Existence and Uniqueness of solutions (pretty )

[lurkre]

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.

 

[HallsofIvy]

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 $|f(x)- f(y)|\le |f'(c)||x- y|$. Since f' is continuous on the closed and bounded set, there is an upper bound, M, for |f'(c)| on the set and $|f(x)- f(y)|\le M|x- y|$.

For example, by the mean value theorem, given any x, y, $|sin(x)- sin(y)|\le |sin(c)||x- y|$ for some c between x and y. In this simple case, $|sin(c)|\le 1$ for all c so $|sin(x)- sin(y)|\le |x-y| and so sin(x) is Lipschitz.<br /> <br /> <br /> ("Lipschitz" is <b>strictly</b> between "continuous" and "continuously differentiable" on a set. There exist functions that are continuous on a set but not Lipschitz and functions that are Lipschitz but not continously differentiable. However, those are very unusual functions.)$