I have developed a differential equation for the variation of a star's semi-major axis with respect to its eccentricity. It is as follows:


Where $y$ is the semi-major axis and $x$ is the eccentricity. The 3-D plots of this equation can be found here

And this is the solution to the above DE: here

My question is this: There is a well defined symmetry for the above equation from the plot. Hence, is it possible to express this in terms of other special function (which have different symmetries). The decay time of stars can be found by solving the following integral: $$T(a_{0},e_{0})=\frac{12(c_{0}^4)}{19\gamma}\int_{0}^{e_0}{\frac{e^{29/19}[1+(121/304)e^2]^{1181/2299}}{(1-e^2)^{3/2}}}de\tag1$$ Where $$\gamma=\frac{64G^3}{5c^5}m_{1}m_{2}(m_{1}+m_{2})$$ For $e_{0}$ close to $1$ the equation becomes: $$T(a_{0},e_{0})\approx\frac{768}{425}T_{f}a_{0}(1-e_{0}^2)^{7/2}\tag2$$ Where $$T_{f}=\frac{a_{0}^4}{4\gamma}$$

I used Appell's hypergeometric functions to solve integral (1), but is there any way in which I can express the solutions in terms of few special functions with simpler symmetries, so that the analysis becomes easier.

Edit: I was suggested that since the powers in the integrand in equation (1) are very non-trivial, probably the hypergeometric function can't be further simplified. But I fail to understand why this might seem to pose a problem. Can't this D.E. be solved by Lie symmetry methods? Or can this solution's field be treated using Frobenius' theorem and the dimensions of it analysed?

  • $\begingroup$ The only symmetry I can see in first derivative is that it is even w.r.t e, which is obvious since negative eccentricity is not really meaningful here. Is that what you are talking about ? $\endgroup$ – Piyush Grover Feb 20 '17 at 15:55
  • $\begingroup$ @PiyushGrover I am talking more about symmetry as a type of invariance, and the application of Lie point symmetry to solve this DE $\endgroup$ – Naveen Balaji Feb 20 '17 at 16:01
  • $\begingroup$ @PiyushGrover In short, can the solution to this DE be expressed in a more elegant manner with the aid of special functions(which have a well defined symmetry)? $\endgroup$ – Naveen Balaji Feb 20 '17 at 16:02
  • $\begingroup$ well your equation for first derivative is invariant under transformation: e-> -e $\endgroup$ – Piyush Grover Feb 20 '17 at 16:03
  • $\begingroup$ Yeah, but what about the latter part of question? $\endgroup$ – Naveen Balaji Feb 20 '17 at 16:04

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