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Maximum interval of the existence

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose f(t,x) is a continuous vector valued function on [tex]\mathbb{R} \times \mathbb{R}^n[/tex]. If f is locally Lipschitz with respect to x with the property that [tex]\|f(t,x)\| \le C \|x\|[/tex] for some positive constant C > 0, then prove that the maximum interval of the existence of the initial value problem x' = f(t,x) with [tex]x(t_0) = x_0[/tex] is equal to [tex](-\infty, \infty)[/tex].

    3. The attempt at a solution

    f is locally Lipschitz with respect to x with the property that [tex]\|f(t,x)\| \le C \|x\|[/tex] for some positive constant C > 0. From the local existence and uniqueness theorem, the system has a unique solution defined on [tex][t_0 - a, t_0 + a][/tex].
    Let
    [tex]\alpha[/tex] = inf{[tex]l < t_0[/tex]: there is a unique solution on [tex][l, t_0][/tex]}
    [tex]\beta[/tex] = sup{[tex]r > t_0[/tex]: there is a unique solution on [tex][t_0, r][/tex]}

    By the definition of [tex]\alpha[/tex] and [tex]\beta[/tex], and the existence-uniqueness theorem, there is a unique solution x(t) defined on [tex](\alpha, \beta)[/tex].

    Assume there is a compact set K [tex]\in \mathbb{R}^n[/tex] s.t. the solution [tex]x(t) \in K[/tex] for all [tex]t \in J[/tex], J = [tex][t_0, \infty)[/tex], that is, the solution is defined for all [tex]t > t_0, t \in \mathbb{R}[/tex].

    Assume J = [tex][t_0, \infty)[/tex] does not hold. Thus, x is defined on [tex][t_0,\beta), \beta < \infty[/tex], but there is no solution on [tex][t_0, \beta][/tex], and [tex]\beta \in \mathbb{R}[/tex].

    Consider the function
    [tex]
    \[ y(t) = \left\{\begin{array}{ll}
    x(t) \,\, if \,\,t \in [t_0, \beta)\\
    y_0 \,\, if \,\,t = \beta \end{array} \right \][/tex]

    where [tex]y_0[/tex] is any element in K. Thus, [tex]\| f(t,y) \| \le C \| y \|[/tex].

    Let g(t) = max{[tex] C \|x\|[/tex]} and [tex]\| f(t,y) \| \le g(t)[/tex]. F(t,y) is bounded by g(t), which is an integrable function on [tex][t_0, \beta][/tex] and is therefore integrable itself.

    Define [tex]x(\beta) = x_0 + \int^\beta_{t_0} f(s, y(s)) \, ds[/tex]

    Since this is the limit of the integral [tex]\int^t_{t_0}[/tex] as [tex]t \to \beta[/tex], and K is closed, [tex]x(\beta) \in K[/tex] and in particular [tex]x(\beta) \in \mathbb{R}^n[/tex]. Also, since f(s, y(s)) = f(s, x(s)) almost everywhere, x satisfies
    [tex]x(t) = x_0 + \int^t_{t_0} f(s, x(s)) \, ds
    = x_0 + \int^t_{t_0} f(s, y(s)) \, ds[/tex]
    on the inteval [tex][t_0, \beta][/tex]. Thus, it is absolutely continuous (from the first equality) and it satisfies the IVP (second equality). We defined an extension to [tex][t_0, \beta][/tex], contradicting maximality.

    Did I make sense?

    Thank you.

    Regards,
    Rayne
     
  2. jcsd
  3. Sep 8, 2007 #2

    EnumaElish

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    I am unfamiliar with some of the concepts you used. How is "maximum interval" defined? How does existence over a compact interval contradict maximality?
     
  4. Sep 9, 2007 #3

    EnumaElish

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    You assume x(t) is in compact K for all t in J. Next you posit that if J was not right, then the solution must be defined over a right-open interval but not over its closure. You then prove that this cannot be the case.

    Then you argue that this contradicts maximality.

    Have I gotten it right so far?

    It makes sense to me up until "contradicting maximality." And that's because I am unfamiliar with the notion of a maximum interval.
     
    Last edited: Sep 9, 2007
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