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.
Best Answer
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:
Note however that the yellow area in general is only a subset of the region of attraction.