Prove eigenvalues of the derivatives of Legendre polynomials >= 0

[lriuui0x0]

Homework Statement

For a particular $\lambda$, we know Legendre polynomials are the solutions to

$$
(1-x^2)\frac{d^2P}{dx^2} - 2x\frac{dP}{dx} + \lambda P = 0
$$

We can show by induction that the k-th derivative of $P$ satisfies

$$
(1-x^2)\frac{d^2P^{(k)}}{dx^2} - 2(k+1)x\frac{dP^{(k)}}{dx} + \lambda_k P^{(k)} = 0
$$

where $\lambda_k$ is fixed once $\lambda$ is fixed.

We want to show if $P^{(k)}$ is not identically zero, then $\lambda_k \ge 0$.

Relevant Equations

Parseval's identity?

The problem has a hint about finding a relationship between $\int_{-1}^1 (P^{(k+1)}(x))^2 f(x) dx$ and $\int_{-1}^1 (P^{(k)}(x))^2 g(x) dx$ for suitable $f, g$. It looks they're the weighting functions in the Sturm-Liouville theory and we may be able to make use of Parseval's identity? However I'm not sure how we exactly do this. I have recast the differential equation to the self-adjoint eigenvalue problem form:

$$
\frac{d}{dx}((1-x^2)^{k+1}\frac{dP^{(k)}}{dx}) = -\lambda_k(1-x^2)^kP^{(k)}
$$

 

[fresh_42]

Can you use an explicit formula of the Legendre Polynomials, e.g. Rodrigues formula? What are $f(x)$ and $g(x)$?

The hint looks like Wirtinger’s inequality for periodic functions.

 

[lriuui0x0]

Can you use an explicit formula of the Legendre Polynomials, e.g. Rodrigues formula? What are $f(x)$ and $g(x)$?

The hint looks like Wirtinger’s inequality for periodic functions.

Maybe we're able to do it using Rodrigues formula but I don't think that's the original intention of the problem. I'd be interested in ways to prove this without explicitly knowing the solution of the Legendre polynomials.

 

[fresh_42]

So we are stuck with a Sturm-Liouville equation.

I'm not an expert in functional analysis, but Wikipedia says, that for the given equation $-(p\psi')'+q\psi=\lambda \omega \psi$ one introduces the linear operator $\mathcal{L}=\dfrac{1}{\omega }\left(-\dfrac{d}{dx}p\dfrac{d}{dx}+q\right)$ and applies spectral theory on $\mathcal{L}\psi=\lambda \psi .$

I also found an asymptotic behavior for the $\lambda_k$ (Weyl asymptotic)
$$
\lambda_k=\pi^2\left(\int_a^b\sqrt{\dfrac{\omega(x)}{p(x)}}\,dx\right)^{-2}k^2+O(k)
$$

Looks as if you can prove Weyl then you can prove your problem.

 

[lriuui0x0]

So we are stuck with a Sturm-Liouville equation.

I'm not an expert in functional analysis, but Wikipedia says, that for the given equation $-(p\psi')'+q\psi=\lambda \omega \psi$ one introduces the linear operator $\mathcal{L}=\dfrac{1}{\omega }\left(-\dfrac{d}{dx}p\dfrac{d}{dx}+q\right)$ and applies spectral theory on $\mathcal{L}\psi=\lambda \psi .$

I also found an asymptotic behavior for the $\lambda_k$ (Weyl asymptotic)
$$
\lambda_k=\pi^2\left(\int_a^b\sqrt{\dfrac{\omega(x)}{p(x)}}\,dx\right)^{-2}k^2+O(k)
$$

Looks as if you can prove Weyl then you can prove your problem.

Emmm... I don't feel Weyl asymptotic is the expected solution either...

 

[lriuui0x0]

Can you use an explicit formula of the Legendre Polynomials, e.g. Rodrigues formula? What are $f(x)$ and $g(x)$?

The hint looks like Wirtinger’s inequality for periodic functions.

The $f(x)$ and $g(x)$ are the functions you're supposed to find. I'm guessing these functions are the weighting function for the weighted inner product. So given the self-adjoint operator, I guess maybe we can try to to find the relationship of the following two integrals:

$$
\begin{aligned}
\int_{-1}^1 [P^{(k)}(x)]^2 (1-x^2)^k dx \\
\int_{-1}^1 [P^{(k+1)}(x)]^2 (1-x^2)^{k+1} dx \\
\end{aligned}
$$

But I don't know how these integrals are related to $\lambda_k$ and $\lambda_{k+1}$.

Also note that if we fix a particular $\lambda$, then we also fix $\lambda_k = \lambda -k-k^2$

 

[fresh_42]

Why is $k=0$ excluded, e.g. for $\lambda =-1$?

 

[fresh_42]

I found the following theorem (with proof but wrong language):

For the eigenvalues $\lambda$ of the Sturm-Liouville problems
$$
\mathcal{L}P=-(f \cdot P')'+g\cdot P = \lambda \cdot \omega \cdot P
$$
over the interval $[-1,1]$ with $f(x)>0$ and $\omega (x)>0$ for all $x\in (-1,1)$ with the specific initial values
$$
P(-1)=0\, , \,P(1)=0\quad \text{(Dirichlet)}\quad\text{ or }\quad P'(-1)=0\, , \,P'(1)=0\quad \text{(Neumann)}
$$
or even $P(-1)=0=P'(1)$ or $P'(-1)=0=P(1)$ we have
$$
\lambda \geq \min_{x\in [-1,1]}\dfrac{g(x)}{\omega(x)}
$$

The proof simply considers
$$
\dfrac{\langle P,\mathcal{L}P \rangle}{\langle P,P \rangle_\omega} =\dfrac{\langle P,\lambda\cdot\omega \cdot P \rangle}{\langle P,P \rangle_\omega} =\lambda
$$
and integrates it.

 

[lriuui0x0]

Why is $k=0$ excluded, e.g. for $\lambda =-1$?

$k=0$ is not excluded? It simply degenerates to the normal Legendre equation in that case? I don't think we will have $\lambda = -1$ as the eigenvalue. Since we know the solution should be the Legendre polynomial and it must be $\lambda = n(n+1)$.

 

[lriuui0x0]

I found the following theorem (with proof but wrong language):

For the eigenvalues $\lambda$ of the Sturm-Liouville problems
$$
\mathcal{L}P=-(f \cdot P')'+g\cdot P = \lambda \cdot \omega \cdot P
$$
over the interval $[-1,1]$ with $f(x)>0$ and $\omega (x)>0$ for all $x\in (-1,1)$ with the specific initial values
$$
P(-1)=0\, , \,P(1)=0\quad \text{(Dirichlet)}\quad\text{ or }\quad P'(-1)=0\, , \,P'(1)=0\quad \text{(Neumann)}
$$
or even $P(-1)=0=P'(1)$ or $P'(-1)=0=P(1)$ we have
$$
\lambda \geq \min_{x\in [-1,1]}\dfrac{g(x)}{\omega(x)}
$$

The proof simply considers
$$
\dfrac{\langle P,\mathcal{L}P \rangle}{\langle P,P \rangle_\omega} =\dfrac{\langle P,\lambda\cdot\omega \cdot P \rangle}{\langle P,P \rangle_\omega} =\lambda
$$
and integrates it.

Thanks for the help! I also got the solution, though much more noisy :) As you said, we basically transform the self-adjoint differential equation:

$$
\begin{aligned}
\frac{d}{dx}((1-x^2)^{k+1}\frac{d}{dx}P^{(k)}) &= -\lambda_k(1-x^2)^kP^{(k)} \\
\frac{d}{dx}((1-x^2)^{k+1}\frac{d}{dx}P^{(k)})P^{(k)} &= -\lambda_k(1-x^2)^k[P^{(k)}]^2 \\
\int_{-1}^1 \frac{d}{dx}((1-x^2)^{k+1}\frac{d}{dx}P^{(k)})P^{(k)}dx &= -\lambda_k \int_{-1}^1(1-x^2)^k[P^{(k)}]^2 dx \\
\biggl[(1-x^2)^{k+1}\frac{d}{dx}P^{(k)}P^{(k)}\biggr]_{-1}^1 - \int_{-1}^1(1-x^2)^{k+1}[\frac{d}{dx}P^{(k)}]^2 dx &=- \lambda_k \int_{-1}^1(1-x^2)^k[P^{(k)}]^2 dx \\
\lambda_k \int_{-1}^1(1-x^2)^k[P^{(k)}]^2 dx &= \int_{-1}^1(1-x^2)^{k+1}[\frac{d}{dx}P^{(k)}]^2 dx
\end{aligned}
$$

Then we have $\lambda_k \ge 0$ given $P^{(k)}$ is not identically zero.

I guess the lesson here is in your compact equation:

$$
\lambda = \frac{\langle P, \mathcal{L}P\rangle}{\langle P, P\rangle_w}
$$

 

[fresh_42]

$k=0$ is not excluded? It simply degenerates to the normal Legendre equation in that case? I don't think we will have $\lambda = -1$ as the eigenvalue. Since we know the solution should be the Legendre polynomial and it must be $\lambda = n(n+1)$.

Neither do I. But if you set $k=0$ in your equation, then the first equation is the result. But you did not specify $\lambda =\lambda_0,$ so negative values would be allowed, in contrast to your claim that $\lambda_k\geq 0.$ I guess $\lambda$ isn't as arbitrary as you said.

 

[lriuui0x0]

Neither do I. But if you set $k=0$ in your equation, then the first equation is the result. But you did not specify $\lambda =\lambda_0,$ so negative values would be allowed, in contrast to your claim that $\lambda_k\geq 0.$ I guess $\lambda$ isn't as arbitrary as you said.

Yeah that's true. By fixing $\lambda$ we still need to make the original diff equation hold, which means $\lambda$ cannot be arbitrary.