# I Legendre Differential Equation

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1. Oct 12, 2016

### TimeRip496

I just started learning Legendre Differential Equation. From what I learn the solutions to it is the Legendre polynomial.

For the legendre DE, what is the l in it? Is it like a variable like y and x, just a different variable instead?

Legendre Differential Equation: $$(1-x^2) \frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + l(l+1)y = 0$$

But what is the differences between the associated Legendre polynomial and unassociated Legendre polynomial? When do we use the associated one or the unassociated one, since both are solutions to the Legendre Differential Equation?

Unassociated Legendre Polynomial: $$P_l(x) = \sum^M_{m=0}(-1)^m\frac{(2l-2m)!}{2^lm!(l-m)!(l-2m)!}x^{l-2m}$$

Associated Legendre Polynomial: $$P^m_l(x) = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}(P_l(x))$$

Lastly, where did the m come from?

2. Oct 12, 2016

### Staff: Mentor

3. Oct 17, 2016

### TimeRip496

How did you get from (8) to (12)?
$$\sum^\infty_{n=0} n(n-1) a_nx^{x-2} ⇒ \sum^\infty_{n=0} (n+2)(n+1) a_{n+2} x^n$$

Shouldnt it be $$\sum^\infty_{n=0} (n+2)(n+1) a_{n+2} x^n + 6a_{-2}x^{-4} +2a_{-1}x^{-3}$$ cause you replace n with n+2?

Besides how do you find the value for a0 and a1 after getting (17)?
Sorry for these questions as I am still new to this.

4. Oct 17, 2016

### Staff: Mentor

The first three terms of the series on the left are $0(-1)a_0x^{-2} + 1(0)a_1x^{-1} + 2(1)a_2x^0$, so you can discard the first two terms, as their coefficients are both zero. The first term of the series on the right is $2(1)a_2x^0$, so now you can adjust the starting index of the series on the left, replacing n by n + 2.

5. Oct 17, 2016

### TimeRip496

Thanks. But how to know the value of a0 and a1? Cause without knowing them, I can't use them to form the solutions (23)&(24).

6. Oct 17, 2016

### Staff: Mentor

These constants are usually determined by the initial conditions (typically y(0) and y'(0)). If they're not given, you can't determine those coefficients of your series.

7. Oct 19, 2016

### TimeRip496

So in this case, a0 and a1 each equals to 1 as the initial conditions right?

As for eqn(23), "If is an even integer, the series reduces to a polynomial of degree with only even powers of and the series diverges.", technically both are diverging right? Is just that one is in the even powers of x while the other is in the odd powers of x.

I also don't understand the representation of (25) in (26)?

8. Oct 19, 2016

### Staff: Mentor

No, I don't see how you can assume that.
One thing that bothers me about this page is their statement, right after equation (2), that "The above form is a special case of the so-called "associated Legendre differential equation" corresponding to the case [PLAIN]http://mathworld.wolfram.com/images/equations/LegendreDifferentialEquation/Inline1.gif." [Broken]
I don't see m mentioned anywhere before that, so it's not clear to me what they're talking about here.

Regarding equation (23), they assert that the series for $y_2(x)$ diverges if l is an even integer, and that $y_1(x)$ diverges if l is an odd integer. These statements omit a lot of work in showing why the two series diverge. I would want to confirm these for myself.

Last edited by a moderator: May 8, 2017
9. Oct 19, 2016

### Staff: Mentor

They're talking about the associated Legendre differential equation. It is that equation that has an $m$, which for $m=0$ reduces to the
Legendre differential equation.

Last edited by a moderator: May 8, 2017
10. Oct 21, 2016

### TimeRip496

Based on eqn(23) & eqn(24), a0 and a1 should be 1 right? If not then I don't know how to find the values of a0 and a1.

11. Oct 21, 2016

### TimeRip496

Sir, do you mind elaborating on the part of equation (23) about y2(x)diverges if l is an even integer, and that y1(x) diverges if l is an odd integer.
I am lost here cause I am new to this.

12. Oct 21, 2016

### Staff: Mentor

In that Wolfram page, it appears that they set both a0 and a1 to 0. These constants normally come from initial conditions such as y(0) and y'(0), but there's no mention of initial conditions on that page.

13. Oct 22, 2016

### TimeRip496

If a0 is 0, then won't a2, a4, ... equal to zero too? $$a_2=-\frac{l(l+1)}{1.2}a_0$$

Likewise for a1, since the odd number of a depends on a1 too.

14. Oct 23, 2016

### Staff: Mentor

Reconsidering, I think I misspoke. For the even solutions it appears that they set $a_0=1$ and $a_1 = 0$. For the odd solutions, it appears they set $a_0=0$ and $a_1 = 1$.

15. Oct 23, 2016

### TimeRip496

Never mind I understand already. But I am not sure whether the equation (26) is correct on the site(http://mathworld.wolfram.com/LegendreDifferentialEquation.html) you suggested.

Shouldn't it be $$_2F_1(-\frac{1}{2}.l, \frac{1}{2}(l+1); \frac{1}{2}; x^2)$$ instead of $$_2F_1(-\frac{1}{2}, \frac{1}{2}(l+1); \frac{1}{2}, x^2)$$ for l even?

16. Oct 24, 2016

### TimeRip496

Thanks a lot for your help but do you know whether the equation (26) is correct on the site(http://mathworld.wolfram.com/LegendreDifferentialEquation.html)?

Shouldn't it be $$_2F_1(-\frac{1}{2}.l, \frac{1}{2}(l+1); \frac{1}{2}; x^2)$$ instead of $$_2F_1(-\frac{1}{2}, \frac{1}{2}(l+1); \frac{1}{2}, x^2)$$ for l even?