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Are there known necessary and sufficient conditions that specify in terms of an algorithm in a real arithmetic model (where real operations, elementary functions, and comparisons are elementary steps) when a first order differential equation $x'(t)=F(x(t),t)$ with a real-valued rational function $F(x,t)$ in real scalars $x$ and $t$ is exactly solvable by elementary functions and finitely many integrations?

Useful (positive and negative) partial results? In particular, are there strong sufficient conditions for rational functions with low numerator and denominator degree?

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    $\begingroup$ You must specify what you mean by both «decide» (and therefore what conditions you put on the coefficients of the differential equation, e.g. belonging to a field where you can decide the equality to $0$ with a halting turing machine) and «exactly solvable» (I assume you mean integrability by quadrature in the sense of Liouville). Even being granted the two defaulted meanings above, I fear that there are no known answer to your question, although it is believed by some that the Poincaré question (solvable by rational first-integral) is undecidable. The key-word is differential Galois theory. $\endgroup$ – Loïc Teyssier Mar 17 '15 at 15:24
  • $\begingroup$ @LoïcTeyssier: Thanks for the link. I made my question more precise. $\endgroup$ – Arnold Neumaier Mar 17 '15 at 16:05
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I don't think a general answer to your question is known. I personnaly doubt that it could be positive.

  1. Partial decidable answer: if the variational linear differential system obtained along a given (algebraic) trajectory is not solvable by quadrature (condition expressed as the virtually solvability of its Picard-Vessiot group) then the equation is not solvable by quadrature. This is a theorem by Casale (an improvement on the works by Morales and Ramis) available here: «Morales-Ramis Theorems via Malgrange pseudogroup, Ann. Institut Fourier 59 n 7 (2009)»

    If I remember correctly such a condition can be algorithmtically checked (although you need to address the "equality-to-$0$" problem).

  2. Partial undecidability answer: if your real arithmetic cannot decide the equality to $0$, then the Poincaré question is undecidable. The question asks for the existence of a rational first-integral of the differential equation. This is easily checked by considering the linear system $$t\dot x=\lambda x$$ with constant $\lambda$: such a rational first-integral exists only if $\lambda$ is rational (which is not decidable under the assumption). There does exist «elementary closed-form» first-integral, though, e.g. $H(t,x)=xt^{-\lambda}$, which are totally acceptable for answering your question (as you point out in the comments). I just thought that mentionning this fact could shed some light on "concrete" diffculties if you try to deal with actual equations.
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  • $\begingroup$ I allow all real operations including comparison of two numbers for equality or sign. Thus the solution $x=ct^\lambda$ of your example differential equation is perfectly valid. What did I formulate wrongly to seemingly exclude this? $\endgroup$ – Arnold Neumaier Mar 17 '15 at 16:29
  • $\begingroup$ Nothing, I understood what you said. I just included the second piece because if you want to achieve effective, concrete algorithms then the issue pops up (although it also does in the first piece). But I understood your concern was at a higher-level of calculability. That being said, I still somehow believe that a lot of undecidability lies down there ;) $\endgroup$ – Loïc Teyssier Mar 17 '15 at 16:55

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