Proving Self-Adjoint ODE for Legendre Polynomials

[ognik]

(I haven't encountered these before, also not in the book prior to this problem or in the near future ...)

Show that the 1st derivatives of the legendre polynomials satisfy a self-adjoint ODE with eigenvalue $\lambda = n(n+1)-2 $

Wiki shows a table of poly's , I don't think this is what the book means ...

Wiki also shows Rodrique's formula which looks more relevant - $ P_n(x) =\frac{1}{2^n}\frac{1}{2!} \d{^n{}}{{x}^n} (x^2-1)^n$
So I differentiated that a couple of times to see, and got $ P_0 =1, P_1 = x, P_2 = \frac{1}{2}(3x^2 - 1) ...$ - and realized I was just generating the polynomials. Basically I don't what I'm supposed to be doing?

 

[ognik]

Got a bit further along, I just couldn't find what one should do with the polys, so I tried to make an ODE out of the 1st 3 and that looks like a good guess (how should I have known?), I get :

$ y''+xy'+\frac{1}{2}(3x^2-1)y=0 ,[ p_0 =1, p_1 = x, p_2 =\frac{1}{2}(3x^2-1)] $ which I can (I hope) use sturm-L theory to to solve. Problem is book is not very clear on S-L theory and couldn't find anything very clear on the web, so please help me get the method right.

I believe I want to get an eigenvalue eqtn of the form: $ \mathcal{L}y=\lambda w(x)y $. First Check if self-adjoint, ie is $p_0' =p_1 $

A) If the eqtn was already self-adjoint, then $p=p_0, q=p_2$ and $ \mathcal{L} = (p\frac{d}{dx})' + q $ ...But where does w(x) normally come from? If I read it off for the original eqtn it = 1. I can't see how it would ever not be 1?

B) if not S-A, multiply the ODE by the integrating factor $= \mu(x)= \frac{1}{p_0} exp\int\frac{p_1}{p_0} dx$. Now $a_0=\mu, a_1=\mu p_1 , a_2=\mu p_2 $ giving a self adjoint ODE= $ a_0 y'' + a_1 y'+ a_2 y=0 $ and I follow A) above.
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Hoping the above is OK, I get $ \mu = e^{\frac{x^2}{2}}, \mathcal{L}y=\lambda y, where \: \mathcal{L}y = (py')'+qy, p=e^{\frac{x^2}{2}}, q=\frac{x}{2} e^{\frac{x^2}{2}} (3x^2-1)$
A hint please on what to do next?

 

[ognik]

I have honestly spent hours trying to make a decent start to this - not getting anywhere. I am not even sure what I'm supposed to doing - assume I'm stupid if necessary just to help get me started. Alternatively if the usual suspects are also stumped, knowing that would be a great comfort, leave a comment!

'Progress' since last post.

Wiki says that legendre polynomials are solutions to the Legendre ODE. I tried $y=p_0=1$ and it was a solution to $(1-x^2)y''-2xy'+n(n-1)=0 $ for n= 0 or 1.
$y=p_1=x$ gave me $2x=n^2+n$ which doesn't seem like a solution?

Tell me if I am mistaken, they want me to differentiate Legendre polynomials and to show they too are solutions of a self-adjoint ODE? If differentiating a poly gave a previous poly, that would answer the question, but they aren't related like that...

1) So - how many of them to differentiate? With a 2nd order ODE I would guess 2, but this doesn't limit it to 2nd order
2) If it was 2 poly's, $p'_0 = 0, p'_1 = 1$ which doesn't seem useful, should I instead start with $p_2= \frac{1}{2}\left(3{x}^{2}-1\right) $?
2) They say 'satisfy a self-adjoint ODE' - which one? I assume Legendres?

 

[Deveno]

I don't know what space you are working on, but let us suppose it is $\mathcal{C}^{\infty}[-1,1]$

Consider the operator:

$H = D\circ(1-x^2)D$

where $D$ is the differentiation operator: $\dfrac{d}{dx}$ (so $D(f) = f'$), and $(1-x^2)D$ refers to "pointwise multiplication" by $(1-x^2)$, so for example, if $f(x) = \cos(x)$, then:

$H(f)(x) = \dfrac{d}{dx}((1-x^2)(-\sin(x)) = (x^2 - 1)\cos(x) + 2x\sin(x)$

Your task is, I believe, to show this operator is Hermitian, and to show $P_n$ is an eigenfunction of this operator, and then to calculate its eigenvalues.

To show $H$ is Hermitian, I believe the inner product you will employ is:

$\langle f,g\rangle = \int\limits_{-1}^1 f(x)g(x)\ dx$ (we can use the Riemann integral since our functions are smooth). This reduces to being "symmetric" (careful, here! some authors do not make this kind of distinction) since we have REAL-VALUED functions. In other words, no need to worry about complex-conjugations.

If I recall correctly (and I may not, it's literally been decades) this is a S-L problem of the form:

$\dfrac{d}{dx}\left[p(x)\dfrac{dy}{dx}\right] + q(x)y = -\lambda w(x)y$

with $p(x) = 1 - x^2, q(x) = 0$ and $w(x) = 1$.

The ODE associated with these polynomials is called Legendre's differential equation, and *that* "diffy q" shows the eigenvalues directly.