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Consider a linear time varying homogeneous system: $$\dot{x}=A(t)x$$ where $x\in\mathbb{R}^n$ and $A(t)$ is a $n\times n$ real symmetric matrix satisfying $A(t)\to -I_n$ as $t\to\infty$. Suppose $A$ is a continuously differentiable function on $[0,\infty)$. Notice that $I_n$ is the $n\times n$ identity matrix. Is it true that $x(t)$ is asymptotically stable? Is it true that $x(t)$ is Lyapunov stable?

For Lyapunov stability, I find a related question. Since $A(t)\to -I_n$, when $t>0$ is large enough, $A(t)$ will be eventually negative definite, so by using the result in the link, it seems we have Lyapunov stability.

For asymptotic stability, it seems to be possible because here we don't have the situation that $A(t)$ will vanish very fast, as shown in several counterexamples in the above link. It is obvious that if $A(t)$ is exactly $-I_n$, it will be asymptotically stable. Also, the fact that $A(t)$ is symmetric may be essential.

William
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1 Answers1

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Let $n(t) = \lambda_1(A(t))$, the $\max$ eigenvalue of $A(t)$. $n$ is continuous and $n(t) \to -1$. Choose $T>0$ such that $n(t) \le -{1 \over 2}$ for $t > T$ and let $B=\sup_{t \in [0,T]} n(t)$.

Pick some $x$ and let $V_x(t) = {1 \over 2} \|x(t)\|^2$, where $x$ is the response of the system to the initial condition $x$. We see that $\dot{V_x}(t) = x(t)^T A(t) x(t) \le 2n(t) V_x(t)$.

On $[0,T]$, $\dot{V_x}(t) \le 2B V_x(t)$ and so $V_x(t) \le V_x(0) e^{2Bt} \le V_x(0) e^{2BT}$ and similarly, for $t >T$, $V_x(t) \le V_x(T) e^{-(t-T)} \le V_x(0) e^{2BT} e^{-(t-T)}$.

Let $\epsilon>0$ and choose $\delta = e^{-BT}$, then if $\|x_0\| < \delta$ we see that $\|x(t)\| < \epsilon$ for $t \ge 0$. Furthermore, $x(t) \to 0$, hence the system is asymptotically stable (in fact exponentially stable).

copper.hat
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