nonlinear systems of differential equations stable points with complex coordinates

Question

I am trying to solve this system of differential equations.

\begin{cases} x_1'= -x_1+2x_1^3+x_2,\\ x_2'= x_1+x_1x_2. \end{cases} *by abusing the notations, I assumed $x(t) = x$

Among fixed points, some of them have complex coordinates. What do these points mean? Should I take them into account when describing the system's behavior?

Answer 1

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Typically $x_1$ and $x_2$ are real variables so they never take on complex values.

Complex values are not stable points for real variables.

$\dot{x_1} = -x_1+2x_1^3+x_2$

$\dot{x_2} = x_1+x_1x_2$

For $\dot{x_1} = 0$ you have $x_2 = x_1 - 2x_1^3$

For $\dot{x_2} = 0$ you have $x_1+x_1x_2 = 0$ either $x_1 = 0$ or $x_2 = -1$

octave:

[x1, x2] = meshgrid(-1.5:0.1:1.5, -1.5:0.1:1.5);
x1dot = -x1 + 2*x1.^3 + x2;
x2dot =  x1 + x1.*x2;
quiver(x1,x2,x1dot, x2dot,"color","blue");
title("x1dot = -x1 + 2*x1^3 + x2;  x2dot =  x1 + x1.*x2;");
hold on;
xx1 = [-1.1:0.1:1.1];
xx2 = xx1 - 2*xx1.^3;

plot(xx1,xx2,"color","red");
line([0 0], [ -1.5 1.5], "color", "green");
line([-1.5 1.5] , [-1 -1] , "color", "green");

legend(" ","x1dot = 0","xdot2  = 0");

The red line is $\dot{x_1} = 0$. The green lines are $\dot{x_2} = 0$.

The steady points are the intercepts at $(x_1,x_2) = $ $(0,0)$ and $(1,-1)$.