Problem 32

Printable View

July 25th 2007, 06:28 AM
ThePerfectHacker

Problem 32

Given a non-negative integer $n$ we define $P_n(x)$ to be a solution* to:
$(1-x^2)y''-2xy'+n(n+1)y=0 \mbox{ on }(-1,1)$ .

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

*)Yes, I realize what I said is not well-defined but it makes no difference. Pick any solution.
July 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).
July 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
July 28th 2007, 05:54 PM
ThePerfectHacker

There is not need to use the Rodrigues formula, nor any identities that exist among Legendre polynomials.
July 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
July 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),
$[(1-x^2)y']'+n(n+1)y=0$

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

Here is a way to get the Rodrigues formula since it was mentioned.

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

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

Take the derivative $n$ times to get,
$(1-x^2)\frac{dz}{dx}+n(n+1)z^{(n-1)}=0$
Take the derivative one last time,
$\left[(1-x^2)\frac{dz}{dx}\right]'+n(n+1)z^n=0$
Hence,
$\frac{d^{n}(x^2-1)^n}{dx^n}$
Is a polynomial which solves the equation.
But we write it as,
$\frac{1}{2^nn!}\cdot \frac{d^n(x^2-1)^n}{dx^n}$
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.
-------------------------------
Warning Complex Analysis Ahead

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

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

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

Evaluating the above integral with this parametrization (and doing some simplification, not shown) we get Laplace's formula.
August 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 $P_n(x)$ are eigenfunctions and thus orthogonal for the $L^2$ inner product - that is, $<br /> \int_{-1}P_n(x)P_m(x) dx = 0<br />$ ; qed!
August 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 $P_n(x)$ are eigenvectors and thus orthogonal for the $L^2$ inner product - that is, $<br /> \int_{-1}P_n(x)P_m(x) dx = 0<br />$ ; 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?
August 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

Doesn't have to be, Sturm-Louiville operators are really flexible about this.
August 2nd 2007, 06:09 PM
mathisfun1

Quote:

Originally Posted by ThePerfectHacker

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

So did you change your signature just for this problem :P?
August 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.