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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.

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    $\begingroup$ Is $A$ continuous, bounded, whatever? $\endgroup$
    – copper.hat
    Commented Nov 19, 2022 at 22:53
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    $\begingroup$ It is a continuously differentiable function in time $t$ on $[0,\infty)$. I have added the assumption. $\endgroup$
    – William
    Commented Nov 19, 2022 at 23:04
  • $\begingroup$ You can use the function $V(x) = {1 \over 2} \|x\|^2$ to show that there is some time $T$ such that $\dot{V}$ is bounded on $[0,T]$ and $ \le -{1 \over 2} \|x\|^2$ for $t>T$. Apply this to a basis. $\endgroup$
    – copper.hat
    Commented Nov 19, 2022 at 23:42
  • $\begingroup$ What do you mean by "a basis"? $\endgroup$
    – William
    Commented Nov 20, 2022 at 0:42

1 Answer 1

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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).

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  • $\begingroup$ Thanks for your kind reply. $\endgroup$
    – William
    Commented Nov 20, 2022 at 3:27
  • $\begingroup$ You are very welcome; in general, it might be better to wait a few days before upvoting in case a better answer arrives. Also, it gives time to uncover any bugs in the above. $\endgroup$
    – copper.hat
    Commented Nov 20, 2022 at 3:29
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    $\begingroup$ Got it. I realized that this is not a tough question, so I upvoted and accepted it. $\endgroup$
    – William
    Commented Nov 20, 2022 at 4:37

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