Prove that even doubly periodic function satisfies a differential equation

Question

Let $f(z)$ be analytic in $\mathbb{C}\setminus\{m+ni:m,n\in\mathbb{Z}\}$. Assume that $f(z)=f(-z)$ , $f(z)=f(z+m+ni)$ and $f$ has a pole of order $2$ at $0$. Prove that there exist numbers $a_0,a_1,a_2,a_3\in\mathbb{C}$ s.t $0$ is a removable singularity of $g(z)=(f'(z))^2-a_3f(z)^3-a_2f(z)^2-a_1f(z)-a_0$ and $g(0)=0$. Hint: use $f$'s Laurent series.

My attempt: since $f$ is even and has a pole of order $2$ at $0$, I can write: $f(z)=\frac{1}{z^2}+h(z)$ with $h$ holomorphic at $0$. Then I can calculate: $$f(z)^2=\frac{1}{z^4}+\frac{2}{z^2}h(z)+h^2(z)$$$$f(z)^3=\frac{1}{z^6}+\frac{3}{z^4}h(z)+\frac{3}{z^2}h^2(z)+h^3(z)$$$$f'(z)=\frac{-2}{z^3}+h'(z)\Rightarrow (f'(z))^2=\frac{4}{z^6}-\frac{4}{z^3}h'(z)+(h'(z))^2 $$

Now it's obvious that to get rid of $\frac{4}{z^6}$, $a_3=4$. Other than that, I wasn't table to make any progress. Any help would be appreciated.

Remark: I am aware that this is a special case of Weirstrass functions, but we have not studied about them yet and I want to use Laurent series to solve it.

Answer 1

The functions $f'^2, f^3, f^2, f$ are all meromorphic in a neighborhood of $z=0$, with poles of order $6, 6, 4, 2$, respectively. These functions are also all even, so that their Laurent expansion have only terms with even exponents.

So you can

• find $a_3$ such that $f'^2 - a_3f^3$ has at most a pole of order $4$ at the origin,
• then $a_2$ such that $f'^2 - a_3f^3-a_2f^2$ has at most a pole of order $2$ at the origin,
• then $a_1$ such that $f'^2 - a_3f^3-a_2f^2-a_1f$ is holomorphic at the origin,
• and finally $a_0$ such that $g = f'^2 - a_3f^3-a_2f^2-a_1f-a_0$ is holomorphic at the origin with $g(0) = 0$.

Then $g$ is an entire doubly periodic function and therefore constant, and since $g(0) = 0$, $g$ is identically zero, i.e. $f$ satisfies the differential equation $$ f'^2 = a_3f^3+a_2f^2+a_1f+a_0 \, . $$