Variational symmetries for the Emden-Fowler equation

[giraffe714]

Homework Statement

The Emden-Fowler equation of astrophysics is $y'' + \frac{2}{x}y' + y^5 = 0$ which arises as the Euler-Lagrange equation to the functional $J(y) = \int_{x_0)^{x_1} \frac{x^2}{2}(y'^2-\frac{1}{3}y^6) dx$. Find the infinitesimal generators that lead to a variational symmetry for this functional and establish the conservation law $$ x^2(y'y + 2x(y'^2+y^5)) = const. $$

Relevant Equations

$$ \xi \frac{\partial f}{\partial x} + \eta \frac{\partial f}{\partial y} + (\eta' - y'\xi')\frac{\partial f}{\partial y'} + \xi' f = 0 $$ where $$ \eta' = \frac{\partial \eta}{\partial x} + \frac{\partial \eta}{\partial y} y' $$ and $$ \xi' = \frac{\partial \xi}{\partial x} + \frac{\partial \xi}{\partial y} y' $$, Noether's theorem $$ \eta \frac{\partial f}{\partial y'} + \xi (f - \frac{\partial f}{\partial y'}y') = const. $$

So firstly I calculated the partial derivatives of f to be:
$$ \frac{\partial f}{\partial x} = \frac{2x}{2} (y'^2 - \frac{1}{3} y^6) + \frac{x^2}{2} (2y'y'' - 2y^5y') = xy'^2 - \frac{1}{3} y^6 + x^2y'(y'' - y^5) $$
$$ \frac{\partial f}{\partial y} = \frac{x^2}{2}*\frac{1}{3}*6y^5 = x^2y^5 $$
$$ \frac{\partial f}{\partial y'} = \frac{x^2}{2}*2y' = x^2y' $$

And then I plugged that into the first equation in the "Relevant equations" (not sure what it's called,) with $\xi_x = \partial \xi / \partial x$ etc. for conciseness:

$$ \xi (xy'^2 - \frac{1}{3}y^6 + x^2y'(y'' - y^5)) - \eta x^2y^5 + (\eta_x + \eta_y y' - \xi_x y' - \xi_y y'^2)x^2 y' + \xi_x (\frac{x^2}{2}(y'^2-\frac{1}{3}y^6) ) + \xi_y (\frac{x^2}{2}(y'^2-\frac{1}{3}y^6)) y' = 0 $$

Which after expanding everything hopefully gives

$$ \xi xy'^2 - \frac{1}{3} \xi y^6 + \xi x^2 y' (y'' - y^5) + \eta x^2 y^5 + \eta_x x^2 y' + \eta_y x^2 y'^2 - \xi_x x^2 y'^2 - \xi_y x^2 y'^3 + \xi_x \frac{x^2}{2} y'^2 - \xi_x \frac{x^2}{6} y^6 + \xi_y \frac{x^2}{2} y'^3 - \eta_y \frac{x^2}{6} y^6 y' = 0 $$

Upon regrouping with $y'^2, y^6, y', y^5, y'^3, y^6y'$ (and this is probably where my mistake lies, but I don't know how to fix it) gives the equations

$$ \xi x + \eta_y x^2 - \eta_x x^2 + \eta_x \frac{x^2}{2} = 0 $$ (from $y'^2$)
$$ -\frac{1}{3} \xi - \xi_x \frac{x^2}{6} = 0 $$ (from $y^6$)
$$ \xi x^2 (y'' - y^5) + \eta_x x^2 = 0 $$ (from $y'$)
$$ \eta x^2 = 0 $$ (from $y^5$)
$$ -\xi_y x^2 + \xi_y \frac{x^2}{2} = 0 $$ (from $y'^3$)
$$ -\xi_y \frac{x^2}{6} = 0 $$ (from $y^6y'$)

But, the equation from $y^5$ implies that $\eta = 0$, and plugging that into both the equation from $y'^2$ and $y'$ implies $\xi = 0$. This however can't be true since this problem has a conservation law by Noether's theorem in the problem statement. I guess what I'm not fully understanding in this problem is how to group the equations together. Is it just be $y'$ and their powers? Is it also by $y$? What about combination terms like $y^6y'$? And if it's the former, what happens to the terms not involving $y'$?

 

[renormalize]

Homework Statement: The Emden-Fowler equation of astrophysics is $y'' + \frac{2}{x}y' + y^5 = 0$ which arises as the Euler-Lagrange equation to the functional $J(y) = \int_{x_0)^{x_1} \frac{x^2}{2}(y'^2-\frac{1}{3}y^6) dx$. Find the infinitesimal generators that lead to a variational symmetry for this functional and establish the conservation law $$ x^2(y'y + 2x(y'^2+y^5)) = const. $$

Please verify that your equation for $J(y)$ that fails to render is:$$J\left(y\right)=\intop_{x_{0}}^{x_{1}}\frac{x^{2}}{2}(y'^{2}-\frac{1}{3}y^{6})dx$$

 

[giraffe714]

Please verify that your equation for $J(y)$ that fails to render is:$$J\left(y\right)=\intop_{x_{0}}^{x_{1}}\frac{x^{2}}{2}(y'^{2}-\frac{1}{3}y^{6})dx$$

Oh, yes, that's correct, my apologies.

 

[renormalize]

Your question is not really in my wheelhouse, but I do suggest that you take a look at this article: https://www.researchgate.net/publication/224831129_The_Lane-Emden_equation_and_its_generalizations. In particular, §2 case 5.1 discusses the Lie point symmetry of your differential equation and §3 case 5.1 covers the Noether point symmetry of your Lagrangian.