# I Legendre's equation

1. Jul 27, 2017

### davidge

Hi
I want to show that $V(x) = (1-x^2)^{m/2} P_l (x)$ is a solution of the equation
$$\frac{d}{dx} \bigg[(1-x^2) \frac{dV(x)}{dx} \bigg] + \bigg[l(l+1) - \frac{m^2}{1-x^2} \bigg]V(x) = 0.$$ Because the equation for $P_l (x)$ is $$\frac{d}{dx} \bigg[(1-x^2) \frac{dP_l (x)}{dx}\bigg] + l(l+1) P_l (x) = 0,$$ my attempts have been consisting of trying to differentiate the last equation $m$ times. I did not realize how to use the Leibnitz formula for general derivative, so I have been trying to differentiate it a small number of times and by induction arguing what it looks like after the $m \text{-th}$ derivative. I'm having no success in by proceeding this way, though.

However, this is just an attempt. I would like equally well any other one, because I just want to show that $V(x)$ given as above is a solution for the equation shown.

Last edited: Jul 27, 2017
2. Jul 30, 2017

A google of the subject shows that $V(x)=(-1)^m (1-x^2)^{m/2} \frac{d^m P_l(x)}{dx^m}$, so that I believe the $V(x)$ as you presented it is incorrect. $\\$ Editing: I think your solution for $P_l(x)$ and its associated equation is simply the case where $m=0$, but then you don't need the $(1-x^2)^{m/2}$ term in $V(x)$, and the result of the $P_l(x)$ equation follows immediately.

Last edited: Jul 30, 2017
3. Jul 30, 2017

### davidge

I think you misunderstood my question. I have actually showed what the solution is, and I said I was interested in showing that that expression (I missed the $(-1)^m$ term) for $V(x)$ is a solution for the equation.

It turns out that after starting this thread I was able to derive both the Legendre's polynomials and the associated Legendre equation (and its solution) with the help of the manuscript on math-phys by @vanhees71.

4. Aug 3, 2017

### Svein

It looks to me like a Sturm-Liouville equation...

Last edited: Aug 3, 2017