# An autonomous system that is asymptotically stable is not also exponentially stable.

I am trying to find a counter example to the following statement:

An autonomous system $$\dot{x}(t)=f(x)$$, that is asymptotically stable is also exponentially stable.

The opposite is true an autonomous system $$\dot{x}(t)=f(x)$$, that is asymptotically stable is also exponentially stable. But I am having a hard time finding a specific counter example to the original statement. Any help would be greatly appreciated.

Notes (or things I know already)

• The definitions of asymptotic stability and exponentially stability I am using.
• If I have a Lyapunov function $$V(x)$$ for $$\dot{x}(t)=f(x)$$, $$\dot{V}(x)<0$$ implies asymptotic stability and $$\dot{V}(x)\leq aV(x)$$ (where $$a$$ is a constant) exponential stability.
• The statement a non-autonomous system $$\dot{x}(t)=f(x,t)$$, that is asymptotically stable is also exponentially stable is false.
• Similar question Aug 20 '19 at 21:59
• $$\dot{x_1}=-x_1^3+\alpha(t)x_2$$ $$\dot{x_2}=-\alpha(t)x_1-x_2^3$$ with $\alpha(t)$ continuous and bounded works with $$V(x) =\frac{1}{2}\left(x_1^2+x_2^2\right)$$ but I don't think is autonomous. Aug 20 '19 at 23:19
• The system is under the label of systems with time dependent parameters (which most consider to be a different family of differential equations). I have papers answering my questions in that regard. Also for clarification I am not asking how to find lyapunov functions or prove stability using Lyapunov functions. Unless there is some lyapunov technique to show a system is only asymptotically stable and cannot be exponentially stable. My understanding is that showing a system is asymptotically stable does not disprove it being exponentially stable.
– AzJ
Aug 20 '19 at 23:53
• It is simple to give (simple) counterexamples, do let us know what you have tried. Aug 21 '19 at 0:27

You know that $$\dot x=-x$$ is exponentially stable, and you would like to find a system where the phase portrait looks the same, $$\longrightarrow 0 \longleftarrow,$$ but where the solutions don't tend to zero so fast. So try making the right-hand side smaller for small $$|x|$$ while still keeping its sign. For example, this might work: $$\dot x = - x^3 .$$ And indeed it does; if you solve this system (by separation of variables), you'll find that the solutions tend to zero roughly like $$1/\sqrt{t}$$ as $$t \to \infty$$, not exponentially.
• Sorry I have an additional question. If we have the Lyapunov function $V(x)=x^2$ the derivative the $\dot{V}=-2x^4=-2 (V(x))^2$ does this imply exponential stability?