- #1

Rasalhague

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

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

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 [itex]\mathbb{R} \times U[/itex] 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?