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Differential Equations behavior for large t?

  1. May 12, 2013 #1
    dy/dt = 2 - 2ty
    y(0) = 1

    I am not asked to solve this (I know it's not easy to solve), but what I am asked is,

    "for large values of t is the solution y(t) greater than, less than, or equal to 1/t"?

    I would think less than because 1/e^(t^2) converges faster than 1/t, but at the same time I solved dy/dt = 0 and got y = 1/t, so I'm not sure what that signifies.
     
  2. jcsd
  3. May 13, 2013 #2
    I am not sure how you got 1/t by solving dy/dt = 0, perhaps there is a typo?

    To look at the behavior for large t, I rescaled variables to t=τ/ε. The differential equation then becomes

    [itex]\epsilon^{2}\frac{dy}{d\tau} = 2\epsilon - 2\tau y[/itex].

    Now large t corresponds to the limit ε→0, so ignoring the second order in ε you would get the behavior [itex]y = \frac{\epsilon}{\tau} = \frac{1}{t}[/itex] for large t.

    As a check I used maple to solve the equation and then plotted it against 1/t. The convergence to 1/t seems pretty rapid actually. By the time you hit t=5 the solution differs from 1/t by only ~0.004.

    I think problems like these come up when you are looking at asymptotic methods for solving differential equations. This is something I am just getting into for a summer research project so I am by no means an authority on the method. I don't really have any book recommendations for this either... but I am reading some sections of Perturbation Methods by Ali Nayfeh. You could probably find more information by searching up asymptotic methods.
     
    Last edited: May 13, 2013
  4. May 13, 2013 #3
    dy/dt = 0 => 2 - 2ty = 0

    2(1 -ty) = 0
    1 - ty = 0
    1 = ty
    y = 1 / t

    This of course assumes can't be zero.

    Thanks for looking into it, just to clarify:

    According to your results from maple, are you saying, it should actually be equal to 1 / t for large t?
     
  5. May 13, 2013 #4
    Yes, for large t it goes like 1/t. Maple was just a check though, I would base it primarily on the epsilon going to zero limit. epsilon^2 goes to zero faster than epsilon which is why I set it to zero, leaving epsilon by itself. If you set both epsilon and epsilon^2 to zero then you get an even cruder approximation for large t, mainly that y tends to 0 for large t, which is also approximately true, but the 1/t approximation is even better.

    What was your reasoning for setting dy/dt = 0?
     
  6. May 13, 2013 #5
    Right, thanks for the help!

    Well usually when we have an autonomous equation dy / dt = f(y) by setting dy/dt = 0 we get the equilibrium solution which is usually a constant, I thought the same idea could be extended to non-autonomous equations dy/dt = f(t,y), but in this case you would get the equilibrium solution as a function t; though I'm not sure if it's mathematically valid, or it may be a case of where it's bad form mathematically, but it still results in the 'correct solution'.
     
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