I am trying to learn about the Lyapunov function. I found an article at:
https://math24.net/method-lyapunov-functions.html, which says that:
Thus, Lyapunov functions allow to determine the stability or instability of a system. The advantage of this method is that we do not need to know the actual solution X(t).
I understand the stability and instability. For instance, in the case of a power system we can have instability due to random delays, and then we apply Lyapunov function to find out controllers which can work in this unstable situation, but how can we find out the new stable values for the controller using Lyapunov function if does not return any solution, I am talking about the following article:
Probability-density-dependent load frequency control of power systems with random delays and cyber-attacks via circuital implementation, available at:
https://ieeexplore.ieee.org/document/9785632
Somebody, please guide me. I want to know what is Lyapunov function and why we don't need a solution in this case, but if we don't get a solution using Lyapunov function how can we get stable values for instance for a controller suffering from random delays? we don't get an answer with the quadrative formula also, if we have 0/0 type of situation, what prevents us from getting a solution in the Lyapunov function? Also, what should we do in that case?
If you find any information lacking in this question, please let me know, I will update my question to add that information.
Zulfi.
Best Answer
You don't need the solution of the differential equation to show stability. This is one of the big advantages of Lyapunov functions.
Lets use a simple example:
$$ \begin{align} \tag{1} \dot{x}(t) &= -x(t) \\ x(0) &= x_0 \end{align} $$
How can we know that $x(t)$ is stable (i.e. converges to the equilibrium without escaping to infinity)?
The "obvious" way is to find the solution $x(t)$ to the initial value problem explicitly and then show this solution is bounded and converges. I.e:
$$ x(t) = x_0 e^{-t} \tag{2} $$
is this solution. Clearly $|x(t)| \leq x_0$ for all time and $\lim_{t\rightarrow \infty}x(t) = 0$ so the system $(1)$ is asymptotically stable. We are done.
However, this method requires you to find the solution of the differential equation which can be very difficult or even impossible if the equations are more complicated (i.e. nonlinear).
With a Lyapunov function this is different. Take $V(x(t))=\frac{1}{2}x(t)^2$ then $\dot{V}(x(t))=-x(t)^2$. Then by the Lyapunov theorems it follows that the system is asymptotically stable.
Note that nowhere did we use the explicit solution $(2)$, we can show stability without knowing the solution of the differential equation $(1)$. This can be a big advantage, although finding such a function can be very difficult as well.
Regarding the linked paper, the sitation is much more complicated if time delays (i.e. random time delays) are involved. So I recommend to study the case without delays first.