Stability Analysis of Linear Time-Delay Systems with Two Incommensurate Delays

Abstract

The contribution focuses on the stability analysis of linear time-delay systems within the framework of the Lyapunov—Krasovskii functionals. The method used is based on the idea to check positive definiteness of the functionals exclusively on the specific Razumikhin-type set of functions. For the systems with incommensurate delays , it is proposed to use the modified functionals depend on the Lyapunov delay matrix related to a nominal system with commensurate delays . The method is applied for the estimation of the stability domains in the parameter space.

Appendices

Appendix 1. Proof of Theorem 4.1

Necessity. This part of the proof is based on the proof of the main result in [4]. The first statement of the theorem holds for the functional $v_0(x_t,U)$ of the form (4.3). To prove the second one, take an arbitrary function $\varphi ~\in ~S$, and denote $\alpha =\Vert \varphi \Vert _h=\Vert \varphi (0)\Vert .$

For the solution of system (4.1), by Gronwall’s lemma, we obtain

$$ \Vert x(t,\varphi )\Vert \leqslant N(t)=\alpha K_1e^{Kt}, $$

where $K=\Vert A_0\Vert +\Vert A_1\Vert +\Vert A_2\Vert ,$ $K_1=1+\Vert A_1\Vert h_1+\Vert A_2\Vert h_2.$ Hence,

$$\Vert \dot{x} (t,\varphi )\Vert \leqslant K N(t)\leqslant K N(\delta ),\quad t\leqslant \delta ,$$

for any $\delta >0$, and $\Vert x(t,\varphi )-x(0,\varphi )\Vert \leqslant K N(\delta ) t$, $\;t\leqslant \delta .$ Choose $\delta$ so that $K N(\delta )=\alpha /(2\delta ).$ Then, $\Vert x(t,\varphi )\Vert \geqslant \Vert \varphi (0)\Vert -\delta K N(\delta )=\Vert \varphi (0)\Vert /2$, $\;t\leqslant \delta .$

System (4.1) is exponentially stable, therefore,

$$\begin{aligned} v_0(\varphi ,U)=\!\int \limits _0^{+\infty } x^T(t,\varphi )W x(t,\varphi )dt \geqslant \lambda _{min }(W)\!\int \limits _0^{\delta } \Vert x(t,\varphi )\Vert ^2dt \geqslant \mu \Vert \varphi (0)\Vert ^2, \end{aligned}$$

where $\mu =\dfrac{\lambda _{min }(W)\delta }{4}>0$, and the proof of the necessity part is complete. Let us note that, in contrast to [4], the constant $\delta$ here does not depend on the initial function $\varphi$.

Sufficiency. Let system (4.1) be not exponentially stable. Then, there exists a sequence $\{t_k\}_{k=1}^\infty$, such that $t_k-t_{k-1}\geqslant h,$ $t_k\xrightarrow [k\rightarrow +\infty ]{}+\infty ,$ $\Vert x(t_k)\Vert \geqslant \beta >0.$

At first suppose the solution x(t) to be uniformly bounded: let there exists $G>0$ such that $\Vert x(t)\Vert \leqslant G,$ $\;t\geqslant -h.$ Hence, $\Vert \dot{x} (t)\Vert \leqslant KG,$ $\;t\geqslant 0$, where $K=\Vert A_0\Vert +\Vert A_1\Vert +\Vert A_2\Vert ,$ and

$$\Vert x(t)-x(t_k)\Vert \leqslant KG(t-t_k)\leqslant KG\tau ,\quad t\in [t_k,t_k+\tau ],\quad \tau >0.$$

Choose $\tau =\min \Bigr \{\dfrac{\beta }{2KG};h\Bigr \}$, then

$$ \Vert x(t)\Vert \geqslant \Vert x(t_k)\Vert -KG\tau \geqslant \dfrac{\beta }{2},\;\; t~\in [t_k,t_k+\tau ], $$

for every k. Let N(t) be the number of intervals $[t_k,t_k+\tau ]\subset [0,t];$ these intervals do not intersect with each other by definition of $\tau ,$ and $N(t)\xrightarrow [t\rightarrow +\infty ]{}+\infty .$ Therefore,

$$\begin{aligned}&\int \limits _0^t x^T(s)Wx(s)ds \geqslant \sum _{k=1}^{N(t)} \int \limits _{t_k}^{t_k+\tau } x^T(s)Wx(s)ds \\&\qquad \geqslant \lambda _{min }(W)\frac{\beta ^2\tau }{4} N(t) \xrightarrow [t\rightarrow +\infty ]{}+\infty . \end{aligned}$$

Since the functional $v_0(x_t,U)$ is bounded when the solution is bounded, we obtain the contradiction:

$$\begin{aligned} v_0(\varphi ,U)=v_0(x_t,U)+\int \limits _0^t x^T(s)W x(s)ds\xrightarrow [t\rightarrow +\infty ]{}+\infty . \end{aligned}$$

Let us now assume that the solution x(t) is not uniformly bounded. It means that the sequence $\{t_k\}_{k=1}^\infty$ can be chosen so that

$$ \Vert x(t_k)\Vert =\max _{\begin{array}{c} t\leqslant t_k \end{array}} \Vert x(t)\Vert \xrightarrow [k\rightarrow +\infty ]{}+\infty . $$

Such a choice results in $x_{t_k}\in S$ for every k, and

$$\begin{aligned} v_0(\varphi ,U)=v_0(x_{t_k},U)+\int \limits _0^{t_k} x^T(s)W x(s)ds \geqslant \mu \Vert x(t_k)\Vert ^2\xrightarrow [k\rightarrow +\infty ]{}+\infty . \end{aligned}$$

We obtain the contradiction that finishes the proof.    $\Box$

Appendix 2. Proof of Theorem 4.2

Necessity. Since system (4.1) satisfies the Lyapunov condition , there exists functional $v_0(x_t,U)$ of the form (4.3) (see [4]), for which the first statement of the theorem is true. Let us prove the second statement.

We first suppose that $\bar{\lambda }=\alpha >0$ is the real eigenvalue of system (4.1). Then, the system has the solution $\tilde{x}(t)=e^{\alpha t}c$, where $c\in \mathbb {R}^n,$ $c\ne \mathbf 0.$ Since $\tilde{x}(t)$ is the increasing function, $\tilde{x}_0\in S.$ On the other hand, we have

$$\begin{aligned} v_0(\tilde{x}_T,U)-v_0(\tilde{x}_0,U)=-\int \limits _0^T \tilde{x}^T(t)W\tilde{x}(t)dt =-\frac{1}{2\alpha }(e^{2\alpha T}-1)\,c^TWc, \end{aligned}$$

(4.16)

where $T=const >0.$ Since $\tilde{x}(T+\theta )=e^{\alpha T}\tilde{x}(\theta ),$ $\theta \in [-h,0],$ it follows that $v_0(\tilde{x}_T,U)=e^{2\alpha T}v_0(\tilde{x}_0,U),$ so (4.16) results in

$$ v_0(\tilde{x}_0,U)=-\frac{1}{2\alpha }c^TWc\leqslant -\frac{\lambda _{min }(W)}{2\alpha }\Vert c\Vert ^2=-\mu \Vert \tilde{x}(0)\Vert ^2, $$

where $\mu =\dfrac{\lambda _{min }(W)}{2\alpha }>0.$ The necessity is proved for $\bar{\lambda }\in R.$

We now turn to the case $\bar{\lambda }=\alpha +i\beta$, where $\alpha >0,$ $\beta \ne 0.$ Let $c=c_1+ic_2$ be the eigenvector corresponding to $\bar{\lambda }$, here $c_1,c_2\in \mathbb {R}^n.$ Choose $T=2\pi /|\beta |$ and consider the T-periodic vector function $\psi (t) = \cos \beta t\,c_1 - \sin \beta t\, c_2.$ Then, $e^{\alpha t}\psi (t)$ is the real part of $e^{\bar{\lambda }t}c$, and, therefore, is the solution of system (4.1). Since the system is time-invariant, function $\tilde{x}(t)=e^{\alpha (t+\bar{t})}\psi (t+\bar{t})$ is also the solution for every $\bar{t}.$ Choose $\bar{t}\in [h,h+T]$ from the condition

$$ \Vert \psi (\bar{t})\Vert = \max \limits _{\begin{array}{c} t\in [0,h+T] \end{array}} \Vert \psi (t)\Vert , $$

such value of $\bar{t}$ exists due to continuity and periodicity of $\psi (t).$ Hence, $\Vert \tilde{x}(\theta )\Vert \leqslant \Vert \tilde{x}(0)\Vert ,$ $\theta \in [-h,0]$, and therefore, $\tilde{x}_0\in S.$ Additionally, as in the first case, $v_0(\tilde{x}_T,U)=e^{2\alpha T}v_0(\tilde{x}_0,U).$

Again consider the first equality in (4.16) and estimate its right-hand side. To this end, first note that $\tilde{x}(t) = e^{\alpha (t+\bar{t})}\bigl (\cos (\beta t)\xi _1 - \sin (\beta t)\xi _2\bigr )$, where $\xi _1=\cos (\beta \bar{t})c_1 - \sin (\beta \bar{t})c_2 = \psi (\bar{t}),$ $\;\xi _2=\sin (\beta \bar{t})c_1 + \cos (\beta \bar{t})c_2.$ Then,

$$\begin{aligned} \int \limits _0^T \tilde{x}^T(t)W\tilde{x}(t)dt&\geqslant \lambda _{min }(W)e^{2\alpha \bar{t}}\biggl [ \int \limits _0^Te^{2\alpha t}\cos ^2(\beta t)dt\,\Vert \xi _1\Vert ^2\\&+\int \limits _0^Te^{2\alpha t}\sin ^2(\beta t)dt\,\Vert \xi _2\Vert ^2-\int \limits _0^Te^{2\alpha t}\sin (2\beta t)dt\,\xi ^T_1\xi _2\biggr ]. \end{aligned}$$

Calculating directly all the integrals, using Cauchy – Bunyakovsky inequality for the term $\xi _1^T\xi _2$ and taking into account the fact that $\Vert \tilde{x}(0)\Vert =e^{\alpha \bar{t}}\Vert \xi _1\Vert ,$ we obtain

$$\begin{aligned} \int \limits _0^T \tilde{x}^T(t)W\tilde{x}(t)dt&\geqslant \lambda _{min }(W)e^{2\alpha \bar{t}}\frac{(e^{2\alpha T}-1)}{4\alpha (\alpha ^2+\beta ^2)} \Bigl [(\alpha ^2+\beta ^2)\Vert \xi _1\Vert ^2\\&+(\alpha \Vert \xi _1\Vert -|\beta |\Vert \xi _2\Vert )^2\Bigr ] \geqslant \mu (e^{2\alpha T}-1)\Vert \tilde{x}(0)\Vert ^2, \end{aligned}$$

where $\mu =\lambda _{min }(W)/4\alpha >0.$ Combining the latter estimate with (4.16), we have $v_0(\tilde{x}_0,U)\leqslant -\mu \Vert \tilde{x}(0)\Vert ^2$ for $\tilde{x}_0\in S,$ as required.

Sufficiency. Let us take the nontrivial initial function $\varphi \in S$ such that $v_0(\varphi ,U)\leqslant -\mu \Vert \varphi (0)\Vert ^2.$ Condition $\varphi \in S$ implies $\Vert \varphi \Vert _h=\Vert \varphi (0)\Vert ,$ so $v_0(\varphi ,U)\leqslant -\mu \Vert \varphi \Vert _h^2.$

Substituting the solution of system (4.1) corresponding to the function $\varphi$ into functional $v_0(x_t,U)$ we obtain

$$\begin{aligned} v_0(x_t(\varphi ),U)=v_0(\varphi ,U)-\int \limits _0^t x^T(s,\varphi )Wx(s,\varphi ) ds \leqslant -\mu \Vert \varphi \Vert _h^2. \end{aligned}$$

(4.17)

Hence, $\mu \Vert \varphi \Vert _h^2\leqslant |v_0(x_t,U)|\leqslant \eta \Vert x_t(\varphi )\Vert _h^2,$ where $\eta =const >0$, and finally,

$$\begin{aligned} \Vert x_t(\varphi )\Vert _h\geqslant \sqrt{\frac{\mu }{\eta }}\Vert \varphi \Vert _h>0, \end{aligned}$$

(4.18)

where the last expression we denote by $\beta .$

Let us prove that the solution $x(t,\varphi )$ is unstable. Conversely, suppose that there exists $G>0$ such that $\Vert x(t,\varphi )\Vert \leqslant G$, $t\geqslant 0.$ Then, $\Vert \dot{x} (t,\varphi )\Vert \leqslant KG$, where $K=\Vert A_0\Vert +\Vert A_1\Vert +\Vert A_2\Vert .$ From (4.18) we have that there exists the sequence $\{t_k\}_{k=1}^{\infty }$, such that $t_k-t_{k-1}\geqslant h,$ $t_k\xrightarrow [k\rightarrow +\infty ]{}+\infty$, and $\Vert x(t_k,\varphi )\Vert \geqslant \beta >0.$ As in the proof of the sufficiency of Theorem 4.1, we can show that

$$\begin{aligned} \int \limits _0^t x^T(s,\varphi )Wx(s,\varphi )ds\xrightarrow [t\rightarrow +\infty ]{}+\infty , \end{aligned}$$

so, according to (4.17), $v_0(x_t,U)\xrightarrow [t\rightarrow +\infty ]{}-\infty$, which contradicts the assumption that the solution $x(t,\varphi )$ is uniformly bounded. The theorem is proved.    $\Box$