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Homework Help: 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


    Staff: Mentor

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


    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
    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
    \frac{dU}{dt} = C_2 U (1 + \log U)
    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
    \lim_{t \to T^{-}} U(t)
    is finite.
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