Regarding of the following system : $$\begin{cases} \dot x=-x^3 \\ \dot y=-x^2y-y-z^2y \\ \dot z=-\sin z \\ \end{cases}$$
I am working to show that the point $(0,0,0)$ is asymptotically stable and find its attractiveness basin.
My attempt: I consider the Lyapunov function: $$ V(x,y,z)=x^4+y^2+(1-\cos z) $$ Note that $V(0,0,0)=0$ and for all $z\neq n\pi$ and $x, y$, $$ V(x,y,z)> x^4+y^2\ge 0 $$
Since $$ \dot V(x,y,z)=-4x^6-2y^2(x^2+z^2+1)-\sin^2 z\le -4x^6-2y^2-\sin^2 z<0, $$ where $\nabla V=[4x^3, 2y, \sin z]^T$, for all $(x,y,z)\neq (0,0,0)$.
Then $V$ is negative definite. We can conclude that the equilibrium point is asymptotically stable in a region around the origin.
However, it seems that it does not have its attractiveness basin. I am confused about this point.
The connectivity component of the origin of the set $\Omega_C=\{ x:V(x)< C \}$ is an estimation of the region of attraction of $\{0\}$ (in the sense that any trajectory starting at some $x_0\in \Omega_C$ tends to the origin without leaving $\Omega_C$) if $\forall x\in\Omega_C\setminus \{0\}\;\; \dot V(x)<0$. This follows from the procedure of proving the Lyapunov’s stability theorem.
In this problem $\dot V$ is negative everywhere except for the points $(0,0,\pi n)$, $n\in\mathbb Z$, so any of the sets $\Omega_C=\{ x:V(x)< C \}$ that does not contain the points $(0,0,\pi n)$ except for $(0,0,0)$, is an estimation of the region of attraction. It is enough here to check the points $(0,0,\pm\pi)$. Since $V(0,0,\pi)=V(0,0,-\pi)=2$, the largest of these sets is $$ \Omega_2= \{ x:V(x)< 2 \}=\{ (x,y,z): x^4+y^2+(1-\cos z)<2\}. $$
Not this entire set is an estimation of the region of attraction, but only the connectivity component of the origin, so the final answer is $$ \Omega_2^c=\{ (x,y,z): x^4+y^2+(1-\cos z)<2\; \wedge \; |z|<\pi\}. $$
Pic.2 The set $\Omega_2^c$.