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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)

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    $\begingroup$ Similar question $\endgroup$
    – Axion004
    Aug 20 '19 at 21:59
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    $\begingroup$ $$\dot{x_1}=-x_1^3+\alpha(t)x_2$$ $$\dot{x_2}=-\alpha(t)x_1-x_2^3$$ 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. $\endgroup$
    – Axion004
    Aug 20 '19 at 23:19
  • $\begingroup$ 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. $\endgroup$
    – AzJ
    Aug 20 '19 at 23:53
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    $\begingroup$ It is simple to give (simple) counterexamples, do let us know what you have tried. $\endgroup$
    – John B
    Aug 21 '19 at 0:27
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Let's try to find a one-dimensional example, to keep things as simple as possible.

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.

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  • $\begingroup$ That you this is what I am looking for. $\endgroup$
    – AzJ
    Aug 21 '19 at 15:30
  • $\begingroup$ 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? $\endgroup$
    – AzJ
    Aug 21 '19 at 16:26
  • $\begingroup$ @AzJ: No, clearly not, as this example shows. $\endgroup$ Aug 21 '19 at 16:46

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