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As a conclusion to a previous question on integrals, I get an answer in terms of inverse Weierstrass elliptic function : $$ f\left(x\right)=\wp^{-1}\left( \beta + \frac{9\beta^2-1}{3(x-\beta)} \right) $$

with the Weierstrass elliptic function associated with the ODE:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad \begin{cases}g_2 &= \frac43\\g_3 &= 4(\beta^3 - \frac{\beta}{3})\end{cases}$$

where :

  • $\beta \in \mathbb{R^{-}_{*}}$
  • $x \in \mathbb{R^{-}_{*}}$

My question is the following : how to express $f\left(x\right)$ in terms of real-valued Jacobi and/or inverse Jacobi functions (or additional special functions available in this list : C++ TR1) ?

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1 Answer 1

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This can be accomplished by the relation between the Weierstrass $\wp$ function and the Jacobi functions as described here. I will describe a possible solution which follows in a straightforward manner from 1.

First notice you have

$$ \wp(f(x); g_2, g_3) = \beta + \frac{9\beta^2-1}{3(x-\beta)}.$$

Then you can use any of (23.6.21-23.6.26) from 1. I will use 23.6.26 so that

$$ \wp(f(x)) = \wp( (f(x) - \omega_3) + \omega_3 ) = e_3 + \left( \frac{K(k) k}{\omega_1}\right)^2 \text{sn}^2\left(\frac{K(k)(f(x)-\omega_3)}{\omega_1}; k\right)\\ = \beta + \frac{9\beta^2-1}{3(x-\beta)} $$

where you can find the definitions for $k, \omega_j, e_j$ in the reference. Then solving this is straightforward:

$$ f(x) = \omega_3 +\left(\frac{\omega_1}{K(k)}\right) \text{sn}^{-1}\left( \frac{\omega_1}{K(k) k}\left( \beta + \frac{9\beta^2-1}{3(x-\beta)} - e_3\right)^{1/2} \right). $$

I hope this helps.

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