General solution of second order differential equaiton.

[E92M3]

Homework Statement

Show that:
$$\theta (\xi) = 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +...$$
is a general solution to the Lane-Emden equation. (assuming that the above equation converges)

Homework Equations

Lane-Emden equation:
$$\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n =0$$

The Attempt at a Solution

$$\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n = \frac{1}{\xi^2} \left( \xi^2 \frac{d^2 \theta}{d \xi^2} + 2 \xi \frac{d\theta}{d \xi}\right )+ \theta^n =\frac{d^2 \theta}{d \xi^2}+ \frac{2}{\xi} \frac{d\theta}{d \xi}+ \theta^n =0$$
$$\frac{d\theta}{d \xi}=\frac{-1}{3}\xi+\frac{n}{30}\xi^3+...$$
$$\frac{d^2 \theta}{d \xi^2}=\frac{-1}{3}+\frac{n}{10}\xi^2+...$$
$$\frac{2}{\xi}\frac{d\theta}{d \xi}=\frac{-2}{3}+\frac{n}{15}\xi^2+...$$
$$\theta^n=\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +...\right )^n$$
Plugging them in we get:
$$\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n \left ( \frac{-1}{3}+\frac{n}{10}\xi^2+...\right ) +\left ( \frac{-2}{3}+\frac{n}{15}\xi^2+...\right ) +\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +...\right )^n$$
Simplifying the 1st 2 terms:
$$=\left ( -1+\frac{n}{6}\xi^2+...\right )+\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +...\right )^n$$
Well, since the series converges I can just look at the 1st 2 terms in the sums (am I correct?). I need to deal with:
$$\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +...\right )^n$$
I don't know how to write it out mathematically but I can try to say what I think. After expanding, the first term I'll get is 1, or more precisely, 1^n. The xi^2 term can only come from the product of the first two terms in the series since the product between any other terms will get me a higher power of xi. I tried it out with the first few n's and find that there will be n of the xi^2 term. Therefore:
$$\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +...\right )^n=1^n-\frac{n}{6}\xi^2+...$$
So theta is indeed the solution (but how is it general? is the question wrong?) Also, how can I write out what I just said more mathematically?

 

[mighty2000]

Thank you for your post regarding the Lane-Emden equation and the solution provided. Your approach to the problem is correct and your solution is also correct. I would like to provide some additional clarification and explanation to your solution.

Firstly, you are correct in saying that the series given in the problem converges, as it is a power series with alternating signs and decreasing terms. This means that we can take the first few terms of the series and still have a good approximation to the actual solution.

Now, to show that the given solution is a general solution, we need to show that it satisfies the Lane-Emden equation for any value of n. In your solution, you have shown this for a specific value of n, but we need to show it for all values of n. This can be done by considering the general term of the series, which is given by:

\frac{d^2 \theta}{d \xi^2}+ \frac{2}{\xi} \frac{d\theta}{d \xi}+ \theta^n = \frac{n(n-1)}{120}\xi^{n-2}+...

As you can see, the term containing n depends on the value of n, which means that for different values of n, we will have different terms in the series. However, the remaining terms in the series are the same for all values of n, which means that the given solution satisfies the Lane-Emden equation for all values of n. Hence, we can say that it is a general solution.

To write this more mathematically, we can say that the given solution is a general solution if it satisfies the Lane-Emden equation for all values of n. In other words, for any value of n, the given solution satisfies:

\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n =0

This is what we have shown in your solution, by considering the general term of the series.

I hope this helps to clarify your understanding of the problem. Keep up the good work!