I am working on a personal math project of mine, and in order for me to continue I need to know the solution to this following system of nonlinear equations.

$$ \frac{d\omega_1}{d\eta} = \omega_2\omega_3 - \omega_1(\omega_2+\omega_3)$$ $$ \frac{d\omega_2}{d\eta} = \omega_1\omega_3 - \omega_2(\omega_1+\omega_3)$$ $$ \frac{d\omega_3}{d\eta} = \omega_1\omega_2 - \omega_3(\omega_1+\omega_2)$$

This equation is referenced in this paper https://arxiv.org/abs/gr-qc/9409025 and the solution is further referenced in F. J. Bureau, Sur des systèmes différentiels du troisième ordre et les équations différentielles associées, Bulletin de la Classe des Sciences LXXIII (1987) 335–353. My native language is Japanese and I can't read a word of French to save my life. In addition I can't even find this book or paper. I looked using two college libraries and Google Scholars. I would rather not need to duplicate effort and re derive the solution of these equations in terms of Hermite modular elliptic functions. If anyone who has worked with these equations before has reference with an English solution that would be nice. Or if anyone knows where I can find the French reference that would help also.

If none can be found then I would appreciate any hints on how to re derive the solution. My project did not intend to run into such formidable looking differential equation, it is a bit outside of my speciality of geometry.

Because this equation is in a mathematical physics paper I believe it is relevant for this stack exchange as well.

Any guidance will be appreciated.


migrated from physics.stackexchange.com Nov 21 '18 at 13:25

This question came from our site for active researchers, academics and students of physics.

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    $\begingroup$ I'm voting to migrate this question to Mathematics SE since it is a math question, even if the topic has physics applications $\endgroup$ – Aaron Stevens Nov 13 '18 at 4:28
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    $\begingroup$ I find it mentioned here and here (search in the docs for Halphen). $\endgroup$ – Keith McClary Nov 13 '18 at 5:54
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    $\begingroup$ And here. $\endgroup$ – Keith McClary Nov 13 '18 at 6:41

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