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Find the Limit?

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Let x be the solution to dx/dt=(pi-x)cos(5x)
    x(0)=4pi/3
    Find the lim x(t) as t goes to infinity.


    2. Relevant equations



    3. The attempt at a solution

    cos(5x)=0 when 5x=n(pi), so when x=n(pi)/5 which = 4(pi)/3 since we have that initial condition.
    so, n=20/3, x=4(pi)/3 ??...I don't know what to do!

    any hints?
     
  2. jcsd
  3. Mar 3, 2009 #2
    Did you try integrating dx/dt?
     
  4. Mar 3, 2009 #3
    I suggest trying a numerical solution to get the answer, and then using analysis to explain it.
     
  5. Mar 4, 2009 #4
    Well, there is no need for numerical solution, unless you want to practice, but like was suggested, all you need to do is integrate with respect to x, and then what you have done is found the solution to that diff .eq.

    Try to do that, and then come back again if you still have trouble.
     
  6. Mar 4, 2009 #5

    gabbagabbahey

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    I'm going to go out on a limb here and guess that neither you or Iubuntu have actually tried to do the integration:wink:....
     
  7. Mar 4, 2009 #6

    gabbagabbahey

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    Really?!....not if n is an integer it doesn't....[itex]cos(0)\neq 0[/itex]

    Huh?! x(0)=4pi/3 is just your initial condition, cos(5x(0)) does not equal zero.

    If [tex]\lim_{t\to\infty} x(t)[/tex] exists, what do you think the slope of the graph of x vs. t will look like for very large t? What does that tell you about dx/dt there?

    What is dx/dt at t=0? is x increasing or decreasing there?
     
    Last edited: Mar 4, 2009
  8. Mar 4, 2009 #7

    lanedance

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    plotting dx/dt is also pretty useful, start at your initial condition and try and imagine how x will move as t as increased (ie for a poistive dt)
     
  9. Mar 4, 2009 #8
    Well, that is sadly correct!

    Looking at it closer, i just realized that it actually is not as easy as i initially thought, since on the RHS we have only x's which are defined implicitly as a function of t. So, we defenitely will run into trouble trying to integrate it, since we need to divide by RHS, and the integral is not going to look that easy, i would guess, since yet, i haven't tried it myself.
     
  10. Mar 4, 2009 #9
    You don't need to find an explicit solution in any event.

    Just work through what happens using logic.

    Initially, x(0) = 4pi/3. Ergo x'(0) = (-1/3 pi) cos(20/3 pi) = -1/6 pi
    ... check my calculation on that...

    So it's heading down. Alright...

    If x decreases a little bit, then the (pi - x) will stay negative, and the cos(5x) will remain positive... so it keeps heading down. In fact, we see that this is true until we get to the point where

    (you fill in the answer!)

    At which point x'(0) = 0 and the thing stops, forever. The end.
     
  11. Mar 4, 2009 #10

    lanedance

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    but will only really approach that point as t approaches infinity
     
  12. Mar 4, 2009 #11
    For a unique set of initial conditions, its asymptotic behavior can be determined. The system is completely deterministic. So yes... it really will.
     
  13. Mar 4, 2009 #12

    lanedance

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    x'(0) is not zero

    hey csprof - agree with everything you said, just making the point that it tends to that asymptotic position but only in the limit, gets as close as you want, but never actually stops
     
  14. Mar 4, 2009 #13
    Oh yeah, I meant that x'(x) = 0, and the thing stops.

    And yes, it never actually makes it there... I guess I should have been more precise and said "even if it did ever make it there, it would then stop and go no further". But since the derivative isn't zero until there, it'll keep getting closer. It can be made more formal by saying the derivative gets smaller and smaller, etc.
     
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