Any suggestions on analyzing the long-term behavior of this system? I've computed the Jacobian, but I'm not sure how to gain much insight about the behavior from this. The eigenvalues appear to be messy, but perhaps that's the right approach? Here's the system:




where $r$ is a non-zero constant.

  • $\begingroup$ In what way are the eigenvalues messy? Don't expect a nice and elegant solution if you have such a non-linear system. $\endgroup$
    – kbau
    Oct 28, 2015 at 12:53
  • $\begingroup$ Maybe this will help $$\dot{z} = -rxz -ryz = -\dot{x} -\dot{y}$$ thus you can get $$\dot{(z+x+y)} = 0$$ so the term in the brackets is conserved. $\endgroup$
    – Chinny84
    Oct 28, 2015 at 13:05
  • $\begingroup$ Could you clarify what it means to be conserved? $\endgroup$
    – statian
    Oct 28, 2015 at 13:07

1 Answer 1


The system $$\begin {cases}0=rxz\\ 0=ryz\\ 0=-rxz-ryz=%-\dot{x}-\dot{y} \end{cases}$$ has solutions $z=0$ and $x=y=0$ which are equilibrium points of the system.

The matrix $$\begin{bmatrix} rz & 0&rx\\ 0&rz&ry\\ -rz&-rz&-rx-ry \end{bmatrix}$$ is sinugular and has an eigenvalues $0$, $r z$, $r (-x - y + z)$.

Since you have a zero eigenvalue you have an unstable saddles at the equilibrium points, that is they are sort of "saddle-sinks" and "saddle-sources".

At $z=0$ the only non zero eigenvalue is $-r (x+ y)$, so whenever $x+y>0$ you have "saddle-sink" if $r>0$ or "saddle-source" if $r<0$. Similarly for $x+y<0$ you get "saddle-source" if $r>0$ or "saddle-sink" if $r<0$.

For $x=y=0$ we have multiple eigenvalue $rz$ which which is "saddle-sink" if $rz<0$ and source otherwise.

You could try to visualize it using Mathematica, but it is not that clear, so you may need to play with options of their plotting function.

VectorPlot3D[{r x z, r y z, -r x z - r y z}, {x,-1,1},{y,-1,1}, {z, -1, 1}]

Also note $$\frac{\dot y}{\dot x}=\frac{dy}{dx}=\frac{ryz}{rxz}=\frac{y}{x}$$ therefore $y=c x$ and $$\dot{z}=-rxz-ryz=-(1+c)rxz=-(1+c)\dot{x}$$ Thus, $$\frac{\dot z}{\dot x}=\frac{dz}{dx}=-(1+c)$$ which gives $z=-(1+c)x$


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