When is a solution $P(f'(x)) = Q(f(x))$ periodic or double periodic?

Question

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\sl}{\operatorname{sl}}\newcommand{\cm}{\operatorname{cm}}\newcommand{\sm}{\operatorname{cm}}$Consider the differential equation

$$P(f '(x)) = Q(f(x))$$

Where $P(x),Q(x)$ are polynomials.

Examples are $f'(x) = 1 + f(x)^2$ where we get a tan solution and $f'(x)^2 = 4 f(x)^3 - g_2 f(x) - g_3$ where we get a Weierstrass elliptic function solution. One of them is periodic , the other double periodic.

In general, When is a solution $P(f'(x)) = Q(f(x))$ periodic or double periodic?

I looked at some famous elliptic functions and most of them are defined with two or three functions like the Dixon elliptic functions with $\cm'(x) = -\sm^2(x), \sm'(x) = \cm^2(x)$ what is related to the Fermat curve $x^3 + y^3 = 1$ and the Eisenstein integers. Or the lemniscate elliptic functions with $\sl'(x) = (1+ \sl^2(x)) \cl(x) , \cl'(x) = -(1 + \cl^2(x)) \sl(x)$. However the lemniscate elliptic function $\sl$ also satisfies a selfreference one : $(\sl'(x))^2 = 1 - \sl(x)^4$ thereby satifying the type of differential equation I was looking for.

It is basically just that solving

$$P(x) + Q(y) = 1$$

for $x$ or $y$ results in at least one function that satisfies :

$$P(f '(x)) = Q(f(x))$$

with the right initial conditions.

Usually what I find is that the degrees of $P$ and $Q$ are between $2$ and $4$.

So, what is going on?

Does every pair of polynomials $P,Q$ with degrees between $2$ and $4$ give double periodic functions? Are degrees above $4$ possible to get double periodic functions?

And when do we get periodic functions that are not double periodic?

Does the Fermat curve

$$x^5 + y^5 = 1 $$

or

$$x^7 + y^7 = 1$$

and their related differential equations give us double periodic functions?

I want to point out that a function of a periodic function is also periodic and the same applies to the double periodic case.

Also we get the trivial case for a polynomial $M(x)$ :

$$ M(P(f'(x))) = M(Q(f(x)))$$

which has as its solutions the same function $f$ as if $M$ was the identity function.

What basicly is an answer to my question of bounded degree, but I am looking for more insightful and general results ofcourse.

Some ideas I had were plugging in a Fourier series with variable coefficients. But I was dealing with infinitely many variables and not sure if my Fourier series was even valid ; was it still analytic and did it still agree with the function it was describing?? Another ideas was an analogue for Fourier series, a series expansion for double periodic functions. But I got stuck there too. Not sure if that was going in the right direction or not. Even if that works, I want to prove it does.

I tried some (complex analysis and geometric function theory ) theorems but they had problems with the poles and analytic continuation around those. The taylor radius was too small and Fourier requires $L^2$ spaces anyways. I might be able to solve a specific case but I want the general idea.

I am ofcourse slightly aware of some basic results such as relating the period(if it exists) with some coefficients ( such as $g_2,g_3$ in the Weierstrass case ) and rewriting the equations as an integral. Or writing functions in terms of eachother. Or some infinite sums. But that does not give me the insight I seek.

I am not an expert at the addition formula's but I also understand that an addition formula implies periodic or double periodic. But again that does not give me what I seek.

How to look at this?

Answer 1

To make this question precise, yo have to specify the analytic nature of solutions that you consider: where they are defined/analytic. When speaking of doubly periodic solutions, they usually require them to be meromorphic in the plane (elliptic). For this case, a complete answer is known: your equation has to satisfy the Fuchs conditions (see for example, E. L. Ince, Ordinary differential equations, Chap XIII.) In general, if an equation $F(y,y')=0$, where $F$ is a polynomial, has a solution meromorphic in the plane, then this solution is either elliptic, or single periodic $R(e^{az}),$ or rational, and the equation itself satisfies Fuchs conditions. The key assumption here is "meromorphic in the plane".

When the Fuchs conditions are not satisfied, solutions are multi-valued (usually infinitely valued), and it is not clear what "periodicity" can mean for them.

For example, your Fermat equations $(y')^n+y^n=1$, for $n\geq 3$, do not have meromorphic solutions (except constant solutions), since they do not satisfy Fuchs's conditions.

It is not difficult to classify all equations of your form which satisfy Fuchs's conditions: in all of them $P(y)=y^m$, and a complete list of such Fuchsian equations, together with general solutions, is given in Ince, section 13.8 (p. 314 of the Dover edition).

When talking about solutions with single period, it is reasonable to assume that it is well defined on the real line, and thus meromorphic in some region containing the real line. For this case we do not have such a complete result, and there is an important distinction, whether we are looking for a single periodic solution, or require that general solution is periodic. But the key observation is that if a solution takes some value $a$ infinitely many times on the real lines (or more generally, more that $\deg P$ times), then we can find a points $x_k$ such that $f(z_k)=f(z_m)$, $f'(z_k)=f'(z_m)$ and therefore $z_k-z_m$ is a period, by uniqueness theorem.