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Let $x \in \mathbb{R}^{n}$ and $A(t) \in \mathbb{R}^{n\times n}$. If $\dot{x}=A(t)x$ and $\dot{x}=cA(t)x$ with $c>1$ are exponentially stable. Is the convergence rate of $x$ to zero of $\dot{x}=cA(t)x$ faster than that of $\dot{x}=A(t)x$?

Here is my initial thought:

For linear time-invariant system, the fact $\dot{x}=A(t)x$ is exponentially stable implies $A(t)=A$ is Hurwitz. It is clear that the $i$th eigenvalue $\lambda_{i}(cA)=c\lambda_{i}(A)$, $i,\ldots,n$, thus, the convergence for $\dot{x}=cAx$ with $c>1$ is faster than that of $\dot{x}=Ax$.

For a linear time-varying system. Let us consider the extreme case where $c=0$, $\dot{x}=cA(t)x=0$, which implies it is no longer exponentially stable.

For $c>1$, can we have the same conclusion for exponentially stable linear time-varying systems, i.e., can we conclude that the convergence is faster when $c>1$?

Update 1: For a scalar time-varying system, i.e., $x\in\mathbb{R}$. We can actually prove this conjecture. In fact, from the uniqueness of the equilibrium point ($x=0$ is the only solution that renders $\dot{x}=0$), the solution of $\dot{x}=ca(t)x$ is a monotone function either strictly increasing or strictly decreasing, depending on its initial condition $x(0)$. Thus, for $\dot{x}=ca(t)x$ where $c>1$, the absolute value of the derivative is larger than that of $\dot{x}=a(t)x$, while for both cases the sign of $\dot{x}(t)$ remains unchanged for all $t\ge0$. Thus, $\dot{x}=ca(t)x$ does converge faster to $x=0$ than $\dot{x}=a(t)x$.

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  • $\begingroup$ I still doubt there might be counterexamples for the time-varying case. $\endgroup$
    – guluzhu
    Aug 13 '18 at 9:41
  • $\begingroup$ But wait, say if $c\lambda_i(A) > 1$ given that $\lambda_(A) < 1$. Then it might be not stable at all, right? $\endgroup$
    – tortue
    Aug 13 '18 at 9:49
  • $\begingroup$ Sorry, I should have added both systems are assumed to be stable. I have edited the question. $\endgroup$
    – guluzhu
    Aug 13 '18 at 9:53
  • $\begingroup$ This is a continuous-time system. For the time-invariant case, if $A$ is Hurwitz, then $cA$ is Hurwitz for any positive $c$. $\endgroup$
    – guluzhu
    Aug 13 '18 at 11:17
  • $\begingroup$ I think this might be false in general case. Sorry my brevity (I’m on my phone). Solving this ODE for general $A(t)$ needs an integration of matrix-function $A(\cdot)$, which may lead to a matrix, which, in turn, may not yield a stable system, letting alone the convergence rates. $\endgroup$
    – tortue
    Aug 13 '18 at 11:38

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