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I want to show the following that there is a bijective correspondence between a certain group action $g$ on $\mathbb{R}$ (I believe that that's called a flow - I don't know much about ODE's, so please keep the explanations at an accessible level) and global solutions of a certain type of differential equations (so talking about $g$ is the same as talking about solving certain ODE's), i.e. I want to show the following two theorems:

"1) Let $g:\mathbb{R}^{n}\times\mathbb{R}\rightarrow\mathbb{R}^{n}$ be a group action of the additive group $\mathbb{R}$ on $\mathbb{R}^{n}$ such that $\partial_{2}g$ exists on the whole domain. Then there exists the ODE $$ x'=\partial_{2}g\,\left(x,0\right)\left(x\right), $$

such that for a solution $l:\mathbb{R}\rightarrow\mathbb{R}^{n}$ with initial value $x_{0}\in\mathbb{R}^{n}$of the above, we have: $l$ is unique and has the form $l\left(t\right):=g\left(x_{0},t\right)$.

(Note, that $\partial_{2}g$ denotes the vector of the partial derivates with respect to the time/the $n+1$'th argument of $\phi$).

2) Conversely if an ODE $$ x'=f\left(x\right) $$ has unique global solutions, then there exists a mapping $g:\mathbb{R}^{n}\times\mathbb{R}\rightarrow\mathbb{R}^{n}$, such that $g$ is a group action ( like described above) and $\partial_{2}g\,\left(x,0\right)=f\left(x\right)$ holds."

My questions are: 1) Have I even stated this proposition correctly ? Is it true what I'm asking ?

2) How can I prove that ? For 1) for example it is easy to show (thanks to the properties of the group action) that if $l$ has the above form, that it is a solution. But taking an arbitrary solution and proving that $l$ has to be of the above form seems to be very difficult.

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There seems to be some confusion here. If you have an ODE with "initial value" $x_0 \in \mathbb{R}^n$, then your solution will need to be a function whose domain is a subset of $\mathbb{R}^n$ not $\mathbb{R}$. –  Bill Cook Mar 28 '12 at 15:00
    
Ignoring your actual question and trying to guess what you might actually be interested in, are you looking for theorems relating group actions and differential equations? If so, $\mathbb{R}$ acting on $\mathbb{R}^n$ in a nice way, yields a $1$-parameter Lie group action. One can study such actions which map solutions to solutions for some ODE, this study is called "symmetry analysis". Looking at these symmetries yields techniques for solving ODEs. There are several good books on thise topic such as Brian Cantwell's "Introduction to Symmetry Analysis" and Peter Hydon's "Symmetry Methods..." –  Bill Cook Mar 28 '12 at 15:05
    
@BillCook Sorry, with "initial value" I meant $l(0)=x_0$, so $l$ has the value $x_0$ for "initial time". I'm not sure if what you describe is what I'm looking for (I don't know what a Lie group even is) - it sounds a bit overkill what you are saying. I don't think that the answer requires so much sophistication. I'm simply interested if there is a theorem that tells me 1)if for a given action of $\mathbb{R}$ on $\mathbb{R}^n$ all solutions of an ODE associated to this action have a certain form (if you would reprove what I stated above that I have already proven, (...) –  resu Mar 28 '12 at 17:08
    
(..) I think you would better see what I would mean. 2) if for a given ODE all solutions have a certain form (i.e. are global) then there exists a group action such that the ODE can be viewed as in 1) to be "induced" by that group action. Together this would be a bijection between group actions and ODEs: For every group action there is an ODE with a certain property and for every ODE there is a group action with certain property. This would be the content of the above theorems stated vaguely in words. Notice that I'm not even sure that this would have to be the case: Maybe (...) –  resu Mar 28 '12 at 17:11
    
(...) there is a simple counterexample, but I can't find it, since I'm very bad at ODEs. –  resu Mar 28 '12 at 17:14

1 Answer 1

up vote 2 down vote accepted

I'm not sure I fully understand your question. Let me rephrase it differently and you can tell me if my rephrasing is actually what you want. To avoid having to think about things that are too pathological, I will assume your $\mathbb R$ action has a continuous time derivative.

Before rephrasing anything, we notice that your condition on the vector fields $f(x)$ in the ODE $\dot x = f(x)$ means that their flows generate a smooth, additive $\mathbb R$ action. Do you want a proof of this claim? It follows from the local existence theorem, coupled with your hypotheses that you have global existence and uniqueness of solutions to the initial value problem and that $f(x)$ does not depend on $t$. (The jargon: $t$ independent vector fields give rise to autonomous differential equations.)

Now I can reformulate what I understand your question to be. Let $\mathcal F$ be the set of differentiable additive $\mathbb R$ actions on $\mathbb R^n$. Let $\mathcal X$ be the set of vector fields on $\mathbb R^n$ that generate flows with the properties you list (we can obtain ``most'' of these by requiring that the vector field be locally Lipschitz and satisfy some growth condition, but this isn't directly relevant).

You have an obvious map $\mathcal X \to \mathcal F$. This map takes a vector field and spits out the flow that comes from it. Your questions are (1) is this map surjective and (2) is this map injective?

Let's start with (2): the map is injective. This follows from the fact that we can recover the vector field by observing that if $g(x,t)$ is your flow (i.e. $\frac{d}{dt} g(x, t) = f( g(x,t) )$, $g(x, 0) = x$), then $\frac{d}{dt}|_{t=0} g(x,t) = f(x)$, recovers the vector field.

The map fails to be surjective, however, because there are ODEs that almost generate flows while not satisfying the global uniqueness property. We will notice first that any smooth, additive $\mathbb R$ action comes from an ODE in a sense. Define $F(x) = \partial_2 g(x, 0)$. This is the candidate for being our vector field. The additive property gives \[ g(x, t+s) = g( g(x, s), t) \] Differentiate both sides with respect to $t$ and evaluate at $t=0$. This gives \[ \partial_2 g(x,s) = (\partial_2 g)(g(x,s), 0) = F(g(x,s)). \] We already have that $g(x,0) = x$, so this indeed looks like a flow generated by the ODE $\dot x = F( x )$.

This doesn't actually show the map is surjective, however... remember that you wanted your vector fields to have global uniqueness. Consider this example on $\mathbb R$: [ \dot x = 3 x^{2/3}. ] The ``flow'' it wants to generate is the $\mathbb R$ action \[ g(x, t) = (t + x^{1/3} )^3. \] Notice that this $g(x,t)$ defines a flow in the sense you wanted. However, the ODE I wrote down above does not satisfy uniqueness. (There are infinitely many solutions satisfying the IC $x(0) = 0$... notably $x(t) = t^3$ and $x(t) = 0$.) Thus, this isn't in the image of the map $\mathcal X \to \mathcal F$.

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sorry for the extremely late response; I've been a little bit ill for some time and afterwards I hadn' time to do anything, so that's why I accepted your answer barely now. I'm sorry you didn't get your bounty - sadly your answer didn't have enough (=2) upvotes when the bountytime expired, otherwise you would have gotten at least have the bounty (even if at that time your answer wasn't yet accepted) - which now is just lost for everyone. –  resu Apr 30 '12 at 13:26

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