4
$\begingroup$

I will begin with some background:

The solutions $\theta$ of $$\cos \theta=x $$ constitute of two families, each of which is an arithmetic progression. Namely, if $\arccos x$ denotes any particular solution then the families are $$\left\{ \arccos x+2 \pi k: k \in \mathbb{Z} \right\} \cup \left\{ -\arccos x+2 \pi k: k \in \mathbb{Z} \right\}. $$ If, instead, we solve the system $$\cos \theta=x \\ \sin \theta=y $$ with $x,y$ satisfying the compatibility condition $$x^2+y^2=1 $$ then the solutions $\theta$ form a single arithmetic progression $$\{\theta^*+2 \pi k: k \in \mathbb{Z} \} $$ where $\theta^*$ is any particular solution. Since $\theta$ is then uniquely defined up to additive integer multiples of $2 \pi$, the differential $\mathrm{d} \theta$ is well-defined, and one can see that $$\mathrm{d} \theta=-y \mathrm{d} x+x \mathrm{d} y. $$


Analogously, one can think of the equation $$\wp(\theta;g_2,g_3)=x $$ as being an underdetermined system, since it has the reflection symmetry $\theta \mapsto -\theta$, as well as the translations $\theta \mapsto \theta+2 \omega_i, \; i=1,2$, where $2\omega_1,2\omega_2$ are generators of the period lattice $$\Lambda=\{2m \omega_1+2n \omega_2:m,n \in \mathbb{Z} \}.$$ If, instead one looks at the system $$\wp(\theta;g_2,g_3)=x \\ \wp'(\theta;g_2,g_3)=y$$ with $x,y$ satisfying the compatibility condition $$y^2=4x^3-g_2 x-g_3 $$ then $\theta$ is uniquely defined only up to translations by points on the period lattice. In fact, one of the branches of $\theta$ is what Mathematica calls InverseWeierstrassP[{x,y},{g2,g3}] or $$\wp^{-1} (x,y;g_2,g_3).$$

My questions are:

  • Is there a nice formula for $\mathrm{d} \theta$ in this case?
  • Similarly, are there simple formulas for the partial derivatives below? $$\frac{\partial}{\partial x} \wp^{-1}(x,y;g_2,4x^3-g_2x-y^2) \\ \frac{\partial}{\partial y} \wp^{-1}(x,y;g_2,4x^3-g_2x-y^2) $$
$\endgroup$
1
  • $\begingroup$ See also how you can construct $\wp(z)$ from its differential equation $\endgroup$
    – reuns
    Aug 19, 2017 at 15:24

2 Answers 2

3
$\begingroup$

Edit and correction (thanks to Nemo for pointing out):

First note that even though $d\theta$ is well defined, it is well defined as a differential of a map on $S^1$ and not on $\mathbb{R}^2$, which means that its representation as a linear sum of $dx$ and $dy$ is not unique. In fact, the tangent space (and the cotangent space) are one-dimensional. Therefore, the choice of map varies the differential (as a differential on $\mathbb{R}^2$) - e.g. the linear combination you have written can be obtained by using $\theta = \arctan(y/x)$.

In our case, the explicit description of this branch of the inverse Weierstrass function is $\wp^{-1}(x,y)=\int_{-\infty}^{x} \frac{1}{\sqrt{4t^3-g_2t-g_3}}dt$. This means that $$ d\theta = y^{-1}dx + (6x^2 - \frac{g_2}{2})^{-1}dy $$

This makes sense regarding the previous answer, where I had forgotten to invert the results in the end.

Old Answer:

Note that the differential $d\theta$ is simply the gradient of the path defined above. In the case of $\wp$ we see that the path is given by $\theta \mapsto (\wp(\theta), \wp'(\theta)) = (x,y)$, hence the gradient is given by $(\wp'(\theta), \wp''(\theta))$. Next, we derive the functional equation $$ (\wp'(\theta))^2 = 4(\wp(\theta))^3 - g_2 \wp(\theta) - g_3 $$ to obtain $$ 2 \wp'(\theta) \wp''(\theta) = 12 (\wp(\theta))^2 \wp'(\theta) - g_2 \wp'(\theta) $$ and hence $$ \wp''(\theta) = 6(\wp(\theta))^2 - \frac{g_2}{2} $$ Recalling that $\wp(\theta) = x$, we see that $\wp''(\theta) = 6x^2-\frac{g_2}{2}$, hence we may write $$ d\theta = ydx + (6x^2 - \frac{g_2}{2})dy $$

Hope that this fully answers your question.

$\endgroup$
2
  • $\begingroup$ It looks like you took $g_3$ as a constant here. Do you know anything about the case where $g_3=4x^3-g_2x-y^2$, as in the end of my question? $\endgroup$
    – user1337
    Aug 3, 2017 at 16:21
  • $\begingroup$ I think this is wrong. $$ydx + (6x^2 - \frac{g_2}{2})dy=y^2d\theta+(6x^2 - \frac{g_2}{2})^2d\theta=\\=\left(4x^3-g_2x-g_3+36x^4-6g_2x^2+g_2^2/4\right)d\theta\neq const\cdot d\theta.$$ However if $g_2=0$ then $ydx-\frac23 xdy=-g_3d\theta$. $\endgroup$ Aug 4, 2017 at 10:57
0
$\begingroup$

For a fixed $g_2$, we will consider a change of coordinates from $(x,y)$ to $(g_3,\theta)$ defined by the equations

$$\begin{align}x&=\wp(\theta;g_2,g_3), \\ y&=\wp'(\theta;g_2,g_3). \end{align} $$

Taking the differential of both equations, one gets

$$\begin{align} \mathrm{d} x&=\wp'(\theta;g_2,g_3) \mathrm{d} \theta+\frac{\partial \wp}{\partial g_3}(\theta;g_2,g_3) \mathrm{d} g_3 \\ \mathrm{d} y&=\wp''(\theta;g_2,g_3) \mathrm{d} \theta+ \frac{\partial \wp'}{\partial g_3}(\theta;g_2,g_3) \mathrm{d} g_3 \end{align} $$

Simplifying, using the 2nd order ODE for $\wp$, and identity 18.6.19 in Abramowitz and Stegun, yields

$$\begin{align} \mathrm{d} x=&y \mathrm{d} \theta+\frac{y \left(6 g_2 \zeta \left(\theta ;g_2,g_3\right)-9 g_3 \theta \right)+12 g_2 x^2-18 g_3 x-2 g_2^2}{2 \left(g_2^3-27 g_3^2\right)} \mathrm{d} g_3 \\ \mathrm{d} y=&\left( 6x^2-\frac{1}{2} g_2 \right) \mathrm{d} \theta+\frac{6 g_2 \left(12 x^2-g_2\right) \zeta \left(\theta ;g_2,g_3\right)-54 g_3 \left(2 \theta x^2+y\right)+9 g_2 \left(g_3 \theta +4 x y\right)}{4 \left(g_2^3-27 g_3^2\right)} \mathrm{d} g_3 \end{align} $$

Now, inverting these relations gives $$\begin{align} d\theta=&\frac{ \left(6 g_2 \left(g_2-12 x^2\right) \zeta \left(\theta ;g_2,g_3\right)+54 g_3 \left(2 \theta x^2+y\right)-9 g_2 \left(g_3 \theta +4 x y\right)\right)}{2 \left(9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3\right)}\mathrm{d} x+\frac{\left(6 g_2 \left(y \zeta \left(\theta ;g_2,g_3\right)+2 x^2\right)-9 g_3 (2 x+\theta y)-2 g_2^2\right)}{9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3} \mathrm{d} y \\ dg_3=&\frac{ \left(g_2^3-27 g_3^2\right) \left(12 x^2-g_2\right)}{9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3}\mathrm{d} x-\frac{2 \left(g_2^3-27 g_3^2\right) y}{9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3} \mathrm{d} y \end{align} $$

Lastly, since $g_3=4x^3-g_2 x-y^2$ the equations take the form

$$ \begin{align} \mathrm{d} \theta=&\frac{ \left(3 g_2 \left(-2 \left(g_2-12 x^2\right) \zeta \left(\theta ;g_2,4 x^3-g_2 x-y^2\right)-3 g_2 \theta x+48 \theta x^3+30 x y-3 \theta y^2\right)-54 \left(4 x^3-y^2\right) \left(2 \theta x^2+y\right)\right)\mathrm{d} x}{108 g_2 x \left(y^2-4 x^3\right)+54 g_2^2 x^2-2 g_2^3+54 \left(y^2-4 x^3\right)^2}+\frac{ \left(g_2 \left(-6 y \zeta \left(\theta ;g_2,4 x^3-g_2 x-y^2\right)+2 g_2-30 x^2-9 \theta x y\right)+9 \left(4 x^3-y^2\right) (2 x+\theta y)\right) \mathrm{d} y}{54 g_2 x \left(y^2-4 x^3\right)+27 g_2^2 x^2-g_2^3+27 \left(y^2-4 x^3\right)^2} \\ \mathrm{d}g_3=&(12x^2-g_2)\mathrm{d} x-2y \mathrm{d} y \end{align}.$$ This finally gives $\frac{\partial \theta}{\partial x},\frac{\partial \theta}{\partial y} .$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.