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Oct 10, 2017 at 12:50 comment added Willie Wong ...needed in the course of the proof. An english translation is available arxiv.org/abs/1604.02484
Oct 10, 2017 at 12:49 comment added Willie Wong Yes, once you have proven existence, uniqueness, and regularity of solution, Taylor's theorem essentially tells you short time preservation of large scale features (among other things). If you have continuous dependence on data you get error estimates from uncertainty. Precise quantification of uncertainty etc. depends on the precise model, and are dealt with on a case-by-case basis. // The proof of wellposedness often requires "fancy stuff". Evans is a good general PDE intro; but for what you are interested in you might as well read Leray's original paper which explains the fancy stuff...
Oct 9, 2017 at 22:56 comment added john mangual I guess this is just Taylor's theorem? I was looking at fancy stuff like wave fronts or existence of solutions to PDE. Any recommendations for intro books? I found Evans.
Oct 9, 2017 at 21:16 comment added Willie Wong Think of $f:[0,1] \times \Sigma \to \mathbb{R}$; then $f^{-1}[1,\infty)$ is a set in $[0,1]\times \Sigma$. You can look at its sections at constant $t$. // Uncertainties in initial value is taken care of by the fact that wellposedness includes "continuous dependence of solution on initial data" as one of the key ideas. If you really are just asking about why qualitative features seem to persist over small periods of time with some sort of uncertainty propagation, then it really is just that the equations are provably locally well-posed.
Oct 9, 2017 at 19:29 comment added john mangual What does "regular" mean in this context? I guess smooth or $C^1$ or whatever we like. Even better question, if $T$ is finite how can you have an inverse image $f^{-1}$? Because you have $[0,T]$ but also $ [0, \infty] $ and yet $ \Sigma $ is compact. Is $[0, \infty] \subset \Sigma $ ? Back to my original question, we don't quite know the initial value, or the PDE or the evolution operator or the final solution but we can guess the vortex should move along some path and the radius of uncertainty should get wider with time.
Oct 9, 2017 at 19:19 comment added Willie Wong I still don't understand your question. Suppose you have a smooth function $f$ defined on $[0,1]\times \Sigma$ where $\Sigma$ is a compact smooth manifold. If you further know that at $t = 0$, the inverse image of $[1,\infty)$ under $f$ is compactly included in some open subset $\Omega\subset \Sigma$, then by continuity there exists some $T$ such that for every $t\in [0,T]$ the inverse image of $[1,\infty)$ is contained in $\Omega$. This is independent of solving any PDEs. (Wellposedness gives you that regular initial data gives regular solutions so the above applies.)
Oct 9, 2017 at 17:36 comment added john mangual The fluid in this case is "air". I have never taken a formal PDE course so this could likely have an elementary answer. However, I ask you if well-posed PDE can be discussed using convex geometry. In certain very simple equations, such as free particle, the time-evolution really is the Minkowski sum of a straight line and a cone in spacetime. @Williewong
Oct 9, 2017 at 17:33 history edited john mangual
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Oct 9, 2017 at 17:12 comment added Willie Wong Why did you tag "dg" and "gt" and not "ap"? And I don't understand your question "What enables us to approximate solutions to PDE in this common-sense way without actually solving anything?" at all. What do you mean by "without actually solving anything?" and by "approximate solutions ... in this common-sense way"? Isn't your question answered by the fact that fluid dynamics have locally well-posed initial value problems?
Oct 9, 2017 at 16:00 history asked john mangual CC BY-SA 3.0