Trajectory intesection with itself

I would like to ask the following. I do know that if all the prerequisites of the Existence and Uniqueness theorem are satisfied, then a trajectory in the Euclidian space-time: \begin{equation} x(t)=(x_1(t), x_2(t),x_3(t)) \end{equation} is not able to intersect itself.

Today I was told that this is possible as long as the derivative $\dot{x}(t)$ of the trajectory differs for the same space point. In other words if $\dot{x}(t_1)\neq \dot{x}(t_2)$, $t_1 \neq t_2$ for the same $x_0=(x_{0_1},x_{0_2},x_{0_3})$ then I can see a trajectory intersecting itself.

Why is that? Am I missing something?

• You should elaborate what you mean by 'intersect itself'. The trajectory of an ideal pendulum intersects itself periodically (in state space and hence in function thereof). – copper.hat Oct 4 '15 at 18:38
• @copper.hat What I do mean by 'intersect itself' is the uniqueness part of the existence and uniqueness theorem. If a trajectory interesects itself then the solution is not unique since I am able to choose more than one trajectories at the intersection point in space. The ideal pendulum nevertheless is a special kind of orbits which are bound to always return to the same points periodically. – Mitscaype Oct 4 '15 at 18:45
• There is a big difference between having two distinct trajectories passing through a given point and a single trajectory intersecting itself. The uniqueness theorem states the the first is impossible, but does not eliminate the second. You asked if a given trajectory can intersect itself, and the answer is it can. A trajectory intersecting itself doesn't contradict uniqueness. – copper.hat Oct 4 '15 at 18:48
• @copper.hat Well, I am not talking about two distinct trajectories passing through a given point. But anyways, let us agree to disagree. I received many opinions and thank you for your help everyone! – Mitscaype Oct 4 '15 at 19:00
• Emm, I'm not disagreeing with anyone. A single trajectory can pass through the same point again and again. – copper.hat Oct 4 '15 at 19:02

By definition, a trajectory $x(t)$ of the vector field $F$ satisfies$$\dot{x}(t)=F(x(t))$$for every $t$. If the trajectory intersects itself, it means that that there are $t_1,t_2$, such that$x(t_1)=x(t_2)$, but $\dot{x}(t_1)\neq\dot{x}(t_2)$, which is impossible, since$$\dot{x}(t_1)=F(x(t_1))=F(x(t_2))=\dot{x}(t_2).$$This is why trajectories of vector fields don't intersect themselves.

However, when we have a time dependent vector field, the above argument does not hold, and there may exist self intersecting trajectories.

• Can you tell, where the argument fails when the vector field is time dependent? Just for my learning! :) – H. R. Oct 4 '15 at 18:16
• @H.R. If the vector field $F$ is time dependent, then $x(t_1)=x(t_2)$ does not imply $F(x(t_1))=F(x(t_2))$. – Amitai Yuval Oct 4 '15 at 18:17
• Periodic solutions intersect themselves all the time. – copper.hat Oct 4 '15 at 18:18
• I found the sentence! In physics, specially electrostatics, they say that field lines doesn't intersect themselves. – H. R. Oct 4 '15 at 18:36

It is not true that the trajectories cannot intersect themselves. (Just to clarify, I mean that there are distinct $t_1,t_2$ such that $x(t_1) = x(t_2)$.)

Take a simple system three dimensional system $\dot{x}_1 = -x_2$, $\dot{x}_2 = x_1$, $\dot{x}_3 = 0$. The solution is periodic, hence must intersect itself.

The theorem states that there is a unique solution passing through any given point in state space. This doesn't imply that it cannot return to the same point (but if it does, it will do so ad nauseum, assuming time invariance).

• @copper.hat I would say that it is a matter of terminology. Nevertheless, I do not talk about the state space graph of a system but for the euclidian space graph. It is true that in state space you can see trajectories intersecting themselves, in my case we call them periodic. But what about the euclidian space? That is the case of the vector fields which I am referring above. – Mitscaype Oct 4 '15 at 18:28
• @H.R.: If a curve is periodic, it must intersect itself. I gave an explicit example in response to your question. – copper.hat Oct 4 '15 at 18:40
• @H.R.: It hinges on what is meant by intersecting. It seems fairly unambiguous to me, but many here seem to disagree. – copper.hat Oct 4 '15 at 19:28
• @H.R.: If you could point me to a single (mathematics) book that defines intersection in that manner I would appreciate it. – copper.hat Oct 4 '15 at 19:44
• @H.R.: This was my first question to the OP above. I find it strange that you think a periodic solution does not intersect itself. – copper.hat Oct 4 '15 at 20:04