Basin of attraction of simple nonlinear coupled ODE

basins-of-attractiondifferential-geometryordinary differential equationssystems of equations

Consider ($\epsilon = 0.1$)

\begin{equation}\label{eq:general eq}
\begin{aligned}
\dot{x}_1(t) &= x_1(x_1-0.5)(x_1+0.5)+\epsilon x_2\\
\dot{x}_2(t) &= x_2(x_2-0.5)(x_2+0.5)+\epsilon x_1
\end{aligned}
\end{equation}

Note that:

  1. If consider the following ODE $$\dot{x}(t) = x(x-0.5)(x+0.5)+\epsilon x,$$ we have three equilibrium points: $\{-0.63,0,0.63\}$. $x=0$ is the stable equilibrium point. The basin of attraction of $x=0$ is $[-0.63,0.63]$.

Go back to the top coupled ODE. Let $$\begin{bmatrix}x_1(0) \\ x_2(0) \end{bmatrix} = \begin{bmatrix}0.63 \\ 0.63 \end{bmatrix}.$$

By doing so (same initial condition of $x_1$ and $x_2$), the top coupled ODE can be reduced to (since both are the same)$$\dot{x}(t) = x(x-0.5)(x+0.5)+\epsilon x,$$ with $x(0) = 0.63$. So $$\begin{bmatrix}x_1(0) \\ x_2(0) \end{bmatrix} = \begin{bmatrix}0.63 \\ 0.63 \end{bmatrix}.$$ is in the basin of attraction of the top coupled ODE.

However, based on the phase plotter, it shows that
$$\begin{bmatrix}x_1(0) \\ x_2(0) \end{bmatrix} = \begin{bmatrix}0.63 \\ 0.63 \end{bmatrix}$$ is not in the basin of attraction of the top coupled ODE.

($\epsilon = 0.1$)

http://kevbase.com/phase/

Can anyone please point out where I am wrong?

enter image description here

Best Answer

Hint.

A good stream plot can help a lot!

In red the equilibrium points for $\epsilon = 0.1$.

enter image description here

The corresponding MATHEMATICA script.

epsilon = 0.1;
str = StreamPlot[{x (x - 0.5) (x + 0.5) + epsilon y, 
    y (y - 0.5) (y + 0.5) + epsilon x}, {x, -1, 1}, {y, -1, 1}, 
   MeshFunctions -> Function[{x, y, vx, vy, n}, n], Mesh -> 55];
sols = Quiet[Solve[{x (x - 0.5) (x + 0.5) + epsilon y == 0, 
     y (y - 0.5) (y + 0.5) + epsilon x == 0}, {x, y}]];
equil = Table[Graphics[{Red, Disk[({x, y} /. sols[[k]]), 0.02]}], {k, 1, 
    Length[sols]}];
Show[str, equil]
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