Prove eigenvalues of the derivatives of Legendre polynomials >= 0

In summary: P\rangle_{\omega}} = \frac{\langle P, \lambda \omega P\rangle}{\langle P,...P\rangle_{\omega}}$$the ##\langle P, \mathcal{L}P\rangle## term is the one that collapses to the differential equation upon integration by parts.In summary, The problem hints at finding a relationship between the integrals ##\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##. The functions ##f(x)## and ##g(x
  • #1
lriuui0x0
101
25
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:

$$
\frac{d}{dx}((1-x^2)^{k+1}\frac{dP^{(k)}}{dx}) = -\lambda_k(1-x^2)^kP^{(k)}
$$
 
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  • #3
fresh_42 said:
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 forumla 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.
 
  • #4
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.
 
  • #5
fresh_42 said:
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...
 
  • #6
fresh_42 said:
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##
 
  • #7
Why is ##k=0## excluded, e.g. for ##\lambda =-1##?
 
  • #8
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.
 
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  • #9
fresh_42 said:
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)##.
 
  • #10
fresh_42 said:
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:

$$
\lambda = \frac{\langle P, \mathcal{L}P\rangle}{\langle P, P\rangle_w}
$$
 
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  • #11
lriuui0x0 said:
##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.
 
  • #12
fresh_42 said:
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.
 

FAQ: Prove eigenvalues of the derivatives of Legendre polynomials >= 0

1. What are eigenvalues and derivatives of Legendre polynomials?

Eigenvalues are special numbers associated with a linear transformation that represent the scaling factor of the corresponding eigenvectors. Derivatives of Legendre polynomials are the rate of change of these polynomials with respect to their independent variable.

2. Why is it important to prove that the eigenvalues of the derivatives of Legendre polynomials are greater than or equal to 0?

It is important because the eigenvalues of the derivatives of Legendre polynomials have physical significance in various fields such as physics, engineering, and mathematics. Proving that they are greater than or equal to 0 helps in understanding the behavior and properties of these polynomials in different applications.

3. How can the eigenvalues of the derivatives of Legendre polynomials be proven to be greater than or equal to 0?

This can be proven using mathematical techniques such as induction, differentiation, and integration. By applying these techniques to the derivatives of Legendre polynomials, we can show that their eigenvalues are always non-negative.

4. What are some real-world applications of the eigenvalues of the derivatives of Legendre polynomials being greater than or equal to 0?

Some examples include using these polynomials to model physical systems, such as the motion of a pendulum or the behavior of a vibrating string. They are also used in solving differential equations and in numerical analysis.

5. Are there any exceptions to the rule that the eigenvalues of the derivatives of Legendre polynomials are greater than or equal to 0?

No, there are no exceptions. This property holds true for all derivatives of Legendre polynomials, regardless of their order or degree.

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