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The Lane-Emden equation is typically written as $$\frac{1}{\xi^2} \frac{d}{d\xi} \left( \xi^2 \frac{d\theta}{d\xi} \right) = -\theta^n,$$ which makes sense given the derivation on the Wikipedia page. But, why not keep going with the simplification?

$$\begin{align*} \frac{1}{\xi^2} \frac{d}{d\xi} \left( \xi^2 \frac{d\theta}{d\xi} \right) &= \frac{1}{\xi^2} \left( 2\xi \frac{d\theta}{d\xi} + \xi^2 \frac{d^2\theta}{d\xi^2} \right) \\ &= \frac{2}{\xi} \frac{d\theta}{d\xi} + \frac{d^2\theta}{d\xi^2} \end{align*}$$

So the Lane-Emden equation could also be written as $$\frac{d^2\theta}{d\xi^2} = -\frac{2}{\xi} \frac{d\theta}{d\xi} - \theta^n,$$ or, seeing as $\theta$ is a function of $\xi$, $$\theta'' = -\frac{2}{\xi}\theta' - \theta^n.$$

In my opinion, this is clearer. When I first came across this, I was confused by the $\frac{d}{d\xi} ( \dots )$ notation and thought, why not multiply out the $\frac{1}{\xi^2}$ and the $\xi^2$ and just have $\frac{d}{d\xi} \frac{d\theta}{d\xi}$? Even as a math major and enthusiast! Furthermore, as Kyle Kanos said in chat,

For $n=1$, this formulation is the way to go because it can be solved with power series

When I first encountered this Lane-Emden equation, I was confused and it took me some time and work to figure out, realize, and understand that it's just a plain ol' second-order ordinary differential equation! So, my question is, why write it the former way and not the latter? Especially since the latter seems to be clearer (to me anyway).

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    $\begingroup$ I can wager a guess: $\frac{1}{\xi^2}\frac{d}{d\xi}\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n$ is quite a convenient form since the radial part of the Laplacian in spherical coordinates, which comes up all the time, is exactly of this form. Therefore, we immediately have some intuition. In any case, it is obvious to someone who has worked with ODE's before that this is just that, and your rewritings don't really achieve anything new... $\endgroup$ – Danu Dec 3 '14 at 20:29
  • $\begingroup$ @Danu: Well, there's my problem! I avoid differential equations, preferring to stick to other branches of math... :P $\endgroup$ – El'endia Starman Dec 3 '14 at 20:32
  • $\begingroup$ As it so happens, most of physics is purely solving (sometimes difficult, nonlinear) PDE's, or the simpler case of ODE's. There's no way around it, so better get used to it ;) $\endgroup$ – Danu Dec 3 '14 at 20:33
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The Lane-Emden is really a non-dimensional form of the Poisson's equation with spherical symmetry: $$ \nabla^2f(r)\equiv\frac1{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right) $$ This should be clear to all that, since we have two factors of $d/dr$ on the right hand side of the above, it must be a 2nd order differential equation. This 2nd-order derivative appears in many places in physics (e.g., heat/diffusion equations), so the fact that it's a commonly appearing relation suggests that familiarity was the reason we keep the Lane-Emden equation as it appears normally.

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  • $\begingroup$ Well if it were really clear to all, there would be no reason for anyone to ask this question ;-) $\endgroup$ – David Z Dec 3 '14 at 20:38
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    $\begingroup$ Familiarity (and consistency) is a pretty common reason to keep certain notations and formulations, so I wasn't surprised that this was the case here as well. $\endgroup$ – El'endia Starman Dec 3 '14 at 20:39

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