# Can a figure 8 be an orbit of $dx /dt =f(x,y)$, $dy/ dt =g(x,y)$ where $f$ and $g$ have continuous partial derivatives with respect to $x$ and $y$?

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 $y$?

• 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?
– chi
Apr 16, 2016 at 16:53
• 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. Dec 8, 2017 at 17:36

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: 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.