Legendre Differential Equation

[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?

 

[DrClaude]

The $l$ is a positive integer. The Legendre differential equation is basically a family of equations, which differ by a parameter $l$.

The Legendre polynomial has no $m$ in it. The associated Legendre polynomial is a solution of the associated Legendre differential equation, which has an additional parameter $m$.

I really recommend that you look at the links to MathWorld I have put in there.

 

[TimeRip496]

The $l$ is a positive integer. The Legendre differential equation is basically a family of equations, which differ by a parameter $l$.

The Legendre polynomial has no $m$ in it. The associated Legendre polynomial is a solution of the associated Legendre differential equation, which has an additional parameter $m$.

I really recommend that you look at the links to MathWorld I have put in there.

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.

 

[Mark44]

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

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.

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.

 

[TimeRip496]

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.

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).

 

[Mark44]

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).

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.

 

[TimeRip496]

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.

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)?

 

[Mark44]

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

No, I don't see how you can assume that.

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)?

I assume you're talking about this page: http://mathworld.wolfram.com/LegendreDifferentialEquation.html
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."
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.

 

[DrClaude]

I assume you're talking about this page: http://mathworld.wolfram.com/LegendreDifferentialEquation.html
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."
I don't see m mentioned anywhere before that, so it's not clear to me what they're talking about here.

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.

 

[TimeRip496]

No, I don't see how you can assume that.

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.

 

[TimeRip496]

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.

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.

 

[Mark44]

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.

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.

 

[TimeRip496]

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.

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.

 

[Mark44]

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.

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 $a_1 = 1$.

 

[TimeRip496]

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.

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?

 

[TimeRip496]

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 $a_1 = 1$.

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?