Completeness of Legendre Polynomials

In summary, the conversation discusses working with Legendre polynomials and their use in the context of Spherical Harmonics. The regular Legendre polynomials are solutions to the differential equation with eigenvalues of the form n(n+1) for integer n. It is questioned why only these values are accepted as valid solutions and what restricts the eigenvalues to this form. The conversation mentions looking into Sturm-Liouville theory for a solution but finding no method to prove that n(n+1) are the only eigenvalues. It is concluded that there is no particular reason to use solutions for other values and the proof for the regular Legendre polynomials being the only solutions is based on pragmatism and aesthetic reasons. The non-polynomial
  • #1
ObsessiveMathsFreak
406
8
I've recently been working with Legendre polynomials, particularly in the context of Spherical Harmonics. For the moment, it's enough to consider the regular L. polynomials which solve the differential equation

[tex][(1-x^2) P_n']'+\lambda P=0[/tex]

However, I've run into a problem. Why in the definition of spherical harmonics are only [tex]\lambda[/tex] of the form [tex]\lambda=n(n+1)[/tex] for integer n, accepted as valid solutions to the problem? In particular, with no boundary value specified, what restricts the eigenvalues to the problem here?

Note that my question is about the eigenvalues. I'm aware that the singular Legendre polynomials of the second kind (Q_n) should be rejected, but what stops a Legendre polynomial of the first kind with a non n(n+1) eigenvalue from being a solution?

I've read excerpts from Sturm-Liouville theory in an effort to find a solution to this problem, but mostly these texts simply repeat the same theorems that the eigenvalues are real, the solutions orthogonal, etc. There seems to be no method in the theory for proving that the eigenvalues [tex]n(n+1)[/tex] are the only eigenvalues.

So my question is this: Why are eigenvalues such as say [tex]\lambda=1[/tex] not permitted as solutions to the Legendre equation? Particularly in the absence of boundary values.
 
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  • #2
If [itex]\lambda[/itex] is not equal to [itex]n(n+1)[/itex], the regular (non-singular) solution to the equation is an infinite series, not a polynomial.

The set of polynomials of degrees 0, 1, 2, ... are obviously linearly independent, and they also have nice orthogonality properties. So there is no particular reason to want to use solutions for other values of [itex]\lambda[/itex] for anything. At least, Legendre didn't have any reason. If somebody else has found a use for them, I don't know about that.

I don't think there is anything more to this than simple pragmatism.
 
  • #3
I understand the pragmatic and aesthetic reasons fro excluding these solutions, and I follow the problem which arises in the infinite series for non integer n.

But where is the proof that the legendre polynomials are the only regular soultions to the differential equation?
 
  • #4
ObsessiveMathsFreak said:
But where is the proof that the legendre polynomials are the only regular soultions to the differential equation?

They aren't the only regular solutions. They are just the pragmatically useful ones.

The non-polynomial regular solutions are called "Legendre functions of the first kind". (The non-regular solutions are the second kind).

http://mathworld.wolfram.com/LegendreDifferentialEquation.html
 
  • #5


The completeness of Legendre polynomials is a well-studied topic in mathematical analysis. The eigenvalues of Legendre polynomials play a crucial role in determining the completeness of these polynomials. The reason why only eigenvalues of the form \lambda=n(n+1) are accepted as valid solutions is because they satisfy the necessary boundary conditions for the problem at hand.

In the context of spherical harmonics, the eigenvalues n(n+1) correspond to the angular momentum quantum numbers, which are well-defined physical quantities in quantum mechanics. These eigenvalues arise naturally from the symmetry of the problem and the boundary conditions imposed. In the absence of boundary values, any non n(n+1) eigenvalue would not satisfy the necessary conditions for the problem and thus would not be a valid solution.

The completeness of Legendre polynomials can be proven using Sturm-Liouville theory, which is a powerful tool for studying eigenvalue problems. This theory ensures that the eigenvalues are real and the solutions are orthogonal, which are crucial properties for the completeness of Legendre polynomials.

In summary, the eigenvalues n(n+1) are the only valid solutions to the Legendre equation in the context of spherical harmonics because they satisfy the necessary boundary conditions and arise naturally from the symmetry of the problem. Sturm-Liouville theory can be used to prove the completeness of Legendre polynomials and the uniqueness of their eigenvalues.
 

Related to Completeness of Legendre Polynomials

1. What are Legendre polynomials?

Legendre polynomials are a set of orthogonal polynomials that are commonly used in mathematics and physics. They are named after French mathematician Adrien-Marie Legendre, who first introduced them in the late 18th century.

2. How are Legendre polynomials used?

Legendre polynomials are used in a variety of mathematical and scientific applications, including solving differential equations, approximation of functions, and solving problems in physics and engineering.

3. What is the completeness of Legendre polynomials?

The completeness of Legendre polynomials refers to their ability to form a complete basis set for functions on a given interval. This means that any function on that interval can be represented as a linear combination of Legendre polynomials, making them useful for solving a wide range of problems.

4. How is the completeness of Legendre polynomials proven?

The completeness of Legendre polynomials can be proven using mathematical techniques, such as the Gram-Schmidt process or the method of induction. These techniques involve showing that the polynomials satisfy certain properties and can be used to construct any function on a given interval.

5. Are Legendre polynomials the only complete basis set for functions?

No, Legendre polynomials are not the only complete basis set for functions. Other examples include Fourier series, Chebyshev polynomials, and Bessel functions. The choice of basis set often depends on the specific problem at hand and the desired properties of the solution.

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