I just consider another Lyapunov function as in question Question about the basin of attraction of the origin of the following ODE
Regarding of the following system : $$\begin{cases} \dot x=-x^3 \\ \dot y=-x^2y-y-z^2y \\ \dot z=-\sin z \\ \end{cases}$$ Show that the point $(0,0,0)$ is asymptotically stable and find its attractiveness basin.
Consider $V(x,y,z)=x^2+y^2+z^2$ and then $$ \dot V(x,y,z)=-2x^4-2y^2(x^2+z^2+1)-2z\sin z $$ which is strictly less than zero for $|z|<\pi$ except for the origin. Hence, it is asymptotically.
This is one example in the page 198 in textbook Morris W. Hirsch, Stephen Smale and Robert L. Devaney (Auth.) - Differential Equations, Dynamical Systems, and an Introduction to Chaos-Academic Press.
It says that
In particular, any solution that begins in the region $|z|<\pi$ must remain trapped in this region for all time. Moreover, we can conclude that the basin of attraction of the origin is the entire region $|z|<\pi$. Any solutions that starts insider a sphere of radius $r<\pi$ must tend to the origin. Outsider of the sphere of radius $\pi$ and between the planes $|z|=\pi$, the function $V$ is still strictly decreasing. Since solutions are trapped between these two planes, it follows that they too must tend to the origin.
For the first part, since $$ \Omega_1=\{(x,y,z): V(x,y,z) \mbox{ and } |z|<\pi\} $$ is positively invariant, then we can prove that "Any solutions that starts insider a sphere of radius $r<\pi$ must tend to the origin.".
Question: But how to prove that the second that solution starting from the outside of the sphere will converge to the origin?
For this question, since for $-\pi+\epsilon<z<0$ (resp. $0<z<\pi-\epsilon$), the function $\sin z$ is increasing (resp. decreasing). Then the solution all converges to the origin.