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Consider a differential equation of the form $$f^{(n)}(z)=P(z,f(z),f'(z),...,f^{(n-1)}(z))$$ where $P$ is a polynomial in $n+1$ variables, with the initial condition (for definitness) $$f^{(k)}(0)=0,\qquad 0\leq k\leq n-1.$$ By the ODE existence theorem, there exists a unique solution $f(z)$ which is holomorphic near the origin of the complex plane. My question is whether it is possible to decide whether $f(z)$ can be analytically continued to the whole complex plane. For example, suppose that the coefficients of P are rational numbers. Is there a finite decision procedure that will take P and determine whether $f(z)$ is entire?

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Don't you need to specify some derivatives of f at zero to get uniqueness? –  S. Carnahan Apr 16 '10 at 22:19
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OK, how about $f''=f+(f')^5-f^5$ with $f(z)=e^z$ then (subtract $1+z^2$ yourself to make it an unrecognizable mess) –  fedja Apr 16 '10 at 23:01
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As a side note, the old question of Borel whether the existence of a polynomial differential equation implies any restriction on the growth rate of an entire function is still wide open. If every function solving a polynomial equation solved a linear one, the answer would be trivially yes, so there is little hope to prove that. But I don't know if there are explicit counterexamples. –  fedja Apr 17 '10 at 0:36
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@GK: I wonder about the reasons of your question, since in such general settings the problem looks really untreatable. If it is not a question for a question, so you have an example in mind, please be more specific. There could be some knowledge in the case $n=1$ which is very special. Otherwise, an example for $f(z)=e^z-z-1$, namely $y''=y^2-{y'}^2+(2z+1)y'-z^2+1$, shows that any perturbation of the coefficients results in the non-entire solution. –  Wadim Zudilin Apr 17 '10 at 11:41
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@Wadim: I understand the general intuition is that in any case where you have an entire solution, an arbitrarily small perturbation will destroy this. This can be stated as the following conjecture: consider the space of polynomials $P$ in $n+2$ variables with degree $d\geq 2$. Give it the natural topology. Then the set of $P$'s for which an entire solution exists is nowhere dense (doesn't contain any open set). –  Guy Katriel Apr 18 '10 at 13:06
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One way to see that the local holomorphic function $f$ is entire is to see if the ODE defined by the polynomial $P$ is integrable. If its not, then you can assure that the solutions are not entire due to the fact that a non-integrable ODE has solutions with singularities. In fact, this is the way to show non-integrability for ODE, to find the singularities of some certain solutions.

For an explicit example, you can search on the vast literature about the integrability question of the Henon-Heiles system.

Hope this is what you were searching.

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Is the criterion you are talking about really an if-and-only-if one? Does the lack of singularities imply integrability in the sense you refer to? –  Guy Katriel Apr 17 '10 at 10:11
    
It's a perfect strategy to decide whether a given algebraic differential equation is integrable. But as far as I know there are no general algorithms (except maybe the 1st order DEs). –  Wadim Zudilin Apr 17 '10 at 11:31
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Thank you kaminoite and Wadim for the very interesting connection with the question of integrability. A quick search led me to a review paper by Martin D.Kruskal, Nalini Joshi, Rod Halburd arxiv.org/abs/solv-int/9710023 There, the authors state that "There is strong evidence [60, 61] that the integrability of a nonlinear sytem is intimately related to the singularity structure admitted by the system in its solutions." It seems from this formulation that the relation is not entirely established. If you can recommend other sources that clarify this, I would be most grateful. –  Guy Katriel Apr 17 '10 at 14:48
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The integrability would (normally) lead to the complete description of singularities of the solutions. But there are many classical non-integrable examples where this task can be done as well. The simplest one is probably the Chazy equation $y'''=2yy''-3y'^2$ and then we have Painleve transcendents,--in these cases the equations possess many symmetries. The moral is that the integrability is helpful when it happens (0% probability for a random polynomial ODE); if it does not happen there is still a chance to have entire solutions. This is not an answer, because I can hardly figure out refs. –  Wadim Zudilin Apr 18 '10 at 0:35
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