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!

Homework Help: Existence and Uniqueness of solutions (pretty )

  1. May 7, 2009 #1
    Existence and Uniqueness of solutions (pretty urgent)

    1. The problem statement, all variables and given/known data

    I need to solve some problems and I've given one as an example.
    The question is if there is existence and uniqueness of solutions to the DE

    2. Relevant equations

    u'(x) = sin(u(x))

    3. The attempt at a solution

    I know that the first step I'm supposed to work out is the satisfaction of the Lipschitz condition... but I don't even know how to begin with that! My text is pretty theoretical... so a worked example would really help. I get the idea of the theorem. I just have trouble applying it.
    Last edited: May 7, 2009
  2. jcsd
  3. May 8, 2009 #2


    User Avatar
    Science Advisor

    Re: Existence and Uniqueness of solutions (pretty urgent)

    The Lischitz condition (on a set) is that, for x and y in the given set, [/itex]|f(x)- f(y)|\le c|x-y|[/itex]. In the case of sin x- sin y you might write sin x and sin y in MacLaurin series as x- (1/3!)x^3+ ..., and y- (1/3!)y^3+ ... and so get sin x- sin y= x-y + terms of order (x-y)^3.

    But here's a simpler way to do it: "Lipschitz" on a compact (closed and bounded) set is between "continuous" on that set and "continuously differentiable" on the set. If f(y) is differentiable on a set, by the mean value theorem, if x and y are in that set then there exist c such that [itex]|f(x)- f(y)|\le |f'(c)||x- y|[/itex]. Since f' is continous on the closed and bounded set, there is an upper bound, M, for |f'(c)| on the set and [itex]|f(x)- f(y)|\le M|x- y|[/itex].

    For example, by the mean value theorem, given any x, y, [itex]|sin(x)- sin(y)|\le |sin(c)||x- y|[/itex] for some c between x and y. In this simple case, [itex]|sin(c)|\le 1[/itex] for all c so [itex]|sin(x)- sin(y)|\le |x-y| and so sin(x) is Lipschitz.

    ("Lipschitz" is strictly between "continuous" and "continuously differentiable" on a set. There exist functions that are continous on a set but not Lipschitz and functions that are Lipschitz but not continously differentiable. However, those are very unusual functions.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook