Trajectory intesection with itself

Question

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?

Answer 1

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.

Answer 2

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).