Weierstrass cubic differential equation

(Gap estimate for the lattice.) Put $A:=\mathrm{\mid }{\omega }_1{\mathrm{\mid }}^2$, $B:=\mathrm{Re}({\omega }_1\overline{{\omega }_2})$, $C:=\mathrm{\mid }{\omega }_2{\mathrm{\mid }}^2$. Expanding with [F2] gives, for real $s,t$, $\mathrm{\mid }s{\omega }_1+t{\omega }_2{\mathrm{\mid }}^2=A{s}^2+2Bst+C{t}^2$; completing the square in the two variables gives $A{s}^2+2Bst+C{t}^2=A(s+\frac{B}{A}t{)}^2+\frac{AC-B^2}{A}{t}^2\ge \frac{AC-B^2}{A}{t}^2$ and symmetrically $A{s}^2+2Bst+C{t}^2=C(t+\frac{B}{C}s{)}^2+\frac{AC-B^2}{C}{s}^2\ge \frac{AC-B^2}{C}{s}^2$, hence $\mathrm{\mid }s{\omega }_1+t{\omega }_2{\mathrm{\mid }}^2\ge \frac{AC-B^2}{\max (A,C)}\max (s^2,t^2)$. Here $A,C>0$ by [F1], and $AC-B^2=(\mathrm{Im}({\omega }_1\overline{{\omega }_2}){)}^2>0$: indeed $AC=\mathrm{\mid }w{\mathrm{\mid }}^2$ and $w\bar{w}=\mathrm{\mid }w{\mathrm{\mid }}^2$ for $w:={\omega }_1\overline{{\omega }_2}$ by [F2], so $AC-B^2=(\mathrm{Im}w{)}^2$, and $\mathrm{Im}w=0$ would make ${\omega }_2\mathrm{/}{\omega }_1=\bar{w}\mathrm{/}\mathrm{\mid }{\omega }_1{\mathrm{\mid }}^2$ real, making ${\omega }_2$ a real multiple of ${\omega }_1$, contrary to [F1]. Therefore $\delta :=\sqrt{(AC-B^2)\mathrm{/}\max (A,C)}$ is positive and $\mathrm{\mid }m{\omega }_1+n{\omega }_2\mathrm{\mid }\ge \delta \max (\mathrm{\mid }m\mathrm{\mid },\mathrm{\mid }n\mathrm{\mid })$ for all integers $m,n$, so every nonzero lattice point has modulus at least $\delta$ and $\mathrm{\Lambda }\cap \{\mathrm{\mid }z\mathrm{\mid }\le R\}$ is finite for every $R$.

(Local uniform convergence and derivatives of the corrected series.) The nonzero lattice points with $\mathrm{\mid }\omega \mathrm{\mid }<1$ are finite by step 1.1. For $k\ge 0$, the shell $2^k\le \mathrm{\mid }\omega \mathrm{\mid }<2^{k+1}$ contains at most $(2^{k+2}\mathrm{/}\delta +1{)}^2$ points, so its contribution to $\sum \mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ is at most $(2^{k+2}\mathrm{/}\delta +1{)}^2{2}^{-3k}=O(2^{-k})$. Thus ${\sum }_{\omega \mathrm{\ne }0}\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}<\mathrm{\infty }$. Put $r:=\delta \mathrm{/}2$. For $\mathrm{\mid }z\mathrm{\mid }\le r$ and $\omega \mathrm{\ne }0$, one has $\mathrm{\mid }z-\omega \mathrm{\mid }\ge \mathrm{\mid }\omega \mathrm{\mid }\mathrm{/}2$ and $\mathrm{\mid }2z\omega -z^2\mathrm{\mid }\le \frac{5}{2}r\mathrm{\mid }\omega \mathrm{\mid }$, so by [F2] and [F2, F3, F9, F11, F10, step 1.1] $\mathrm{\mid }{f}_{\omega }(z)\mathrm{\mid }=\frac{\mathrm{\mid }2z\omega -z^2\mathrm{\mid }}{\mathrm{\mid }z-\omega {\mathrm{\mid }}^2\mathrm{\mid }\omega {\mathrm{\mid }}^2}\le \frac{10r}{\mathrm{\mid }\omega {\mathrm{\mid }}^3}\mathrm{.}$ Hence the finite-subset net of corrected summands converges uniformly on $\mathrm{\mid }z\mathrm{\mid }\le r$; call its sum $S$. By [F9], $S$ is holomorphic on $\mathrm{\mid }z\mathrm{\mid }<r$, and for $0<\mathrm{\mid }z\mathrm{\mid }<r$ the defining formula gives $S(z)=\mathrm{\wp }(z)-z^{-2}$. The lattice is countable by [F11], so choose an enumeration $({\omega }_j)$; its partial sums converge locally uniformly to $S$. Applying [F9] gives $S^{(k)}(0)={\sum }_j{f}_{{\omega }_j}^{(k)}(0)$. For $k\ge 1$, [F10] gives $f_{\omega }^{(k)}(0)=(k+1)!{\omega }^{-k-2}$, while $f_{\omega }(0)=0$ by [F3]. Therefore $S^{(k)}(0)\mathrm{/}k!=(k+1)G_{k+2}$ for $k\ge 1$ and $S(0)=0$.

(Absolute convergence of $G_4$ and $G_6$.) By step 2.1, $\sum \mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ converges. For $\mathrm{\mid }\omega \mathrm{\mid }\ge 1$ and $k\ge 3$, $\mathrm{\mid }\omega {\mathrm{\mid }}^{-k}\le \mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$, while the lattice points with $0<\mathrm{\mid }\omega \mathrm{\mid }<1$ are finite by step 1.1. Thus the families $({\omega }^{-k})$ are absolutely summable for $k=3,4,5,6$. In particular $G_4$ and $G_6$ are well-defined complex numbers, with convergent finite-subset sums.

(Odd sums vanish.) The map $\omega \mapsto -\omega$ is a bijection of $\mathrm{\Lambda }\setminus \{0\}$ and the families $({\omega }^{-3})$ and $({\omega }^{-5})$ are absolutely summable by step 3.1, so reindexing gives $G_3={\sum }_{\omega }(-\omega {)}^{-3}=-{\sum }_{\omega }{\omega }^{-3}=-G_3$ and $G_5=-G_5$; hence $G_3=G_5=0$.

(Taylor expansion of $\mathrm{\wp }$ at the origin.) By [F8] applied to the holomorphic function $S$ on $\mathrm{\mid }z\mathrm{\mid }<r$, $S(z)={\sum }_{k\ge 0}\frac{S^{(k)}(0)}{k!}{z}^k$ for $\mathrm{\mid }z\mathrm{\mid }<r$, so by step 2.1 [F8, step 2.1, step 4.1] $S(z)={\sum }_{k\ge 1}(k+1)G_{k+2}{z}^k=2G_{3}z+3G_4{z}^2+4G_5{z}^3+5G_6{z}^4+{\sum }_{k\ge 6}(k+1)G_{k+2}{z}^k,$ and since $G_3=G_5=0$ by step 4.1 the last sum is $z^{6}U(z)$ for a holomorphic $U$ near $0$. Thus, on $0<\mathrm{\mid }z\mathrm{\mid }<r$, $\mathrm{\wp }(z)=z^{-2}+3G_4{z}^2+5G_6{z}^4+z^{6}U(z)\mathrm{.}$

(Derivative expansion.) Differentiating the identity of step 5.1 termwise, which is legitimate for the locally uniformly convergent power series by [F9], gives [F9, F10, step 5.1] ${\mathrm{\wp }}^{\mathrm{\prime }}(z)=-2z^{-3}+6G_{4}z+20G_6{z}^3+z^{5}V(z)$ for a holomorphic $V$ near $0$ (indeed $\frac{d}{dz}(z^{6}U(z))=z^5(6U(z)+z{U}^{\mathrm{\prime }}(z))$ by the product rule of [F10]).

(Expansions of $({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2$ and ${\mathrm{\wp }}^3$.) Write steps 5.1 and 6.1 as $\mathrm{\wp }(z)=z^{-2}(1+3G_4{z}^4+5G_6{z}^6+z^{8}U(z))$ and ${\mathrm{\wp }}^{\mathrm{\prime }}(z)=-2z^{-3}(1-3G_4{z}^4-10G_6{z}^6+z^8{V}_1(z))$ with $V_1$ holomorphic near $0$ (one has $z^{5}V(z)=-2z^{-3}\cdot z^8{V}_1(z)$ with $V_1=-\frac{1}{2}V$, and the constant term is absorbed since $V$ is holomorphic). Squaring and cubing the brackets with the product rule of [F10] gives [F10, step 5.1, step 6.1] $({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2=4z^{-6}(1-6G_4{z}^4-20G_6{z}^6+z^8{W}_1(z))=4z^{-6}-24G_4{z}^{-2}-80G_6+z^{2}W(z),$ $4{\mathrm{\wp }}^3=4z^{-6}(1+9G_4{z}^4+15G_6{z}^6+z^8{X}_1(z))=4z^{-6}+36G_4{z}^{-2}+60G_6+z^{2}X(z)$ with $W,X$ holomorphic near $0$; all displayed coefficients are read off by expanding the products $(1+a+b+c{)}^2$ and $(1+a+b+c{)}^3$ and using that the resulting remainders are holomorphic.

(The difference is bounded at the origin.) Put $g_2:=60G_4$, $g_3:=140G_6$ and $F:=({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2-4{\mathrm{\wp }}^3+g_2\mathrm{\wp }+g_3$, a meromorphic function on $\mathbb{C}\setminus \mathrm{\Lambda }$. Using step 7.1 and $\mathrm{\wp }(z)=z^{-2}+O(z^2)$ from step 5.1 [F7, step 7.1, step 5.1] $F(z)=(-24-36+60)G_4{z}^{-2}+(-80-60+140)G_6+z^{2}Y(z)=z^{2}Y(z)$ for a holomorphic $Y$ near $0$; in particular $F(z)\to 0$ as $z\to 0$, so $F$ is bounded on a punctured neighbourhood of $0$. By [F7] the singularity of $F$ at $0$ is removable and the extension has $F(0)={\lim }_{z\to 0}F(z)=0$.

($F$ is elliptic and pole-free.) By [F4], $\mathrm{\wp }$ and ${\mathrm{\wp }}^{\mathrm{\prime }}$ are $\mathrm{\Lambda }$-elliptic; by [F5] constants are elliptic and sums, products and constant multiples of elliptic functions are elliptic, so $F=({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2-4{\mathrm{\wp }}^3+g_2\mathrm{\wp }+g_3$ is a $\mathrm{\Lambda }$-elliptic function. Its poles can only occur where $\mathrm{\wp }$ or ${\mathrm{\wp }}^{\mathrm{\prime }}$ has a pole, i.e. at lattice points, by [F4]; but $F$ extends holomorphically at $0$ by step 8.1 and $F$ is $\mathrm{\Lambda }$-periodic, so near every $\lambda \in \mathrm{\Lambda }$ one has $F(\lambda +u)=F(u)$ for small $u\mathrm{\ne }0$, and the holomorphy at $0$ passes to $\lambda$. Hence $F$ is holomorphic on all of $\mathbb{C}$: a $\mathrm{\Lambda }$-elliptic function without poles.

(Conclusion.) By [F6] the pole-free elliptic function $F$ is constant, and the constant is $F(0)=0$ by step 8.1. Hence $({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2-4{\mathrm{\wp }}^3+g_2\mathrm{\wp }+g_3=0$ on $\mathbb{C}\setminus \mathrm{\Lambda }$, that is $({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2=4{\mathrm{\wp }}^3-g_2\mathrm{\wp }-g_3$; and the absolute convergence of $G_4$ and $G_6$ is step 3.1.