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Thread: Terminating Motion

  1. #1
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    Terminating Motion

    For the dynamical system described by


    dx/dt =x(x−1)(x+1)



    1. (a) show that the motion is not terminating for theinitial condition x(0) = x0 = 1/2
    2. (b) show that the motion is terminating for the initialcondition x(0) = x0 = 2
    3. (c) for the initial condition given in (b) find an upperbound on the time required to reach +∞
    4. (d) for the initial condition in (b) calculate the timerequired to reach +∞ exactly.







    Please help I have really no idea where to even begin. I'm not sure what I'm even supposed to be showing! Can anyone give me a method?
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  2. #2
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    Re: Terminating Motion

    Then where in the world did you get this problem? Do you know what "x" is? Do you know what "t" is? The motion "terminates" when dx/dt= 0. Where does that happen?
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  3. #3
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    Re: Terminating Motion

    further from toying with it a bit the problem has got it wrong.

    initial positions of $|x|<1$ lead to terminal motion.

    positions of $|x|>1$ lead to non-terminal motion.
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    Re: Terminating Motion

    Okay so it terminates at the points (0,0) (-1, 0) (1,0)
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  5. #5
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    Re: Terminating Motion

    It looks like you still don't understand the problem. For any x between 0 and 1: x is positive, x- 1 is negative, and x+ 1 is positive. Therefore, dx/dt= x(x- 1)(x+ 1) is negative. The solution with x(0)= 1/2 has dx/dt negative so x decreasing to x= 0 where it terminates so 1 (a) is incorrect- it does termininate. For any x larger than 1, all three terms, x, x- 1, and x+ 1 are positive so dy/dx= x(x+ 1)(x- 1) is positive. The solution with x(0)= 2 is increasing and does not terminate, x goes to infinity as the last two parts indicate. To answer the last two parts you will need to actually find x(t) by integrating, using "partial fractions".
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