Lyapunov function of perturbed linear time-varying ODE system

Question

I am considering an ODE system as follows: $$\dot{x}(t) = A(t)x(t) + g(t,x),$$ where $A(t)\in \mathbb{R}^{n\times n}$ for any $t\geqslant t_0$, and $g(t,x)$ is a vanishing perturbing term (i.e., $g(t,0)=0$ for all $t\geqslant t_0$) satisfying $\Vert g(t,x)\Vert\leqslant \gamma \Vert x\Vert$ (edit: Here $\gamma \rightarrow 0$ when $\Vert x\Vert \rightarrow 0$ ). I want to find a Lyapunov function that meets two conditions: (1) $\alpha_1(\Vert x\Vert)\leqslant V(t,x)\leqslant \alpha_2(\Vert x\Vert)$; and (2) $\frac{d}{dt} V(t,x(t)) = \frac{\partial V}{\partial t}+\frac{\partial V}{\partial x}\dot{x}(t)\leqslant -\alpha_3(\Vert x\Vert)$, where functions $\alpha_1, \alpha_2, \alpha_3$ are class-$\mathcal{K}$ function. Choosing the Lyapunov function of its nominal system $\dot{x}(t) = A(t)x(t)$ ,$V = x^T P(t) x$ where $P(t)$ satisfies matrix differential equation $-\dot{P}(t) = P(t)A(t)+A^T(t)P(t)+Q(t)$, seems quite reasonable. Here $Q(t)$ can be any continuous positive-definite time-varying bounded matrix. We can assume $0<c_1I\leqslant Q(t)\leqslant c_2I$. The solution of above matrix differential equation is given by $P(t) = \int_{t}^{\infty} \Phi^T(\tau,t)Q(\tau)\Phi(\tau,t)d\tau$, where $\Phi(\tau,t)=e^{\int_{t}^{\tau} A(s)ds}$ is the state transition matrix of the nominal system.

My attempt: It is easy to verify that this Lyapunov function meets condition (2), by properly bounding the perturbing term when taking derivative of it. If assuming that the $0$ equilibrium point of nominal system is exponentially stable (implying that $\Vert \Phi(\tau,t)\Vert\leqslant ke^{-\lambda (\tau-t)}$), then the upper bound of $V$ can be obtained by $$V(t,x)\leqslant \Vert x\Vert^2 \cdot c_2\int_{t}^{\infty}\Vert \Phi(\tau,t)\Vert^2d\tau \leqslant c_3 \Vert x\Vert^2.$$ However, I just cannot bound the $V$ from below. I have already known how lower bound is obtained in the linear time-varying situation, by assuming that $\Vert A(t)\Vert \leqslant L$, and the state-transition property that $\Phi(\tau,t)x(t;t_0,x_0) = x(\tau; t,x(t))$. But it is not the same in the perturbed system. I have chosen the $\Phi(\tau,t)$ as the state transition matrix of the nominal system, and the solution of original system doesn't have state-transition property in terms of $\Phi(\tau,t)$.

My question: Is the $0$ equilibrium point of this perturbed system reach stability for sure? If not, is there counter-example? If so, can anyone bound the $V$ from below in the form of a class-$\mathcal{K}$ function? Any reasonable extra assumption is allowed to made, and any other form of Lyapunov function is welcomed.

Answer 1

Assume that $A(t)$ is such that the origin of the system $$\dot{x}(t)=A(t)x(t),\ x(t_0)=x_0$$ is globally uniformly exponentially stable. This is equivalent to saying that there exists a differentiable matrix-valued function $P(t)$ such that $$0\prec\alpha_1I\preceq P(t)\preceq\alpha_2I$$ and $$ \dot{P}(t)+A(t)^TP(t)+P(t)A(t)\preceq -\alpha_3I $$ for all $t\ge t_0$ and all $t_0\ge0$. An explicit solution for it is, for instance, $$ P(t)=\int_t^\infty\Phi(s,t)^TQ(s)\Phi(s,t)ds $$ where $\Phi(t,s)$ is the state-transition matrix defined as $x(t)=\Phi(t,s)x(s),t\ge s\ge t_0$ and satisfying $\Phi(t,t)=I$ for all $t\ge t_0$. The matrix-valued function $Q(t)$ is coercive and bounded.

Now consider the Lyapunov function $V(t,z)=z^TP(t)z$ for the perturbed system $$ \dot{z}(t)=A(t)z(t)+g(t,z(t)),\ z(t_0)=z_0 $$ where $||g(t,z)||\le\gamma ||z||$ for all $t\ge0$.

Computing the derivative of $V$ along the trajedctory of the perturbed system yields $$ \dfrac{d}{dt}V(t,z(t))=z(t)^T(\dot P(t)+A(t)^TP(t)+P(t)A(t))z(t)+2z(t)^TP(t)g(t,z(t)). $$

We have

$$ 2z(t)^TP(t)g(t,z(t))\le \epsilon z(t)^TP(t)z(t)+\epsilon^{-1}g(t,z(t))^TP(t)g(t,z(t)) $$ for all $\epsilon>0$. This implies that $$ 2z(t)^TP(t)g(t,z(t))\le \epsilon\alpha_2||z(t)||^2+\epsilon^{-1}\alpha_2\gamma^2||z(t)||^2. $$ Therefore, we have $$ \dfrac{d}{dt}V(t,z(t))\le (-\alpha_3+\epsilon\alpha_2+\epsilon^{-1}\alpha_2\gamma^2)||z(t)||^2. $$ The value for $\epsilon$ that minimizes the right hand side is $\epsilon=\gamma$ which yields the upper-bound $$ \dfrac{d}{dt}V(t,z(t))\le (-\alpha_3+2\alpha_2\gamma)||z(t)||^2. $$

This implies that the origin of the perturbed system is globally uniformly exponentially stable if $$\gamma<\dfrac{\alpha_3}{2\alpha_2}.$$

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Edit. For the lower bound for $P(t)$, just observe that there exists a sufficiently large $%c>0$ such that

$$ \dfrac{d}{ds}\Phi(s,t)^TQ(s)\Phi(s,t)\succeq -c \Phi(s,t)^TQ(s)\Phi(s,t). $$ This follows from the boundedness of $\Phi(t,s)$ and $Q(t)$. Integrating both sides from $t$ to $\infty$ yields $$ -Q(t)\succeq -c P(t) $$ which yields $$ P(t)\succeq Q(t)/c. $$ The conclusion follows from the fact that $Q(t)$ is coercive.