# Finding the region of attraction using a Lyapunov function

I'm trying to find an estimate for the region of attraction of an equilibrium point. The notes from Nonlinear Control by Khalil suggest that defining $$V(x) = x^TPx,$$ where $$P$$ is the solution of $$PA+A^TP=-I,$$ will yield the best results for an estimate. It also assures that the estimate can be found from these types of Lyapunov functions for exponentially stable equilibrium points.

The system I am studying is defined by $$\begin{gathered} \dot{x}_1 = x_1 - x_1^3 + x_2 \\ \dot{x}_2 = x_1 - 3x_2. \end{gathered}$$ The linealization matrix around the equilibrium point $$x^* = \{\frac{2}{\sqrt{3}},\frac{2}{3\sqrt{3}}\}$$, is $$A = \begin{bmatrix} -3 & 1 \\ 1 & -3\\ \end{bmatrix}$$, and so, $$P = \begin{bmatrix} 3/16 & 1/16 \\ 1/16 & 3/16\\ \end{bmatrix}$$.

Given the previous values, I took the derivative of V(x) w.r.t. time and substituted the system in the equation (did it in Mathematica to try and simplify it as much as possible), giving $$\dot{V}(x) = \frac{1}{8}\left(4x_1^2-3x_1^4+4x_1x_2-x_1^3x_2-8x_2^2\right).$$ Now I need to find a region where $$\dot{V}(x)$$ is negative definite, but when I try to use inequalities with the norm of $$x$$ I get only positive norms raised to a power, and so $$\dot{V}(x)$$ can never be negative definite.

Using $$|x_1|\leq||x||, \quad |x_1x_2|\leq\frac{1}{2}||x||^2,$$ I arrived at $$\begin{gathered} \dot{V}(x) \leq \frac{1}{8}\left(4||x||^2+3||x||^4+2||x||^2+\frac{1}{2}||x||^4+8||x||^2\right) \\ \leq \frac{1}{8}\left(14||x||^2+\frac{7}{2}||x||^4\right). \end{gathered}$$ As you can see, it never is negative.

Am I being overly aggresive with converting everything to norms? How can I find the region where $$\dot{V}(x)$$ is negative?

• Hello and welcome to math.stackexchange. The overall strategy and the choice of $P$ are good, but you should then use the Lyapunov function $y^TPy$, where $y = x - x^\ast$, since the equilibrium under study is not the origin. Commented May 19, 2021 at 1:18
• @HansEngler Thanks a lot!, I see, then I'd have a $\dot{V}(y)$ with $y + x*$ instead of $x$ in the function? With a different system that had an equilibrium outside the origin I skipped the change of variable and still could find the basin, so I thought it'd work for every equilibrium. Commented May 19, 2021 at 1:25
• You can either write down the system for $y(t) = x(t) - x^\ast$ and work with $V_0= y^TPy$, or work with $V_1(x) = (x-x^\ast)^TP(x-x^\ast)$ and use the original system. Commented May 20, 2021 at 0:22

Correcting an error done.

In my previous answer (this is the edition of changes) I considered

$$\dot V = (x,y)P(x-x^3+y,x-3y) \ \ \text{translated to}\ \ (x^*, y^*)\ \ \text{which is wrong}$$

The correct procedure is to consider

$$\dot V = (x-x^*,y-y^*)P(x-x^3+y,x-3y)$$

thus giving

$$\dot V = -\frac{3 x^4}{8}-\frac{x^3 y}{8}+\frac{5 x^3}{6 \sqrt{3}}+\frac{x^2}{2}+\frac{x y}{2}-\frac{4 x}{3 \sqrt{3}}-y^2+\frac{2 y}{3 \sqrt{3}}$$

thus we have in light blue the region with $$\dot V\le 0$$ and in black the contour lines for $$V$$. As we can observe, for the choose Lyapounov function, the small attraction basin is shown by the tangent inner level curve.

• Hi, thanks for your response! So, the region where $\dot{V}(x) \leq 0$ is enclosed by those curves? Commented May 22, 2021 at 19:17
• $\dot V(x) \le 0$ is OUT those curves. Commented May 22, 2021 at 22:35
• Then would that mean that this equilibrium isn't even locally asymptotically stable? Linearization suggests that it is exponentially stable near this equilibrium, but if one were to take a ball around that point, then it would always have some places where $\dot{V}(x) > 0$. Commented May 23, 2021 at 1:48
• Corrected a mistake in the processing for $\dot V$ Commented May 23, 2021 at 13:32
• I see, I was disregarding the translation and it threw off all of my graphs and analysis. Thanks a lot! Commented May 24, 2021 at 2:20

I get the following image for region of attraction:

Yellow: $$\dot{V}(x) > 0$$

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

Blue: $$\dot{V}(x) \leq 0$$ and $$V(x) \leq 0.244$$

Black cross: $$x^* = (\frac{2}{\sqrt{3}}, \frac{2}{3\sqrt{3}})$$

The Lyapunov function is $$V(x) = (x - x^*)^T\begin{pmatrix} \frac{3}{16}&\frac{1}{16}\\ \frac{1}{16} & \frac{3}{16} \end{pmatrix}(x - x^*)$$ for the system

\begin{align} \dot{x}_1 &= x_1 - x_1^3 + x_2 \\ \dot{x}_2 &= x_1 - 3 x_2 \end{align}

The equilibrium is stable because the linearized system at $$x^*$$ has eigenvalues $$-2$$ and $$-4$$.

So $$V(x) \leq 0.244$$ is sufficient that $$x \rightarrow x^*$$.

• Thanks!, I didn't take into account the translation... How did you get the value of the upper bound for V(x)? Commented May 24, 2021 at 2:22
• @Octavio Just check the point where blue hits yellow. Commented May 25, 2021 at 15:33