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Consider the curve \begin{equation} \label{eq:curve} y^2 = R(x) = ax^4+bx^3+cx^2+dx+e \end{equation} and the associated differential equation \begin{equation} \label{eq:fde} f'^2 = R(f) \end{equation} Previously we solved \eqref{eq:fde} by computing the Möbius transform mapping this curve to Weierstrass form. Here is a more direct method which can be generalised to other curves.

Even in this relatively simple case these calculations require computing eight terms of the power series and solving a system of equations in eight unknowns. As a result they are best carried out using CAS.

Power Series Solution

By repeatedly differentiating \eqref{eq:fde} it is easy to compute the first few terms of the power series solution with boundary condition $f(0)=x_0$ and $f'(0)=y_0$ as follows \begin{multline} \label{eq:fseries} f(z) = x_0 + y_0 z + \tfrac 1 4 R_1 z^2 + \tfrac 1 6 y_0 R_2 z^3 + \tfrac 1 {48} \left(6 R_0 R_3 + R_1 R_2\right) z^4 + \tfrac 1 {240} y_0 \left(24 R_0 R_4 + 9 R_1 R_3 + 2 R_2^2\right) z^5 + \\ \tfrac 1 {2880} \left(144 R_0 R_1 R_4 + 60 R_0 R_2 R_3 + 9 R_1^2 R_3 + 2 R_1 R_2^2\right) z^6 + \tfrac 1 {5040} y_0 \left(132 R_0 R_2 R_4 + 45 R_0 R_3^2 + 45 R_1^2 R_4 + 27 R_1 R_2 R_3 + R_2^3\right) z^7 + \bigO(z^8) \end{multline} where $R_i$ are the coefficients of the Taylor expansion $R$ evaluated at $x_0$ ie. $\displaystyle R_i= \frac 1 {i!} R^{[i]}(x_0)$. In a similar way when $x_0=\infty$ we get the Laurent series \begin{multline} \label{eq:flaurent} f(z) = {a^{-1/2}} z^{-1} - \tfrac 1 4 b a^{-1} - \tfrac 1 {48} \left(8ac - 3b^2\right) a^{-3/2} z - \tfrac 1 {64} \left(8a^2d - 4abc + b^3\right) a^{-2} z^2 -\tfrac 1 {11520} \left(1152a^3e - 288a^2bd - 224a^2c^2 + 240ab^2c - 45b^4\right)a^{-5/2}z^3 + \\ \tfrac 1 {3072} \left(8a^2d - 4abc + b^3\right)\left(8ac - 3b^2\right)a^{-3}z^4 +\bigO(z^5) \end{multline} We can similarly compute the power series for $\wp(z,g_2,g_3)$ as follows \begin{equation} \label{eq:pseries} \wp(z) = z^{-2} + \tfrac 1 {20} g_2 z^2 + \tfrac 1 {28} g_3 z^4 + \tfrac 1 {1200} g_2^2 z^6 + \tfrac 3 {6160} g_2 g_3 z^8 + \bigO(z^{10}) \end{equation}

Now assuming $f(z)$ has two simple poles at $z_1,z_2 \neq 0$ with residues $\alpha,-\alpha$ we can use the order 2 version of this formula to write $f(z)$ as \begin{equation} \label{eq:solution0} f(z) = f(0) + \tfrac 1 2 \alpha \left[ \frac {\wp'(z) + \wp'(z_1)} {\wp(z) - \wp(z_1)} -\frac {\wp'(z) + \wp'(z_2)} {\wp(z) - \wp(z_2)} \right] \end{equation} Attempting to solve this for the unknown coefficients $\alpha,\wp(z_1),\wp(z_2)$ leads to nonlinear equations. However rewriting it as \begin{equation} \label{eq:solution} f(z) = \frac {A_1 \wp'(z) + A_2\wp(z)^2 + A_3\wp(z) + A_4} {\wp(z)^2 + A_5\wp(z) + A_6} \end{equation} then moving the denominator to the left hand side of \eqref{eq:solution}, substituting in \eqref{eq:fseries} and \eqref{eq:pseries}, and finally equating the coefficients of the first eight $z^i$ terms, gives six linear equations in the six unknown coefficients $A_1,A_2,A_3,A_4,A_5,A_6$ and a further two linear equations in $g_2,g_3$. This set of eight equations is easily solved to yield: \begin{align*} A_1 &= -\tfrac 1 2 y_0 \\ A_2 &= x_0 \\ A_3 &= \tfrac 1 {12} \Big[ 3b x_0^2 + 4c x_0 + 3d \Big] \\ A_4 &= \tfrac 1 {144} \Big[ (18ad - 3bc) x_0^2 + (36ae + 9bd - 5c^2) x_0 + 18be - 3cd \Big] \\ A_5 &= -\tfrac 1 6 \Big[ 6a x_0^2 + 3b x_0 + c \Big] \\ A_6 &= \tfrac 1 {144} \Big[ (9b^2 - 24ac) x_0^2 + (6bc - 36ad) x_0 - 36ae + c^2 \Big] \\ g_2 &= \tfrac {1} {12} \Big[ 12ae - 3bd + c^2 \Big] \\ g_3 &= \tfrac {1} {432} \Big[ 72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3 \Big] \end{align*} In this way $f(z)$ can be expressed as a rational functional of $\wp(z),\wp'(z),a,b,c,d,e$ and $x_0,y_0$. And we have expressed $g_2,g_3$ as polynomials in $a,b,c,d,e$.

Reverse Solution

A similar procedure can be applied to express $\wp(z)$ as a rational function of $f(z),f'(z),a,b,c,d,e$ and $x_0,y_0$. \begin{equation} \label{eq:reverse} \wp(z) = \frac {B_1 f'(z) + B_2 f(z)^2 + B_3 f(z) + B_4} {f(z)^2 + B_5 f(z) + B_6} \end{equation} where \begin{align*} B_1 &= \tfrac 1 2 y_0 \\ B_2 &= \tfrac 1 {12} \Big[ 6a x_0^2 + 3b x_0 + c \Big] \\ B_3 &= \tfrac 1 {12} \Big[ 3b x_0^2 + 4c x_0 + 3d \Big] \\ B_4 &= \tfrac 1 {12} \Big[ c x_0^2 + 3d x_0 + 6e \Big] \\ B_5 &= -2 x_0 \\ B_6 &= x_0^2 \\ \end{align*}

Is this a general algorithm ?

This algorithm could be applied to any genus 1 curve $F(x,y)=0$ with a known holomorphic differential $\displaystyle dz = \frac {Q(x,y)} {F_y(x,y)} dx = -\frac {Q(x,y)} {F_x(x,y)} dy$.

1. Choose an arbitrary point on the curve $x_0,y_0$ where $F(x_0,y_0)=0$.
2. Compute the power series solution for the uniformising elliptic functions $f(z),\space g(z)$ with boundary condition $f(0)=x_0,\space g(0)=y_0$. This is done by repeatedly differentiating the differential equations $f' = F_y(f,g) / Q(f,g)$ and $g' = -F_x(f,g) / Q(f,g)$ derived from the holomorphic differential. The coefficients will be rational functions^† of the coefficients of $F$ and $Q$, say $a_i$, and $x_0,y_0$.
3. Equate $f(z)$ to an appropriate rational function of $\wp(z),\wp'(z)$ with unknown coefficients $A_i$ and rearrange it into a linear equation for the $A_i$
4. Substitute the power series into this equation and equate the coefficients to zero, giving a set of linear equations for the $A_i$ and a further two relations for $g_2,g_3$. Solve these equations for $A_i$ and $g_2,g_3$
5. This yields $f(z)$ as a rational function of the coefficients of $a_i$ and $x_0,y_0$ and $\wp(z),\wp'(z)$ and also formulae expressing $g_2,g_3$ as a rational function in the coefficients of $a_i$.

Special Case 1

When $x_0 \rightarrow \infty$ we have $y_0/x_0^2 \rightarrow \sqrt{a}$, one of the poles say $z_2 \rightarrow 0$ and \eqref{eq:solution} collapses to \begin{equation} \label{eq:solution3} f(z) = \tfrac 1 2 \alpha \frac {\wp'(z) + \wp'(z_1)} {\wp(z) - \wp(z_1)} + \text{const.} = \frac {\pm 24\sqrt{a}\thinspace\wp'(z) - 12b\thinspace\wp(z) - 6ad + bc} {48a\thinspace\wp(z) + 8ac - 3b^2} \end{equation} So there are two solutions with $f(0)=\infty$ and $f'(0)=\infty$ and the solution is no longer rational.

Special Case 2

When $R(x_0)=0$ we have $y_0=0$ and \eqref{eq:solution} collapses to the Möbius transform computed in a previous section. \begin{equation} \label{eq:solutionL} f(z) = L(\wp(z)) = x_0 + \frac {B_1} {\wp(z) + B_2} \end{equation} where \begin{align*} B_1 &= \tfrac 1 {4} \Big[4 a x_0^3 + 3 b x_0^2 + 2 c x_0 + d\Big] \\ B_2 &= -\tfrac 1 {12} \Big[6 a x_0^2 + 3 b x_0 + c\Big] \\ \end{align*} The reduction is not obvious but follows from the fact that the numerator and denominator of \eqref{eq:solution} have a common factor when $R_0=0$ because \begin{equation*} \resultant(A_2 x^2 + A_3 x + A_4, x^2 + A_5 x + A_6) = \tfrac 1 {64} R_0 (R_0 R_3^2 - R_1^2 R_4) = 0 \end{equation*}

Rational Points

If $a,b,c,d,e$ are rational and $(x_0, y_0)$ is a rational point on curve \eqref{eq:curve} then \eqref{eq:solution} has rational coefficients. It puts the rational points on \eqref{eq:curve} in one to one correspondence with the rational points on the corresponding Weierstrass curve. In particular the point $(x_0, y_0)$ corresponds to the point $(\infty, \infty)$ on the Weierstrass curve.

Using \eqref{eq:solution} we can compute the birational map of the Jacobi curve \begin{equation} \label{eq:jacobi} y^2 = (1-x^2)(1-k^2x^2) \end{equation} to the Weierstrass curve \begin{equation} v^2 = 4u^3 - g_2u - g_3 \end{equation} We obtain \begin{equation} g_2=\tfrac 1 {12}(k^4 + 14k^2 + 1),\qquad g_3=\tfrac 1 {216}(k^6 - 33k^4 - 33k^2 + 1) \end{equation} and for eight obvious points on \eqref{eq:jacobi}

Addition Formula

When we use this algorithm to compute a rational mapping of a curve onto itself we are also computing the addition formulae for the uniformising elliptic functions. For example in \eqref{eq:solution} substitute the coefficients for the Weierstrass curve $a=0,b=4,c=0,d=-g_2,e=-g_3$. Then $f(z)=\wp(z+z_0)$ where $\wp(z_0)=x_0$ and $\wp'(z_0)=y_0$ and we have \begin{equation} \wp(z+z_0) = \frac {-\tfrac 1 2 \wp'(z_0)\wp'(z) + \wp(z_0)\wp(z)^2 + \wp(z_0)^2\wp(z) - \tfrac 1 4 g_2\wp(z) + \tfrac 1 4 g_2\wp(z_0) -\tfrac 1 2 g_3} {\wp(z)^2 - 2 \wp(z_0)\wp(z) + \wp(z_0)^2} \end{equation} which is equivalent to the standard addition formula \begin{equation} \wp(x+y) = \tfrac 1 4 \left[\frac {\wp'(x) - \wp'(y)} {\wp(x) - \wp(y)}\right]^2 - \wp(x) - \wp(y) \end{equation}