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Phase Space - does each point have a unique time associated with it?

  1. Oct 16, 2008 #1
    Phase Space -- does each point have a unique time associated with it?

    Hi all,

    If I have an autonomous system:


    The k-dimensional state vector x lives in a k-dimensional phase space. Does each point in the k-dimensional phase space have a UNIQUE time associated with it?

    I don't think it does, and this is why. Lets say we choose some point x0, we can iterate the solution forward using a runge-kutta routine. (I'm going to drop the bold vector notation.) That is we find x1 = O(x0, j), where O() is an operator that evolves the solution (our runge-kutta), and j is the step. In our case we use j=1, so we iterate one time step.

    Now, if we let the initial condition x0 exist at time t=0, then when we iterate forward via O() we get x1 which exists at time t=0+1*step. Now, we could just as easily have chosen x0 to exist at time t=100. So it seems to me, that "we" pick a reference time, and tracing out the trajectory returns phase space points from our initial (chosen) reference time!

    So is it just that all points in phase space are unique, like we don't have intersecting trajectories? BUT, we could have a point x0 at time t=0, a point x1 at time t=0, x2 at time t=0, etc... where we picked the t=0 point.

    Could someone explain this to me? I'm trying to read and understand, but I think I'm confusing myself. Thank you!

    P.S. Sorry for the sloppy notation. I hope it can be understood.
  2. jcsd
  3. Oct 16, 2008 #2


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    Re: Phase Space -- does each point have a unique time associated with it?

    For example, the differential equation
    [tex]\frac{d\left(\begin{array}{c}x \\ y\end{array}\left)}{dt}= \left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right)[/tex]
    has general solution
    [tex]\left(\begin{array}{c}C_1cos(t)+ C_2sin(t) \\ -C_1sin(t)+ C_2cos(t)\end{array}\right)[/tex]

    The trajectories in the xy-plane (phase plane) are circles. Each point corresponds to an infinite number of t-values, differing by multiples of [itex]2\pi[/itex]. In general, a closed path in the phase plane indicates a periodic solution with every point on the path corresponding to infinitely many t-values.

    And, of course, there is no "t" on the xy-plane. We can choose the same (x,y) point to correspond to any value of t and see the same trajectory followed for t greater than that starting value.

    The fact that we are free to choose "t= 0" as we please in an autonomous equation is quite different from the existance of periodic solutions but I wasn't sure which you meant.

    Finally, let me point out that we can have intersecting trajectories where f(x) is 0.
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