I am studying the book Nonlinear Systems from H.K. Khalil and I am stuck with a proof.
I have a Lyapunov function V that satisfies the condition $$c_1\|y\|^2\leq V(t,x,y)\leq c_2\|y\|^2,$$ for all $y\in\{\|y\|\leq\rho_0\}$. Moreover, I have the following estimate on $\dot{V}$: $$\dot{V}\leq\frac{-c_3}{2\epsilon}\|y\|^2+c_4L_3\|y\|,$$ for some $\epsilon$ sufficiently small.
From here, Khalil states that if at some time $t^*\geq t_0$, we have that $$\|y(t^*,\epsilon)\|\leq\rho_0\sqrt{c_1/c_2}=:\mu,$$
then the solution $y(t,\epsilon)$ satisfies the exponentially decaying bound $$\|y(t,\epsilon)\|\leq\mu\sqrt{c_2/c_1}e^{\frac{-\alpha(t-t^*)}{\epsilon}}+\epsilon\delta, \quad \forall \;\; t\geq t^*,$$
where $\alpha=c_3/4c_2$ and $\delta=2c_2c_4L_3/c_1c_3$.
I have the feeling that he is using the comparison lemma, however I am missing the details of how to get such a bound.
Any help is really appreciated.
EDIT
I figured out where the exponential bound is coming from.
Consider for a moment only $$\dot{V}\leq\frac{-c_3}{2\epsilon}\|y\|^2,$$ then from the right-hand side on the top inequality, we get $$\dot{V}\leq\frac{-c_3}{2\epsilon c_2}V.$$
Applying the comparison Lemma (Khalil, Lemma 3.4), we have $$V(t,x(t),y(t))\leq V(t^*,x(t^*),y(t^*))e^{-A(t-t^*)},$$ where $A:=\frac{-c_3}{2\epsilon c_2}$. At this point, we use the left-hand side of the top inequality and get: $$\|y(t)\|\leq \bigg[\frac{V(t,x(t),y(t))}{c_1}\bigg]^\frac{1}{2}\leq\bigg[\frac{V(t^*,x(t^*),y(t^*))e^{-A(t-t^*)}}{c_1}\bigg]^\frac{1}{2}$$ and now using the right-hand side of the top inequality it becomes $$\bigg[\frac{V(t^*,x(t^*),y(t^*))e^{-A(t-t^*)}}{c_1}\bigg]^\frac{1}{2}\leq\bigg[\frac{c_2\|y(t^*)\|^2e^{-A(t-t^*)}}{c_1}\bigg]^\frac{1}{2}.$$ Finally, substituting $\mu$, $A$ and solving the square root, we get the exponential bound.
However, I am still struggling with the bound $\epsilon\delta$...