In linear control theory, a system is stable if and only if if satisfies the Routh–Hurwitz stability criterion, so we can use this to solve for the limits of stability. E.g. you can find the maximum gain that will allow for a stable system.

However, for nonlinear systems, if we can find a candidate function that satisfies the Lyapunov conditions, the system is stable, but the failure of a candidate function to satisfy the conditions doesn't mean it's unstable.

What is an example where one can find a Lyapunov candidate function that proves stability for a range of parameters, but where the system is still stable outside that range? I.e. where the stability bounds suggested by the Lyapunov function aren't the actual stability bounds?

  • 1
    $\begingroup$ I am not aware of a scenario where this occurs, practically. Lyapunov is extensively used and is rigorous. You may want to read up on the different definitions of stability. In practical scenarios, like robotics, we often use uniform ultimate boundedness. This is a local concept. For example, it might say that if the error in the system starts in a region less than some value, it will eventually be bounded. Note that this is different than global uniform ultimate boundedness because we cannot make any guarantees due to disturbances present in the system. So it is complicated ;) $\endgroup$ Oct 2, 2018 at 3:38
  • $\begingroup$ Not sure about the last paragraph, but I can answer your first question: indeed, the lack of a Lyapunov function does mean that the system is unstable (google "converse Lyapunov Theorem"), i.e., we also have an if and only if condition. $\endgroup$
    – Nukular
    Oct 2, 2018 at 21:30
  • $\begingroup$ Nukular, I'll revise the question to clarify that I meant that that if a candidate function doesn't meet the Lyapunov condition it doesn't mean the system is unstable, not that if the Lyapunov function doesn't exist. $\endgroup$
    – tkw954
    Oct 4, 2018 at 15:11

1 Answer 1


If I understand your question correctly, you are looking for an example where a Lyapunov candidate function proves the stability of a nonlinear system inside a region, while the system is still stable outside that region too. Well, it is safe to say that most nonlinear systems and Lyapunov candidate functions are like this. Just search for the terms region of attraction or basin of attraction to see hundreds of results and papers on this topic.

Consider e.g. the Vanderpol oscillator:

$$\frac{d^2\theta}{dt^2}-\mu(1-\theta^2)\frac{d\theta}{dt}+\theta=0$$ which is converted to the standard form of $$\dot x_1=x_2\\ \dot x_2=\mu(1-x_1^2)x_2-x_1$$ with $x=[x_1\;x_2]^T$ as the state vector. The trivial equilibrium point of this system is stable for $\mu<0$. This can be verified by setting the Lyapunov candidate as $$V(x)=\frac 12x^Tx=\frac 12(x_1^2+x_2^2)$$ which results in $$\dot V=x_1\dot x_1+x_2\dot x_2=\mu(1-x_1^2)x_2^2$$ As you see, the gradient of $V$ is negative when $\mu<0$ and $|x_1|<1$. In other words, the basin of attraction of this Lyapunov candidate is the region where $|x_1|<1$. But as you see in the below phase portrait (where $\mu=-1$), there are points outside that region which belong to the stable area of the system.

phase portrait

  • $\begingroup$ This is almost exactly what I was looking for. Do you have an example of this with respect to parameters, rather than state space? E.g. in your example where mu was >1 but still stable? $\endgroup$
    – tkw954
    Oct 9, 2018 at 15:39
  • $\begingroup$ @tkw954 You may already know about Chetaev instability theorem which says that if $V$ is a Lyapunov candidate and $\dot V>0$ in a neighborhood of $0$, then the system is surely unstable. In other words, if $V$ is dependent on a parameter and $\dot V$ becomes positive in a range of that parameter, then it would be unstable in that range. $\endgroup$
    – polfosol
    Oct 9, 2018 at 15:58
  • $\begingroup$ The Chetaev Instability Theorem is exactly what I was looking for. Accepted your answer. $\endgroup$
    – tkw954
    Oct 9, 2018 at 16:43

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