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?
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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 $(x_0,y_0)$.
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.