Problem 32

ThePerfectHacker · #1 $Old$ 07-25-2007, 08:28 AM

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.

mathisfun1 · #2 $Old$ 07-27-2007, 04:57 PM

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).

topsquark · #3 $Old$ 07-27-2007, 06:49 PM

Quote:

Originally Posted by mathisfun1 $View Post$

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

ThePerfectHacker · #4 $Old$ 07-28-2007, 07:54 PM

There is not need to use the Rodrigues formula, nor any identities that exist among Legendre polynomials.

topsquark · #5 $Old$ 07-29-2007, 06:17 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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

ThePerfectHacker · #6 $Old$ 07-29-2007, 03:30 PM

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.

ThePerfectHacker · #7 $Old$ 07-29-2007, 04:35 PM

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.

Rebesques · #8 $Old$ 08-02-2007, 11:26 AM

Quote:

The trick is to write the differencial equation as (in Sturm-Louiville form)

I claim this is cheating!

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, $\int_{-1}P_n(x)P_m(x) dx = 0$ ; qed!

ThePerfectHacker · #9 $Old$ 08-02-2007, 12:24 PM

Quote:

Originally Posted by Rebesques $View Post$

I claim this is cheating!

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, $\int_{-1}P_n(x)P_m(x) dx = 0$ ; 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?

Rebesques · #10 $Old$ 08-02-2007, 01:23 PM

Quote:

Change "eigenvectors" to "eigenfunctions" to fix your post

thnx!

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.