Timeline for Entire solutions of polynomial ODE's
Current License: CC BY-SA 2.5
17 events
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| May 4, 2014 at 17:46 | comment | added | Ali Taghavi | @GuyKatriel Could you please give a reference for this conjecture?thanks | |
| Apr 18, 2010 at 13:06 | comment | added | Guy Katriel | @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). | |
| Apr 18, 2010 at 0:22 | comment | added | Wadim Zudilin | Formal proofs are really hard in such cases! My "proof" is completely experimental in nature (plus some personal experience with algebraic ODEs). The case of linear ODEs in your settings is easy because their coefficients lie in $\mathbb C[z]$, hence no singular points for their solutions (except at infinity). In the nonlinear case no control of the singularities is available: they "move" as singularities of the correpsonding DE. I know the existence of general theorems of global behaviour of solutions if $n=1$ but nothing general for $n>1$. | |
| Apr 17, 2010 at 14:34 | comment | added | Guy Katriel | @Wadim: My question was not prompted by a specific example. Rather, I was thinking about the fact that writing a polynomial ODE with initial conditions is an explicit way to define a holomorphic function, so that it is natural to wonder what properties of the function can be deduced from this representation. It seems that indeed it is harder than I suspected. Can you give a hint as to how you show that any perturbation of the example you gave leads to a non-entire solution? | |
| Apr 17, 2010 at 11:41 | comment | added | Wadim Zudilin | @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. | |
| Apr 17, 2010 at 9:32 | answer | added | Kaminoite | timeline score: 1 | |
| Apr 17, 2010 at 7:27 | history | edited | Guy Katriel | CC BY-SA 2.5 |
$0\leq k\leq n-1$ instead of $1\leq k\leq n-1$
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| Apr 17, 2010 at 7:08 | comment | added | Guy Katriel | Could it be that the Borel problem you mentioned relates to the more general equation $P(z,f(z),f'(z),...f^{(n)}(z))=0$, and has a known solution in the case that the equation is of the form $f^{(n)}(z)=P(z,f(z),f'(z),...f^{(n-1)}(z))$? Just wondering.. (of course my original question can also be formulated for the more general form of the differential equation). | |
| Apr 17, 2010 at 0:40 | history | edited | Harald Hanche-Olsen | CC BY-SA 2.5 |
Corrected the statement as per the OP's response in the comments
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| Apr 17, 2010 at 0:36 | comment | added | fedja | 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. | |
| Apr 17, 2010 at 0:30 | comment | added | Guy Katriel | It's also possible to formulate a somewhat more general question for which the linear equation issue will (probably) not be relevant, by allowing $P$ to be a rational function, in which case there are probably many examples of entire functions which are not solutions of linear equations. E.g. $f(z)=e^{e^z}$ solves $f''(z)=\frac{(f'(z))^2}{f(z)} +f'(z)$, but I would guess it doesn't solve any linear differential equation with coefficients that are polynomials in $z$ (proof?). | |
| Apr 17, 2010 at 0:28 | comment | added | Guy Katriel | Leonid's question and fedja's response brings to mind the question whether any entire solution of such a differential equation is also a solution of linear equation. I don't know the answer. Even if it is positive, it will lead to the solution of the decision problem only if it can be decided whether the solution of such a nonlinear equation is also the solution of a linear one. | |
| Apr 16, 2010 at 23:29 | comment | added | Guy Katriel | Scott: Thankyou for the correction. I should have written $f^{(k)}(0)=0,\;\;1\leq k\leq n-1$ | |
| Apr 16, 2010 at 23:01 | comment | added | fedja | 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) | |
| Apr 16, 2010 at 22:41 | comment | added | fedja | @Leonid. Sure: $f''=f+(f')^5-f^4$ with the solution $f(z)=0$. This example is rather stupid, of course, but how exactly do you suggest to rule such things out? | |
| Apr 16, 2010 at 22:19 | comment | added | S. Carnahan♦ | Don't you need to specify some derivatives of f at zero to get uniqueness? | |
| Apr 16, 2010 at 21:48 | history | asked | Guy Katriel | CC BY-SA 2.5 |