# One- parameter groups

1. Aug 6, 2007

### WWGD

Hi, everyone. I am new here, so I hope I am follow the protocols. Please
let me know otherwise. Also, I apologize for not knowing Latex yet, tho
I hope to learn it soon.

am trying to show that the vector field:

X^2(del/delx)+del/dely

Is not a complete vector field. I think this is
from John Lee's book, but I am not sure (it was in my
class notes.)

From what I understand, we need to find the
integral curves for the vector field first, i.e
we need to solve the system:

dx/dt=[x(t)]^2

and

dy/dt=1

I found the solutions to be given by (1/(x+c),y+c')

c,c' real constants.

In my notes ( 2-yrs old, unfortunately) , there is a solution:

Phi(x,t)=(1/(1-tx), y+t)

somehow in function of (x,t)

In addition, there is a statement that Phi(x,t)
satisfies:

Phi(x,t+s)=Phi(x,t)oPhi(x,t) (o = composition)

and that Phi satisfies certain initial conditions
(which were not given explicitly for the problem).

I suspect this has to see with one-parameter groups,
but I am not sure of it, and I don't understand them
that well, nor the relation with complete V.Fields.

I would appreciate any explanation or help.

2. Aug 26, 2007

### Chris Hillman

First of all, just to be clear, are you asking whether the vector field
$$x^2 \, \partial_x + \partial_y$$
is complete on $R^2$? (Click on the graphic to see the latex code I used to produce this expression.)

(For other readers: a complete vector field, thought of as a homogeneous linear first order differential operator, generates a globally defined flow, i.e. a global action by R on our manifold.)

Good book by the way, one of the few which stresses the crucial global versus local distinction.

3. Aug 27, 2007

### WWGD

Actually, yes, that is what I was asking. Sorry, I tried to use the quote option,
but it came out garbled.

" (For other readers: a complete vector field, thought of as a homogeneous linear first order differential operator, generates a globally defined flow, i.e. a global action by R on our manifold.)

Good book by the way, one of the few which stresses the crucial global versus local distinction.[/QUOTE]"

Thanks.

4. Aug 31, 2007

### Chris Hillman

Hi, WWGD,

Well, first you should try to find the integral curves of the vector field $x^2 \, \partial_x + \partial_y$; good, that's what you did. Good, you got
$$x(s) = \frac{x_0}{1 - s \, x_0}, \; y(s) = y_0+s$$
(That's the unique integral curve, parameterized by s, which passes through $(x_0,y_0)$.)

There are basically two things to look for:
1. some of the integral curves "run off to infinity" in finite lapse of parameter,
2. some point in R^2 is a singularity of the vector field.
See p. 440 of your textbook.

About one parameter groups (aka unidimensional subgroups of the group of diffeomorphisms on R^2, aka uniparametric subgroups), yes, a globally defined flow on R^2 is an action by R on R^2, so one way of thinking about this is to ask whether, when we let s by any real number above, our integral curves define an action by R on R^2 (think of R as the group of real numbers under addition). See p. 448 of your textbook.

Last edited: Aug 31, 2007