Estimating the region of attraction of a non-linear system with a Lyapunov Function I'm trying to estimate the region of attraction of the following system:
$$
\begin{gathered}
\dot{x}_1 = \sin(x_2) \\
\dot{x}_2 = -x_1 - \sin(x_2). 
\end{gathered}
$$
From Khalil, I know that if I define V as follows:
$$
V(x) = x^TPx,
$$
where $P$ is the solution of
$$
PA+A^TP=-I,
$$
will yield the best results for an estimate.
The unique equilibrium point of the system is in the origin, and it is asymptotically stable. Linearizing around the origin, I obtain the following:
$$A =
\begin{bmatrix} 
0 & 1 \\
-1 & -1\\ 
\end{bmatrix} 
\text{ and }
P = 
\begin{bmatrix} 
3/2 & 1/2 \\
 1/2 & 1\\ 
\end{bmatrix}
.$$From which
$$
V(x) = (3x_1^2)/2 + (x_1x_2) + x_2^2
$$
taking its derivative and substituting the initial system:
$$
\dot{V}(x) = -sin(x_2)(x_2-2x_1) - x_1^2 - 2x_1x_2 
$$
Introducing the following relations:
$$
|x_1|\leq\|x\|, \quad |x_1x_2|\leq\frac{1}{2}\|x\|^2, \quad |x_2 - 2x_1|\leq\sqrt5\|x\|, \quad |sinx|\leq1
$$
It is possible to write:
$$
\dot{V}(x) \leq-\sqrt5\|x\| - 2\|x\|^2
$$
This expression is less than $0$ only for:
$$
\|x\|\leq-\sqrt5/2 \
\text{ and } \
\|x\|\geq0
$$
The first one is clearly impossible because the $\|x\|$ is always positive. If I choose the second one, I should select $r = 0$, and so the region of attraction should be null too because from Khalil:
$$
V(x) < c = \lambda_{min}(P)*r^2
$$
This sound quite strange to me because the origin is asymptotically stable, so the region of attraction should be not null. Any help or suggestions are highly appreciated.
 A: You have the system
$$
\begin{align}
\dot{x}_1&=\sin(x_2)\\
\dot{x}_2&=-x_1-\sin(x_2)
\end{align}
$$
which has the linearization matrix
$$
A(x)=\begin{pmatrix}0&\cos{x_2}\\-1&-\cos{x_2}\end{pmatrix}
$$
which at the origin is
$$
A=A(0)=\begin{pmatrix}0&1\\-1&-1\end{pmatrix}
$$
Now taking $Q=I$ you can solve
$$
PA+A^TP=-Q
$$
for $P>0$ and get $P=\begin{pmatrix}3/2&1/2\\1/2&1\end{pmatrix}$ so that
$$
V(x)=x^TPx=3 x_1^2/2 + x_1 x_2 + x_2^2
$$
Now finding a region of attraction can be done by solving
$$
k=\min_x V(x) \text{ s.t. } \dot{V}(x)=0, x\neq0
$$
This is a constrained minimization problem:
$$
\begin{align}
k=&\min_x\quad 3 x_1^2/2 + x_1 x_2 + x_2^2 \\
&\text{ s.t. } \quad 2 x_1\sin(x_2) - 2 x_1 x_2 - x_2 \sin(x_2) - x_1^2=0\\
&\phantom{\text{ s.t. }}\quad x\neq0
\end{align}
$$
This is not easy to solve analytically but we can solve it numerically. We can solve $\dot{V}(x)=0$ for $x_2$ and get two different solutions which we can substitute in $V(x)$ to get:
$$
V_1(x_2)=\frac{\left(2\,x_{2}-3\,\sin\left(x_{2}\right)\right)\,\left(3\,x_{2}-2\,\sin\left(x_{2}\right)+2\,\sqrt{{x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2}\right)}{2}
$$
and
$$
V_2(x_2)=-\frac{\left(2\,x_{2}-3\,\sin\left(x_{2}\right)\right)\,\left(2\,\sin\left(x_{2}\right)-3\,x_{2}+2\,\sqrt{{x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2}\right)}{2}
$$
both assuming ${x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2\geq 0.$ Now we can plot both functions:

In this image, only the parts of $V_1$ and $V_2$ are plotted where ${x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2\geq 0.$ We can now see that we have two minima, which we can use to compute $x_1$. We end up with two solutions:
$$
\begin{align}
(x_1,x_2)&=(0.9885, -2.2013)\\
(x_1,x_2)&=(-0.9885, 2.2013)
\end{align}
$$
The first corresponds to the minimum of the blue graph, the second to the minimum of the red graph. In both cases we have
$$
k=V(0.9885,-2.2013)=V(-0.9885,2.2013)=4.1355
$$
which is the level you are looking for. We can confirm this by checking another plot:


*

*Teal: $\dot{V}(x)\leq 0$

*Dark blue: $\dot{V}(x)> 0$

*Yellow: $\dot{V}(x)\leq 0$ and $V(x)\leq k$

*Black cross: origin

*Blue dot: first solution $(x_1,x_2)=(0.9885, -2.2013)$

*Red dot: second solution $(x_1,x_2)=(-0.9885, 2.2013)$
Note however that the yellow area in general is only a subset of the region of attraction.
