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This is a follow-up to my previous question: Are there periodic functions satisfying a quadratic differential equation?

Let $P(u,v)$ be a bivariate polynomial. Then its zero set $\{(u,v): P(u,v)=0\}$ is an algebraic curve. It might make it easier to assume in what follows that the curve has no singularities.

If we substitute $y$ and $y'$ into $P(u,v)$ to take the places of $u$ and $v$ respectively, then we get a possibly non-linear first-order ODE. I am curious about how many ODEs of this type have solutions.

1. How many/which first order ODE's formed by substituting into a bivariate polynomial have solutions?

2. Does any solution to such an ODE parametrize the algebraic curve that characterizes the defining ODE? EDIT: I realize now that the answer to this question is obviously yes, because it follows directly from the fact that $y$ and $y'$ are both functions of the same single variable (say $t$) and that they satisfy the equation $P(y,y')=0$, so if they exist they must parametrize the algebraic curve, as a result of their definition.

Some examples might help. For second-degree algebraic curves, we have that the equation of the unit circle corresponds to a first order ODE which has both sine and cosine as solutions, and that these two functions parametrize the unit circle as well: $$(y')^2 + y^2 -1=0 $$ Likewise, the hyperbolic sine and hyperbolic cosine parametrize the unit hyperbola, and both correspond to solutions of a first order ODE whose form is that of the unit hyperbola: $$(y')^2 - y^2 -1=0$$ Finally, smooth cubic curves in Weierstrass normal form can be parametrized by the Weierstrass $\wp$ functions, and the (complex) differential equations characterizing the Weierstrass $\wp$ functions have the form of a smooth cuvic curve in Weierstrass normal form: $$(\wp')^2 = 4[\wp^2] -g_2 \wp -g_3 $$ Then my question is essentially: how far does this go? For examples, if I were to choose the algebraic curve $$ v^4 - v^3 = u^5 +u^2 -7$$ would the first-order ODE $$(y')^4 - (y')^3 = y^5 + y^2 -7$$ have a solution? And would the solution to this ODE (if it existed) parametrize the original algebraic curve $v^4 - v^3 = u^5 + u^2 -7$? This example is completely arbitrary, but hopefully it makes clear to some extent the level of generality I am interested in.

Note: I am not sure how to properly tag this question.

This question is related but addresses second order ODE's -- however, I am not interested in second order ODE's, only first order: Algebraic Curves and Second Order Differential Equations

I think that this question is probably the most related, assuming that epicycloids can be defined by first-order ODEs (I don't know either way). Proof that Epicycloids are Algebraic Curves?

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Chill2Macht
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    All differential equations $P(y, y') = 0$ have solutions (at least locally) wherever $ \dfrac{\partial}{\partial v} P(u,v) \ne 0$. – Robert Israel Sep 17 '16 at 00:41
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    In particular, the differential equation is separable, so you have solutions in the implicit form $\int \dfrac{dy}{G(y)} = x + c$ where $P(y, G(y)) = 0$. – Robert Israel Sep 17 '16 at 00:49
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    Are you asking whether there is a "closed-form" solution? – Robert Israel Sep 17 '16 at 00:57
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    @RobertIsrael I imagine that a closed-form solution would probably be too ambitious, since even for cubic curves the Weierstrass $\wp$ functions lack one. So if we assume that the algebraic curve is smooth, ie if at least one of $\frac{\partial}{\partial u}P(u,v)\not=0$ or $\frac{\partial}{\partial v}P(u,v)\not=0$ holds at every point, then at least locally we can use the fact that the ODE is separable to prove the existence of a solution? Can we say anything qualitative about the solutions, e.g. if they are periodic, like for quadratic or cubic curves? Do their singularities correspond to the – Chill2Macht Sep 17 '16 at 11:07
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    genus of the algebraic curve, or their periodicity, the way that the Weierstrass $\wp$ functions can be defined on a torus, which corresponds to the genus of smooth cubic curves being one? I am curious, because it seems like solutions might give a whole class of special functions generalizing the trigonometric functions and which could be put into one-to-one correspondence with algebraic curves, whose defining equations are simple to describe. – Chill2Macht Sep 17 '16 at 11:08

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This isn't really so much of an answer as a compendium of relevant links I have found about the subject on the internet -- most seem to take a much more abstract approach than what I had in mind, which was just to generalize the trigonometric and Weierstrass $\wp$ functions.

Chill2Macht
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What you're looking for is, probably, the following theorems: (Theorems A and B of "A Malmquist–Yosida type of theorem for the second-order algebraic differential equations" by Liangwen Liao)

Theorem (Malmquist): Let $R(z,f)$ be birational. If a differential equation of the form $$f'(z)=R(z,f)$$ admits a transcendental meromorphic solution, then the equation can be reduced into a Riccati differential equation $$f'(z)=a(z)+b(z)f(z)+c(z)f^2(z),$$ where $a(z), b(z)$ and $c(z)$ are rational functions.

That is, if your polynomial $P(y,y')$ has degree $1$ in $y'$, it must have degree at most $2$ in $y$.

Theorem (Steinmetz): Let $R(z,f)$ be birational. If the following differential equation $$f'(z)^n=R(z,f),$$ admits a transcendental meromorphic solution, then the differential equation can be reduced into $$f'(z)^n=\sum_{i=0}^{2n}a_i(z)f(z)^i,$$ where the $a_i(z)(i=0,1,...,n)$ are rational functions and at least one of then does not vanish.

So if your equation has degree $n$ in $y'$, then it must have degree at most $2n$ in $y$.

The general case with more then on power of $y'$ is adressed in the "Fusch conditions" that you can find in section $1$ of "Meromorphic solutions of algebraic differentlal equations" by Eremenko.

Diego Santos
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