# Proof of orthogonality of associated Legendre polynomial

1. Dec 25, 2005

### maverick6664

Hi,
I'm trying to prove the orthogonality of associated Legendre polynomial which is called to "be easily proved":

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

And prove

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

Though it should be easily proved, I don't know how. When $$m = m' = 0,$$ 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.

Last edited: Dec 25, 2005
2. Dec 25, 2005

### maverick6664

When $$m = m' = 0$$, assume $$l \leqq l'$$ and we have

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

and partial integral ($$l'$$ times) yields,

$$= \frac {(-1)^{l'}} {2^{l+l'} l! l'!} \int_{-1}^{1} (\frac {d^{l+l'}} {dx^{l+l'}} (x^2-1)^l) (x^2-1)^{l'} dx$$

When $$l < l'$$ this is 0. When $$l = l'$$,

$$\frac {(-1)^{l}} {2^{2l} (l!)^2} \int_{-1}^1 (x^2-1)^{l} \frac {d^{2l}} {dx^{2l}} (x^2-1)^{l} dx = \frac {(-1)^{l} (2l)!} {2^{2l} (l!)^2} \int_{-1}^1 (x^2-1)^{l} dx$$

Then replace $$x = 2u-1$$ and obtain

$$\frac {(-1)^{l} 2 (2l)!} {(l!)^2} \int_0^1 u^{l} (u-1)^{l} du = \frac {2} {2l+1}$$

So we have

$$\frac{2}{2l+1} \delta_{ll'}$$

Last edited: Dec 25, 2005
3. Oct 22, 2006

### wendybosy

could you explain the last step i dont see why they are =

4. Feb 8, 2008

### dhk

a little more detail...

A bit more about maverick6664's reasoning above: for the case $$l=l'$$, after you get

$$\frac{(-1)^l 2 (2l)!}{(l!)^2} \int_0^1 u^l (u-1)^l du,$$

integrate by parts $$l$$ more times to get

$$2 \int_0^1 y^{2l} dy = \frac{2}{2l+1}.$$

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. Jul 17, 2008

### krome

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.

#### Attached Files:

• ###### legendre_orthogonality.pdf
File size:
47.6 KB
Views:
1,601
6. May 18, 2009

### fderoo

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. Oct 20, 2009

### beth_p

krome-- thank you so much for posting that pdf. it was exactly what i needed for a big hw problem which has been giving me so much trouble. thank you thank you!!

8. Nov 2, 2009

### lavster

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 doesnt equal 0? its prob really obv but ive been stuck at it for days!!

9. Nov 16, 2009

### timewalker

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.

$$P_{l}^{m} (x) = (1-x^2 ) ^{\frac{m}{2}} \frac{d^m}{dx^m} P_l (x)$$​

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

$$P_l (x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^2 -1 )^l$$​

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

$$I = \int_{-1}^{+1} P_l^m (x) P_l'^m (x) dx$$​

By the trichotomy property of real numbers, $$l$$ and $$l'$$ must have a following relation.

$$l < l'$$ or $$l' < l$$ or $$l = l'$$​

Without loss of generality, here we assume $$l \leq l'$$.

Now expand the integral by following procedure.

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

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

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

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

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

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

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

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

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

$$P_l (x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^2 -1)^l$$​

From this, we can find that

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

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

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

then we have,

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

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

$$(-1)^m \frac{1}{2^l l!} \frac{(2l)!}{l!} \frac{l!}{(l-m)!} \frac{d^m}{dx^m} x^l+m$$​

and do another differentiations,

$$(-1)^m \frac{1}{2^l l!} \frac{(2l)!}{l!} \frac{l!}{(l-m)!} \frac{(l+m)!}{l!} x^l$$​

But we know that the highest degree term of $$l$$-th degree Legendre polynomial is

$$\frac{1}{2^l l!} \frac{(2l)!}{l!} x^l$$​

So,

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

Therefore, remind the first form of our integral.

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

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

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

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

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

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

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

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. Nov 17, 2009

### lavster

thanks so much for that!!! i completely understand it now!!!!! :):):)

11. Apr 12, 2011

### colinjohnstoe

The relation that was asked for in the first post is not correct. It is not the case that

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

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

$$\int_{-1}^1 P_l^m(x)P_{l'}^{m'} (x)dx$$

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

$$-\int_{-1}^1 (1-x^2)^{1/2}dx$$

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

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook