5
$\begingroup$

To me it seems that is not always the case that a discrete dynamical system of the form $$ x_{k+1} = f(x_k) $$ can be seen as the discretization of a continuous one like $$ \dot{x}(t) = f(x), $$ while we can always go in the opposite direction.

My question is: is there some keyword I can search in the literature to understand better this connection or do you have any reference to suggest?

I would like to understand if there are some sufficient conditions guaranteeing the possibility of doing this transition, I don't even need a constructive way to build this continuous version, "just" existence would already be a great thing to understand.

To be clear, I would appreciate even results on this sort of "continuous prolongation" based on increasing the dimension of the phase space, i.e. for example for a discrete system of $\mathbb{R}^2$ if there is a dimension $d\geq 2$ such that on it there is a well defined continuous system whose discretization at certain time instants correspond to the discrete one. This question should be easier since there is not the constraint in the dimension.

I know it is quite a broad question, but I need at least some terminology to look for.

Up to now I have read about embeddability of homeomorphisms into flows and this seems to answer to my question in some cases, so I just wanted to know if the embeddability into flows is the right thing to study or if there are other perspectives.

P.S. About getting an approximation of them I am finding very interesting this question Links between difference and differential equations? at the moment.

$\endgroup$
9
  • $\begingroup$ I don't understand why you say Clearly, it is not always the case that a discrete dynamical system of the form. Why isn't $ x_{k+1} = f(x_k) $ a discretization of $\dot{x}(t) = f(x) -x$ by writing $\dot{x}(t) \approx x_{k+1} - x_k$? $\endgroup$ Dec 30, 2020 at 17:49
  • 1
    $\begingroup$ This is pure conjecture but I am not sure if I agree with the first statement. I think you can always go from discrete to continuous if you don't insist that f be a nice function. Especially, if you allow f to have time dependency. $\endgroup$
    – Calculon
    Dec 30, 2020 at 17:50
  • $\begingroup$ Ok nice, my thought was based on the fact that with discrete dynamical systems you can change the topology of the sets you evolve, while for example you can have some obstructions in continuous (even non-autonomous) systems because you can not have trajectories intersecting at the same time instant. By the way I remove clearly, it is not that clear at this point and it might even be a wrong fact $\endgroup$
    – Dadeslam
    Dec 30, 2020 at 17:52
  • 1
    $\begingroup$ I am not a specialist in dynamical systems, but I once heard a very prominent one claiming that the difficulty in studying dynamical systems progresses as follows: (1) flows in dimension $n$, (2) diffeomorphisms in dimension $n$, (3) flows in dimension $n+1$. The reason is that from (1) you get an example of (2) by considering the time 1 map of the flow, and from (2) you get an example of (3) by considering the flow on the mapping torus. $\endgroup$
    – Ruy
    Dec 30, 2020 at 17:59
  • $\begingroup$ En passant, it is indeed not true that every diffeomorphism of $\mathbb R^n$ is the time 1 map of an autonomous flow, since the latter always preserves orientation. For example the map $f(x)=-x$ on $\mathbb R$ doesn't. $\endgroup$
    – Ruy
    Dec 30, 2020 at 18:09

1 Answer 1

5
$\begingroup$

As you allude to in your question, there is indeed a construction which inputs a discrete dynamical system in the form of a self-diffeomorphism $f : M \to M$ of a smooth manifold $M$, and outputs a continuous dynamical system called the suspension flow of $f$. This suspension flow is defined on a manifold $M_f$ of one dimension higher than $f$, namely the quotient of $M \times \mathbb R$ under the equivalence relation generated by $(x,t) \sim (f(x),t-1)$. This suspension flow is also generated in the usual fashion by a vector field on $M_f$, namely the quotient vector field of the $\frac{d}{dt}$ vector field on $M \times \mathbb R$ equipped with coordinates $(x,t)$, $x \in M$, $t \in \mathbb R$. Thus, $f$ can be realized as the time $1$ map of the suspension flow, restricted to the image of $M$ embedded in $M_f$.

There are further generalizations of this construction beyond the realm of self-diffeomorphisms.

In the case of a self-homeomorphism $f : M \to M$ where $M$ is just a topological space, the construction of the suspension flow on $M_f$ works exactly as stated, producing an action of $\mathbb R$ on $M_f$. However, $M_f$ is not a smooth manifold and so its rather tricky to express the suspension flow as being generated by a vector field.

There is a still further generalization to any continuous self-map $f : M \to M$ where $M$ is again a general topological space. There is again a "suspension flow" construction, defined on the quotient of $M \times [0,\infty)$ by the equivalence relation $(x,t) \sim (f(x),t-1)$ for $x \in M$, $t \in [1,\infty)$. The reason for restricting to positive time is that $f$ itself can only be iterated with positive exponents; there is no inverse map $f^{-1}$.

$\endgroup$
5
  • 1
    $\begingroup$ That is exactly what I called the mapping torus of a diffeomorphism in my comment above. $\endgroup$
    – Ruy
    Dec 30, 2020 at 21:14
  • $\begingroup$ In some sense, this is just evading the posed problem using a rather trivial construction. The question asked for a vector field on $M$ itself. $\endgroup$ Dec 30, 2020 at 21:30
  • 3
    $\begingroup$ @LutzLehmann: It's less clear to me than to you that the OP was asking for a vector field on $M$ itself. The post does not clearly elucidate the phase space in any situation, and it does indeed allude to "embeddability of homeomorphisms into flows" as a potential answer. $\endgroup$
    – Lee Mosher
    Dec 30, 2020 at 21:58
  • 1
    $\begingroup$ $+1$ This is of course the canon. $\endgroup$
    – John B
    Jan 1, 2021 at 16:35
  • 1
    $\begingroup$ Thank you, very clear. $\endgroup$
    – Dadeslam
    Jan 3, 2021 at 16:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .