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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?

Bazinga
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  • 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 04 '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. – Bazinga Oct 04 '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 04 '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! – Bazinga Oct 04 '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 04 '15 at 19:02

2 Answers2

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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.

Amitai Yuval
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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
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  • Well, I'd say this is a matter of terminology. According to the definitions I know, a periodic solution wouldn't be called self intersecting. – Amitai Yuval Oct 04 '15 at 18:19
  • What do you mean? The curve $\left{ \matrix{ {x_1} = \cos t \hfill \cr {x_2} = \sin t \hfill \cr} \right.$ is periodic but doesn't intersect itself. Am I wrong? – Hosein Rahnama Oct 04 '15 at 18:21
  • $x(0) = x(2 \pi)$, and $x(t) \neq x(0)$ for all $t \notin 2 \pi \mathbb{Z}$ (assuming $x(0) \neq 0$). – copper.hat Oct 04 '15 at 18:22
  • @AmitaiYuval: How do you define self intersecting? – copper.hat Oct 04 '15 at 18:23
  • Especially in this context, I understand "self intersecting" as "there exist $t_1,t_2$, such that $x(t_1)=x(t_2)$, but $\dot{x}(t_1)\neq\dot{x}(t_2)$". – Amitai Yuval Oct 04 '15 at 18:26
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    @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. – Bazinga Oct 04 '15 at 18:28
  • I didn't get you! You said that every periodic curve intersect itself but I gave an example of a periodic curve which doesn't intersect itself! – Hosein Rahnama Oct 04 '15 at 18:29
  • @Mitscaype: It is all a matter of terminology. I am not clear what you mean by Euclidean space as opposed to state space. If the state space trajectory intersects, then so must any projection of that state space (which is what I presume you must mean by Euclidean space?). – copper.hat Oct 04 '15 at 18:36
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    @H.R.: If a curve is periodic, it must intersect itself. I gave an explicit example in response to your question. – copper.hat Oct 04 '15 at 18:40
  • Your response isn't comprehensive and I have some lack of knowledge to understand it! :) – Hosein Rahnama Oct 04 '15 at 19:24
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    @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 04 '15 at 19:28
  • By intersection we mean that for some ${t_2} \ne {t_1}$ we have $\left{ \matrix{ x({t_2}) = x({t_1}) \hfill \cr \dot x({t_2}) \ne \dot x({t_1}) \hfill \cr} \right.$ – Hosein Rahnama Oct 04 '15 at 19:42
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    @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 04 '15 at 19:44
  • I will look it up for you but that's strange for me that how you cannot imagine a self-intersecting curve in 3 dimensional euclidean space. Use some intuitions from analytic geometry. – Hosein Rahnama Oct 04 '15 at 19:52
  • @H.R.: Huh? Who said I could not imagine such a thing??? This is getting silly. – copper.hat Oct 04 '15 at 19:54
  • This is what is meant here. What will be your definition of intersection for such a curve? – Hosein Rahnama Oct 04 '15 at 19:55
  • So, it's a matter of terminology here! Your definition of intersection is somehow strange! :) However, with that definition you are right! – Hosein Rahnama Oct 04 '15 at 19:59
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    @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 04 '15 at 20:04
  • Can you please take a look at my question too

    http://math.stackexchange.com/questions/1447956/how-to-solve-the-following-biharmonic-eigenvalue-problem

    I don't know how to tell you this on chat.

     – Hosein Rahnama Oct 04 '15 at 20:06
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    @H.R.: That question is out of my comfort zone :-). – copper.hat Oct 04 '15 at 20:31
  • Thanks anyway. :) Where is this chat thing? How can I find it? – Hosein Rahnama Oct 04 '15 at 20:33
  • @H.R.: On a web browser, click on Stack Exchange at the top left, you should see 'chat' to the right of 'Mathematics'. – copper.hat Oct 04 '15 at 20:34
  • Rather late to this argument, but I think copper.hat is thinking about intersection as one would think about set intersection. Applied to a parametrized curve in Euclidean space, the corresponding notion of self-intersection would simply require the same point be reached at two different times. However, this notion would deem tangency to be an intersection. Amitai uses a more stringent (and dare I say, more common) notion of (self-)intersection which requires not only coincidence at a point at two different times, but also distinction in the derivatives at the two times. – daegan Sep 17 '20 at 02:02