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If I would ask for $\phi'(x) = f( \phi(x))$ and $\phi(0)=f(0)$, I would get that the inverse of $\phi$ is forced to be of the form:

$$\phi^{-1}(z) = \int_{0}^z \frac{1}{f(x)} d x.$$

Now it is natural to ask whether there is something similiar for the problem $\phi''(x) =f( \phi(x),\phi'(x))$?

Or a little bit mor restrictive: Given a (homogenous) polynomial $f_j(X_0, \dots X_n) \in \mathbb{C}[X_0, \dots, X_n]$, where $j=1, \dots, m$, such that the variety $V(f_1, \dots, f_m)$ is a non singular variety over $\mathbb{C}$.

Q1: Is it known in general, if there exists a function $\phi: \Omega \rightarrow \mathbb{C}$, such that $$f_j( \phi(s), \phi'(s), \dots, \phi^{(n)}(s))=0$$ or perhaps the easier problem with isolated variables $$\left( \phi^{(n+1)}(s) \right)^{\mathrm{deg} (f_j)}= f_j( \phi(s), \phi'(s), \dots, \phi^{(n)}(s)).$$ for all $s \in \Omega$? Here $\Omega$ is any open subset of $\mathbb{C}$. Are there global meromorphic solutions? What is the algebraic invariant, which counts the linear independent solutions here?

Q2: (see Denis Serre's answer): What is known in the case $m=1$? Or what if require all $f_j$'s to have the same degree?

Q3: Is there a general theory/algorithm, how to compute the solutions in special cases?

Q4: Are there some easy examples, which give an intuition, what we can expect to be true and what not?

I make this community wiki, since I do not know wheter my question is naive. I have little intuition about this stuff. Please comment if you do some changes. As a motivation, the Weierstrass $\wp $ function does the job for elliptic curves.

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The answer is NO in general because your differential system is overdetermined: it has several equations and only one unknown function. Take the polynomials $\omega^2 X_0-X_2$ and $X_2^2-X_1^2-X_0^2$ with $\omega^4\ne1$. The solutions of the first equation are $ae^{\omega s}+be^{-\omega s}$, but none of them satisfy the second equation, but $\phi\equiv0$.

A more interesting question should be which varieties $V$ yield non-trivial solutions. The answer must be of geometrical nature. Overdetermined linear systems have been studied for a long time (dating back to Goursat and E. Cartan), but the situation for nonlinear ones might be widely open.

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  • $\begingroup$ Thanks for this point. Since the system you define, can be simplified to $X_0^2 (\omega^4-1)- X_1^2$. So this solution in this case will be $\exp ( x \sqrt{ \omega^4 -1})$. So how tho modify the question that it makes suitable sense? Should I require just one polynomial equation. Please forgive me that I will change the question accordingly. $\endgroup$
    – Marc Palm
    Commented Nov 17, 2010 at 8:10
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    $\begingroup$ When $m=1$, it is reasonnable to think that there are solutions. Two difficulties arise. 1- that $\partial f/\partial X_n$ vanishes at some points of $V$. This makes the ODE singular, and you have to go through Fuchs theory to see which kind of singularities are present in the solution. In some cases, the singularity is fake, as for the ODE $y'^2=y^3+ay+b$ (Weirstrass functions). 2- since the equation is non-linear, it is never obvious whether the solutions are global or not. Perhaps this is not an issue since you accept poles (meromorphic solutions). $\endgroup$
    – Denis Serre
    Commented Nov 17, 2010 at 8:55
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    $\begingroup$ I'm not sure which work of Cartan is being referred to, but it seems to me that the theory now known as Cartan-Kahler theory developed for analyzing overdetermined systems of PDE's could be applied to this question. This theory works for nonlinear systems. Cartan may have some additional work specifically on ODE's, but I am unfamiliar with this. $\endgroup$
    – Deane Yang
    Commented Nov 17, 2010 at 14:17
  • $\begingroup$ Dear Deane Yang, can you please give me a better reference here? Wikipedia is not to precise here: en.wikipedia.org/wiki/Cartan-Kähler_theorem. But thanks for the hint in which direction I have to look! Do they give a constructive method or do they just consider uniqueness/existence questions? $\endgroup$
    – Marc Palm
    Commented Nov 18, 2010 at 8:13

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