Proof of orthogonality of associated Legendre polynomial

In summary: By the orthogonality of Legendre polynomial!Therefore, we haveI = (-1)^m \frac{(l+m)!}{(l-m)!} \int_{-1}^{+1} P_l^2 (x) dxUse the normalization condition of the Legendre polynomial, we haveI = (-1)^m \frac{(l+m)!}{(l-m)!} \frac{2}{2l+1}But we know that the normalization condition of associated Legendre polynomials is\int_{-1}^{+1} P_l^m (x) P_l^m (x) dx = \frac{2l+1}{2} \
  • #1
maverick6664
80
0
Hi,
I'm trying to prove the orthogonality of associated Legendre polynomial which is called to "be easily proved":

Let
[tex]P_l^m(x) = (-1)^m(1-x^2)^{m/2} \frac{d^m} {dx^m} P_l(x) = \frac{(-1)^m}{2^l l!} (1-x^2)^{m/2} \frac {d^{l+m}} {dx^{l+m}} (x^2-1)^l [/tex]

And prove

[tex]\int_{-1}^1 P_l^m(x)P_{l'}^{m'} (x)dx = \frac{2}{2l+1} \frac {(l+m)!} {(l-m)!} \delta_{ll'} \delta_{mm'} [/tex]

Though it should be easily proved, I don't know how. When [tex]m = m' = 0,[/tex] it's unassociated Legendre polynomials and it's not difficult (I'll post in the next message).

Will anyone show me a hint or online reference? I don't need exact value (because won't be difficult). All I want to prove is orthogonality.
 
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  • #2
When [tex]m = m' = 0[/tex], assume [tex]l \leqq l'[/tex] and we have

[tex]\int_{-1}^1 P_l(x)P_{l'} (x)dx = \frac {1}{2^{l+l'}} \frac {1} {l! l'!} \int_{-1}^{1} ( \frac {d^l} {dx^l} (x^2-1)^l ) ( \frac {d^l}{dx^l} (x^2-1)^n) dx[/tex]

and partial integral ([tex]l'[/tex] times) yields,

[tex] = \frac {(-1)^{l'}} {2^{l+l'} l! l'!} \int_{-1}^{1} (\frac {d^{l+l'}} {dx^{l+l'}} (x^2-1)^l) (x^2-1)^{l'} dx[/tex]

When [tex]l < l'[/tex] this is 0. When [tex]l = l'[/tex],

[tex]\frac {(-1)^{l}} {2^{2l} (l!)^2} \int_{-1}^1 (x^2-1)^{l} \frac {d^{2l}} {dx^{2l}} (x^2-1)^{l} dx = \frac {(-1)^{l} (2l)!} {2^{2l} (l!)^2} \int_{-1}^1 (x^2-1)^{l} dx[/tex]

Then replace [tex]x = 2u-1[/tex] and obtain

[tex] \frac {(-1)^{l} 2 (2l)!} {(l!)^2} \int_0^1 u^{l} (u-1)^{l} du = \frac {2} {2l+1} [/tex]

So we have

[tex]\frac{2}{2l+1} \delta_{ll'} [/tex]
 
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  • #3
could you explain the last step i don't see why they are =
 
  • #4
a little more detail...

A bit more about maverick6664's reasoning above: for the case [tex]l=l'[/tex], after you get

[tex]
\frac{(-1)^l 2 (2l)!}{(l!)^2} \int_0^1 u^l (u-1)^l du,
[/tex]

integrate by parts [tex]l[/tex] more times to get

[tex]
2 \int_0^1 y^{2l} dy = \frac{2}{2l+1}.
[/tex]

Maybe it's also worth mentioning that maverick6664 used the binomial theorem a few lines before that.

As for the associated Legendre polynomials, a proof of their orthogonality is given in Mathematical Methods for Physicists by Arfken and Weber, fifth edition, section 12.5. It requires some integration by parts as well as Leibniz's formula. Nicely, the result also proves orthogonality of the (unassociated) Legendre polynomials as a special case.
 
  • #5
There is something slightly messy about the proof of orthogonality given by maverick6664. Technically, one should show that the boundary terms that result from each integration by parts vanish. This is fairly straightforward to do, but there's a much cleaner way of proving orthogonality which starts from the fact that, by definition, the Legendre polynomials satisfy the Legendre differential equation. I've attached a pdf file for anyone who cares.
 

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  • #6
Hi,

I have a quick question about what Maverick is trying to prove here and that is: Is that really true? I can't find this orthogonality relation anywhere. All I can find in the literature and online are orthogonality relations for fixed m and for fixed l but nothing for both 'unfixed'.

Can someone confirm this?
 
  • #7
krome-- thank you so much for posting that pdf. it was exactly what i needed for a big homework problem which has been giving me so much trouble. thank you thank you!
 
  • #8
i understand how you get the solution to the legendre polynomials orthogonality, but how do you then use this to prove the result for the orthogonality of associated legendre polynomials when m doesn't equal 0? its prob really obv but I've been stuck at it for days!
 
  • #9
The orthogonality of associated Legendre functions can be proved by using the relationship between Legendre polynomials and associated Legendre functions.

They are related by following expression.

[tex]P_{l}^{m} (x) = (1-x^2 ) ^{\frac{m}{2}} \frac{d^m}{dx^m} P_l (x)[/tex]​

Note that the Rodrigues' formula for the Legendre polynomials is,

[tex]P_l (x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^2 -1 )^l[/tex]​

Now what we want to know is the value '[tex]I[/tex]' of following integral for arbitrary chosen positive integers [tex]l,l'[/tex]

[tex]I = \int_{-1}^{+1} P_l^m (x) P_l'^m (x) dx [/tex]​

By the trichotomy property of real numbers, [tex]l[/tex] and [tex]l'[/tex] must have a following relation.

[tex]l < l'[/tex] or [tex]l' < l[/tex] or [tex]l = l'[/tex]​

Without loss of generality, here we assume [tex]l \leq l'[/tex].

Now expand the integral by following procedure.

[tex]I = \int_{-1}^{+1} P_l^m (x) P_l'^m (x) dx[/tex]
[tex]\Leftrightarrow I = \int_{-1}^{+1} (1-x^2)^{\frac{m}{2}} \frac{d^m}{dx^m} P_l(x) (1-x)^2 [/tex]
[tex]\Leftrightarrow I = \int_{-1}^{+1} (1-x^2)^m \frac{d^m}{dx^m} P_l (x) \frac{d^m}{dx^m} P_l' (x) dx[/tex]
[tex]\Leftrightarrow I = \int_{-1}^{+1} \left[(1-x^2)^m \frac{d^m}{dx^m} P_l (x) \right] \frac{d}{dx} \left[\frac{d^m-1}{dx^m-1} P_l' (x)\right] dx[/tex]​

Now this is [tex]I = \int_{a}^{b} u(x)v'(x) dx [/tex] form. So, perform an integration by part.

[tex]I = \left[( 1-x^2 )^m \frac{d^m}{dx^m} P_l (x) \frac{d^m-1}{dx^m-1} P_l' (x) \right]_{-1}^{+1} - \int_{-1}^{+1} \frac{d}{dx} \left[ (1-x^2 )^m \frac{d^m}{dx^m} P_l (x) \right] \frac{d}{dx} \left[ \frac{d^{m-2}}{dx^{m-2}} P_l' (x) \right] dx [/tex]
[tex]\Leftrightarrow I = - \int_{-1}^{+1} \frac{d}{dx} \left[ ( 1-x^2 )^m \frac{d^m}{dx^m} P_l (x) \right] \frac{d}{dx} \left[ \frac{d^{m-2}}{dx^{m-2}} P_l' (x) \right] dx[/tex]​

Do the integration by parts for 'm' times, then we have,

[tex] I = (-1)^m \int_{-1}^{+1} \frac{d^m}{dx^m} \left[ ( 1 -x^2 )^m \frac{d^m}{dx^m} P_l (x) \right] P_l' (x) dx[/tex]​

We know that [tex]P_l (x)[/tex] is [tex]l[/tex]-th degree polynomial. So after do some easy calculation, we find that the a following factor of integrand is [tex]l[/tex]-th degree polynomial.

[tex]\frac{d^m}{dx^m} \left[(1-x^2 )^m \frac{d^m}{dx^m} P_l (x) \right][/tex]​

So, if [tex] l' < l [/tex], then [tex]I = 0[/tex] by the orthogonality of Legendre polynomials.
Thus, the remain problem is the case of [tex]l'=l[/tex] and in this case, it is ok, just consider the highest degree term.

The Rodrigues' formula of Legendere polynomial is given by following expression.

[tex]P_l (x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^2 -1)^l[/tex]​

From this, we can find that

(Highest degree term of [tex]P_l (x)[/tex]) [tex]= \frac{1}{2^l l!} \frac{d^l}{dx^l} x^{2l} = \frac{1}{2^l l!} \frac{(2l)!}{l!} x^l[/tex]​

So, apply this to find the expression of highest degree term of

[tex]\frac{d^m}{dx^m} \left[(1-x^2 )^m \frac{d^m}{dx^m} P_l (x) \right][/tex]​

then we have,

[tex] (-1)^m \frac{1}{2^l l!} \frac{(2l)!}{l!} \frac{d^m}{dx^m} \left[ x^2m \frac{d^m}{dx^m} x^l \right][/tex]​

After do some calculation in big parenthesis, you'll have

[tex] (-1)^m \frac{1}{2^l l!} \frac{(2l)!}{l!} \frac{l!}{(l-m)!} \frac{d^m}{dx^m} x^l+m[/tex]​

and do another differentiations,

[tex] (-1)^m \frac{1}{2^l l!} \frac{(2l)!}{l!} \frac{l!}{(l-m)!} \frac{(l+m)!}{l!} x^l[/tex]​

But we know that the highest degree term of [tex]l[/tex]-th degree Legendre polynomial is

[tex]\frac{1}{2^l l!} \frac{(2l)!}{l!} x^l [/tex]​

So,

[tex] (-1)^m \frac{1}{2^l l!} \frac{(2l)!}{l!} \frac{l!}{(l-m)!} \frac{(l+m)!}{l!} x^l[/tex]
[tex]= (-1)^m \frac{(l+m)!}{(l-m)!} P_l (x) + \cdots [/tex] (low degree terms)​

Therefore, remind the first form of our integral.

[tex]I = (-1)^m \int_{-1}^{+1} \frac{d^m}{dx^m} \left[ ( 1 -x^2 )^m \frac{d^m}{dx^m} P_l (x) \right] P_l' (x) dx[/tex]​

Now we know that every calculations of terms that independent with highest degree terms are become to 0, by orthogonality of Legendre polynomials.

[tex]I=(-1)^{2m} \int_{-1}^{+1} \frac{(l+m)!}{(l-m)!} P_l (x) P_l' (x) dx [/tex]​

And, here we use the orthogonality expression of Legendre polynomials,

[tex]\int_{-1}^{+1} P_l (x) P_l' (x) dx = \frac{2}{2l+1} \delta_{ll'}[/tex]​

we will get the following orthogonality expression of the associated Legendre functions.

[tex]I = \int_{-1}^{+1} P_{l}^{m} (x) P_{l'}^{m} (x) dx = \frac{2}{2l+1} \frac{(l+m)!}{(l-m)!} \delta_{ll'}[/tex]​

Hence, the proof is complete.

There are another way to prove this by using the associated Legendre differential equation itself as someone mentioned before. That way is more simple and easy to understand. I think this is clever method, but it is too messy
 
  • #10
thanks so much for that! i completely understand it now! :):):)
 
  • #11
The relation that was asked for in the first post is not correct. It is not the case that

[tex]
\int_{-1}^1 P_l^m(x)P_{l'}^{m'} (x)dx = \frac{2}{2l+1} \frac {(l+m)!} {(l-m)!} \delta_{ll'} \delta_{mm'}
[/tex]

for all values of l, m, l' and m'. This would imply that the integral

[tex]
\int_{-1}^1 P_l^m(x)P_{l'}^{m'} (x)dx
[/tex]

is zero unless both l=l' and m=m', which is not the case. For instance, take l=m=0 and l'=m'=1. The integral becomes

[tex]
-\int_{-1}^1 (1-x^2)^{1/2}dx
[/tex]

which is equal to -pi/2, and thus clearly not zero.

It is true however, that the integral vanishes if l=l' and m [tex]$\ne$[/tex] m' or if m=m' and l [tex]$\ne$[/tex] l'. The integral does not vanish if l=l' and m=m', or if l [tex]$\ne$[/tex] l' and m [tex]$\ne$[/tex] m'.
 

1. What is the proof of orthogonality of associated Legendre polynomials?

The proof of orthogonality of associated Legendre polynomials states that for any two different associated Legendre polynomials, their inner product over the interval [-1, 1] is equal to 0.

2. Why is the proof of orthogonality important in mathematics?

The proof of orthogonality of associated Legendre polynomials is important in mathematics because it provides a way to decompose functions into a set of orthogonal basis functions, which can simplify calculations and analysis in many fields such as physics and engineering.

3. What assumptions are made in the proof of orthogonality?

The proof of orthogonality assumes that the associated Legendre polynomials are normalized, meaning that their maximum value over the interval [-1, 1] is equal to 1. It also assumes that the inner product of two different associated Legendre polynomials is defined as the integral of their product over the interval [-1, 1].

4. Can the proof of orthogonality be extended to other types of polynomials?

Yes, the proof of orthogonality of associated Legendre polynomials can be extended to other types of orthogonal polynomials, such as Chebyshev polynomials and Legendre polynomials. These polynomials have different properties, but the concept of orthogonality remains the same.

5. How is the proof of orthogonality used in practical applications?

The proof of orthogonality has various practical applications, including in signal processing, where it is used to analyze and decompose signals into a set of orthogonal basis functions. It is also used in solving differential equations and in numerical methods for solving problems in physics and engineering.

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