本人为研二小白,在看论文的过程中记录一下自己的学习过程和想法。

在之前转载的Lyapunov-Krasovskii泛函二重积分项求导_原理文章的基础上,这里给出Lyapunov-Krasovskii泛函三重积分项求导简单的计算过程及不等式放缩引理,主要是在文章的基础上进行一个总结。

1 Lyapunov-Krasovskii泛函三重积分项举例

这里给出研究时滞系统、网络化控制系统时常出现的Lyapunov-Krasovskii泛函三重积分项,主要参考的是以下三篇文章。

  1. P. Park, W. Lee, and S. Lee, “Auxiliary function-based integral inequalities for quadratic functions and their applications to time-delay systems,” Journal of the Franklin Institute, vol. 352, no. 4, pp. 1378-1396, Apr. 2015.
  2. J, Zhang and Y. Ma, “Event-triggered dissipative double asynchronous controller for interval type-2 fuzzy semi-Markov jump systems with state quantization and actuator failure,” ISA Transactions, vol. 138, pp. 226-242, Jul. 2023.
  3. D. Zhang, Z. Ye, G. Feng and H. Li, “Intelligent event-based fuzzy dynamic positioning control of nonlinear unmanned marine vehicles under DoS attack,” IEEE Transactions on Cybernetics, vol. 52, no. 12, pp. 13486-13499, Dec. 2022.

定义时变时延 h ( t ) h(t) h(t)满足 0 ≤ h 1 ≤ h ( t ) ≤ h 2 0\leq h_1\leq h(t)\leq h_2 0h1h(t)h2 h 12 = h 2 − h 1 h_{12}=h_2-h_1 h12=h2h1。给出如下几种常见的Lyapunov-Krasovskii泛函三重积分项。
V 1 ( t ) = ∫ − h 1 0 ∫ − h 1 γ ∫ t + β t x ˙ T ( α ) Z 1 x ˙ ( α ) d α d β d γ V 2 ( t ) = ∫ − h 2 − h 1 ∫ γ − h 1 ∫ t + β t x ˙ T ( α ) Z 2 x ˙ ( α ) d α d β d γ V 3 ( t ) = ∫ − h 2 − h 1 ∫ − h 2 γ ∫ t + β t x ˙ T ( α ) Z 3 x ˙ ( α ) d α d β d γ (1) \begin{aligned}V_1(t)&=\int_{-h_1}^{0}\int_{-h_1}^{\gamma}\int_{t+\beta}^{t}\dot{x}^T(\alpha)Z_1\dot{x}(\alpha)d\alpha d\beta d\gamma\\ V_2(t)&=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}\int_{t+\beta}^{t}\dot{x}^T(\alpha)Z_2\dot{x}(\alpha)d\alpha d\beta d\gamma\\ V_3(t)&=\int_{-h_2}^{-h_1}\int_{-h_2}^{\gamma}\int_{t+\beta}^{t}\dot{x}^T(\alpha)Z_3\dot{x}(\alpha)d\alpha d\beta d\gamma\end{aligned}\tag{1} V1(t)V2(t)V3(t)=h10h1γt+βtx˙T(α)Z1x˙(α)dαdβdγ=h2h1γh1t+βtx˙T(α)Z2x˙(α)dαdβdγ=h2h1h2γt+βtx˙T(α)Z3x˙(α)dαdβdγ(1)

定义时变时延 τ ( t ) \tau(t) τ(t)满足 0 ≤ τ 1 ≤ τ ( t ) ≤ τ 2 0\leq \tau_1\leq \tau(t)\leq \tau_2 0τ1τ(t)τ2 τ 12 = τ 2 − τ 1 \tau_{12}=\tau_2-\tau_1 τ12=τ2τ1。给出如下几种常见的Lyapunov-Krasovskii泛函三重积分项。
V 1 ( t ) = ∫ − τ 2 − τ 1 ∫ β − τ 1 ∫ t + u t e 2 α ( s − u ) x ˙ T ( s ) R 1 x ˙ ( s ) d s d u d β V 2 ( t ) = ∫ − τ 2 − τ 1 ∫ − τ 2 β ∫ t + u t e 2 α ( s − u ) x ˙ T ( s ) R 2 x ˙ ( s ) d s d u d β (2) \begin{aligned}V_1(t)&=\int_{-\tau_2}^{-\tau_1}\int_{\beta}^{-\tau_1}\int_{t+u}^{t}e^{2\alpha(s-u)}\dot{x}^T(s)R_1\dot{x}(s)ds dud\beta\\ V_2(t)&=\int_{-\tau_2}^{-\tau_1}\int_{-\tau_2}^{\beta}\int_{t+u}^{t}e^{2\alpha(s-u)}\dot{x}^T(s)R_2\dot{x}(s)ds dud\beta\end{aligned}\tag{2} V1(t)V2(t)=τ2τ1βτ1t+ute2α(su)x˙T(s)R1x˙(s)dsdudβ=τ2τ1τ2βt+ute2α(su)x˙T(s)R2x˙(s)dsdudβ(2)

定义时变时延 d ( t ) d(t) d(t)满足 0 ≤ d m ≤ d ( t ) ≤ d M 0\leq d_m\leq d(t)\leq d_M 0dmd(t)dM d 12 = d M − d m d_{12}=d_M-d_m d12=dMdm。给出Lyapunov-Krasovskii泛函三重积分项。
V 1 ( t ) = ∫ − d M − d m ∫ v − d m ∫ t + θ t e 2 α ( t − u ) x ˙ T ( u ) Q 1 x ˙ ( u ) d u d θ d v (3) V_1(t)=\int_{-d_M}^{-d_m}\int_{v}^{-d_m}\int_{t+\theta}^{t}e^{2\alpha(t-u)}\dot{x}^T(u)Q_1\dot{x}(u)du d\theta dv\tag{3} V1(t)=dMdmvdmt+θte2α(tu)x˙T(u)Q1x˙(u)dudθdv(3)

2 Lyapunov-Krasovskii泛函三重积分项求导

对于Lyapunov-Krasovskii泛函二重积分项,参考之前的文章Lyapunov-Krasovskii泛函二重积分项求导_原理,可以得到如下的计算过程和结果。
V ( t ) = ∫ − h 1 0 ∫ t + β t x ˙ T ( α ) M 1 x ˙ ( α ) d α d β d V ( t ) d t = ∫ − h 1 0 [ x ˙ T ( t ) M 1 x ˙ ( t ) − x ˙ T ( t + β ) M 1 x ˙ ( t + β ) ] d β = ∫ − h 1 0 x ˙ T ( t ) M 1 x ˙ ( t ) d β − ∫ − h 1 0 x ˙ T ( t + β ) M 1 x ˙ ( t + β ) d β = x ˙ T ( t ) M 1 x ˙ ( t ) − ∫ t − h 1 t x ˙ T ( β ) M 1 x ˙ ( β ) d β (5) \begin{aligned}V(t)&=\int_{-h_1}^{0}\int_{t+\beta}^{t}\dot{x}^T(\alpha)M_1\dot{x}(\alpha)d\alpha d\beta\\ \frac{dV(t)}{dt}&=\int_{-h_1}^{0}[\dot{x}^T(t)M_1\dot{x}(t)-\dot{x}^T(t+\beta)M_1\dot{x}(t+\beta)]d\beta\\ &=\int_{-h_1}^{0}\dot{x}^T(t)M_1\dot{x}(t)d\beta-\int_{-h_1}^{0}\dot{x}^T(t+\beta)M_1\dot{x}(t+\beta)d\beta\\ &=\dot{x}^T(t)M_1\dot{x}(t)-\int_{t-h_1}^{t}\dot{x}^T(\beta)M_1\dot{x}(\beta)d\beta\end{aligned}\tag{5} V(t)dtdV(t)=h10t+βtx˙T(α)M1x˙(α)dαdβ=h10[x˙T(t)M1x˙(t)x˙T(t+β)M1x˙(t+β)]dβ=h10x˙T(t)M1x˙(t)dβh10x˙T(t+β)M1x˙(t+β)dβ=x˙T(t)M1x˙(t)th1tx˙T(β)M1x˙(β)dβ(5)

同样的原理,对于Lyapunov-Krasovskii泛函三重积分项,我们可以看成是二重积分项和一重积分项的嵌套,那么,我们可以按照以下的步骤进行Lyapunov-Krasovskii泛函三重积分项求导的计算。
V ( t ) = ∫ − h 2 − h 1 ∫ γ − h 1 ∫ t + β t x ˙ T ( α ) M 2 x ˙ ( α ) d α d β d γ d V ( t ) d t = ∫ − h 2 − h 1 ∫ γ − h 1 [ x ˙ T ( t ) M 2 x ˙ ( t ) − x ˙ T ( t + β ) M 2 x ˙ ( t + β ) ] d β d γ = ∫ − h 2 − h 1 ∫ γ − h 1 x ˙ T ( t ) M 2 x ˙ ( t ) d β d γ − ∫ − h 2 − h 1 ∫ t + γ t − h 1 x ˙ T ( β ) M 2 x ˙ ( β ) d β d γ (6) \begin{aligned}V(t)&=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}\int_{t+\beta}^{t}\dot{x}^T(\alpha)M_2\dot{x}(\alpha)d\alpha d\beta d\gamma\\ \frac{dV(t)}{dt}&=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}[\dot{x}^T(t)M_2\dot{x}(t)-\dot{x}^T(t+\beta)M_2\dot{x}(t+\beta)]d\beta d\gamma\\ &=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}\dot{x}^T(t)M_2\dot{x}(t)d\beta d\gamma-\int_{-h_2}^{-h_1}\int_{t+\gamma}^{t-h_1}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta d\gamma\end{aligned}\tag{6} V(t)dtdV(t)=h2h1γh1t+βtx˙T(α)M2x˙(α)dαdβdγ=h2h1γh1[x˙T(t)M2x˙(t)x˙T(t+β)M2x˙(t+β)]dβdγ=h2h1γh1x˙T(t)M2x˙(t)dβdγh2h1t+γth1x˙T(β)M2x˙(β)dβdγ(6)

到这一步,我们看到,得到的结果和Lyapunov-Krasovskii泛函二重积分项求导的结果很类似,对于第一项,由于被积函数 x ˙ T ( t ) M 2 x ˙ ( t ) \dot{x}^T(t)M_2\dot{x}(t) x˙T(t)M2x˙(t) 不含有积分变量 β \beta β γ \gamma γ,因此,我们可以将此项进一步处理,得到如下结果。
∫ − h 2 − h 1 ∫ γ − h 1 x ˙ T ( t ) M 2 x ˙ ( t ) d β d γ = ∫ − h 2 − h 1 ∫ γ − h 1 x ˙ T ( t ) M 2 x ˙ ( t ) d γ = ∫ − h 2 − h 1 ( − h 1 − γ ) x ˙ T ( t ) M 2 x ˙ ( t ) d γ = h 12 2 2 x ˙ T ( t ) M 2 x ˙ ( t ) (7) \begin{aligned}\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}\dot{x}^T(t)M_2\dot{x}(t)d\beta d\gamma&=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}\dot{x}^T(t)M_2\dot{x}(t)d\gamma\\ &=\int_{-h_2}^{-h_1}(-h_1-\gamma)\dot{x}^T(t)M_2\dot{x}(t)d\gamma\\ &=\frac{h^2_{12}}{2}\dot{x}^T(t)M_2\dot{x}(t)\end{aligned}\tag{7} h2h1γh1x˙T(t)M2x˙(t)dβdγ=h2h1γh1x˙T(t)M2x˙(t)dγ=h2h1(h1γ)x˙T(t)M2x˙(t)dγ=2h122x˙T(t)M2x˙(t)(7)

因此,结合公式 (6) 和 (7),可得到如下结果。
d V ( t ) d t = ∫ − h 2 − h 1 ∫ γ − h 1 [ x ˙ T ( t ) M 2 x ˙ ( t ) − x ˙ T ( t + β ) M 2 x ˙ ( t + β ) ] d β d γ = ∫ − h 2 − h 1 ∫ γ − h 1 x ˙ T ( t ) M 2 x ˙ ( t ) d β d γ − ∫ − h 2 − h 1 ∫ t + γ t − h 1 x ˙ T ( β ) M 2 x ˙ ( β ) d β d γ = h 12 2 2 x ˙ T ( t ) M 2 x ˙ ( t ) − ∫ − h 2 − h 1 ∫ t + γ t − h 1 x ˙ T ( β ) M 2 x ˙ ( β ) d β d γ = h 12 2 2 x ˙ T ( t ) M 2 x ˙ ( t ) − ∫ − h ( t ) − h 1 ∫ t + γ t − h 1 x ˙ T ( β ) M 2 x ˙ ( β ) d β d γ − ∫ − h 2 − h ( t ) ∫ t + γ t − h ( t ) x ˙ T ( β ) M 2 x ˙ ( β ) d β d γ − ( h 2 − h ( t ) ) ∫ t − h ( t ) t − h 1 x ˙ T ( β ) M 2 x ˙ ( β ) d β (8) \begin{aligned}\frac{dV(t)}{dt}&=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}[\dot{x}^T(t)M_2\dot{x}(t)-\dot{x}^T(t+\beta)M_2\dot{x}(t+\beta)]d\beta d\gamma\\ &=\int_{-h_2}^{-h_1}\int_{\gamma}^{-h_1}\dot{x}^T(t)M_2\dot{x}(t)d\beta d\gamma-\int_{-h_2}^{-h_1}\int_{t+\gamma}^{t-h_1}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta d\gamma\\ &=\frac{h^2_{12}}{2}\dot{x}^T(t)M_2\dot{x}(t)-\int_{-h_2}^{-h_1}\int_{t+\gamma}^{t-h_1}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta d\gamma\\ &=\frac{h^2_{12}}{2}\dot{x}^T(t)M_2\dot{x}(t)-\int_{-h(t)}^{-h_1}\int_{t+\gamma}^{t-h_1}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta d\gamma\\ &-\int_{-h_2}^{-h(t)}\int_{t+\gamma}^{t-h(t)}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta d\gamma-(h_2-h(t))\int_{t-h(t)}^{t-h_1}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta\end{aligned}\tag{8} dtdV(t)=h2h1γh1[x˙T(t)M2x˙(t)x˙T(t+β)M2x˙(t+β)]dβdγ=h2h1γh1x˙T(t)M2x˙(t)dβdγh2h1t+γth1x˙T(β)M2x˙(β)dβdγ=2h122x˙T(t)M2x˙(t)h2h1t+γth1x˙T(β)M2x˙(β)dβdγ=2h122x˙T(t)M2x˙(t)h(t)h1t+γth1x˙T(β)M2x˙(β)dβdγh2h(t)t+γth(t)x˙T(β)M2x˙(β)dβdγ(h2h(t))th(t)th1x˙T(β)M2x˙(β)dβ(8)

注意到,最后结果对项 ∫ − h 2 − h 1 ∫ t + γ t − h 1 x ˙ T ( β ) M 2 x ˙ ( β ) d β d γ \int_{-h_2}^{-h_1}\int_{t+\gamma}^{t-h_1}\dot{x}^T(\beta)M_2\dot{x}(\beta)d\beta d\gamma h2h1t+γth1x˙T(β)M2x˙(β)dβdγ 拆分成了三项,这是为了让更多的时滞信息得到利用,降低设计的保守性。对于结果中的一重积分项和二重积分项的处理,即通过一些不等式关系来放缩,将会在下面两节来介绍。

对于带有 e e e 指数的Lyapunov-Krasovskii泛函三重积分项,可以看出是常规的Lyapunov-Krasovskii泛函三重积分和 e e e 指数乘在一起构成的,直接运用乘积求导法则即可得到求导结果,这里不再给出详细的推导过程。

3 基于辅助函数的一重积分不等式

参考文章 “Auxiliary function-based integral inequalities for quadratic functions and their applications to time-delay systems”,可以得到比常规的 Jensen 不等式更紧的放缩结果,引理如下:

引理1. 对于可积函数 { x ( s ) ∣ s ∈ [ a , b ] } \{x(s)|s\in[a,b]\} {x(s)s[a,b]} 和 任意的正定矩阵 R > 0 R>0 R>0,下列不等式成立:
∫ a b x T ( s ) R x ( s ) d s ≥ 1 b − a ( ∫ a b x ( s ) d s ) T R ( ∫ a b x ( s ) d s ) + 3 b − a Ω 1 T R Ω 1 ∫ a b x T ( s ) R x ( s ) d s ≥ 1 b − a ( ∫ a b x ( s ) d s ) T R ( ∫ a b x ( s ) d s ) + 5 b − a Ω 2 T R Ω 2 ∫ a b x T ( s ) R x ( s ) d s ≥ 1 b − a ( ∫ a b x ( s ) d s ) T R ( ∫ a b x ( s ) d s ) + 3 b − a Ω 1 T R Ω 1 + 5 b − a Ω 2 T R Ω 2 \begin{aligned}\int_{a}^{b}x^{T}(s)Rx(s)ds&\geq \frac{1}{b-a}\bigg(\int_{a}^{b}x(s)ds\bigg)^{T}R\bigg(\int_{a}^{b}x(s)ds\bigg)+\frac{3}{b-a}\Omega^{T}_{1}R\Omega_{1}\\ \int_{a}^{b}x^{T}(s)Rx(s)ds&\geq \frac{1}{b-a}\bigg(\int_{a}^{b}x(s)ds\bigg)^{T}R\bigg(\int_{a}^{b}x(s)ds\bigg)+\frac{5}{b-a}\Omega^{T}_{2}R\Omega_{2}\\ \int_{a}^{b}x^{T}(s)Rx(s)ds&\geq \frac{1}{b-a}\bigg(\int_{a}^{b}x(s)ds\bigg)^{T}R\bigg(\int_{a}^{b}x(s)ds\bigg)+\frac{3}{b-a}\Omega^{T}_{1}R\Omega_{1}+\frac{5}{b-a}\Omega^{T}_{2}R\Omega_{2} \end{aligned} abxT(s)Rx(s)dsabxT(s)Rx(s)dsabxT(s)Rx(s)dsba1(abx(s)ds)TR(abx(s)ds)+ba3Ω1TRΩ1ba1(abx(s)ds)TR(abx(s)ds)+ba5Ω2TRΩ2ba1(abx(s)ds)TR(abx(s)ds)+ba3Ω1TRΩ1+ba5Ω2TRΩ2

其中
Ω 1 = ∫ a b x ( s ) d s − 2 ( b − a ) ∫ a b ∫ u b x ( s ) d s d u Ω 2 = ∫ a b x ( s ) d s − 6 ( b − a ) ∫ a b ∫ u b x ( s ) d s d u + 12 ( b − a ) 2 ∫ a b ∫ u b ∫ β b x ( s ) d s d u d β \begin{aligned}\Omega_{1}&=\int_{a}^{b}x(s)ds-\frac{2}{(b-a)}\int_{a}^{b}\int_{u}^{b}x(s)dsdu\\ \Omega_{2}&=\int_{a}^{b}x(s)ds-\frac{6}{(b-a)}\int_{a}^{b}\int_{u}^{b}x(s)dsdu +\frac{12}{(b-a)^2}\int_{a}^{b}\int_{u}^{b}\int_{\beta}^{b}x(s)dsdud\beta\end{aligned} Ω1Ω2=abx(s)ds(ba)2abubx(s)dsdu=abx(s)ds(ba)6abubx(s)dsdu+(ba)212abubβbx(s)dsdudβ

引理2. 对于可积函数 { x ( s ) ∣ s ∈ [ a , b ] } \{x(s)|s\in[a,b]\} {x(s)s[a,b]} 和 任意的正定矩阵 R > 0 R>0 R>0,下列不等式成立:
∫ a b x ˙ T ( s ) R x ˙ ( s ) d s ≥ 1 b − a Ω 3 T R Ω 3 + 3 b − a Ω 4 T R Ω 4 ∫ a b x ˙ T ( s ) R x ˙ ( s ) d s ≥ 1 b − a Ω 3 T R Ω 3 + 3 b − a Ω 4 T R Ω 4 + 5 b − a Ω 5 T R Ω 5 \begin{aligned}\int_{a}^{b}\dot{x}^{T}(s)R\dot{x}(s)ds&\geq\frac{1}{b-a}\Omega^{T}_{3}R\Omega_{3}+\frac{3}{b-a}\Omega^{T}_{4}R\Omega_{4}\\ \int_{a}^{b}\dot{x}^{T}(s)R\dot{x}(s)ds&\geq\frac{1}{b-a}\Omega^{T}_{3}R\Omega_{3}+\frac{3}{b-a}\Omega^{T}_{4}R\Omega_{4}+\frac{5}{b-a}\Omega^{T}_{5}R\Omega_{5} \end{aligned} abx˙T(s)Rx˙(s)dsabx˙T(s)Rx˙(s)dsba1Ω3TRΩ3+ba3Ω4TRΩ4ba1Ω3TRΩ3+ba3Ω4TRΩ4+ba5Ω5TRΩ5

其中
Ω 3 = x ( b ) − x ( a ) Ω 4 = x ( b ) − x ( a ) − 2 ( b − a ) ∫ a b x ( s ) d s Ω 5 = x ( b ) − x ( a ) + 6 ( b − a ) ∫ a b x ( s ) d s − 12 ( b − a ) 2 ∫ a b ∫ u b x ( s ) d s d u \begin{aligned}\Omega_{3}&=x(b)-x(a)\\ \Omega_{4}&=x(b)-x(a)-\frac{2}{(b-a)}\int_{a}^{b}x(s)ds\\ \Omega_{5}&=x(b)-x(a)+\frac{6}{(b-a)}\int_{a}^{b}x(s)ds-\frac{12}{(b-a)^2}\int_{a}^{b}\int_{u}^{b}x(s)dsdu\end{aligned} Ω3Ω4Ω5=x(b)x(a)=x(b)x(a)(ba)2abx(s)ds=x(b)x(a)+(ba)6abx(s)ds(ba)212abubx(s)dsdu

4 基于辅助函数的二重积分不等式

参考文章 “Auxiliary function-based integral inequalities for quadratic functions and their applications to time-delay systems”,给出基于辅助函数的二重积分不等式放缩结果,引理如下:

引理2. 对于可积函数 { x ( s ) ∣ s ∈ [ a , b ] } \{x(s)|s\in[a,b]\} {x(s)s[a,b]} 和 任意的正定矩阵 R > 0 R>0 R>0,下列不等式成立:
∫ a b ∫ u b x ˙ T ( s ) R x ˙ ( s ) d s d u ≥ 2 Ω 6 T R Ω 6 + 4 Ω 7 T R Ω 7 ∫ a b ∫ u b x ˙ T ( s ) R x ˙ ( s ) d s d u ≥ 2 Ω 8 T R Ω 8 + 4 Ω 9 T R Ω 9 \begin{aligned}\int_{a}^{b}\int_{u}^{b}\dot{x}^{T}(s)R\dot{x}(s)dsdu&\geq2\Omega^{T}_{6}R\Omega_{6}+4\Omega^{T}_{7}R\Omega_{7}\\ \int_{a}^{b}\int_{u}^{b}\dot{x}^{T}(s)R\dot{x}(s)dsdu&\geq2\Omega^{T}_{8}R\Omega_{8}+4\Omega^{T}_{9}R\Omega_{9} \end{aligned} abubx˙T(s)Rx˙(s)dsduabubx˙T(s)Rx˙(s)dsdu2Ω6TRΩ6+4Ω7TRΩ72Ω8TRΩ8+4Ω9TRΩ9

其中
Ω 6 = x ( b ) − 1 ( b − a ) ∫ a b x ( s ) d s Ω 7 = x ( b ) + 2 ( b − a ) ∫ a b x ( s ) d s − 6 ( b − a ) 2 ∫ a b ∫ u b x ( s ) d s d u Ω 8 = x ( a ) − 1 ( b − a ) ∫ a b x ( s ) d s Ω 9 = x ( a ) − 4 ( b − a ) ∫ a b x ( s ) d s + 6 ( b − a ) 2 ∫ a b ∫ u b x ( s ) d s d u \begin{aligned}\Omega_{6}&=x(b)-\frac{1}{(b-a)}\int_{a}^{b}x(s)ds\\ \Omega_{7}&=x(b)+\frac{2}{(b-a)}\int_{a}^{b}x(s)ds-\frac{6}{(b-a)^2}\int_{a}^{b}\int_{u}^{b}x(s)dsdu\\ \Omega_{8}&=x(a)-\frac{1}{(b-a)}\int_{a}^{b}x(s)ds\\ \Omega_{9}&=x(a)-\frac{4}{(b-a)}\int_{a}^{b}x(s)ds+\frac{6}{(b-a)^2}\int_{a}^{b}\int_{u}^{b}x(s)dsdu\end{aligned} Ω6Ω7Ω8Ω9=x(b)(ba)1abx(s)ds=x(b)+(ba)2abx(s)ds(ba)26abubx(s)dsdu=x(a)(ba)1abx(s)ds=x(a)(ba)4abx(s)ds+(ba)26abubx(s)dsdu

注:3、4节中得到的结果比常规的Jenson不等式更紧,可以在一定程度上利用更多的系统时滞信息,降低设计结果的保守性。这里只是简单列举出了放缩结果,若对放缩结果有所疑问,在文章“Auxiliary function-based integral inequalities for quadratic functions and their applications to time-delay systems”中有非常详细的推导和证明过程,这里不再详细介绍。


创作不易,希望大家支持,多多点赞收藏!!!!非常感谢!!!!

参考文献:

Auxiliary function-based integral inequalities for quadratic functions and their applications to time-delay systems.
Event-triggered dissipative double asynchronous controller for interval type-2 fuzzy semi-Markov jump systems with state quantization and actuator failure.
Intelligent event-based fuzzy dynamic positioning control of nonlinear unmanned marine vehicles under DoS attack.

点赞(0) 打赏

评论列表 共有 0 条评论

暂无评论

微信公众账号

微信扫一扫加关注

发表
评论
返回
顶部