Consider
$\dot{x} = \beta x - x \left( x^2 + y^2 \right) - xy^2$
$\dot{y} = \beta y - y \left( x^2 + y^2 \right) - yx^2$
Show that the line $x = y$ separates two distinct domains of attractions.
I was curious if my solution was correct.
I calculated some of my equilibrium points to be the following:
$E_2 = (0, \sqrt{\beta})$,
$E_3 = (0, -\sqrt{\beta})$,
$E_4 = (\sqrt{\beta}, 0)$, and
$E_5 = (-\sqrt{\beta},0)$
Let $y = kx$. Then we have
$\frac{\dot{y}}{\dot{x}} = \frac{\beta k x - k x \left(x^2 + k^2 x^2 \right) - k x^3}{\beta x - x \left(x^2 + k^2 x^2 \right) - k^2 x^3}$
Let $\frac{\dot{y}}{\dot{x}} = k$. If $x \neq 0$ and $y \neq 0$, then $k = \pm 1$. Therefore $y = x$ and $y = -x$ are invariant straight line.
When $x < y$, $E_2$ and $E_5$ are attractors.
When $x > y$, $E_3$ and $E_5$ are attractors.
So $x = y$ separates two distinct domain of attraction.