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I am taking a nonlinear control class. As we were discussing stability, we came across 'weaker' and 'stronger' stability conditions. For example asymptotic stability (AS) is stronger than stability in the sense of lyapunov (SISL), and exponential stability (ES) is stronger than AS. Then we across local and global definitions of stability.

The question then arises, if for some system we have a global form of some stability, say AS, and a local form of another 'stronger' stability for the same system, say ES, then which statement is actually stronger. Or does it so happen that global AS + local ES => global ES?

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Nope. A system can have better local stability than global. As an extreme example, a system can be stable within some boundary in the state space, and unstable beyond it.

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  • $\begingroup$ 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? $\endgroup$
    – k_j
    May 2 '19 at 3:04
  • $\begingroup$ @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? $\endgroup$
    – fibonatic
    May 2 '19 at 11:55
  • $\begingroup$ @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|| $\endgroup$
    – k_j
    May 3 '19 at 18:23

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