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    The Lipschitz condition seems states that l f(x,u)-f(x,v) l ≤ A l u-v l with a few conditions which you know if you know what Lipschitz is...

    However what if you have f(x,y)=x√y on the rectangle

    Does this satisfy Lipschitz?

    You have that
    lf(x,u)-f(x,v)l ≤ h l √u-√v l

    but surely this doesnt mean Lipschitz is satsfied yet I'm sure I need Lipschitz to be satisfied so that I can do the rest of the question...

    I'm doing the same worksheet as you it seems... I haven't any idea either. And it was due in hours ago. Sorry.

    Notice that f is differentiable wrt y everywhere except y=0. Differentiable => Lipschitz so f satisfies that Lipschitz condition everywhere except y=0.

    Basically, that condition puts a bound on the gradient as y increases. And as you get close to the line y=0 the gradient shoots off to infinity. So there is no bound.

    I'll add more. y' = x√y, y(0)=b > 0.
    You want to find a rectangle R = {(x,y) : |x|≤ h, |y-b|≤ k, h,k>0} that you can apply Picard's theorem on.
    You need f to satisfy the Lipschitz condition on R so R cannot contain the x axis, that is, 0<k<b.
    Next you want the rectangle to satisfy M≤(k/h) where M is the maximum of f on R. Well it's fairly clear that the largest value x√y takes on R is at x=h, y=b+k. So take M=h√(b+k). That gives h²≤ k / √(b+k). So as long as h and k satisfy those conditions you can apply Picard's to R to get a unique solution on [-h , h]
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Updated: October 16, 2006

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