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.
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.