# Region of attraction of simple ODE with perturbation

Consider the following simplest example:

$$\dot{x} = x(x-1)(x+1)$$ $[-1,1]$ is the ROA.

Now consider the two dimensional case:

\begin{aligned} &\dot{x} = x(x-1)(x+1)\\ &\dot{y} = y(y-1)(y+1) \end{aligned}Obviously, ROA is a square. However, if I consider the following coupled ODE:

\begin{aligned} &\dot{x} = x(x-1)(x+1) + \epsilon (y-x)\\ &\dot{y} = y(y-1)(y+1) + \epsilon (x-y) \end{aligned} where $\epsilon$ is a very small number. Or

\begin{aligned} &\dot{x} = x(x-1)(x+1) + \epsilon (-y+x)\\ &\dot{y} = y(y-1)(y+1) + \epsilon (-x+y) \end{aligned} Then I have the following ROAs: (blue line-case three, black line-case two, red line-case one)

My questions are:

1. There are two different tilt directions for case two and three. I know this is because of the slope of the coupling term (for case two, the slope of $x$ and $y$ in the coupling terms are $-1$). But how could I analyze this formally?

2. Is it a good way to analyze 1. by perturbation method? (observe the sign of the leading order term of the solution obtained from perturbation method?) and how could I proceed it for the coupling term?

3. Are there any reference about my questions?

Note: It is simple to check that if you just use the linearization method to find the Jacobian matrix (w.r.t the point $(0,0)$), the ROA will be the whole $\mathbb{R}^2$ , which is not correct.