Solving Legendre's Equation with Leibnitz Formula

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Discussion Overview

The discussion centers around demonstrating that a specific function, ##V(x) = (1-x^2)^{m/2} P_l (x)##, is a solution to a differential equation related to Legendre's equation. Participants explore the use of Leibnitz's formula for differentiation and the implications of different values of ##m##.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to show that ##V(x)## is a solution to a differential equation involving Legendre polynomials, expressing difficulty in applying Leibnitz's formula for higher derivatives.
  • Another participant suggests that the proposed form of ##V(x)## is incorrect and provides an alternative expression, indicating that the original formulation may only apply when ##m=0##.
  • A later reply clarifies that the original poster has derived the Legendre polynomials and associated equations, acknowledging a missed term in their expression for ##V(x)##.
  • One participant notes that the equation resembles a Sturm-Liouville equation, suggesting a potential classification of the problem.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the proposed function ##V(x)## and its relation to the differential equation. There is no consensus on the validity of the original formulation or the necessity of the ##(1-x^2)^{m/2}## term.

Contextual Notes

Some assumptions regarding the values of ##m## and the applicability of Leibnitz's formula remain unresolved. The discussion does not clarify the implications of these assumptions on the overall problem.

davidge
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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.
 
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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.
 
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Charles Link said:
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.
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. :smile:
 
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It looks to me like a Sturm-Liouville equation...
 
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