# 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.

• Are you sure that the derivative of $V$ is correct?
– KBS
Commented Dec 24, 2022 at 12:03
• I think when you compute $\dot{V}$, you need to plug in the linearized equations, not the fully nonlinear ones. Commented Dec 24, 2022 at 12:14
• @KBS I substituted the initial system in the time derivative, I have just checked again and it is correct. Commented Dec 24, 2022 at 13:28
• The error is that $V(x)=x^TPx=3x_1^2/2+x_1x_2+x_2^2$.
– KBS
Commented Dec 24, 2022 at 14:54
• This $1/2$ is pretty useless and it is better to skip it here because it leads to simpler expressions.
– KBS
Commented Dec 24, 2022 at 15:32

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.

• Thank you for this brilliant solution. I'm wondering if the same solution could be reached even with my procedure, so utilizing the formula I found on Khalil ($c = \lambda_{min}(P)*r^2$) and exploiting relations with the $||x||$. Commented Dec 27, 2022 at 11:18