Given a state equation X'=AX . The equilibrium point (x0,y0) is a stable point if we can find a Lyapunov function V(x,y) about the equilibrium point.
Most problems on stability using lyapunov functions, the stability is guarantee if the lyapunov function is positive definite and its derivative along solution path is negative. Many functions often have these properties,hence lyapunov may not be unique in general.
Prof Hacine BENCHOUBANE
The Lyapunov funcction is not unique, for the same non linear system, you can find one that gives you stability and another one that can give you asymptotic stability.
Thanks to both Professors. This is not a good news to me because I'm trying to obtain a Lyapunov function through a numerical method not analytical. How about pointwise convergence? Is it also not unique? (or does it makes sense my question here????)
No, Lyapunov function is not unique at all. Moreover, it is not meant to be unique. According to Lyapunov approach you are to find any function satisfying the necessary properties, mentioned by prof. oyelami Oyediran.
Consider a simple example:
x' = -x
A possible Lyapunov function is V = x^2. Indeed, it is positive and
V' = (x^2)' = 2 x' x = -2x^2,
e.g. it decreases along any trajectory.
Now take V = x^4
V' = 4x'x^3 = -4x^4,
hence it is a Lyapunov function as well.
Moreover, for every n>0 a function V = x^(2n) would be a feasible Lyapunov function. Similarly, polynomial
k_(2n) x^(2n) + k_(2n-2) x^(2n-2) + ... defines a Lyapunov function for the considered system if all k are positive.
So, the Lyapunov function is not just non-unique, but they form an infinite dimensional subspace in the space of all functions.
Thanks everyone. No wonder I have more unknowns than equations in my numerical methods. I guess I do not need to search for more equations. V(x0,y0)=0, V(x,y) > 0 and V'(x,y) < 0 are what is required.
Dear Amran
Of Course Laypunov functions are not unique , in fact one can determine families of
such functions to study the equilibrium state for the given Dynamical System.
but it depends on your given system of equations, it seems to me that your system in the form X'=AX which is linear and the study of the eigen values of the matrix A is enough to test stability unless some real parts of the eigenvalues are zeros and others are negative . there are direct methods to construct Laypunov functions in quadratic form.
best
issam kaddoura
Just to add to previous contributions, Lyapunouv function can't be unique since you choose any function which satisfy the Lyaponouv stability condition. Despite non uniqueness of these Lyapunouv functions, they exhibit common characteristics: one, satisfying Lyapunouv stability criteria and to large extent determine the bound for an unstable system. Probably, the function which gives a satisfactory bound can be considered depending on the focus of the problem
Dear Hussin,
Generally, The Lyapunouv function is a sufficient but not necessary condition to analyze the stability. in other words:
If you find a lyapunouv function for the system stability, its done. but if you dont, we cant consider any idea about the system stability at all.
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