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?
Best Answer
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.