# Legendre Differential Equation

• I
• TimeRip496
In summary, the Legendre Differential Equation is a family of equations with a parameter l, and its solution is the Legendre polynomial. The associated Legendre polynomial and unassociated Legendre polynomial are both solutions to the Legendre Differential Equation, with the former having an additional parameter m. The value of a0 and a1 in the series solution for the Legendre Differential Equation is determined by initial conditions. The series for y2(x) diverges if l is an even integer, while y1(x) diverges if l is an odd integer. The associated Legendre differential equation has an m parameter, which reduces to the Legendre differential equation when m = 0.
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 said:
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.

TimeRip496 said:
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.
TimeRip496 said:
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 said:
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).

TimeRip496 said:
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.

Mark44 said:
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)?

TimeRip496 said:
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.
TimeRip496 said:
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)?
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.

Last edited by a moderator:
Mark44 said:
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.

Last edited by a moderator:
Mark44 said:
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.

DrClaude said:
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.

TimeRip496 said:
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.

Mark44 said:
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.

TimeRip496 said:
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##.

DrClaude said:
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?

Mark44 said:
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?

## 1. What is a Legendre Differential Equation?

A Legendre Differential Equation is a second-order ordinary differential equation named after the French mathematician Adrien-Marie Legendre. It is used to describe the behavior of certain physical systems, such as the motion of a pendulum or the flow of heat in a solid.

## 2. What is the general form of a Legendre Differential Equation?

The general form of a Legendre Differential Equation is (1 - x2)y'' - 2xy' + n(n+1)y = 0, where n is a constant. This form is also known as the Legendre's canonical form.

## 3. What is the significance of Legendre Differential Equations?

Legendre Differential Equations are important in many areas of physics and engineering. They are used to solve problems in electrostatics, quantum mechanics, and fluid dynamics. They also have applications in other fields such as image processing and signal analysis.

## 4. How are Legendre Differential Equations solved?

There are several methods for solving Legendre Differential Equations, including the Frobenius method, the power series method, and the method of variation of parameters. The specific method used depends on the coefficients and initial conditions of the equation.

## 5. What are the special solutions of Legendre Differential Equations?

Legendre Differential Equations have two types of special solutions: Legendre polynomials and Legendre functions. Legendre polynomials are used to solve problems in electrostatics and quantum mechanics, while Legendre functions are used to solve problems in fluid dynamics and heat transfer.

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