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One- parameter groups

  1. Aug 6, 2007 #1

    WWGD

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    Gold Member

    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. jcsd
  3. Aug 26, 2007 #2

    Chris Hillman

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    First of all, just to be clear, are you asking whether the vector field
    [tex]x^2 \, \partial_x + \partial_y[/tex]
    is complete on [itex]R^2[/itex]? (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.
     
  4. Aug 27, 2007 #3

    WWGD

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    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.
     
  5. Aug 31, 2007 #4

    Chris Hillman

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    Hi, WWGD,

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

    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
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