Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Finding simpler symmetries to differential equations

+ 0 like - 0 dislike
1598 views

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:

$$\frac{dy}{dx}=\frac{12}{19}\frac{y\left(1+\left(\frac{73}{24}x^2\right)+\left(\frac{37}{26}x^4\right)\right)}{x\left(1+\left(\frac{121}{304}x^2\right)\right)}$$
 

Where $y$ is the semi-major axis and $x$ is the eccentricity.


Where $y$ is the semi-major axis and $x$ is the eccentricity. The 3-D plots of this equation can be found [here](http://www.wolframalpha.com/input/?i=3D+plot++%5Cfrac%7B12%7D%7B19%7D%5Cfrac%7By%5Cleft(1%2B%5Cleft(%5Cfrac%7B73%7D%7B24%7Dx%5E2%5Cright)%2B%5Cleft(%5Cfrac%7B37%7D%7B26%7Dx%5E4%5Cright)%5Cright)%7D%7Bx%5Cleft(1%2B%5Cleft(%5Cfrac%7B121%7D%7B304%7Dx%5E2%5Cright)%5Cright)%7D)

And this is the solution to the above DE [here](http://www.wolframalpha.com/input/?i=solve+y'%3D+%5Cfrac%7B12%7D%7B19%7D%5Cfrac%7By%5Cleft(1%2B%5Cleft(%5Cfrac%7B73%7D%7B24%7Dx%5E2%5Cright)%2B%5Cleft(%5Cfrac%7B37%7D%7B26%7Dx%5E4%5Cright)%5Cright)%7D%7Bx%5Cleft(1%2B%5Cleft(%5Cfrac%7B121%7D%7B304%7Dx%5E2%5Cright)%5Cright)%7D)

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.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).

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?

asked Feb 19, 2017 in Astronomy by Naveen (85 points) [ revision history ]
edited Feb 21, 2017 by Naveen

Powers in the integrand in your equation (1) are very non-trivial, so I don't think that hypergeometric function can be further simplified.

The links don't work. Please try the corrected ones independently of their generation before posting the corrections. Use the chain symbol in the editor to hide the URL under the [here].

@Arnold Neumaier I have edited the links. They work now

The links now do work, but content of the second link is still uninformative. 

I'm sorry about that I am unable to pin info via Desmos.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...