Can a figure 8 ever be an orbit of

\begin{align} \frac{dx}{dt} & =f(x,y), \\[10pt] \frac{dy}{dt} & =g(x,y) \end{align} where $f$ and $g$ have continuous partial derivatives with respect to $x$ and $y$?

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    $\begingroup$ Hint: Let $(x,y)$ be the self-intersection of the 8. Put $(x(0),y(0)) = (x,y)$ as initial conditions. What would the solutions be? Do we know that the solution is unique? $\endgroup$
    – chi
    Apr 16, 2016 at 16:53
  • $\begingroup$ Don't know if you find this relevant, but it can be the so-called $\omega$-limit of an ode, i.e. the set of accumulation points of an orbit. Can develop if you think it is of interest. $\endgroup$
    – H. H. Rugh
    Dec 8, 2017 at 17:36

1 Answer 1


As was already stated by @chi, a figure 8 can't be an orbit because of the uniqueness of the solution at the self-intersection point; but it can be a composition of 3 trajectories: two parts of the figure and a self-intersection point. For instance, the phase portrait of the system $$ \dot x=xy,\quad \dot y=y^2-x^4 $$ is as shown below: enter image description here The trajectories in the right and left half-planes are separate solutions tending to zero at $t\to\pm\infty$, so they do not intersect.

(This example is from the book N. N. Bautin and E. A. Leontovich, “Methods and Ways of Qualitative Study of Dynamic Systems on a Plane,” In Russian, Nauka, Moscow, 1976. )


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