一类新的二重积分不等式在时滞系统中的应用

A Novel Double Integral Inequalities Applied to Time-Delay Systems

摘要: 本文主要研究一类新的二重积分不等式并用于时滞系统的稳定性分析。首先，通过引入零等式得到新的二重积分不等式。然后，利用增广向量构造李雅普诺夫泛函(LKF)，再通过不等式对泛函导数中的积分项进行处理，得到保守性更低的稳定性条件。接着用更加宽松的二次函数负决定引理得到新的稳定条件。最后，通过一个数值例子验证所得结果的有效性和优越性。

Abstract: This article is concerned with a novel double integral inequalities applied to time-delay systems. Firstly, two zero-value equations have been introduced to estimate the upper bound of double in-tegral inequality. Secondly, augmented vectors are used to construct Lyapunov function, and the inequality is utilized to estimate the derivative of functional, and then relax quadratic function negative-determination lemma is used to obtain stability criterion. Finally, examples are given to show the effectiveness of the obtained result.

1. 引言

时滞是现实世界中不可避免的一种自然现象。我们在很多领域都会遇到时滞，例如信号处理，模式识别，组合优化等等 [1] [2] [3]。很多相关研究表明时滞会导致不良的动态行为，这些行为会使得系统的性能下降甚至不稳定 [4]，因此研究时滞系统的稳定性是非常有意义的。

到目前为止，研究具有低保守性稳定性标准的时滞系统依旧是个热门话题。学者们针对时滞系统的稳定性研究主要采用李雅普诺夫函数方法。通过利用李雅普诺夫稳定性理论 [5]，构造恰当的LKF以及尽可能紧密地估计泛函的导数来得到低保守性的稳定性标准。对于构造LKF方法，目前比较常见的是构造带有增广的LKF [6]，以及利用时滞分割方法构造泛函 [7] 和构造带有多重积分项的泛函 [8]。对于估计泛函的导数，主要是估计求导后产生的积分项，力求获得求导后更严格的下界。为了处理这些积分，学者们研究了许多估计技术，例如Jensen不等式 [9]，基于Wirtinger的积分不等式 [10]，基于自由矩阵的积分不等式 [11]，改进的自由矩阵积分不等式 [12] 以及互凸不等式 [13] 等等。虽然这些不等式有效地降低系统的保守性，但是还有改进的空间，文献 [14] 中通过引入增广的LKF并应用于wirtinger不等式中，得到了保守性较低的时滞相关稳定性标准，文献 [15] 利用零等式改进具有增广的一重积分不等式，与直接使用Jensen不等式放缩相比，运用零等式可以巧妙加入自由矩阵，从而提供更大的自由度进而得到更紧的时滞上界。然而，在目前的工作中利用零等式构造具有增广的二重积分不等式，目前尚未被考虑。事实上，高阶积分项对降低时滞系统的保守性扮演着十分重要的作用，例如构造带有三重积分的LKF，在对这个泛函求导后会产生二重积分项，估计这个二重积分项会将一重积分状态信息以及系统的状态信息引入并使它们与二重积分状态信息相互交叉，因此建立一个新的带有自由矩阵的具有增广的二重积分不等式对降低时滞系统的保守性具有深远的意义。此外，值得一提的是在增广向量中引入s相关项能有效克服系统的保守性，因为它将一重积分放入增广向量中并把这个增广向量放入带有一重积分的LKF中，在对这个泛函求导后能够获得更多状态信息。因此，本文拟通过新的积分不等式处理技巧结合更符合系统的LKF，获得保守性更低的稳定性条件。

为了简化表述，我们有必要做如下的符号说明： $R^n$ 表示n维欧式空间； $R^{n\times m}$ 代表 $n\times m$ 的矩阵值； $\text{sym}\left\{X\right\}=X+X^{\text{T}}$ ； $\text{diag}\{\cdots \}$ 表示对角矩阵； $\text{col}\{\cdots \}$ 表示列向量； $X^{\text{T}}$ 和 $X^{-1}$ 分别矩阵的转置和逆矩阵； $*$ 表示矩阵的对称项。

2 预备知识

2.1. 系统描述

考虑一般的带有时变时滞的线性系统

$\{\begin{array}{l}\stackrel{\dot{}}{x}\left(t\right)=Ax\left(t\right)+Bx\left(t-h\left(t\right)\right),\\ x\left(t\right)=\phi \left(t\right),t\in \left[-h,0\right]\end{array}$ (1)

其中 $x\left(t\right)\in R^n$ 是系统的状态向量； $A\in R^{n\times n}$ 和 $B\in R^{n\times n}$ 是已知的实数矩阵；初始条件 $\phi \left(t\right)$ 是连续的微分函数，h和u是已知的常数，且满足如下条件：

$0<h\left(t\right)<h$， $\stackrel{\dot{}}{h}\left(t\right)<u$. (2)

2.2. 引理

在分析系统(1)的时滞依赖稳定性判据之前，本节首先引入如下一些重要的引理：

引理1 [16] 给定一个二次函数 $f\left(y\right)=a_2{y}^2+a_{1}y+a_0$，当 $\beta \in \left[0,1\right]$ 时，如果有

$T_1=f\left(h_i\right)<0\left(i=1,2\right)$， $T_2=-{\beta }^2{h}_{12}^2{a}_2+f\left(h_1\right)<0$， $T_3=-{\left(1-\beta \right)}^2{h}_{12}^2{a}_2+f\left(h_2\right)<0$，(3)

那么 $f\left(y\right)<0\left(h_1\le y\le h_2\right)$ 成立，其中 $h_{12}=h_2-h_1$。

引理2 [12] $w\left(t\right)$ 是一个可微函数： $\left[a,b\right]\to R^n$，若存在对称正定矩阵 $H\in R^{n\times n}$，对于任意矩阵L，M，则下列的不等式成立：

${\displaystyle {\int }_a^b{w}^{\text{T}}\left(s\right)Hw\left(s\right)\text{d}s}\ge -\text{sym}\left\{{\sigma }_0^{\text{T}}L{\sigma }_1+{\sigma }_0^{\text{T}}M{\sigma }_2\right\}-\left(b-a\right){\sigma }_0^{\text{T}}\left(\frac{3L{H}^{-1}{L}^{\text{T}}+M{H}^{-1}{M}^{\text{T}}}{3}\right){\sigma }_0$ (4)

其中 ${\sigma }_1=\left[\begin{array}{c}{\displaystyle {\int }_a^{b}x\left(s\right)\text{d}s}\\ x\left(b\right)-x\left(a\right)\end{array}\right]$， ${\sigma }_2=\left[\begin{array}{c}-{\displaystyle {\int }_a^{b}x\left(s\right)\text{d}s}+\frac{2}{b-a}{\displaystyle {\int }_a^b{\displaystyle {\int }_s^{b}x\left(u\right)\text{d}u\text{d}s}}\\ x\left(b\right)+x\left(a\right)-\frac{2}{b-a}{\displaystyle {\int }_a^{b}x\left(s\right)\text{d}s}\end{array}\right]$， ${\sigma }_0$ 是任意向量。

引理3 [9] 若对称正定矩阵 $R>0$， $x\left(t\right)$ 是一个可微函数： $\left[a,b\right]\to R^n$，则下列不等式成立

${\displaystyle {\int }_a^b{x}^{\text{T}}\left(s\right)\Omega x\left(s\right)\text{d}s}\ge \frac{1}{b-a}{\left({\displaystyle {\int }_a^{b}x\left(s\right)\text{d}s}\right)}^{\text{T}}R\left({\displaystyle {\int }_a^{b}x\left(s\right)\text{d}s}\right)+\frac{3}{b-a}{\Omega }_1^{\text{T}}R{\Omega }_1$，

${\displaystyle {\int }_a^b{\displaystyle {\int }_{\beta }^b{x}^{\text{T}}\left(s\right)Rx\left(s\right)\text{d}s\text{d}\beta }}\ge \frac{2}{{\left(b-a\right)}^2}{\left({\displaystyle {\int }_a^b{\displaystyle {\int }_{\beta }^{b}x\left(s\right)\text{d}s\text{d}\beta }}\right)}^{\text{T}}R\left({\displaystyle {\int }_a^b{\displaystyle {\int }_{\beta }^{b}x\left(s\right)\text{d}s\text{d}\beta }}\right)$，

其中 ${\Omega }_1={\displaystyle {\int }_a^{b}x\left(s\right)\text{d}s}-\frac{2}{b-a}{\displaystyle {\int }_a^b{\displaystyle {\int }_{\beta }^{b}x\left(s\right)\text{d}s\text{d}\beta }}$。

引理4 若正定对称矩阵 $Y_i\in R^{n\times n}\left(i=1,3\right)$， $X_i\in R^{n\times n}\left(i=1,2\right)$，任意矩阵 $Y_2\in R^{n\times n}$， $x\left(t\right)$ 是一个可微函数： $\left[a,b\right]\to R^n$，满足：

${\Omega }_0=\left[\begin{array}{cc}{Y}_1& Y_2\\ *& Y_3\end{array}\right]\ge 0$， ${\Omega }_1=\left[\begin{array}{cc}{Y}_1& Y_2+X_1\\ *& Y_3\end{array}\right]\ge 0$， ${\Omega }_2=\left[\begin{array}{cc}{Y}_1& Y_2+X_2\\ *& Y_3\end{array}\right]\ge 0$，

那么，对任意向量 $c\in \left[a,b\right]$，下面的不等式成立

$-{\displaystyle {\int }_a^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}\le {\xi }^{\text{T}}\Omega \xi$ (5)

其中

$W\left(s\right)=\text{col}\left\{\stackrel{\dot{}}{x}\left(s\right),x\left(s\right)\right\}$， $\xi =\text{col}\left\{x\left(b\right),x\left(c\right),x\left(a\right),w_1,w_2,v_1,v_2\right\}$， $w_1=\frac{1}{b-c}{\displaystyle {\int }_c^{b}x\left(s\right)\text{d}s}$，

$w_2=\frac{1}{c-a}{\displaystyle {\int }_a^{c}x\left(s\right)\text{d}s}$， $v_1=\frac{1}{{\left(b-c\right)}^2}{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^{b}x\left(s\right)\text{d}s\text{d}\theta }}$， $v_2=\frac{1}{{\left(c-a\right)}^2}{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^{c}x\left(s\right)\text{d}s\text{d}\theta }}$，

$\begin{array}{c}\Omega =\left(c-a\right)e_2^{\text{T}}{X}_2{e}_2+\left(b-c\right)e_1^{\text{T}}{X}_1{e}_1-\left(b-c\right)e_4^{\text{T}}{X}_1{e}_4-\left(c-a\right)e_5^{\text{T}}{X}_2{e}_5-\left(c-a\right)\left(b-a\right)e_4^{\text{T}}{Y}_1{e}_4\\ -\text{sym}\{{\left(b-c\right)}^2{e}_6^{\text{T}}{Y}_1{e}_6+2\left(b-a\right)e_6^{\text{T}}\left(Y_1+X_1\right)\left(e_1-e_4\right)+{\left(e_1-e_4\right)}^{\text{T}}{Y}_3\left(e_1-e_4\right)+{\left(c-a\right)}^2{e}_7^{\text{T}}{Y}_1{e}_7\\ +2\left(c-a\right)e_7^{\text{T}}\left(Y_2+X_2\right)\left(e_2-e_5\right)+{\left(e_2-e_5\right)}^{\text{T}}{Y}_3\left(e_2-e_5\right)+\left(c-a\right)e_4^{\text{T}}{Y}_2\left(e_1-e_2\right)\}\\ -\frac{c-a}{b-a}{\left(e_1-e_2\right)}^{\text{T}}{Y}_3\left(e_1-e_2\right),\end{array}$

$e_i=\left[\begin{array}{ccc}{0}_{n,\left(i-1\right)n}& I_n& 0_{n,\left(7-i\right)n}\end{array}\right]\left(i=1,2,\cdots ,7\right)$。

证明：对于对称半正定矩阵 $X_i\in R^{n\times n}\left(i=1,2\right)$，有下面的零等式成立

$0=-2{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^b{x}^{\text{T}}}}\left(s\right)X_1\stackrel{\dot{}}{x}\left(s\right)\text{d}s\text{d}\theta +\left(b-c\right)x^{\text{T}}(b)X_{1}x\left(b\right)-{\displaystyle {\int }_c^b{x}^{\text{T}}\left(s\right)X_{1}x\left(s\right)\text{d}s}$，

$0=-2{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^c{x}^{\text{T}}}}\left(s\right)X_2\stackrel{\dot{}}{x}\left(s\right)\text{d}s\text{d}\theta +\left(c-a\right)x^{\text{T}}\left(c\right)X_{2}x\left(c\right)-{\displaystyle {\int }_a^c{x}^{\text{T}}\left(s\right)X_{2}x\left(s\right)\text{d}s}$。

将上面两个零等式加入 $-{\displaystyle {\int }_a^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}$。

可得

$\begin{array}{c}\Phi =-{\displaystyle {\int }_a^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}-2{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^b{x}^{\text{T}}}}\left(s\right)X_1\stackrel{\dot{}}{x}\left(s\right)\text{d}s\text{d}\theta +\left(b-c\right)x^{\text{T}}\left(b\right)X_{1}x\left(b\right)-{\displaystyle {\int }_c^b{x}^{\text{T}}\left(s\right)X_{1}x\left(s\right)\text{d}s}\\ -2{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^c{x}^{\text{T}}}}\left(s\right)X_2\stackrel{\dot{}}{x}\left(s\right)\text{d}s\text{d}\theta +\left(c-a\right)x^{\text{T}}\left(c\right)X_{2}x\left(c\right)-{\displaystyle {\int }_a^c{x}^{\text{T}}\left(s\right)X_{2}x\left(s\right)\text{d}s}\end{array}$

其中

$-{\displaystyle {\int }_a^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}=-{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}-{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}$

$-{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}=-{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^{\text{c}}{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}-{\displaystyle {\int }_a^c{\displaystyle {\int }_c^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}$

$-{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}=-{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^c{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}-{\displaystyle {\int }_c^b{\displaystyle {\int }_c^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s\text{d}\theta }}$

合并同类项后可以得到

$\begin{array}{c}\Phi =-{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{1}W\left(s\right)\text{d}s\text{d}\theta }}-{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^c{W}^{\text{T}}\left(s\right){\Omega }_{2}W\left(s\right)\text{d}s\text{d}\theta }}-\left(c-a\right){\displaystyle {\int }_c^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s}\\ -{\displaystyle {\int }_c^b{x}^{\text{T}}\left(s\right)X_{1}x\left(s\right)\text{d}s}-{\displaystyle {\int }_a^c{x}^{\text{T}}\left(s\right)X_{2}x\left(s\right)\text{d}s}+\left(b-c\right)x^{\text{T}}\left(b\right)X_{1}x\left(b\right)+\left(c-a\right)x^{\text{T}}\left(c\right)X_{2}x\left(c\right).\end{array}$

利用Jensen不等式可得

$-{\displaystyle {\int }_c^b{x}^{\text{T}}\left(s\right)X_{1}x\left(s\right)\text{d}s}\le -\left(b-c\right)w_1^{\text{T}}{X}_1{w}_1$

$-{\displaystyle {\int }_a^c{x}^{\text{T}}\left(s\right)X_{2}x\left(s\right)\text{d}s}\le -\left(c-a\right)w_2^{\text{T}}{X}_2{w}_2$

由引理3，可以得到

$-{\displaystyle {\int }_c^b{\displaystyle {\int }_{\theta }^b{W}^{\text{T}}\left(s\right){\Omega }_{1}W\left(s\right)\text{d}s\text{d}\theta }}\le -2{\left[\begin{array}{c}\left(b-c\right)v_1\\ x\left(b\right)-w_1\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2+X_1\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(b-c\right)v_1\\ x\left(b\right)-w_1\end{array}\right]$

$-{\displaystyle {\int }_a^c{\displaystyle {\int }_{\theta }^c{W}^{\text{T}}\left(s\right){\Omega }_{2}W\left(s\right)\text{d}s\text{d}\theta }}\le -2{\left[\begin{array}{c}\left(c-a\right)v_2\\ x\left(c\right)-w_2\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2+X_2\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(c-a\right)v_2\\ x\left(c\right)-w_2\end{array}\right]$

$\begin{array}{c}-\left(c-a\right){\displaystyle {\int }_c^b{W}^{\text{T}}\left(s\right){\Omega }_{0}W\left(s\right)\text{d}s}\le -\left(c-a\right)\{\frac{1}{b-c}{\left[\begin{array}{c}\left(b-c\right)w_1\\ x\left(b\right)-x\left(c\right)\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(b-c\right)w_1\\ x\left(b\right)-x\left(c\right)\end{array}\right]\\ +\frac{3}{b-c}{\left[\begin{array}{c}\left(b-c\right)\left(w_1-2v_1\right)\\ 2w_1-x\left(b\right)-x\left(c\right)\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(b-c\right)\left(w_1-2v_1\right)\\ 2w_1-x\left(b\right)-x\left(c\right)\end{array}\right]\}\end{array}$

整理可得

$\begin{array}{c}\Phi \le -2{\left[\begin{array}{c}\left(b-c\right)v_1\\ x\left(b\right)-w_1\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2+X_1\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(b-c\right)v_1\\ x\left(b\right)-w_1\end{array}\right]-2{\left[\begin{array}{c}\left(c-a\right)v_2\\ x\left(c\right)-w_2\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2+X_2\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(c-a\right)v_2\\ x\left(c\right)-w_2\end{array}\right]\\ -\left(c-a\right)\{\frac{1}{b-c}{\left[\begin{array}{c}\left(b-c\right)w_1\\ x\left(b\right)-x\left(c\right)\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(b-c\right)w_1\\ x\left(b\right)-x\left(c\right)\end{array}\right]\\ +\frac{3}{b-c}{\left[\begin{array}{c}\left(b-c\right)\left(w_1-2v_1\right)\\ 2w_1-x\left(b\right)-x\left(c\right)\end{array}\right]}^{\text{T}}\left[\begin{array}{cc}{Y}_1& Y_2\\ *& Y_3\end{array}\right]\left[\begin{array}{c}\left(b-c\right)\left(w_1-2v_1\right)\\ 2w_1-x\left(b\right)-x\left(c\right)\end{array}\right]\}\\ +\left(b-c\right)x^{\text{T}}\left(b\right)X_{1}x\left(b\right)+\left(c-a\right)x^{\text{T}}\left(c\right)X_{2}x\left(c\right)-\left(b-c\right)w_1^{\text{T}}{X}_1{w}_1-\left(c-a\right)w_2^{\text{T}}{X}_2{w}_2.\end{array}$

同时加上和减去 ${\left(x\left(c\right)-x\left(a\right)\right)}^{\text{T}}{Y}_3\left(x\left(c\right)-x\left(a\right)\right)+3{\left(2w_2-x\left(c\right)-x\left(a\right)\right)}^{\text{T}}{Y}_3\left(2w_2-x\left(c\right)-x\left(a\right)\right)$，运用互凸组合可以得到

$\begin{array}{c}\Phi \le -2{\left(b-c\right)}^2{v}_1^{\text{T}}{Y}_1{v}_1-4\left(b-c\right)v_1^{\text{T}}\left(Y_1+X_2\right)\left(x\left(b\right)-w_1\right)-2{\left(x\left(b\right)-w_1\right)}^{\text{T}}{Y}_3\left(x\left(b\right)-w_1\right)\\ -2{\left(c-a\right)}^2{v}_2^{\text{T}}{Y}_1{v}_2-4\left(c-a\right)v_2^{\text{T}}\left(Y_2+X_2\right)\left(x\left(c\right)-w_2\right)-2{\left(x\left(c\right)-w_2\right)}^{\text{T}}{Y}_3\left(x\left(c\right)-w_2\right)\\ -\left(c-a\right)\left(b-c\right)w_1^{\text{T}}{Y}_1{w}_1-2\left(c-a\right)w_1^{\text{T}}{Y}_2\left(x\left(b\right)-x\left(c\right)\right)-\frac{c-a}{b-a}{\left(x\left(b\right)-x\left(c\right)\right)}^{\text{T}}{Y}_3\left(x\left(b\right)-x\left(c\right)\right)\\ -3\left(c-a\right)\left(b-c\right){\left(w_1-2v_1\right)}^{\text{T}}{Y}_1\left(w_1-2v_1\right)-6\left(c-a\right){\left(w_1-2v_1\right)}^{\text{T}}{Y}_2\left(2w_1-x\left(b\right)-x\left(c\right)\right)\\ -\frac{3\left(c-a\right)}{b-c}{\left(2w_1-x\left(b\right)-x\left(c\right)\right)}^{\text{T}}{Y}_3\left(2w_1-x\left(b\right)-x\left(c\right)\right)+\left(b-c\right)x^{\text{T}}\left(b\right)X_{1}x\left(b\right)\\ +\left(c-a\right)x^{\text{T}}\left(c\right)X_{2}x\left(c\right)-\left(b-c\right)w_1^{\text{T}}{X}_1{w}_1-\left(c-a\right)w_2^{\text{T}}{X}_2{w}_2\end{array}$

因此，结论成立，证毕。

目前已有的文献中，处理带增广的二重积分不等式都是直接用Jenson不等式处理，带有很强的保守性。在本篇文章中，结合前人研究以及所学知识，通过引入两个零等式来估计积分的上限，引入自由矩阵让这个积分拥有更多的自由度。

为了便于后面的研究，需要对带有分母的积分项进行处理，因此在这个不等式中，引入 ${\left(x\left(c\right)-x\left(a\right)\right)}^{\text{T}}{Y}_3\left(x\left(c\right)-x\left(a\right)\right)+3{\left(2w_2-x\left(c\right)-x\left(a\right)\right)}^{\text{T}}{Y}_3\left(2w_2-x\left(c\right)-x\left(a\right)\right)$ 让它与后面这一项组合 $-\left(c-a\right)\left\{\frac{1}{b-a}{\left(x\left(b\right)-x\left(c\right)\right)}^{\text{T}}{Y}_3\left(x\left(b\right)-x\left(c\right)\right)+\frac{3}{b-c}{\left(2w_1-x\left(b\right)-x\left(c\right)\right)}^{\text{T}}{Y}_3\left(2w_1-x\left(b\right)-x\left(c\right)\right)\right\}$，然后巧妙地 运用互凸组合进行化简和放缩。通过这一系列的处理后，我们可以得到一个拥有多个松弛矩阵和大量状态向量交叉信息的具有增广的二重积分不等式。利用这个不等式对二重积分进行放缩，在降低保守性具有非常良好的效果。

3. 理论结果及证明

在这个部分当中，基于上面提到的二重积分不等式引理，我们通过恰当地构造李雅普诺夫泛函得到关于系统(1)的新的稳定标准。

为了简化向量和矩阵表示给出以下术语：

${\xi }_1\left(t\right)=\text{col}\left\{x\left(t\right),{\displaystyle {\int }_{t-h}^{t}x\left(s\right)\text{d}s}\right\}$， ${\xi }_2\left(t\right)=\text{col}\left\{x\left(t\right),\stackrel{\dot{}}{x}\left(t\right)\right\}$， ${\xi }_3\left(t\right)=\text{col}\left\{x\left(s\right),{\displaystyle {\int }_s^{t}x\left(u\right)\text{d}u}\right\}$，

${\xi }_4\left(t\right)=\text{col}\left\{x\left(t\right)-x\left(t-h\left(t\right)\right),x\left(t\right)+x\left(t-h\left(t\right)\right)-2w_1,x\left(t\right)-x\left(t-h\left(t\right)\right)+6w_1-12v_1\right\}$，

${\xi }_5\left(t\right)=\text{col}\left\{x\left(t-h\left(t\right)\right)-x\left(t-h\right),x\left(t-h\left(t\right)\right)+x\left(t-h\right)-2w_2,x\left(t-h\left(t\right)\right)-x\left(t-h\right)+6w_2-12v_2\right\}$，

$r_1=\frac{1}{h\left(t\right)}{\displaystyle {\int }_{t-h\left(t\right)}^{t}x\left(s\right)\text{d}s}$， $r_2=\frac{1}{t-h\left(t\right)}{\displaystyle {\int }_{t-h}^{t-h\left(t\right)}x\left(s\right)\text{d}s}$，

$t_1=\frac{1}{h{\left(t\right)}^2}{\displaystyle {\int }_{t-h\left(t\right)}^t{\displaystyle {\int }_{\theta }^{t}x\left(s\right)\text{d}s\text{d}\theta }}$， $t_2=\frac{1}{{\left(t-h\left(t\right)\right)}^2}{\displaystyle {\int }_{t-h}^{t-h\left(t\right)}{\displaystyle {\int }_{\theta }^{t-h\left(t\right)}x\left(s\right)\text{d}s\text{d}\theta }}$

定理1 对于给定的实数h和u，若存在对称正定矩阵 $U\in R^{2n\times 2n}$， $Q_i\in R^{2n\times 2n}\left(i=1,2,3,4\right)$， $R\in R^{n\times n}$， $P\in R^{2n\times 2n}$， $Z_i\in R^{n\times n}\left(i=1,2,3\right)$， $Z=\left[\begin{array}{cc}{Z}_1& Z_2\\ *& Z_3\end{array}\right]\in R^{2n\times 2n}$， $M_i\in R^{n\times n}\left(i=1,2\right)$，任意矩阵 $S\in R^{3n\times 3n}$， $L_i\in R^{9n\times 2n}\left(i=1,2\right)$， $M_j\in R^{9n\times 2n}\left(j=3,4\right)$，使得下列线性矩阵不等式成立：

${\Gamma }_1=\left[\begin{array}{cc}\bar{R}& S\\ S& \bar{R}\end{array}\right]>0$ (5)

${\Gamma }_2=\left[\begin{array}{cc}{Z}_1& Z_2+M_i\\ *& Z_3\end{array}\right]>0\left(i=1,2\right)$ (6)

${\Gamma }_3\left(h\left(t\right)=0\right)={\left({\delta }^{\perp }\right)}^{\text{T}}\left[\begin{array}{ccccc}\Xi \left(h\left(t\right)\right)& \sqrt{h\left(t\right)}{L}_1& \sqrt{h\left(t\right)}{M}_3& \sqrt{h-h\left(t\right)}{L}_2& \sqrt{h-h\left(t\right)}{M}_4\\ *& -P& 0& 0& 0\\ *& *& -\frac{1}{3}P& 0& 0\\ *& *& *& -P& 0\\ *& *& *& *& -\frac{1}{3}P\end{array}\right]\left({\delta }^{\perp }\right)<0$ (7)

${\Gamma }_4\left(h\left(t\right)=h\right)={\left({\delta }^{\perp }\right)}^{\text{T}}\left[\begin{array}{ccccc}\Xi \left(h\left(t\right)\right)& \sqrt{h\left(t\right)}{L}_1& \sqrt{h\left(t\right)}{M}_3& \sqrt{h-h\left(t\right)}{L}_2& \sqrt{h-h\left(t\right)}{M}_4\\ *& -P& 0& 0& 0\\ *& *& -\frac{1}{3}P& 0& 0\\ *& *& *& -P& 0\\ *& *& *& *& -\frac{1}{3}P\end{array}\right]\left({\delta }^{\perp }\right)<0$ (8)

${\Gamma }_5={\left({\delta }^{\perp }\right)}^{\text{T}}\left(-{\beta }^2{h}_2^2{J}_2+{\Gamma }_3\right)\left({\delta }^{\perp }\right)<0$ (9)

${\Gamma }_6={\left({\delta }^{\perp }\right)}^{\text{T}}\left(-{\left(1-\beta \right)}^2{h}_2^2{J}_2+{\Gamma }_4\right)\left({\delta }^{\perp }\right)<0$ (10)

则时变时滞系统(1)是渐近稳定的，其中

$\Xi \left(h\left(t\right)\right)={\displaystyle \underset{n=1}{\overset{6}{\sum }}{\Pi }_n}$， $\xi \left(t\right)=\text{col}\left\{x\left(t\right),x\left(t-h\left(t\right)\right),x\left(t-h\right),w_1,w_2,v_1,v_2,\stackrel{\dot{}}{x}\left(t-h\left(t\right)\right),\stackrel{\dot{}}{x}\left(t-h\right)\right\}$，

$\bar{R}=\text{diag}\left\{R,3R,5R\right\}$， $e_i=\left[\begin{array}{ccc}{0}_{n,\left(i-1\right)n}& I_n& 0_{n,\left(9-i\right)n}\end{array}\right]\left(i=1,2,\cdots ,9\right)$， $f_0=A{e}_1+B{e}_2$， $G_4=\text{col}\left\{e_2,e_8\right\}$，

$G_1=\text{col}\left\{e_1,h\left(t\right)e_4+\left(h-h\left(t\right)\right)e_5\right\}$， $G_2=\text{col}\left\{f_0,e_1-e_3\right\}$， $G_3=\text{col}\left\{e_1,f_0\right\}$， $G_{19}=\text{col}\left\{e_0,e_4\right\}$，

$G_5=\text{col}\left\{e_3,e_9\right\}$， $G_6=\text{col}\left\{e_1,e_0\right\}$， $G_7=\text{col}\left\{e_2,h^2\left(t\right)e_4\right\}$， $G_8=\text{col}\left\{e_3,h\left(t\right)e_4+\left(h-h\left(t\right)\right)e_5\right\}$，

$G_9=\text{col}\left\{e_0,e_1\right\}$， $G_{10}=\text{col}\left\{h\left(t\right)e_4,h\left(t\right)e_6\right\}$， $G_{18}=\text{col}\left\{-\left(h-h\left(t\right)\right)e_5+2\left(h-h\left(t\right)\right)e_7,e_2+e_3-2e_5\right\}$，

$G_{11}=\text{col}\left\{h\left(t\right)e_4+\left(h-h\left(t\right)\right)e_5,h^2\left(t\right)e_6+{\left(h-h\left(t\right)\right)}^2{e}_7+h\left(t\right)\left(h-h\left(t\right)\right)e_4\right\}$， $G_{20}=\text{col}\left\{e_0,e_4-e_5\right\}$，

$G_{12}=\text{col}\left\{e_1-e_2,e_1+e_2-2e_4,e_1-e_2+6e_4-12e_6\right\}$， $G_{17}=\text{col}\left\{\left(h-h\left(t\right)\right)e_5,e_2-e_3\right\}$， $e_0=0_{n\times 9n}$，

$G_{13}=\text{col}\left\{e_2-e_3,e_2+e_3-2e_5,e_2-e_3+6e_5-12e_7\right\}$， $G_{16}=\text{col}\left\{-h\left(t\right)e_4+2h\left(t\right)e_6,e_1+e_2-2e_4\right\}$，

$G_{14}=\text{col}\left\{e_1,e_2,e_3,e_4,e_5,e_6,e_7,e_8,e_9\right\}$， $G_{15}=\text{col}\left\{h\left(t\right)e_4,e_1-e_2\right\}$， $G_{21}=\text{col}\left\{e_0,e_6+e_7-e_4\right\}$，

${\Pi }_1=\text{sym}\left\{G_{1}U{G}_2\right\}$， ${\Pi }_2=G_3^{\text{T}}{Q}_1{G}_3-\left(1-u\right)G_4^{\text{T}}{Q}_1{G}_4+G_3^{\text{T}}{Q}_2{G}_3-G_5^{\text{T}}{Q}_2{G}_5$， ${\Pi }_6=\frac{h^2}{2}{G}_3^{\text{T}}Z{G}_3+{\Upsilon }_1$，

${\Pi }_3=G_6^{\text{T}}{Q}_3{G}_6-\left(1-u\right)G_7^{\text{T}}{Q}_3{G}_7+G_6^{\text{T}}{Q}_4{G}_6-G_8^{\text{T}}{Q}_4{G}_8+\text{sym}\left\{G_{10}^{\text{T}}{Q}_3{G}_9+G_{11}^{\text{T}}{Q}_4{G}_9\right\}$，

${\Pi }_4=h^2{f}_{0}R{f}_0-{\left[\begin{array}{c}{G}_{12}\\ G_{13}\end{array}\right]}^{\text{T}}{\Gamma }_1\left[\begin{array}{c}{G}_{12}\\ G_{13}\end{array}\right]$， ${\Pi }_5=h{G}_3^{\text{T}}P{G}_3+\text{sym}\left\{{\eta }_0^{\text{T}}{L}_1{\eta }_1+{\eta }_0^{\text{T}}{M}_3{\eta }_2+{\eta }_0^{\text{T}}{L}_2{\eta }_3+{\eta }_0^{\text{T}}{M}_4{\eta }_4\right\}$，

$\begin{array}{c}{\Upsilon }_1=\left(h-h\left(t\right)\right)e_2^{\text{T}}{M}_2{e}_2+h\left(t\right)e_1^{\text{T}}{M}_1{e}_1-h\left(t\right)e_4^{\text{T}}{M}_1{e}_4-\left(h-h\left(t\right)\right)e_5^{\text{T}}{M}_2{e}_5-h\left(t\right)\left(h-h\left(t\right)\right)e_4^{\text{T}}{Z}_1{e}_4\\ -\text{sym}\{h{\left(t\right)}^2{e}_6^{\text{T}}{Z}_1{e}_6+2h\left(t\right)e_6^{\text{T}}\left(Z_1+M_1\right)\left(e_1-e_4\right)+{\left(e_1-e_4\right)}^{\text{T}}{Z}_3\left(e_1-e_4\right)+{\left(h-h\left(t\right)\right)}^2{e}_7^{\text{T}}{Z}_1{e}_7\\ +2\left(h-h\left(t\right)\right)e_7^{\text{T}}\left(Z_2+M_2\right)\left(e_2-e_5\right)+{\left(e_2-e_5\right)}^{\text{T}}{Z}_3\left(e_2-e_5\right)+\left(c-a\right)e_4^{\text{T}}{Z}_2\left(e_1-e_2\right)\}\\ -\frac{h-h\left(t\right)}{h}{\left(e_1-e_2\right)}^{\text{T}}{Z}_3\left(e_1-e_2\right),\end{array}$

${\Upsilon }_2=\text{sym}\left\{h\left(t\right)G_{14}^{\text{T}}\left(\frac{3L_1{P}^{-1}{L}_1+M_3{P}^{-1}{M}_3}{3}\right)G_{14}+\left(h-h\left(t\right)\right)G_{14}^{\text{T}}\left(\frac{3L_2{P}^{-1}{L}_2+M_4{P}^{-1}{M}_4}{3}\right)G_{14}\right\}$ (11)

证明：构造如下李雅普诺夫泛函

$V\left(t\right)={\displaystyle \underset{d=1}{\overset{6}{\sum }}{V}_d}$

$V_1\left(t\right)={\xi }_1^{\text{T}}\left(t\right)U{\xi }_1(\; t\; )$

$V_2\left(t\right)={\displaystyle {\int }_{t-h\left(t\right)}^t{\xi }_2^{\text{T}}\left(s\right)Q_1}{\xi }_2\left(s\right)\text{d}s+{\displaystyle {\int }_{t-h}^t{\xi }_2^{\text{T}}\left(s\right)Q_2}{\xi }_2\left(s\right)\text{d}s$

$V_3={\displaystyle {\int }_{t-h\left(t\right)}^t{\xi }_3^{\text{T}}\left(t,s\right)Q_3}{\xi }_3\left(t,s\right)\text{d}s+{\displaystyle {\int }_{t-h}^t{\xi }_3^{\text{T}}\left(t,s\right)Q_4}{\xi }_3\left(t,s\right)\text{d}s$， $V_4=h{\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_{\theta }^t\stackrel{\dot{}}{x}\left(s\right)R\stackrel{\dot{}}{x}\left(s\right)\text{d}s\text{d}\theta }}$，

$V_5={\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_{\theta }^t{\xi }_2^{\text{T}}\left(s\right)P{\xi }_2\left(s\right)\text{d}s\text{d}\theta }}$， $V_6={\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_{{\theta }_2}^t{\displaystyle {\int }_{{\theta }_1}^t{\xi }_2^{\text{T}}\left(s\right)Z{\xi }_2\left(s\right)\text{d}s\text{d}{\theta }_1\text{d}{\theta }_2}}}$

对 $V\left(t\right)$ 求导，可得

${\stackrel{\dot{}}{V}}_1\left(t\right)=2{\xi }^{\text{T}}\left(t\right)G_1^{\text{T}}U{G}_2\xi (\; t\; )$

${\stackrel{\dot{}}{V}}_2\le {\xi }_2^{\text{T}}\left(t\right)Q_1{\xi }_2\left(t\right)-\left(1-u\right){\xi }_2^{\text{T}}\left(t-h\left(t\right)\right)Q_1{\xi }_2\left(t-h\left(t\right)\right)+{\xi }_2^{\text{T}}\left(t\right)Q_2{\xi }_2\left(t\right)-{\xi }_2^{\text{T}}\left(t-h\right)Q_2{\xi }_2\left(t-h\right)$

$\begin{array}{c}{\stackrel{\dot{}}{V}}_3\le {\xi }_3^{\text{T}}\left(t,t\right)Q_3{\xi }_3\left(t,t\right)-\left(1-u\right){\xi }_3^{\text{T}}\left(t,t-h\left(t\right)\right)Q_3{\xi }_3\left(t,t-h\left(t\right)\right)+{\xi }_3^{\text{T}}\left(t,t\right)Q_4{\xi }_3\left(t,t\right)-{\xi }_3^{\text{T}}\left(t,t-h\right)Q_4{\xi }_3\left(t,t-h\right)\\ +\text{sym}\left\{\left[\begin{array}{c}{\displaystyle {\int }_{t-h\left(t\right)}^{t}x\left(s\right)\text{d}s}\\ {\displaystyle {\int }_{t-h\left(t\right)}^t{\displaystyle {\int }_s^{t}x\left(u\right)\text{d}u\text{d}s}}\end{array}\right]Q_3\left[\begin{array}{c}0\\ x\left(t\right)\end{array}\right]+\left[\begin{array}{c}{\displaystyle {\int }_{t-h}^{t}x\left(s\right)\text{d}s}\\ {\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_s^{t}x\left(u\right)\text{d}u\text{d}s}}\end{array}\right]Q_4\left[\begin{array}{c}0\\ x\left(t\right)\end{array}\right]\right\},\end{array}$

${\stackrel{\dot{}}{V}}_4=h^2\stackrel{\dot{}}{x}\left(t\right)R\stackrel{\dot{}}{x}\left(t\right)-h{\displaystyle {\int }_{t-h}^t\stackrel{\dot{}}{x}\left(s\right)R\stackrel{\dot{}}{x}\left(s\right)\text{d}s}$， ${\stackrel{\dot{}}{V}}_5=h{\xi }_2^{\text{T}}\left(t\right)P{\xi }_2\left(t\right)-{\displaystyle {\int }_{t-h}^t{\xi }_2^{\text{T}}\left(s\right)P{\xi }_2\left(s\right)\text{d}s}$

${\stackrel{\dot{}}{V}}_6=\frac{h^2}{2}{\xi }_2^{\text{T}}\left(t\right)Z{\xi }_2\left(t\right)-{\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_{{\theta }_2}^t{\xi }_2^{\text{T}}\left(s\right)Z{\xi }_2\left(s\right)\text{d}s\text{d}{\theta }_2}}$

运用Jensen不等式以及互凸组合对 ${\stackrel{\dot{}}{V}}_4\left(t\right)$ 中的积分项进行放缩，可以得到

$\begin{array}{c}-h{\displaystyle {\int }_{t-h}^t\stackrel{\dot{}}{x}\left(s\right)R\stackrel{\dot{}}{x}\left(s\right)\text{d}s}=-h{\displaystyle {\int }_{t-h\left(t\right)}^t\stackrel{\dot{}}{x}\left(s\right)R\stackrel{\dot{}}{x}\left(s\right)\text{d}s}-h{\displaystyle {\int }_{t-h}^{t-h\left(t\right)}\stackrel{\dot{}}{x}\left(s\right)R\stackrel{\dot{}}{x}\left(s\right)\text{d}s}\\ \le -\frac{h}{h\left(t\right)}{\xi }_4^{\text{T}}\left(t\right)\bar{R}{\xi }_4\left(t\right)-\frac{h}{h-h\left(t\right)}{\xi }_5^{\text{T}}\left(t\right)\bar{R}{\xi }_5\left(t\right)\\ \le -{\left[\begin{array}{c}{\xi }_4\\ {\xi }_5\end{array}\right]}^{\text{T}}{\Gamma }_1\left[\begin{array}{c}{\xi }_4\\ {\xi }_5\end{array}\right].\end{array}$

为了考虑更多的积分信息，利用时滞分割方法将一重积分区间 $\left[0,h\right]$ 划分成 $\left[0,h\left(t\right)\right]$ 和 $\left[h\left(t\right),h\right]$ 两个部分，并运用引理2中的自由矩阵不等式， ${\stackrel{\dot{}}{V}}_5\left(t\right)$ 中的积分项进行放缩可以得到

$-{\displaystyle {\int }_{t-h}^t{\xi }_2^{\text{T}}\left(s\right)P{\xi }_2\left(s\right)\text{d}s}=-{\displaystyle {\int }_{t-h\left(t\right)}^t{\xi }_2^{\text{T}}\left(s\right)P{\xi }_2\left(s\right)\text{d}s}-{\displaystyle {\int }_{t-h}^{t-h\left(t\right)}{\xi }_2^{\text{T}}\left(s\right)P{\xi }_2\left(s\right)\text{d}s}$

$-{\displaystyle {\int }_{t-h\left(t\right)}^t{\xi }_2^{\text{T}}\left(s\right)P{\xi }_2\left(s\right)\text{d}s}\le \text{sym}\left\{{\eta }_0^{\text{T}}{L}_1{\eta }_1+{\eta }_0^{\text{T}}{M}_3{\eta }_2\right\}+h\left(t\right){\eta }_0^{\text{T}}\left(\frac{3L_1{P}^{-1}{L}_1+M_3{P}^{-1}{M}_3}{3}\right){\eta }_0$

其中

${\eta }_0=G_{14}\xi \left(t\right)$， ${\eta }_1=\left[\begin{array}{c}h\left(t\right)r_1\\ x\left(t\right)-x\left(t-h\left(t\right)\right)\end{array}\right]$， ${\eta }_2=\left[\begin{array}{c}-h\left(t\right)r_1+2h\left(t\right)t_1\\ x\left(t\right)+x\left(t-h\left(t\right)\right)-2r_1\end{array}\right]$

${\eta }_3=\left[\begin{array}{c}\left(h-h\left(t\right)\right)r_2\\ x\left(t-h\left(t\right)\right)-x\left(t-h\right)\end{array}\right]$， ${\eta }_4=\left[\begin{array}{c}-\left(h-h\left(t\right)\right)r_2+2\left(h-h\left(t\right)\right)t_2\\ x\left(t-h\left(t\right)\right)+x\left(t-h\right)-2r_2\end{array}\right]$

运用引理4提出的新二重积分不等式，对 ${\stackrel{\dot{}}{V}}_6\left(t\right)$ 中的积分放缩得

$-{\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_{{\theta }_2}^t{\xi }_2^{\text{T}}\left(s\right)Z{\xi }_2\left(s\right)\text{d}s\text{d}{\theta }_2}}\le {\xi }^{\text{T}}\left(s\right){\Upsilon }_1\xi (\; s\; )$

取 $\Psi \left(h\left(t\right)\right)=\Xi \left(h\left(t\right)\right)+{\Upsilon }_2$，由于 $h\left(t\right)$ 在是二次的，因此可以改写成 $\Psi \left(h\left(t\right)\right)=J_2{h}^2\left(t\right)+J_{1}h\left(t\right)+J_0$，其中 $J_0$， $J_1$， $J_2$ 和都是常数矩阵，且

$J_2=e_4^{\text{T}}{Z}_1{e}_4-\left(1-u\right)G_{19}^{\text{T}}{Q}_3{G}_{19}-G_{20}^{\text{T}}{Q}_3{G}_{20}-\text{sym}\left\{e_6^{\text{T}}{Z}_1{e}_6+e_7^{\text{T}}{Z}_1{e}_7+e_{21}^{\text{T}}{Q}_3{e}_9\right\}$

利用Finsler’s引理 [17]，让 $\delta \xi \left(t\right)=0$，如果(5)~(10)的条件满足，我们可以得到 ${\left({\delta }^{\perp }\right)}^{\text{T}}\Psi \left(h\left(t\right)\right)\left({\delta }^{\perp }\right)<0$ 成立，这就暗示 ${\xi }^{\text{T}}\left(t\right)\Psi \left(h\left(t\right)\right)\xi \left(t\right)<0$ 成立。因此，如果(7)~(10)的条件满足，那么通过引理1，可以得到 $\Psi \left(h\left(t\right)\right)<0$ 成立。从而定理1的结论成立，证毕。

需要说明的是， $V_3\left(t\right)$ 是在增广向量中引入S相关的单重积分项 ${\displaystyle {\int }_s^{t}x\left(u\right)\text{d}u}$，对 $V_3\left(t\right)$ 求导后能产生二重积分 ${\displaystyle {\int }_{t-h\left(t\right)}^t{\displaystyle {\int }_s^{t}x\left(u\right)\text{d}u}}\text{d}s$ 和 ${\displaystyle {\int }_{t-h}^t{\displaystyle {\int }_s^{t}x\left(u\right)\text{d}u}}\text{d}s$，可以看到更多的时滞信息和相互交叉的状态向量，这些都可以为降低保守性提供帮助。

$V_6\left(t\right)$ 求导后产生的二重积分就是利用引理4的积分不等式进行放缩的，值得注意的是利用互凸组合巧妙的化解分母为零的现象，并且可以看到利用新的积分不等式能产生更多的时滞信息和自由矩阵。

4. 数值实例

我们通过了下面的例子证明本文所提方法的有效性和优越性。

$A=\left[\begin{array}{cc}0& 1\\ -1& -2\end{array}\right]$， $B=\left[\begin{array}{cc}0& 0\\ -1& 1\end{array}\right]$

Table 1. Upper bound of the maximum allowable delay of h when u takes different values

表1. 当u取不同值时h最大允许的时滞上界

Figure 1. State response diagram

图1. 状态响应图

这个例子主要用来验证时滞系统的稳定性和计算最大允许时滞上界h，通过图1我们可以看到，当取 $\beta =0.5$， $u=0.2$ 时，令 $h=5.4514$，取 $h\left(t\right)=10.9028\sin \left(t/2\right)$，我们可以清楚看到系统是渐近稳定的，这证明我们所提方法的有效性。运用定理1通过表1中的数据对比可以的出我们所提方法的确具有优越性，能有效降低时滞系统的保守性，且可以通过控制调节参数 $\beta$ 找到最优解。当 $\beta =0.7$， $u=0.2$ 时，运用定理1可以得到最大允许时滞上界h的值是6.1771，它比文献 [7] [10] [11] [12] [14] 以及 [15] 分别大68.18%，39.96%，38.31%，36.73%，36.43%，33.18%，11.7%。

5. 结语

本文探讨了时滞系统的稳定性问题，首先通过两个零等式引入自由矩阵改进具有增广的二重积分不等式，然后在充分考虑各个状态向量的基础上构造带有增广的李雅普诺夫泛函，并利用提出的二重积分不等式以及广义的自由矩阵积分不等式处理泛函的导数。最后，通过一个数值例子来说明所得结论的有效性和优越性。本文的方法还可以扩展到具有马尔科夫跳跃的时滞神经网络系统的稳定性以及控制问题的研究中。

基金项目

国家自然科学基金(12061088)。