Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is it not possible to have a period-2 orbit(or greater) for a 2D autonomous system whereas the same is possible for a non-autonomous system?

I can explain the existence of period-2 orbits in non-autonomous systems by saying that the forcing function changes the phase and hence the point is not essentially the same point when the orbit returns back to the original point, but ,I am not able to come-up with a justification for a autonomous system.Can anyone? enter image description here

share|cite|improve this question

Consider the system $(x,y)\to((\cos a) x-(\sin a) y,(\sin a) x+(\cos a) y)$. If $a=2\pi/k$ for some integer $k$, this autonomous 2D system has period-$k$ orbits.

For example, $(x,y)\to(-x,-y)$ has period-2 orbits and $(x,y)\to(-y,x)$ has period-4 orbits.

Edit: The edited version of the question seems to ask about trajectories of dynamical systems intersecting themselves. This is an entirely different problem, and the mention of "period-2 orbit" may seem unfortunate in this context. Thus, consider a dynamical system $x'(t)=f(x(t),y(t))$, $y'(t)=g(x(t),y(t))$. The configuration of the picture is impossible since the point $(x^*,y^*)$ at the intersection would be such that $(f(x^*,y^*),g(x^*,y^*))$ takes two different values at the same time. In fact, if $(x(t+s),y(t+s))=(x(t),y(t))$ for some given $t$ and $s\gt0$, one has $(x(t+s+u),y(t+s+u))=(x(t+u),y(t+u))$ for every $u\geqslant0$, that is, the trajectory is a cycle.

share|cite|improve this answer
I meant system of continuous differential equations with their solution curves represented in 2D phase space. I remember reading somewhere that trajectories can never cross in a autonomous system(except at an equilibrium point or a fixed point in discrete system case). Am I clear with the question now?I edited the original question with a diagram. – Sunny Marella Apr 4 '13 at 9:48

No, the intersection like you shown in the picture can never happen. The reason is that if your differential equations have a Lipschitz continuous r.h.s, the solution to the ODE with a given initial condition are unique. Let us say the point of intersection is $(x_0,y_0)$ at time $t_0$. Now if you integrate your ODE's with initial condition $(x_0,y_0)$, your picture shows that it could go in two different directions, which means the solution is NOT unique. Hence the contradiction. Note that at the point of intersection ,there are 2 different directions on the curve along which the time is increasing. So how will the ODE know which one to take, since ODE solution at time $t_0+\delta$ depends solely on $(x_0,y_0)$.

share|cite|improve this answer
Yes,I understand now that the Lipschitz condition $$|f(t,u)-f(t,v)| \leq L*|u-v| $$ and the Lipschitz Lemma that the partial derivative $$ \frac{\partial f}{\partial y} \leq K $$ are sufficient conditions to check uniqueness of solution. But,my question is: 'Is there anyway that the solution curve intersects itself(like in diag.) in a non-autonomous Diff system?' I mean,won't "phase change" be a suitable enough condition to say that when the solution curve intersects itself(like in diag.), it is actually unique in the sense that the phase has changed from the original. Is there a fn. example? – Sunny Marella Apr 8 '13 at 6:12
More curious with counter-arguments, is it possible that the domain D for the Lipschitz condition is different for non-autonomous systems? I'm not aware so much whether domains are defined differently for non-autonomous systems. – Sunny Marella Apr 8 '13 at 6:57
i.e could you give an example where the Lipschitz condition does not hold,like when there is a periodic forcing function? Hope I have made my query clear. – Sunny Marella Apr 8 '13 at 13:05
Any non-autonomous system can be written as autonomous by taking time as the (n+1)th coordinate. So 'phase change' is equivalent to saying that the (n+1)th coordinate is different, hence the trajectories are not crossing themselves, but only look like crossing because you are taking a projection on a lower-dimensional manifold. – nonlinearism Apr 8 '13 at 14:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.