Let's say I have a Lyapunov candidate of the following form: $$V(x,y)=\frac{a_1}{2}x(t)^2+\frac{a_2}{2}y(t)^2$$ Where both $a_1$ and $a_2$ are known, positive, non-zero constants.
Let's now assume that the solution to this Lyapunov function takes on the following exponential form (no need to go through the math):
$$V\leq (a_2+V(t_i))e^{\Delta t}-a_2 \leq V_m$$
Knowing that this function can be bounded by $V_m=\frac{a_1}{2}x_{max}^2+\frac{a_2}{2}y_{max}^2$ I can then solve for a $\Delta t$: $$\Delta t\leq ln(\frac{V_m+a_2}{V(t_i)+a_2})$$
After looking at the characteristics of the above inequality I notice that I can maximize $\Delta t$ by making $a_2$ arbitrarily small.
The question is now, "How Small"? Can I take the right-hand limit of this inequality as $a_2\rightarrow0$? $$\lim_{a_2 \to 0^+} \Delta t=ln(\frac{x_{max}^2}{x(t_i)})$$
By doing this can I still claim that the Lyapunov function is Positive Definite?
I argue that I can since the definition of the right-hand limit is just telling me what happens when $a_2$ approaches $0$ which is not the same as $a_2$ being zero. However, I am very hesitant to make this claim.
What over conclusions can I make from this? Can I argue that the above inequality is not dependent on $y(t)$? That just sounds insane to me but it seems that the math allows for it.
Some guidance and understanding would be greatly appreciated.