Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relationship between two solutions

  1. Sep 10, 2011 #1
    Let [itex] t \in \mathbb{R} [/itex] be large.
    Let [itex] f [/itex] be a function over [itex] [0,t] [/itex] satisfying [itex] f(0) = 1 [/itex] and [itex] f'(x) = e^{-f(x)} [/itex] for all [itex] x[/itex].
    Let [itex] g [/itex] be a function over [itex] [0,t] [/itex] satisfying [itex] g(0) = 1 [/itex] and [itex] g'(x) = (1 - g(x)/t^2)^{t^2} [/itex] for all [itex] x[/itex]. Note that [itex] g'(x) \sim e^{-g(x)}[/itex].

    Without solving the two differential equations and finding out [itex] f [/itex] and [itex] g [/itex] (which can be done at least approximately), can the fact that [itex] f(0)=g(0)=1 [/itex] and [itex] g'(x) \sim e^{-g(x)} [/itex] be used to show that [itex] f(x) \sim g(x) [/itex] for all [itex] x[/itex] in [itex] [0,t][/itex] ?
    Last edited: Sep 10, 2011
  2. jcsd
  3. Sep 11, 2011 #2


    User Avatar
    Science Advisor

    This makes no sense. If g is a function of x only, its derivative cannot depend upon both x and t. Did you mean [itex]g'(x) = (1 - g(x)/x^2)^{x^2} [/itex]?

  4. Sep 11, 2011 #3
    HallsofIvy, I meant for [itex] g(x) [/itex] to depend on [itex]x[/itex] and [itex]t[/itex]. You should think of [itex]t[/itex] as being a large constant, say, [itex] t=10^{10} [/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook