Lyapunov-Krasovskii泛函三重积分项求导_原理
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。
- 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.
- 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.
- 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.
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h_{12}=h_2-h_1
h12=h2−h1。给出如下几种常见的Lyapunov-Krasovskii泛函三重积分项。
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\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)=∫−h10∫−h1γ∫t+βtx˙T(α)Z1x˙(α)dαdβdγ=∫−h2−h1∫γ−h1∫t+βtx˙T(α)Z2x˙(α)dαdβdγ=∫−h2−h1∫−h2γ∫t+βtx˙T(α)Z3x˙(α)dαdβdγ(1)
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τ12=τ2−τ1。给出如下几种常见的Lyapunov-Krasovskii泛函三重积分项。
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\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)=∫−τ2−τ1∫β−τ1∫t+ute2α(s−u)x˙T(s)R1x˙(s)dsdudβ=∫−τ2−τ1∫−τ2β∫t+ute2α(s−u)x˙T(s)R2x˙(s)dsdudβ(2)
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d12=dM−dm。给出Lyapunov-Krasovskii泛函三重积分项。
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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)=∫−dM−dm∫v−dm∫t+θte2α(t−u)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。
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\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)=∫−h10∫t+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)dβ−∫−h10x˙T(t+β)M1x˙(t+β)dβ=x˙T(t)M1x˙(t)−∫t−h1tx˙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。
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\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)=∫−h2−h1∫γ−h1∫t+βtx˙T(α)M2x˙(α)dαdβdγ=∫−h2−h1∫γ−h1[x˙T(t)M2x˙(t)−x˙T(t+β)M2x˙(t+β)]dβdγ=∫−h2−h1∫γ−h1x˙T(t)M2x˙(t)dβdγ−∫−h2−h1∫t+γt−h1x˙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
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\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}
∫−h2−h1∫γ−h1x˙T(t)M2x˙(t)dβdγ=∫−h2−h1∫γ−h1x˙T(t)M2x˙(t)dγ=∫−h2−h1(−h1−γ)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。
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\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)=∫−h2−h1∫γ−h1[x˙T(t)M2x˙(t)−x˙T(t+β)M2x˙(t+β)]dβdγ=∫−h2−h1∫γ−h1x˙T(t)M2x˙(t)dβdγ−∫−h2−h1∫t+γt−h1x˙T(β)M2x˙(β)dβdγ=2h122x˙T(t)M2x˙(t)−∫−h2−h1∫t+γt−h1x˙T(β)M2x˙(β)dβdγ=2h122x˙T(t)M2x˙(t)−∫−h(t)−h1∫t+γt−h1x˙T(β)M2x˙(β)dβdγ−∫−h2−h(t)∫t+γt−h(t)x˙T(β)M2x˙(β)dβdγ−(h2−h(t))∫t−h(t)t−h1x˙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 ∫−h2−h1∫t+γt−h1x˙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)ds∫abxT(s)Rx(s)ds∫abxT(s)Rx(s)ds≥b−a1(∫abx(s)ds)TR(∫abx(s)ds)+b−a3Ω1TRΩ1≥b−a1(∫abx(s)ds)TR(∫abx(s)ds)+b−a5Ω2TRΩ2≥b−a1(∫abx(s)ds)TR(∫abx(s)ds)+b−a3Ω1TRΩ1+b−a5Ω2TRΩ2
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\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−(b−a)2∫ab∫ubx(s)dsdu=∫abx(s)ds−(b−a)6∫ab∫ubx(s)dsdu+(b−a)212∫ab∫ub∫βbx(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)ds∫abx˙T(s)Rx˙(s)ds≥b−a1Ω3TRΩ3+b−a3Ω4TRΩ4≥b−a1Ω3TRΩ3+b−a3Ω4TRΩ4+b−a5Ω5TRΩ5
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\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)−(b−a)2∫abx(s)ds=x(b)−x(a)+(b−a)6∫abx(s)ds−(b−a)212∫ab∫ubx(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} ∫ab∫ubx˙T(s)Rx˙(s)dsdu∫ab∫ubx˙T(s)Rx˙(s)dsdu≥2Ω6TRΩ6+4Ω7TRΩ7≥2Ω8TRΩ8+4Ω9TRΩ9
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\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)−(b−a)1∫abx(s)ds=x(b)+(b−a)2∫abx(s)ds−(b−a)26∫ab∫ubx(s)dsdu=x(a)−(b−a)1∫abx(s)ds=x(a)−(b−a)4∫abx(s)ds+(b−a)26∫ab∫ubx(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。
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.
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