Finding the region of attraction using a Lyapunov function

Question

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?

Thanks in advance.

Answer 1

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.

Answer 2

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 $x \rightarrow x^*$.