0
$\begingroup$

I'm only aware of ODE uniqueness on open subsets of $R^n$, which is useful when we want to prove uniqueness of integral curves on a smooth manifold.

Suppose we have an interal curve $f:[0,\delta]\to M$ that starts on the boundary, how can we show it is unique?

$\endgroup$
  • $\begingroup$ Welcome to MSE. What is the context of your question? What are your assumptions on the vector field? What have you done? Where are you stuck? $\endgroup$ – user539887 Apr 22 at 7:45
  • $\begingroup$ @user539887 I thought this is legitimate as a stand alone question. The real context is I'm self learning Lee, and there is a problem that asks you to understand flowouts from the boundary of a manifold with a boundary. I have a good sketch of proof apart from this part I'm not sure about (I believe it's true, but not sure). The naive attempt says we take a boundary chart $A$, so that we have a smooth map $A \to R^n$ representing the vector field, so that we can extend this to a map from a nbr of $A$, but the problem is that the extension has nontrivial freedom. $\endgroup$ – user135743 Apr 22 at 17:37
  • $\begingroup$ If at each point of the boundary the vector field is transverse to it then integral curves are defined (at least for $t>0$ or for $t<0$, sufficiently close to $0$), and the necessity to extend the field "outside" $M$ makes no difficulty. But if the field is at some points of the boundary tangent and at others transversal, the situation becomes much more complicated. Such problems have been considered by control theory people, see Bony–Brezis theorem. $\endgroup$ – user539887 Apr 23 at 8:29
  • $\begingroup$ @user539887 Thanks, we may assume that at each point of the vector field it's inwardly pointing. I don't think I follow your argument; if we have two integral curves $a,b$ of a smooth inward pointing vector field $V$ starting on $p$ (a point on the boundary) that are different in arbitarily small nbr of $p$, I am not able to obtain a contradiction from the following; if we take a boundary chart $U$ of $p$, and consider an extension of $V$ to this $U$, then there is a unique integral curve passing through $p$, but it doesn't have to be either of $a,b$, who are defined on an half interval. $\endgroup$ – user135743 Apr 23 at 18:44
  • $\begingroup$ The integral curve starting at a point on the boundary is independent, for $t>0$, of an extension of the vector field. $\endgroup$ – user539887 Apr 23 at 20:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.