1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove that x(t) is bounded

  1. Nov 10, 2013 #1
    Hi, I want to answer the following question:

    x=x(t) is continuous on [0,T) and satisfies

    1 ≤ x(t) ≤ C[itex]_{1}[/itex] + C[itex]_{2}[/itex]∫[itex]^{t}_{0}[/itex] x(s)(1+logx(s)) ds

    for 0 ≤ t < T. Prove x(t) is bounded on [0,T].


    Using Gronwall's inequality I get to

    x(t) ≤ C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex] ∫[itex]^{t}_{0}[/itex] (1+logx(s)) ds )

    ≤ C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex]t + C[itex]_{2}[/itex]∫[itex]^{t}_{0}[/itex] logx(s) ds )

    Can I say that this is less than C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex]T + ∫[itex]^{T}_{0}[/itex] logx(s) ds ) ?

    I'm not too sure where to proceed from here. Would it be helpful to use x(s) > logx(s)?

    Any help is appreciated!
     
    Last edited: Nov 10, 2013
  2. jcsd
  3. Nov 15, 2013 #2
    Can any body help at all? I'm not sure how to bound the integral of logx(s).
     
  4. Nov 15, 2013 #3

    Mark44

    Staff: Mentor

    What does logx(s) mean?
     
  5. Nov 15, 2013 #4

    pasmith

    User Avatar
    Homework Helper

    Gronwall's inequality will not help you here; your bound is not of the correct form.

    You have
    [tex]1 \leq x(t) \leq U(t) = C_1 + C_2 \int_0^t x(s)(1 + \log x(s))\,ds.[/tex]

    The worst case scenario is [itex]x(t) = U(t)[/itex], which gives
    [tex]
    U(t) = C_1 + C_2 \int_0^t U(s)(1 + \log U(s))\,ds.[/tex]

    The right hand side is differentiable, so we obtain
    [tex]
    \frac{dU}{dt} = C_2 U (1 + \log U)
    [/tex]
    This can be solved subject to the initial condition [itex]U(0) = C_1[/itex] to obtain [itex]U(t)[/itex], and one can then check whether
    [tex]
    \lim_{t \to T^{-}} U(t)
    [/tex]
    is finite.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted