Long term Behavior of Dynamical System

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Given the following dynamical system:

$ \dot x = -6x^2+yz+x-1 $

$ \dot y = 4xz-3y^2+y-2 $

$ \dot z = 9xy-2z^2+z-3 $

What can you say about its long term behavior?

Attempt:

First, finding the fixed points.

There is only one real solution to $x'=0$, $y'=0$ and $z'=0$ and this is at the point $(1,2,3)$. At this point, the eigenvalues of the Jacobian matrix are $\lambda = 1,-17 $ Because of the positive eigenvalue, this fixed point is unstable.

I have also run into fixed points for the above system but they are complex. Do these complex fixed points have an influence on the dynamics of the system?

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The change of variables $x \to \xi$, $y \to 2 \eta$, $z \to 3 \zeta$ yields the system \begin{align} \dot{\xi} &= f(\xi) + 6 \eta \zeta, \\ \dot{\eta} &= f(\eta) + 6 \xi \zeta, \tag{1}\\ \dot{\zeta} &= f(\zeta) + 6 \xi \eta, \end{align} with \begin{equation} f(x) = -6 x^2 +x-1. \end{equation} Not only is the vector field of $(1)$ conservative (its curl vanishes), it is also invariant under all permutations of the triple $(\xi,\eta,\zeta)$.

This type of systems has been studied extensively by Martin Golubitsky. A good source is

M. Golubitsky, I. Stewart, The Symmetry Perspective, Birkhäuser, Basel, 2002.

Of particular interest is section 3.4 in chapter 3, on rings of cells.

Addition: The symmetry suggests another coordinate change. Note that the linear combination $\xi+\eta+\zeta$ is invariant under permutations of $(\xi,\eta,\zeta)$. Therefore, introducing the (suitably normalised) coordinates \begin{align} X &= \frac{\xi - \eta}{\sqrt{2}},\\ Y &= \frac{\xi+\eta - 2 \zeta}{\sqrt{6}},\\ Z &= \frac{\xi+\eta+\zeta}{\sqrt{3}}, \end{align} yields \begin{align} \dot{X} &= X(1 - 6 \sqrt{3} Z),\\ \dot{Y} &= Y(1 - 6 \sqrt{3} Z),\\ \dot{Z} &= Z-\sqrt{3} - 3 \sqrt{3}(X^2+Y^2). \end{align} The equation for $Z$ suggests the introduction of polar coordinates $X = R \cos \theta$, $Y = R \sin \theta$, yielding \begin{align} \dot{R} &= R(1 - 6 \sqrt{3} Z),\\ \dot{\theta} &= 0\\ \dot{Z} &= Z-\sqrt{3} - 3 \sqrt{3}R^2. \end{align} So, the three-dimensional system is reduced to a planar system.