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The intuition:

Consider a single variable polynomial differential equation with integer-polynomial coefficients, for example

$$ y '' = -y $$

Then consider a pair of algebraic varieties (that is solutions to systems of polynomial equations with integer coefficients) on $\mathbb{R}^2$ for example $\lbrace (x,y) | x^2+y^2 =1 \rbrace$ and $\lbrace (x,y) | x = 0\rbrace $

Now observe that our algebraic varities in $\mathbb{R}^2$ can also be viewed as sets in $\mathbb{C}^2$, as the unit circle and real-line respectively. Now recall the famous fact that curiously enough ONE of the solutions to that differential equation $y = e^{ix}$ happens to map the real-line to the unit circle

This is is a pretty surprising result. But theres a family of such results

The curves $re^{ix}$ and $re^{-ix}$ both map the real line to the circle of radius r.

And in general the solution $ae^{ix} + be^{-ix}$ maps the real line to an ellipse given by $(x(t),y(t) = ((a+b)\cos(t), (a-b)\sin(t))$ (introducing complex numbers for $a,b$ allows us to tilt our ellipses around).

So generally speaking the maps between ${x=0}$ to any ellipse centered at 0, are induced by the solutions of $y'' = -y$.

The Question:

Given a polynomial differential equation $D$, with integer-polynomial coefficients, how can one find the set of all [if any] pairs of algebraic varieties $U,V$ that get mapped to each other under solutions of $D$ viewed as complex functions.

For example given $y'' = -y$ one could could generate the pair $\lbrace (x,y), x = 0 \rbrace, \lbrace (x,y) | x^2 + y^2 = 1 \rbrace$

But if we just consider some arbitrarily harder equation such as:

$$ 2yy'' + 3y' - 5x^2 y = 0$$

It becomes extremely hard to figure how to do algorithmically do this. Your best luck is to solve the equation, and hope that you recognize some nice properties about the functions used in the general solution.

Note that since the definition of variety is quite general, I want cut straight to the interesting stuff by insisting the two varieties should have covering dimension $1$ (that is they are line-like).

Some Results:

Going the other way from algebraic variety to differential equation is surprisingly easier. You can always parametrize algebraic curves and once you have a parametric function of 2 variables that can be transformed into a complex function of 2 variables and from here there is a lot more hope at being able to find a differential equation [although insisting the differential equation has integral/polynomial coefficients or polynomial form might make matters much harder]

Some stray thoughts:

Instead of asking about algebraic varieties you could ask about topological manifolds [forgetting the algebraic part] and as the original intuition suggests, a more natural question is to ask for infinite families of varieties and/or families of manifolds (often indexed by real or complex parameters)

You could also restrict the form of the differential equations [linear, constant coefficient, etc...]

So you end up with very large space of similar questions. But for the purposes of this site I decided to pick one such specific question feel free to answer any of the forms if one of them appears to be more well known/promising.

Some potentially related questions/ideas:

Algebraic Curves and Second Order Differential Equations

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    This is an interesting question I would like to spend more time thinking about. For very simple dynamical systems there can be solutions. For example, consider $A \in \mathbb{R}^{n\times n}.$ If $A = -A^\top$ then you can compute the algebraic variety induced by the linear time invariant dynamics $\dot{x} = A x.$ Pick $p_1(x) = \rho - x^\top x$ for any $\rho>0$ and it is easy to verify that this set is invariant under the dynamics. Moreover, for $n$ odd we have rank deficiency in $A$ producing another invariant $p_2(x) = x^\top v $ where $v$ is in the nullspace of $A.$ – Rollen S. D'Souza May 21 '20 at 01:00

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I think that the complete answer to this question lies in a deep study of singularities of solutions to each equation. Let's see a simple example. Consider the first order linear ODE: $$aty'-y=0.$$ If $a=n\in \Bbb{N}$, then the general solution is $y=C\sqrt[n]t$. So it's a map from the Riemann surface $w=z^n$ to the Riemann sphere. But if $a\notin\Bbb{Q}$ the general solution is transcendental.

Now one step further. For second order linear ODE's the Fusch theorem guarantees a solution in form of a Frobenius series around a regular singular point. Here the situation gets harder. For example: if the indicial equation has a irrational root or roots differing by an interger, the solutions will be transcendental or contain a logarithm therm. So in order to solutions be a map from an algebraic Riemann surface to the Riemann sphere, we need a Fuschian equation such that all exponents of all singularities be, pairwise, rationals and non differing by an interger. The construction of such a Riemann surface wold be a careful 'cut and paste' along some branch cuts.

For nonlinear equations all become even more complicated because we can have movable singularities that are not, in principle, evident in the equation as for linear ones. All this leads to the study of Painlevé property for ODE's.

Diego Santos
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