13
$\begingroup$

What justification can you give for the fact that "most ODEs do not have an explicit solution"?

$\endgroup$
  • 4
    $\begingroup$ First of all, one would need to say what is meant by "explicit". Could you please say where you saw this quote or phrase, and in what context? $\endgroup$ – Yemon Choi Sep 22 '14 at 14:42
  • 1
    $\begingroup$ I added the tag "soft question" as it is purposefully vague. $\endgroup$ – Ryan Budney Sep 22 '14 at 17:32
  • 3
    $\begingroup$ Solving an ODE has to be harder than integrating or solving an equation for an unknown number, and these assignments are already typically hopeless as far as explicit solutions are concerned. $\endgroup$ – Christian Remling Sep 22 '14 at 17:44
  • 4
    $\begingroup$ As is often the case, "explicit" is a misnomer for "elementary", where the latter more literally refers to polynomials, exponentials-and-logs, trig functions, roots, ... As in is implicit in Loic Teyssier's answer, the coefficients of a differential equation can be elementary, while the solutions are demonstrably not. $\endgroup$ – paul garrett Sep 22 '14 at 19:21
31
$\begingroup$

If the ODE is linear --and the notion of «explicit» refers to Liouvillian solutions (towers of iterated quadrature and exponential of meromorphic functions)-- then its differential Galois group (Picard-Vessiot theory) must be a solvable algebraic subgroup of $GL_n (\mathbb C)$. Such subgroups are rare: they define a proper algebraic subvariety. The defining equations correspond to trivial commutations relations, e.g $[G,G]=0$ in the Abelian case, encoding the tower of normal subgroups intervening in the definition of solvable group.

In the non-linear context the intuition is the same: nice «transverse structures» are rare.

Edit: for a statement regarding the non-linear case, see my paper http://fr.arxiv.org/abs/1308.6371v2, Corollary C.

$\endgroup$
  • $\begingroup$ In other words, it is much like the case for solutions to polynomials in one variable over $\mathbf{Q}$. Just as an aside, sometimes the notion of "explicit" permits the intervention of algebraic functions, corresponding to the finite component group of the differential Galois group being non-trivial; that finite group might also be non-solvable, but that is a separate can of worms. :) $\endgroup$ – user27920 Sep 22 '14 at 17:04
  • 1
    $\begingroup$ Yes, I was being purposefully vague while linking to a more precise formulation. $\endgroup$ – Loïc Teyssier Sep 22 '14 at 17:14
10
$\begingroup$

On the non-linear side, there is a theorem of Hudai-Verenov MR0147699, which says that for a generic (open dense set) equation $y'=R(x,y)$ the graph of a generic solution is dense. See also Ilyashenko, MR0247176. This implies that this equations do not have first integrals, so they are not "solvable" in the sense that we teach in the elementary courses.

This is an example of a general theorem on the subject. But of course it is well known from the last 300 years of experience with differential equations. For example, all integrable cases of the rotating top in uniform gravity field are explicitly known, and they are exceptional. There are many other examples like this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.