Lyapunov Stability class K functions

Question

I'm reading the book on Nonlinear System Analysis by M. Vidyasagar.

I see they define functions of class K as continuous strictly increasing functions such that $\phi(0)=0$ and from there, they define locally positive definite functions as a continuous function $V(t,x)$ if there exist $r>0$ such that $V(t,0)=0$ for all t and $\phi(\|x\|) \leq V(t,x)$ for all $x \in B_r$ and all $t$.

From there they prove the Lyapunov theorems, all this seems fine. I learned in my differential equations class where we worked only with autonomous systems, the definition of positive definites of a function was a lot easier without the need for a K class function. It was basically defined as a continuous function $V(x)$ such that $V(0)=0$ and it was positive for all other $x \in B_r$.

I was wondering wether all the definition of class K (and class L or KL) functions is really needed or we can just work it out with a simpler definition on the locally positive definite functions which does not imply those other functions.

Such a definition would be simply a continuous function $V(t,x)$ if there exist $r>0$ such that $V(t,0)=0$ for all t and $0 < V(t,x)$ for all $x \in B_r$ and all $t$. Am I missing something important about this other functions?

Answer 1

The point of those special functions are to place bounds on the time-varying behaviour of the candidate $V(t,x).$ For instance, you'd probably agree that the system $$ \dot{x}(t) = 0, $$ does not exhibit asymptotic stability. But consider the candidate, $$ V(t,x) = \frac{1}{t + 1}\,x^2 $$ and observe that, $$ \dot{V}(t,x) = -\frac{1}{(t+1)^2}\,x^2. $$ Observe that,

• $V(t,x)$ is positive for all $x\neq 0$ and $t>0$
• $V(t, 0) = 0$ for all $t > 0.$
• $\dot{V}(t, x) < 0$ for all $x\neq0$ and $t>0.$

Essentially the different classes of function introduced are to place restrictions on how time-varying these Lyapunov candidates are allowed to be, ensuring they don't degenerate in strange ways like the way above. It does so in a way that doesn't require us to diverge too far away from the traditional results of Lyapunov ($V$ is greater than something, $\dot V$ is less than something). If you don't have a time-varying Lyapunov candidate, of course these functions are not required for precisely the reason you've identified (the definitions all decay into "something is positive [semi-]definite").