Analytically estimate the region of attraction of a non-linear system I know this exercise has already been posted, but here I'm reformulating it to ask for help to find an analytical solution. I want to estimate the region of attraction of the following system:
$$
\begin{gathered}
\dot{x}_1 = \sin(x_2) \\
\dot{x}_2 = -x_1 - \sin(x_2). 
\end{gathered}
$$
Defining V as:
$$
V(x) = x^TPx,
$$
where $P$ is the solution of:
$$
PA+A^TP=-I,
$$
From which, after linearizing the system in the neighborhood of the origin, I get:
$$A =
\begin{bmatrix} 
0 & 1 \\
-1 & -1\\ 
\end{bmatrix} 
\text{ and }
P = 
\begin{bmatrix} 
3/2 & 1/2 \\
 1/2 & 1\\ 
\end{bmatrix}
.$$
$$
V(x) = (3x_1^2)/2 + (x_1x_2) + x_2^2
$$
Taking the time derivative:
$$
\dot{V}(x) = -sin(x_2)(x_2-2x_1) - x_1^2 - 2x_1x_2 
$$
I'm aware of the following inequalities:
$$
|x_1|\leq\|x\|, \quad |x_1x_2|\leq\frac{1}{2}\|x\|^2, \quad |x_2 - 2x_1|\leq\sqrt5\|x\|, \quad |sinx|\leq1
$$
But I don't know how to introduce these to find a function $\rho(||x||)$ such that:
$$
\dot{V}(x) \leq \rho(||x||) \;, \text{where}\; \rho(||x||) < 0 \;\; \forall ||x|| \ge r
$$To then estimate the region of attraction as:
$$
V(x) < c = \lambda_{min}(P)*r^2
$$
Any help would be much appreciated.
 A: First, take note that the given system has multiple equilibrium points at $x_{1} = 0$ and $x_{2} = \pi n$, where $n \in \mathbb{Z}$ due to the trigonometric function. I'm unsure if you can try the following. Let
$$ y = \sin(x_{2}) \iff \sin^{-1}\left(y\right) = x_{2} $$
and
$$ \dot{y} = \cos(x_{2}) \dot{x_{2}} = \cos(\sin^{-1}\left(y\right)) \left(- x_{1} - y\right) $$
which can be expressed as
$$ \dot{y} = \sqrt{1 - y^{2}} \left(- x_{1} - y\right) $$
Thus, the dynamical model can be transformed to
$$ \left[\matrix{\dot{x}_{1} \cr \dot{y}}\right] = \left[\matrix{y \cr \sqrt{1 - y^{2}} \left(- x_{1} - y\right)}\right] $$
Since the square root of any real number cannot be negative, the stability analysis is performed for $|y| < 1$. Consider the Lyapunov function candidate:
$$ V = \frac{1}{2} x_{1}^{2} + 1 - \sqrt{1 - y^{2}} > 0 $$
The derivative of V along the trajectories of the system is given by
$$ \dot{V} = x_{1} \dot{x}_{1} + \frac{y}{\sqrt{1 - y^{2}}} \dot{y} $$
$$ \dot{V} = x_{1} y + \frac{y}{\sqrt{1 - y^{2}}} \sqrt{1 - y^{2}} \left(- x_{1} - y\right) $$
$$ \dot{V} = x_{1} y - x_{1} y - y^{2} $$
$$ \dot{V} = - y^{2} \leq 0 $$
Since $\dot{V}$ is negative semidefinite, by applying LaSalle’s invariance principle, it can be shown that that no trajectory can stay identically
at points where $V = 0$, except at the origin, and the origin is asymptotically stable for $|y| < 1$.
From
$$ \dot{x}_{2} = - x_{1} - \sin(x_{2}) $$
and $|\sin(x_{2})| < 1$ for $- \frac{\pi}{2} < x_{2} < \frac{\pi}{2}$, one can guarantee that $\lim_{t \to \infty} x_{1}(t) = 0$ and $\lim_{t \to \infty} x_{2}(t) = 0$ hold in the region of attraction of the sets $- 1 \leq x_{1} \leq 1$ and $- \frac{\pi}{2} < x_{2} < \frac{\pi}{2}$. However, if $|x_{1}| > 1$, then $x_{2}$ may converge to $\lim_{t \to \infty} x_{2}(t) = 2 \pi n$, where $n \in \mathbb{Z}$.
