# Problem 32

• Jul 25th 2007, 06:28 AM
ThePerfectHacker
Problem 32
Given a non-negative integer $\displaystyle n$ we define $\displaystyle P_n(x)$ to be a solution* to:
$\displaystyle (1-x^2)y''-2xy'+n(n+1)y=0 \mbox{ on }(-1,1)$.

Show that:
$\displaystyle \int_{-1}^1P_n(x)P_m(x) dx = 0 \mbox{ for }n\not = m$.

*)Yes, I realize what I said is not well-defined but it makes no difference. Pick any solution.
• Jul 27th 2007, 02:57 PM
mathisfun1
The given equation is the Legendre equation and its solutions are the Legendre orthogonal polynomials (Legendre polynomials - Wikipedia, the free encyclopedia). I'm not sure how to show the orthogonality of the solutions from scratch though (i.e. w/o Rodrigues's formula).
• Jul 27th 2007, 04:49 PM
topsquark
Quote:

Originally Posted by mathisfun1
The given equation is the Legendre equation and its solutions are the Legendre orthogonal polynomials (Legendre polynomials - Wikipedia, the free encyclopedia). I'm not sure how to show the orthogonality of the solutions from scratch though (i.e. w/o Rodrigues's formula).

But Rodrigues' formula is the simplest way to go with it!

-Dan
• Jul 28th 2007, 05:54 PM
ThePerfectHacker
There is not need to use the Rodrigues formula, nor any identities that exist among Legendre polynomials.
• Jul 29th 2007, 04:17 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
There is not need to use the Rodrigues formula, nor any identities tha exist among Legendre polynomials.

Well, I suppose you could plug the differential equation into the integral and get a diffeo-integral equation and work from there, but I've never seen it done.

-Dan
• Jul 29th 2007, 01:30 PM
ThePerfectHacker
As mentioned this problem can be solved with Legendre Polynomials and their identities, but there is no need. There are many nice identities among the Legendre Polynomials and not a single one is used here.

The trick is to write the differencial equation as (in Sturm-Louiville form),
$\displaystyle [(1-x^2)y']'+n(n+1)y=0$

Thus,
$\displaystyle [(1-x^2)P_n'(x)]'+n(n+1)P_n(x)=0$
Multiply by $\displaystyle P_m(x)$ to get,
$\displaystyle [(1-x^2)P_n'(x)]'P_m(x)+n(n+1)P_n(x)P_m(x)=0$
And integrate,
$\displaystyle \int_{-1}^1 [(1-x^2)P_n'(x)]'P_m(x) dx + n(n+1)\int_{-1}^1 P_n(x)P_m(x)dx=0$
Use integration by part on the first integral,
$\displaystyle (1-x^2)P_n'(x)P_m(x)\big|_{-1}^1 - \int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+n(n+1)\int_{-1}^1P_n(x)P_m(x)dx=0$
The first term vanishes,
$\displaystyle - \int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+n(n+1)\int_{-1}^1P_n(x)P_m(x)dx=0$
Doing the similar steps with $\displaystyle P_m(x)$ instead we have,
$\displaystyle -\int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+m(m+1)\int_{-1}^1P_n(x)P_m(x)dx=0$
Now subtract these two equations to get,
$\displaystyle [n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x) dx = 0$
Since $\displaystyle n\not = m$ and they are integers we have,
$\displaystyle \int_{-1}P_n(x)P_m(x) dx = 0$.
Q.E.D.
• Jul 29th 2007, 02:35 PM
ThePerfectHacker
Here is a way to get the Rodrigues formula since it was mentioned.

Let $\displaystyle z=(x^2-1)^n$.

We see that,
$\displaystyle (x^2-1)\frac{dz}{dx} - 2nx z = 0$

Take the derivative $\displaystyle n$ times to get,
$\displaystyle (1-x^2)\frac{dz}{dx}+n(n+1)z^{(n-1)}=0$
Take the derivative one last time,
$\displaystyle \left[(1-x^2)\frac{dz}{dx}\right]'+n(n+1)z^n=0$
Hence,
$\displaystyle \frac{d^{n}(x^2-1)^n}{dx^n}$
Is a polynomial which solves the equation.
But we write it as,
$\displaystyle \frac{1}{2^nn!}\cdot \frac{d^n(x^2-1)^n}{dx^n}$
Because if we solve the Legendre equation with a power series we find that the leading coefficient of the polynomial starts with that factor. So to make Rodrigues formula agree with an infinite series solution we make the leading term equal to that factor.
-------------------------------

Here is a fabulous identity due to Laplace.
$\displaystyle P_n(x) = \frac{1}{\pi}\int_0^{\pi} (x+\sqrt{x^2-1}\cos \phi)^n d\phi$

We begin by noting that,
$\displaystyle P_n(x) = \frac{1}{2\pi i}\oint_{\gamma} \frac{1}{2^n}\frac{(z^2-1)^n}{(z-x)^{n+1}}dz$
Where $\displaystyle \gamma$ is any peicewise smooth simple closed curve containing $\displaystyle x$.
This is immediately true by Cauchy's Integral Formula.

Let $\displaystyle x>1$ define $\displaystyle \gamma = x+\sqrt{x^2-1}e^{i\phi}$ for $\displaystyle 0\leq \phi \leq 2\pi$. That is a circle centered at $\displaystyle x$ with radius $\displaystyle \sqrt{x^2-1}$.

Evaluating the above integral with this parametrization (and doing some simplification, not shown) we get Laplace's formula.
• Aug 2nd 2007, 09:26 AM
Rebesques
Quote:

The trick is to write the differencial equation as (in Sturm-Louiville form)
I claim this is cheating! :p

Since, by identifying a Sturm-Liouville operator here, the $\displaystyle P_n(x)$ are eigenfunctions and thus orthogonal for the $\displaystyle L^2$ inner product - that is, $\displaystyle \int_{-1}P_n(x)P_m(x) dx = 0$; qed!
• Aug 2nd 2007, 10:24 AM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
I claim this is cheating! :p

Since, by identifying a Sturm-Liouville operator here, the $\displaystyle P_n(x)$ are eigenvectors and thus orthogonal for the $\displaystyle L^2$ inner product - that is, $\displaystyle \int_{-1}P_n(x)P_m(x) dx = 0$; qed!

Yes, that is cheating. It takes all the fun away from the problem if it can be done.*

(Change "eigenvectors" to "eigenfunctions" to fix your post.)

*)I say "if it can be done" because this is not a boundary value problem. Thus, Sturm-Louiville Theory does not apply. Are you sure?
• Aug 2nd 2007, 11:23 AM
Rebesques
Quote:

Change "eigenvectors" to "eigenfunctions" to fix your post
thnx! :o

Quote:

this is not a boundary value problem. Thus, Sturm-Louiville Theory does not apply
• Aug 2nd 2007, 06:09 PM
mathisfun1
Quote:

Originally Posted by ThePerfectHacker

The trick is to write the differencial equation as (in Sturm-Louiville form),
$\displaystyle [(1-x^2)y']'+n(n+1)y=0$

So did you change your signature just for this problem :P?
• Aug 2nd 2007, 06:15 PM
ThePerfectHacker
Quote:

Originally Posted by mathisfun1
So did you change your signature just for this problem :P?

You mean my Sturm-Louiville signature? No, it is just I sometimes post equations that I love. And sometimes that which I just learned. So for instance, the Sturm-Louiville signature was posted when I was doing boundary value problem from my Partial Differencial Equations book.