Timeline for If a system is globally asymptotically stable AND locally exponentially stable, can we say claim that it is globally exponentially stable?
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| May 3, 2019 at 18:23 | comment | added | k_j | @fibonatic Thanks for the counter-example. The system that you present is indeed NOT globally ES. Here is an analysis when the initial condition x(t)=x_0 at t=0. W.l.o.g. let us assume that x_0 is positive. Analysis for x_0>1: Then the state reaches x=1 at time T=x_0-1. After this time, it exponentially decays to zero. So the state evolution for t>T looks like $$x(t)=e^{-(t-T)}$$ which simplifies to $$x(t)=e^{-t)e^{x_0-1}$$ The state is upper bounded by the above equation for all t>0. This however is not considered to be exponentially stable, as there is no linear dependence ||x_0|| | |
| May 2, 2019 at 11:55 | comment | added | fibonatic | @k_j Take this system for example $$\dot{x} = \left\{ \begin{array}{ll} -1 & \text{if}\,x>1 \\ -x & \text{if}\,-1\leq x\leq 1 \\ 1 & \text{if}\,-1>x \end{array} \right.$$ which is locally ES and globally AS. Can you show that this also globally ES? | |
| May 2, 2019 at 3:04 | comment | added | k_j | Sure, but my question asks that if a system with a single equilibrium point at the origin does have global asymptotic stability (the 'weak' kind) and local exponential stability, then could the combined effect of these two assumptions lead to global exponential stability? | |
| May 2, 2019 at 1:32 | history | answered | TimWescott | CC BY-SA 4.0 |