# Is the convergence of $\dot{x}=2A(t)x$ faster than that of $\dot{x}=A(t)x$?

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 $\dot{x}=A(t)x$?

This question is initially asked on Mathematics: https://math.stackexchange.com/questions/2881136/is-the-convergence-of-dotx-2atx-faster-than-that-of-dotx-atx

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 $\dot{x}=a(t)x$.

• Math.SE user AVK provides a possible counterexample here, did you check it? Also, even in dimension one highly oscillatory coefficients of the form $A(t)=c \sin(e^t)$ may lead to exponentially fast convergence to a non-zero value $e^{c\pi/2} x(0)$, with convergence rate independent of $c$, but I suppose this is not what you had in mind. Sep 11 '18 at 11:20
• The counterexample considered $c\in(0, 0.55)$, while the questions focus on $c\in(1,\infty)$. The assumption exponential stability of $\dot{x}=A(t)x$ implies that $x(t)$ will convege to the origin as $t$ tends to infty. Thus, $A(t)=c\sin(e^{t})$ does not satisfy the above assumption. Sep 11 '18 at 11:28
• I think AVK's counterexample may work if you substitute $\tfrac{1}{4}A(t)$ for $A(t)$. Still, apparently there is a simpler counterexample that I describe in my answer. Sep 11 '18 at 19:55

A simple counterexample can be constructed as follows: take $$A(t) = \pmatrix{0&\tfrac{\pi}{2}\\-\tfrac{\pi}{2}&0} \qquad \text{for } t \in [2n, 2n+1) ,$$ and $$A(t) = \pmatrix{-\alpha&0\\0&-\beta} \qquad \text{for } t \in [2n+1, 2n+2) ,$$ where $0 < \alpha \leqslant \beta$.
The equation $x'(t) = c A(t) x(t)$ describes then the following evolution: during each of the intervals $[2n, 2n+1)$ the particle is rotated by an angle $\tfrac{c \pi}{2}$ counter-clockwise, while during the interval $[2n, 2n+1)$ each of the coordinates converges exponentially to zero at rate $c \alpha$ or $c \beta$.
For $c = 1$ (or, more generally, for odd values of $c$), rotation essentially swaps the coordinates (if we disregard signs), so effectively $$x(4n) = (-1)^n e^{-c(\alpha + \beta) n} x(0).$$ The rate of convergence is thus $\tfrac{1}{4} c (\alpha + \beta)$.
For $c = 2$ (or, more generally, for even values of $c$), rotation merely changes the sign, so we end up with $$x(4n) = \pmatrix{e^{-2c\alpha n}&0\\0&e^{-2c\beta n}} x(0).$$ Rate of convergence is therefore $\tfrac{1}{2} c \alpha$.
Choose $\alpha = 1$ and $\beta = 7$. Then solutions of $x'(t) = A(t) x(t)$ converge to zero at rate $\tfrac{1}{4} \times 1 \times (1 + 7) = 2$, while solutions of $x'(t) = 2 A(t) x(t)$ converge to zero at slower rate $\tfrac{1}{2} \times 2 \times 1 = 1$.
• Thank you for the counterexample. My follow-up questions is: Does the conjecture hold for sufficiently large $c$? Sep 12 '18 at 1:05