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May 8, 2020 at 11:13 vote accept Ishan Deo
Apr 14, 2020 at 14:55 comment added Willie Wong In regards to my previous comment: I think in at least the case of the Peano theorem (where the vf is only continuous), the usual proof by looking at the integral formulation and using compactness should go through.
Apr 14, 2020 at 14:43 comment added Willie Wong When we lose uniqueness, maybe we can do with an "existence" formulation of continuous dependence? Something along the lines of "for every bounded open $\Omega\subset\mathbb{R}^N$ there exists a time interval $(-T,T)$ and a continuous function $f: \Omega\times(-T,T) \to \mathbb{R}^N$ such that $\frac{d}{dt} f(\alpha,t) = F(f(\alpha,t))$ with initial data $f(\alpha,0) = \alpha$"?
Apr 14, 2020 at 1:39 answer added demolishka timeline score: 5
Apr 12, 2020 at 14:34 comment added Alexandre Eremenko mathoverflow.net/questions/234183/…
Apr 12, 2020 at 14:27 comment added Alexandre Eremenko For non-Lipschitz differentil equations, there is no uniqueness theorem. So we cannot speak of "dependence" on initial conditions at all. There are some generalizations of uniqueness theorem to more general functions, but it is something very close to the Lipschitz condition.
Apr 12, 2020 at 14:17 comment added Jochen Glueck Welcome to MathOverflow! Hmm, if the vector field is not Lipschitz, the solution will not be unique, in general. So it seems to me that it is not even clear what one means by "continuous dependance on the initial parameter", since there could be many solutions for each initial value. (+1 anyway, since I find the question to be an interesting thought.)
Apr 12, 2020 at 7:50 history edited YCor CC BY-SA 4.0
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Apr 12, 2020 at 10:33
Apr 12, 2020 at 5:47 history asked Ishan Deo CC BY-SA 4.0