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The concepts of overdetermined and underdetermined PDE systems are well known. However, all sources I have so far looked into appear to avoid giving any name to PDE systems which are neither overdetermined nor underdetermined. Is there any reasonably commonly used name for such PDE systems? If possible, please provide the references where the name from your answer is used.

EDIT: It was suggested by Igor Khavkine and Robert Bryant that one should consider formally integrable systems (which are neither over- nor underdetermined and have no nontrivial compatibility conditions). I like the idea but in my case this term would appear within the discussion of systems which are completely integrable (via the inverse scattering transform), and this might confuse the non-expert readers. Is there any sensible way out of this conundrum?

Thanks in advance!

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    $\begingroup$ Well-posed? Formally integrable? I guess it would depend if you are concerned with purely local or also with global properties. $\endgroup$ Jun 22, 2011 at 15:50
  • $\begingroup$ @Igor: I am interested in purely local properties. $\endgroup$ Jun 22, 2011 at 16:23
  • $\begingroup$ @Igor: Thanks for the suggestions, formally integrable sounds interesting. $\endgroup$ Jun 22, 2011 at 16:28
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    $\begingroup$ I would be a little wary of the definitions given for "overdetermined" and "underdetermined". Although they work for most examples that arise, they do not in general. $\endgroup$
    – Deane Yang
    Jun 22, 2011 at 17:47
  • $\begingroup$ I'm confused: why formally integrable? May overdetermined or underdetermined systems not also be formally integrabel? $\endgroup$ Jun 22, 2011 at 19:39

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determined: Bryant et. al, Exterior Differential Systems, p. 189

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    $\begingroup$ The notion of counting 'number of unknowns' versus 'number of equations' was always meant to be a first crude property of the PDE, it wasn't meant to tell you everything. In fact, this count is only a very rough guide. Even 'determined' systems can fail to have good properties. For example, the equation curl U = F where F is a known vector field in 3-space and U is the unknown vector field, is determined (3 equations for 3 unknowns), but it is not formally integrable (in the usual sense) unless div F = 0 $\endgroup$ Jun 22, 2011 at 17:06
  • $\begingroup$ @Robert Bryant: Dear Professor Bryant, thanks a lot for your comment. May I ask what you would call a system for which BOTH of the following properties hold: a) the number of unknown functions equals the number of equations; b) there are no nontrivial compatibility conditions? Is it formally integrable, determined, or what? $\endgroup$ Jun 22, 2011 at 17:16
  • $\begingroup$ A small problem that I have with the term formally integrable is that in the context I want to use it will appear together with the term completely integrable (through existence of a Lax pair and the inverse scattering transform) which might confuse the non-expert readers. Which is the best way to handle this? Many thanks in advance once again! $\endgroup$ Jun 22, 2011 at 17:30
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    $\begingroup$ I thought to suggest involutive determined, but that isn't quite perfect either. If the integrability conditions are satisfied, then the system will be formally involutive, and also determined, but not quite what you want. $\endgroup$
    – Ben McKay
    Jun 22, 2011 at 20:12
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    $\begingroup$ Well, the salient thing for a 'determined' system, which Deane was alluding to with his definition in the first order quasi-linear case, is whether there are non-characteristic covectors at every point. This is the necessary and sufficient condition that (locally) the system can be expressed as a system in Cauchy-Kowaleski form (not necessarily of first order). Thus, another possible phrase you could use is 'determined, of CK-type'. It's not graceful, but it is clear. I don't think it's a good idea to try to change the accepted definition of 'determined', even if it is naïve. $\endgroup$ Jun 23, 2011 at 15:29

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