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Mar 16, 2022 at 21:13 history closed Diego Santos
Todd Trimble
Needs details or clarity
Mar 16, 2022 at 19:41 review Close votes
Mar 16, 2022 at 21:17
Mar 16, 2022 at 19:22 history edited Diego Santos CC BY-SA 4.0
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Nov 13, 2021 at 7:45 answer added Phil Harmsworth timeline score: 2
Nov 10, 2021 at 21:26 comment added Diego Santos @AlexandreEremenko, you are right. Thinking with more caution, in the book that I'm reading says "sufficiently many first integrals are known globally", but I didn't see a precise definition for "globally" and just assumed that it means "on a OPEN dense subset".
Nov 10, 2021 at 21:13 comment added Alexandre Eremenko "Analytic, defined on a dense set?" For example on rational points?
Nov 10, 2021 at 21:01 history edited Diego Santos CC BY-SA 4.0
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Nov 10, 2021 at 20:36 comment added Loïc Teyssier @DiegoSantos: I was aiming at the main question, but obviously got myself mixed up in the degrees. But, yes, you got my point ;)
Nov 10, 2021 at 20:33 comment added Diego Santos @LoïcTeyssier, are your comment/question referent to initial question or main question? If I didn't get it wrong, you are helping me see that I need add "minimal" before "set of functions", right?
Nov 10, 2021 at 20:20 comment added Loïc Teyssier @Diego Santos: you could simply take the function space spanned by all analytic solutions of polynomial 1st order polynomial ODEs. If you want something more meaningful, you'd probably like to quantify on the kind of "determinacy" of the sets you're after...?
Nov 10, 2021 at 17:52 comment added Diego Santos The function $H$ must be analytic. Yes, I can see that, without sufficient specification, the sentences makes no sense. Reading yours questions I'm realizing that many important details had escaped from my attention.
Nov 10, 2021 at 17:40 comment added Alexandre Eremenko According to your definition, EVERY equation $y'=P(x,y)$, where $P$ is a polynomial, or even a continuous function satisfying Lipschitz condition will be integrable. This makes no sense.
Nov 10, 2021 at 17:34 comment added Alexandre Eremenko What kind of function $H$? Not even continuous??
Nov 10, 2021 at 17:27 comment added Diego Santos For this equation, is the existence of a function $H(x,y)$ such that: i) $H(x,y)=C$ is a solution for the equation, and ii) $H$ is defined on a dense subset of $R^2$.
Nov 10, 2021 at 17:14 comment added Alexandre Eremenko From y point of view, 1-st order equation is $F(y',y,x)=0$, where $F$ is a polynomial. Can you spell EXACTLY what does it mean that THIS equation is integrable?
Nov 10, 2021 at 14:08 history edited Diego Santos CC BY-SA 4.0
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Nov 10, 2021 at 13:27 comment added Diego Santos @AlexandreEremenko, by "integrable" I mean the Goriely's "suffciently many first integrals are known globally". I will edit my question and would be very glad if you could reread.
Nov 10, 2021 at 13:06 review Close votes
Nov 15, 2021 at 3:04
Nov 10, 2021 at 12:49 comment added Alexandre Eremenko The answer to your "initial question" depends on the precise definition of "integrable" and "elementaty function". There are various definitions. Elliptic functions are not elementary according to most of them. The answer to your second question is negative: most linear differential equations with polynomial coefficients cannot be solved in terms of any "special functions". But again the precise answer depends on what you mean by "generalized hypergeometric functions" (Several definitions exist in the literature).
S Nov 10, 2021 at 0:34 review First questions
Nov 10, 2021 at 3:55
S Nov 10, 2021 at 0:34 history asked Diego Santos CC BY-SA 4.0