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Math Help - alpha and omega limit sets

  1. #1
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    alpha and omega limit sets

    Consider the solution to x''= -x with the initial conditions x(0) = 0, x'(0) = 1. Find the alpha and omega limit sets for this solution.
    (Then rewrite as a 2x2 system)
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  2. #2
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    Quote Originally Posted by Borkborkmath View Post
    Consider the solution to x''= -x with the initial conditions x(0) = 0, x'(0) = 1. Find the alpha and omega limit sets for this solution.
    (Then rewrite as a 2x2 system)
    If you solve the ODE you get

    x(t)=\sin(t)

    The definition of an omega limit point is if there exists a sequence that goes to infinity such that

    a_n \to \infty \implies x(a_n)=r, r \in \mathbb{R}

    Since there exists such a sequence for any

    r \in [-1,1]

    This is the omega set.

    The alpha set is the same except the limit must go to negative infinity.

    for the next part let

    y(t) = x'(t) \implies y'(t)=x''(t)

    Now use the ODE to write this as a first order system.
    Last edited by TheEmptySet; May 23rd 2011 at 09:43 AM. Reason: typo
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  3. #3
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    Ty for your fast response.
    I also got the x(t) = sin(t) after staring at the problem for a little.

    But, I think that the way you wrote the limit set and what I got are different.
    \alpha-limit set = {sin(t) : t \in [-2pi, 0]}
    \omega-limit set = {sin(t) : t \in [0, 2pi]}
    Unless, what you wrote and what I wrote mean the same thing.

    Also, for the last part, I seemed to have forgotten a word. "Rewrite the equation as a 2x2 system". I am not sure which equation it is referring to.
    But would it be:
    (cos(t) sint(t)
    -sin(t) cos(t))
    Last edited by Borkborkmath; May 24th 2011 at 10:29 AM.
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  4. #4
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    Quote Originally Posted by Borkborkmath View Post
    Ty for your fast response.
    I also got the x(t) = sin(t) after staring at the problem for a little.

    But, I think that the way you wrote the limit set and what I got are different.
    /alpha-limit set = {sin(t) : t /in [-2pi, 0]}
    /omega-limit set = {sin(t) : t /in [0, 2pi]}
    Unless, what you wrote and what I wrote mean the same thing.
    sin(t) and cos(t) take on all values between -1 and 1.

    Also, for the last part, I seemed to have forgotten a word. "Rewrite the equation as a 2x2 system". I am not sure which equation it is referring to.
    But would it be:
    (cos(t) sint(t)
    -sin(t) cos(t))
    There is only one equation given: x''= -x. As TheEmptySet suggested, let y= x' so your equation becomes y'= -x and you have the system of equations x'= y, and y'= -x.
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