymbol{Z} & \boldsymbol{S}\\ * & \boldsymbol{Z} \end{array}} \right] > 0$，$\boldsymbol{\varXi } + \boldsymbol{\delta }{\boldsymbol{N}^{ - 1}}{\boldsymbol{\delta }^{\rm{T}}} + {\boldsymbol{E}^{\rm{T}}}\boldsymbol{NE} < 0$，${\boldsymbol{Q}_1} = {\boldsymbol{Q}_1}^{\rm{T}}, $${\boldsymbol{Q}_2}, {\boldsymbol{Q}_3} = {\boldsymbol{Q}_3}^{\rm{T}}$，${\boldsymbol{R}_1} = {\boldsymbol{R}_1}^{\rm{T}}, {\boldsymbol{R}_2}, {\boldsymbol{R}_3} = {\boldsymbol{R}_3}^{\rm{T}}$，对角矩阵L_i > 0，其中i=1, 2，$\cdots$, 4，${\boldsymbol{D}_1} = {\rm{diag}}\left\{ {{d_{11}}, {d_{12}}, \! \cdots \!, {d_{1n}}} \right\} > 0$，${\boldsymbol{D}_2} = {\rm{diag}}\left\{ {{d_{21}}, {d_{22}}, } \right.\left. {\! \cdots \!, {d_{2n}}} \right\} > 0$，标量γ$ \leqslant $0神经网络(1)是鲁棒无源的.

$ \left[{\begin{array}{*{20}{c}} \! {{\boldsymbol{\varPsi}}_1} + {\varepsilon _1}\boldsymbol{E}_1^{\rm{T}}{\boldsymbol{E}_1} & {{\boldsymbol{\varPsi}}_3} & {{\boldsymbol{\varPsi}}_4} & {{\boldsymbol{\varPsi}}_2}S & S & 0 & {{\boldsymbol{\varPsi}}_5} &-\bar \tau {\boldsymbol{AZ}} & (\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2}-{\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{H} \! \\ \! * & {{\boldsymbol{\varPsi}}_7} &-{\boldsymbol{F}_2}{\boldsymbol{L}_4} & 0 & 0 & 0 & {{\boldsymbol{\varPsi}}_8} & - \bar \tau {\boldsymbol{W}^{\rm{T}}}Z & ({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\boldsymbol{H} \! \\ \! * & {{\boldsymbol{\varPsi}}_6} + {\varepsilon _2}\boldsymbol{E}_2^{\rm{T}}{\boldsymbol{E}_2} & {{\boldsymbol{\varPsi}}_9} + {\varepsilon _3}\boldsymbol{E}_3^{\rm{T}}{\boldsymbol{E}_3} & {{\boldsymbol{\varPsi}}_{10}} & 0 & 0 & 0 & \bar \tau \boldsymbol{W}_1^{\rm{T}}\boldsymbol{Z} & 0 \! \\ \! * & * & * & {{\boldsymbol{\varPsi}}_{11}} & \boldsymbol{Z} - \boldsymbol{S} & 0 & 0 & 0 & 0 \! \\ \! * & * & * & * & - {\boldsymbol{R}_1} - \boldsymbol{Z} & - {\boldsymbol{R}_2} & 0 & 0 & 0 \! \\ \! * & * & * & * & * & - {\boldsymbol{R}_3} & 0 & 0 & 0 \! \\ \! * & * & * & * & * & * & - \gamma \boldsymbol{I} & \bar \tau \boldsymbol{Z} & 0 \! \\ \! * & * & * & * & * & * & * & - \boldsymbol{Z} & \bar \tau {\boldsymbol{ZH}} \! \\ \! * & * & * & * & * & * & * & * & - \boldsymbol{N} \! \end{array}} \right] \leqslant 0, $

(29)

其中

$ \begin{aligned} & {\boldsymbol{\sigma }_1} = - \Delta \boldsymbol{A}{(t)^{\rm{T}}}{(\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1})^{\rm{T}}} - ({\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1} +\\ & \;\;\;\;\;\;\; \boldsymbol{P})\Delta \boldsymbol{A}(t), \\ & {\boldsymbol{\sigma }_2} = (\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1})\Delta \boldsymbol{W}(t) - \Delta \boldsymbol{A}{(t)^{\rm{T}}}({\boldsymbol{D}_1} - {\boldsymbol{D}_2}), \\ & {\boldsymbol{\sigma }_3} = (\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - {\boldsymbol{K}_2}{\boldsymbol{D}_1})\Delta {\boldsymbol{W}_1}(t), \\ & {\boldsymbol{\sigma }_4} = ({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\Delta \boldsymbol{W}(t) + \Delta {\boldsymbol{W}^{\rm{T}}}(t)({\boldsymbol{D}_1} - {\boldsymbol{D}_2}), \\ & {\boldsymbol{\sigma }_5} = ({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\Delta {\boldsymbol{W}_1}(t). \end{aligned} $

由定理1可得

$ \begin{aligned} & \boldsymbol{\dot V}(\boldsymbol{x}(t)) - \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - 2\boldsymbol{y}{(t)^{\rm{T}}}\boldsymbol{u}(t) \leqslant 2\boldsymbol{x}{(t)^{\rm{T}}}(\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2} - \\ & {\boldsymbol{K}_2}{\boldsymbol{D}_1})\boldsymbol{\dot x}(t) + 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}({\boldsymbol{D}_1} - {\boldsymbol{D}_2})\boldsymbol{\dot x}(t) + \boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{Q\eta }(t) - \\ & \left( {1 - \mu } \right)\boldsymbol{\eta }{(t - \tau (t))^{\rm{T}}}\boldsymbol{Q\eta }(t - \tau (t)) + \boldsymbol{\eta }{(t)^{\rm{T}}}\boldsymbol{R\eta }(t) - \\ & \boldsymbol{\eta }{(t - \bar \tau )^{\rm{T}}}\boldsymbol{R\eta }(t - \bar \tau ) + {{\bar \tau }^2}\boldsymbol{\dot x}{(t)^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(t) - \\ & \bar \tau \int\limits_{t - \bar \tau }^t {\boldsymbol{\dot x}{{(s)}^{\rm{T}}}\boldsymbol{{\rm Z}\dot x}(s)ds} - \gamma \boldsymbol{u}{(t)^{\rm{T}}}\boldsymbol{u}(t) - \\ & 2\boldsymbol{f}{(\boldsymbol{x}(t))^{\rm{T}}}\boldsymbol{u}(t) < \varPhi . \end{aligned} $

设$\delta = \left[{\boldsymbol{P} + {\boldsymbol{K}_1}{\boldsymbol{D}_2}-{\boldsymbol{K}_2}\boldsymbol{D}\;{\boldsymbol{D}_1}-{\boldsymbol{D}_2}\;0\;0\;0\;0\;0\;\bar \tau \boldsymbol{Z}} \right]\boldsymbol{H}$，其中H=[H₁H₂H₃]，$\boldsymbol{E}=\left[\begin{aligned} {\boldsymbol{E}_1}\;\;\;0\;\;\;0\;\;\;0\;\;\;0\;\;\;0\;\;\;0\;\;\;0\\ 0\;\;\;{\boldsymbol{E}_2}\;\;\;0\;\;\;0\;\;\;0\;\;\;0\;\;\;0\;\;\;0\\ 0\;\;\;0\;\;\;{\boldsymbol{E}_3}\;\;\;0\;\;\;0\;\;\;0\;\;\;0\;\;\;0 \end{aligned} \right]$，可得$\boldsymbol{\varPhi } = \boldsymbol{\varXi } + \boldsymbol{\delta FE} + {(\boldsymbol{\delta FE})^{\rm{T}}}$，$\boldsymbol{F} = {\rm{diag}}\left\{ {({\boldsymbol{F}_1}(t), {\boldsymbol{F}_2}(t), } \right.$$\left. {{\boldsymbol{F}_3}(t))} \right\}$.

由引理4得$\boldsymbol{\varPhi } = \boldsymbol{\varXi } + \boldsymbol{\delta FE} + {(\boldsymbol{\delta FE})^{\rm{T}}} \leqslant \boldsymbol{\varXi } + \boldsymbol{\delta }{\boldsymbol{N}^{ - 1}}{\boldsymbol{\delta }^{\rm{T}}} +$$ {\boldsymbol{E}^{\rm{T}}}\boldsymbol{{N}E}$，$\boldsymbol{N} = {\rm diag}\left\{ {{\varepsilon _1}\boldsymbol{I}, {\varepsilon _2}\boldsymbol{I}, {\varepsilon _3}\boldsymbol{I}} \right\}.$

综合引理3可得$\boldsymbol{\varXi } + \boldsymbol{\delta }{\boldsymbol{{N}}^{ - 1}}{\boldsymbol{\delta }^{\rm{T}}} + {\boldsymbol{E}^{\rm{T}}}\boldsymbol{{N}E} < 0$，即当Φ < 0则神经网络(1)鲁棒无源.证毕.

3 仿真实例

考虑神经网络(12)，选取连接权值矩阵为

$ \begin{aligned} & \boldsymbol{A} = \left[{\begin{array}{*{20}{c}} {0.113\;7} & 0 \\ 0 & {0.152\;8} \end{array}} \right], \boldsymbol{W} = \left[{\begin{array}{*{20}{c}} {0.627\;7} & {-0.774\;4} \\ {0.926\;6} & {1.405\;0} \end{array}} \right], \\ & {\boldsymbol{W}_1} = \left[{\begin{array}{*{20}{c}} {-0.531\;4} & {1.194\;8} \\ {-0.617\;6} & {-1.617\;0} \end{array}} \right], \end{aligned} $

假定激活函数为

$ {f_i}({x_i}) = 0.5(\left| {{x_i} + 1} \right| - \left| {{x_i} - 1} \right|), i = 1, 2. $

选取满足激活函数的参数$F_1^ - = 0.05, F_2^ - = 0.01$，$F_1^ + = 0.15, F_2^ + = 0.5$，通过以上参数可计算出满足定理1在定值μ下τ最大值，对不同μ值所得结果如表 1所示。

在引入新颖李雅普诺夫泛函以及应用更广泛的不等式(6)后，通过把本文得出的结果与文献[17]的结果相比较，得到神经网络(12)无源条件保守性更小.同时假定τ=0.6，μ=0.5，计算满足定理1的矩阵Q与R得：

$\begin{array}{l} \boldsymbol{Q} = \left[{\begin{array}{*{20}{c}} {{\boldsymbol{Q}_1}}&{{\boldsymbol{Q}_2}}\\ {\boldsymbol{Q}_2^{\rm{T}}}&{{\boldsymbol{Q}_3}} \end{array}} \right] = \left[{\begin{array}{*{20}{c}} {0.168\;3}&{0.021\;8}&{-0.306\;6}&{-0.090\;3}\\ {*}&{0.142\;3}&{-0.051\;1}&{ - 0.276\;4}\\ {*}&{*}&{0.046\;7}&{ - 0.130\;3}\\ {*}&{*}&{*}&{1.027\;5} \end{array}} \right], \\ \boldsymbol{R} = \left[{\begin{array}{*{20}{c}} {{\boldsymbol{R}_1}}&{{\boldsymbol{R}_2}}\\ {\boldsymbol{R}_2^{\rm{T}}}&{{\boldsymbol{R}_3}} \end{array}} \right] = \left[{\begin{array}{*{20}{c}} {0.021\;9}&{0.025\;9}&{-0.419\;7}&{0.016\;6}\\ {*}&{-0.110\;3}&{-0.061\;5}&{ - 0.123\;8}\\ {*}&{*}&{3.883\;4}&{ - 0.338\;7}\\ {*}&{*}&{*}&{0.640\;2} \end{array}} \right]. \end{array}$

计算矩阵Q的特征值为-0.236 4，0.071 3，0.427 3，1.122 6，R的特征值为-0.135 8，-0.021 8，0.629 2，3.963 6，可得出矩阵Q与R不是正定的，但Q与R确保了李雅普诺夫泛函(16)是正定的，同时满足神经网络(12)无源.

4 结论

本文对时滞神经网络及范数有界参数不确定的时滞神经网络的无源性进行了研究，利用LMI方法，构造了不需要所有对阵矩阵是正定的李雅普诺夫泛函，通过引入Jensen不等式、Schur补引理及自由权矩阵等方法，分别得到时滞神经网络无源的条件和参数不确定的时滞神经网络的鲁棒无源条件.仿真实例表明，本文满足神经网络无源性的结果比文献[17]的结果的保守性更小.