Understanding Legendre differential equation

[chwala]

TL;DR

Analysis of $(1-x^2)y^{"}-2x y'+ n(n+1)y=0$.

I am looking at this now, $(1-x^2)y^{"}-2x y' + n(n+1)y=0$, my interest being on $P_n(x)$, i will go straight to the point,

... steps leading to this are clear, i.e

we have the recursive relationship,

$a_{k+2} = \dfrac{(k-n)(k+n+1)}{(k+2)(k+1)}$

For even solutions, we set $a_1 =0$ and $a_0=1$ Is this always the case?, of course i get the even and odd part... using symmetry,

Now in my understanding, we shall end up with,

$k=0, a_2=-3$

$k=2, a_4 = 0$

...

$y(x)=a_0x^0 + a_2x^2 + a_4x^4$ (which terminates at $k=2$).

$y(x)=1-3x^2= P_2(x)$ my step here is clear, now my confusion is here or misunderstanding, arises from the following lines,

$y(x)=a_0(1-3x^2)$ where is this $a_0$ coming from where? or its set again for conformity? and then further, $P_2(x)= \dfrac{1}{2}(3x^2-1)$, how?

...i read scale factor or something, i may need a clear explanation...cheers !

 

[anuttarasammyak]

https://en.wikipedia.org/wiki/Legendre_polynomials

Does your sequence coincide with this ?

 

[pasmith]

TL;DR Summary: Analysis of $(1-x^2)y^{"}-2x y'+ n(n+1)y=0$.

I am looking at this now, $(1-x^2)y^{"}-2x y' + n(n+1)y=0$, my interest being on $P_n(x)$, i will go straight to the point,

... steps leading to this are clear, i.e

we have the recursive relationship,

$a_{k+2} = \dfrac{(k-n)(k+n+1)}{(k+2)(k+1)}$

This is the correct recurrence. If you se $l = k - 2$ you will get it in the form posted by @anuttarasammyak.

For even solutions, we set $a_1 =0$ and $a_0=1$ Is this always the case?

Note that the ODE is linear, so a constant multiple of a solution is a solution. The same is true of the recurrence relation for the $a_k$. But the recurrence has to start somewhere, so $a_0$ and $a_1$ are arbitrary constants.

The recurrence is of the form $a_{k+2} = f(k)a_k$. It follows that if $a_1 = 0$ and $a_0 \neq 0$ then $y$ will be even, and if $a_0 = 0$ and $a_1 \neq 0$ then $y$ will be odd.

The Legendre polynomials are normalized such that $P_n(1) = 1$. This means that we don't get $P_{2m}$ by setting $(a_0,a_1) = (1,0)$; instead we have to leave $a_0$ arbitrary until we know the value of $y(1)$ in terms of $a_0$.