Existence of genus 0 solution for linear ordinary differential equation

Question

This question is about the linear differential equations with polynomial coefficients. I am interested in the necessary and sufficient conditions for the existence of genus 0 for linear differential equations, and the understand the geometry underlying them. Let me give few examples that prompt me to ask the questions.

Let us analyse what kind of polynomial coefficient will give rise to the rational solution. Let's begin with analysing the linear ODE of order 1 with polynomial coefficients of the form, where $a(x), b(x)$ polynomials in $x$. $$ a(x)\partial_{x} + b(x) y(x) =0 $$ It's easy to see that that for $y(x)$ to have algebraic solution we must have $\frac{b(x)}{a(x)} = \sum_{{i}} \frac{c_{i}}{(x-a_{i})}$

I have noticed the in many examples where zeros of $ a(x)$ are all distinct and rational and $b(x)$ is constant and rational the solution will be algebraic solutions. In case $b(x)$ is integers I got most of my solution to be genus 0.

${\bf{Example-1}}$ Let take the ODE $$ \left( 1-4\,x \right) \left( 1-x \right) {\frac {\rm d}{{\rm d}x}}y \left( x \right) =-2\,y \left( x \right) $$ which have solution $$ y \left( x \right) = {C_1 \left( -1+4\,x \right) ^{2/3} }{ \left( x-1 \right) ^{-2/3}} $$ where $C_{1}$ is a constant. Taking $C_{1}$ to be 1 the algebraic curve is given by $$ \left( x-1 \right) ^{2}{x}^{3}{y}^{3}- \left( -1+4\,x \right) ^{2}=0 $$ This above curve has genus 0

$\bf{Example -2} $ Let take the ODE $$ \left( 1-4\,x \right) x{\frac {\rm d}{{\rm d}x}}y \left( x \right) =- 2\,y \left( x \right) $$ which have the solution $$ y \left( x \right) =C_1\left( 16-8\,{x}^{-1}+{x}^{-2} \right) $$ Taking $C_{1}$ as constant, the algebraic curve has genus 0

$\bf{Example-3}$\ ${\bf{Genus 1}}$ $$ {\frac { \left( 1-x \right) \left( x+1 \right) {\frac {\rm d}{{\rm d} x}}y \left( x \right) }{x}}=-1/2\,y \left( x \right) $$ The solution is $$ y \left( x \right) ={C_1}\,\sqrt [4]{x-1}\sqrt [4]{x+1} $$ taking $C_{1} =1 $ the curve is $$ {y}^{4}{x}^{4}- \left( x-1 \right) \left( x+1 \right) $$ The above curve has genus 1 What first question is how to general linear ordinary differential equation what will be the condition for the existence of genus 0 solutions? I recently browse through the following paper claiming to have the condition for general algebraic ODE http://mmrc.iss.ac.cn/~ryfeng/papers/p155-feng.pdf unfortunately it's a bit difficult for me to follow. So I am expecting the condition will be much milder for linear ODE.

My second part of the question is if I get a genus 0 solution I can parameterise the with respect to coordinate say $z$ now so now the curve is given by $\left(x(z), y(z)\right)$. Now the change of coordinate will make the linear ODE and the solution is given in the form of $\frac{P(z)}{Q(z)}$, where $P, Q$ are polynomials. Due to the change of coordinate in $z$ $$ \frac{\partial}{\partial_x} = \frac{1}{x^{'}(z)}\frac{\partial}{\partial_z} $$ the zeros of $Q(z)$ contain the zeros of $x^{'}(z)$. Is there any condition that it will only contain the zeros of $x^{'}(z)$?

Answer 1

This question gets way more difficult even for second order ODE's. Note that, by applying the Frobenius method, even if you have a Fuchsian equation you can obtain irrational exponents on some singularities. This would lead to transcendental solutions. Note also that in your examples all $a_i$'s and $c_i$'s are rational. Take for example the simple Euler equation: $$t^2y''+ty'+y=0.$$ It's general solution is $$y(t)=C_1\cos(\ln(t))+C_2\sin(\ln(t)).$$ So I think that the complete answer to your question would impose severe constraints in the singularity structure of the equations.