I am studying stability for control systems, and I have written in the notes of my professor that the Lyapunov Criterion only gives sufficient conditions for stability, and not necessary and sufficient.
Can someone explain to me why?
Why does the Lyapunov criterion only gives sufficient conditions for stability
control theorylyapunov-functionsstability-theory
Related Solutions
Assymptotically stable does indeed mean that a solutions assymptotically convergences to the equilibrium point. However this does not mean that it has to happen in infinite time. Namely if $x(\tau)=x^*$ for some finite $\tau$ then it still satisfies the limit that $x(t) \to x^*$ as $t \to \infty$, because $x(t)$ should remain equal to $x^*$ for all $t\geq\tau$.
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.
Best Answer
The reason why the conditions are only sufficient is that failure OF A GIVEN FUNCTION to be a Lyapunov function for a system does not mean that no other Lyapunov function exists. This is valid for both linear and nonlinear systems.
There exist converse Lyapunov theorems (the original ones due to Massera, if memory serves me right) establishing that, if a system is stable, then a Lyapunov function exists. The proof in the nonlinear case is neither trivial not completely constructive.
In the case of linear system, there exists a recipe for constructing a Lyapunov function, which consists in solving the Lyapunov matrix equation. Thus if the equation does not provide a Lyapunov function, none exists, and the system is not stable. Apart from existence of the constructive method, the situations for linear and nonlinear systems are quite similar, technicalities aside.