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Existence and Uniqueness of solutions (pretty )

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Existence and Uniqueness of solutions (pretty urgent)

Homework Statement



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

Homework Equations



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

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.
 
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Answers and Replies

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HallsofIvy
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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.)
 

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