Estimating the region of attraction of a non-linear system with a Lyapunov Function

Question

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.

Answer 1

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.