The system is: \begin{align} \dot{x_1} &= -x_1 \\ \dot{x_2} &= (x_1x_2-1)x_2^3 +(x_1x_2 - 1 +x_1^2)x_2 \end{align}
I'm trying to prove the following set is positive invariant: \begin{equation} \Omega_c = \{ x \in \mathbb{R^2} \mid x_1x_2 \ge 2\} \end{equation}
A positive invariant set is such that: \begin{equation} \Omega_c = \{ x \in \mathbb{R^2} \mid V(x) \le c \} \;\; with \;\; \dot{V} \le 0 \;\; in \;\; \Omega_c \end{equation}
I analyzed the derivative of the Lyapunov function $V(x_1, x_2) = x_1x_2$: \begin{equation} \dot{V} = x_1x_2(x_1^2 + x_1x_2^3 + x_1x_2 - x_2^2 - 2) \end{equation}
But it seems like for $x_1x_2 \ge 2$ we have $\dot{V} > 0$.
I would appreciate some help!
The goal of a positive invariant set $\Omega_c$ is that once $x(t)\in\Omega_c$ we have that for all $t'\geq t \Rightarrow x(t')\in\Omega_c$.
If $\varphi$ is a real valued function such that $\frac d{dt}(x(t))>0$ and $\Omega_c=\{x:\varphi(x)\geq c\}$ we can show that $\Omega_c$ is such a set. Switching the direction of the inequalities will yield an equivalent definition.
$\dot{x}_1 x_2 = -x_1x_2$ and $x_1\dot{x}_2=x_1x_2[(x_1x_2-1)(x_2^2+1)+x_1^2]$. Thus:
$$\dot V = x_1x_2[(x_1x_2-1)(x_2^2+1)+x_1^2-1]\geq 2[x_2^2+x_1^2]\geq 4x_1x_2\geq 8>0$$