Show that the dynamical system contains a closed orbit

$\dot x = xf(x,y)+yg(x,y)$ and $\dot y = yf(x,y)-xg(x,y)$

Given Information:

f(x,y) and g(x,y) are single valued functions and differentiable at the point (0,0).

A simple bounded curve C is given by the region $ R_C = \{(x,y) | f(x,y)=0\}$ with the origin as an interior point.

Additionally, the interior region and exterior region of $R_C$ is given by:

$ R_I = \{(x,y) | f(x,y)>0\}$ and $ R_E = \{(x,y) | f(x,y)<0\}$


(0,0) is clearly a fixed point as $\dot x$ and $\dot y$ equal zero at the origin

I converted the dynamical system to polar coordinates using:

$\dot rr = x\dot x + y\dot y$ and $r^2\dot \theta = x\dot y - y\dot x$

This gives:

$\dot r = r(f(x,y) = r[f(rcos\theta,rsin\theta)]$ and $\dot \theta = -g(x,y) = -g(rcos\theta,rsin\theta)$

My next step is to find a trapping region and prove by Poincare Bendixson theorem that there is a closed orbit for the system. Looking at the region $R_C$, taking values of r locally around the fixed point, $\dot r > 0$ since in $R_I , f(x,y) >0$ which means the vector field points outward in this region. Using the same argument for $R_E$, $\dot r <0$ so the vector field points inward at the boundary of $R_C$. Hence, if another small simply closed curve is chosen to surround the fixed point, a trapping region is formed. This trapping region has no fixed points so trajectories that enter this region will either be a closed orbit or spiral towards one. Is this analysis enough to show that a closed orbit exists? Can linear analysis be employed to classify the fixed point?


Yes. The Poicare-Bendixon Theorem says that for a dynamical system defined in an open subset of $\mathbb R^2$, every compact, nonempty $\omega$ limit set of an orbit is either

  • a periodic orbit
  • a fixed point
  • a countable number of homoclinic orbits connecting finitely many fixed points.

You found an open subset of $\mathbb R^2$ which is forward invariant with respect to your system, so the system is defined on an open subset of $\mathbb R^2$. This particular subset is important because it contains no fixed points. Therefore, the second and third options are out; you must have a periodic orbit. Since this set is bounded, every $\omega$ limit set is compact and nonempty.

  • $\begingroup$ Is there a way to show that the origin is an unstable spiral for the system? $\endgroup$ – Jay Apr 5 '16 at 1:34
  • $\begingroup$ It's unstable because $\dot r>0$ near the origin. $\endgroup$ – Alex S Apr 5 '16 at 1:44

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