Why is Arnold's extended phase space a strip rather than a rectangle?

[Rasalhague]

I'm reading Arnold: Ordinary Differential Equations, Chapter 1. In section 1.2, an integral curve was defined as the graph, in the extended phase space, \mathbb{R} \times M, of the motion \phi : \mathbb{R} \rightarrow M of a phase point in M. In 2.2, an integral curve is defined as the graph of a solution, \phi : I \rightarrow U, to a differential equation \dot{x} = \mathbf{v}(x), where I and U are open intervals.

Now the extended phase space is said to be "a strip \mathbb{R} \times U in the direct product of the t-axis and the x-axis". Why is it not a rectangle I \times U? What if I \neq \mathbb{R}?

I see the Wikipedia article Dynamical systems, in defining a dynamical system in general, makes the domain of the evolution function a subset of what Arnold calls the "extended phase space", and suggests that I(x) is not necessarily equal to T (in the notation of this page). Is I(x) always equal to T = R for a real dynamical system, a.k.a. flow? And is that why Arnold's extended phase space has to be \mathbb{R} \times U rather than IxU?

Is "the integral curve of a differential equation" (being the graph of a solution) not necessarily defined for all of the extended phase space of the equation, and therefore not an integral curve in the sense of Arnold Ch. 1, section 1.2?

 

[mathwonk]

you have identified a crucial property in differential equations, namely when is the solution defined for all "time"? I agree with your reading that arnol'd has designated the words "phase space", or "one parameter group" for the case where the solution IS defined for all real numbers. Check out sections 3.5 and 3.6 of chapter 1, where he discusses when this may not happen. as i recall it holds when the manifold is compact, and maybe the equation is linear?? I am not an expert, but it is usual for different authors to make their own conventions as to the use of language. Whatever they call it, it is important to know when the solution is defined for all t.

 

[Rasalhague]

Thanks for the pointer, mathwonk. In sections 3.4 and 3.5, he says that not every differential equation on the line has an associated one-parameter group (=phase flow). In 3.6, he says the reason there is no phase flow in the case of the example in 3.5 is that the t-advance mappings g^t are not defined for all x, that is, I think, the domain is not the whole of R for all (any?) of them. And yes, he says in 3.6, that "every differentiable velocity field on a compact manifold is the phase velocity field of a one-parameter group of diffeomorphisms. The example he gave in 3.5 was nonlinear in x. 3.3 talks about linearity--but I'll hold off paraphrasing for now till I've got some of these definitions straight in my head.

A couple of incidental ponderings on glancing ahead to section 3:

In section 3, "the phase flow associated with a differential equation" is the family (i.e. set?) of mappings {g^t : t is in R} : M --> M, where M is a set called the phase space. In section 1, a phase flow was a tuple (M,{g^t : t is in R}). I thought when I read section 1 that this definition seemed a bit superfluous, since M is already part of the definition of each g^t. Maybe this is why the M has been dropped in section 3.

Wikipedia defines a "one-parameter group" as a continuous group homomorphism from R (with addition) to G, where G is the underlying set of another topological group. It says that a "one-parameter group", so defined, is not a group. But Arnold's one-parameter group maps a set (the phase space) to itself, and composition follows the same rules as addition of real numbers, which makes me think that Arnold's one-parameter group is indeed a group.