Why Does the Integral of Legendre Polynomials Yield a Kronecker Delta?

In summary, the integral of a Laplace equation in spherical coordinates is zero if L ≠ L'. However, if L = L', then the integral is equal to 2/(2L+1)*δ. using Rodrigues' formula or the generating function, the Legendre Polynomials can be solved to give the value of δ.
  • #1
Don'tKnowMuch
21
0
I am doing a Laplace's equation in spherical coordinates and have come to a part of the problem that has the integral...

∫ P(sub L)*(x) * P(sub L')*(x) dx (-1<x<1)

The answer to this integral is given by a Kronecker delta function (δ)...

= 0 if L ≠ L'

OR...
= 2/(2L+1)*δ if L = L' (where δ = 1)

I believe the reason why the integral is equal to zero when L ≠ L' is because of the orthogonality of Legendre polynomials, however i cannot figure out how the integral is equal 2/(2L+1). If L = L' then the integral would equal...

∫ [P(sub L)*(x)]^2 dx

I tried to use a simple power rule of integration to solve the above integral, but i am afraid my method is flawed. Any suggestions?

Pre-emptive thanks to whomever takes the time to help!
 
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  • #2
For clarity i should add that...

P(sub L)(x)

is a Legendre polynomial
 
  • #3
I think i have to use Rodrigues' formula
 
  • #4
This is one possibility. Another is to use the generating function, which is used to define in the msot convenient way the Legendre polynomials.

It's given by
[tex]\Phi(r,u)=\frac{1}{\sqrt{1-2u r+r^2}}=\sum_{l=0}^{\infty} P_l(u) r^l.[/tex]
From this you find
[tex]P_l(u)=\frac{1}{l!} \left .\frac{\partial^l}{\partial r^l} \Phi(r,u) \right|_{r=0}.[/tex]
This series expansion is obviously valid for [itex]|r|<1[/itex] if [itex]|u|<1[/itex].

We know that the Legendre Polynomials are orthogonal to each other since they are solutions of the eigen-value equation
[tex]\frac{\mathrm{d}}{\mathrm{d} u} \left [(1-u^2) \frac{\mathrm{d} P_l}{\mathrm{d} u} \right ]=-l(l+1) P_l,[/tex]
where the differential operator is self-adjoint on [itex]L^2([-1,1])[/itex]. Thus we have
[tex]\int_{-1}^{1} \mathrm{d} u P_l(u) P_{l'}(u)=N_l \delta_{ll'}.[/tex]
To evaluate the normalization factor we take the following integral:
[tex]I(r)=\int_{-1}^{1} \mathrm{d} u \Phi^2(r,u)=\int_{-1}^{1} \mathrm{d} u \frac{1}{1-2 u r +r^2}=-\left [\frac{1}{2r} \ln(1-2 u r+r^2) \right]_{u=-1}^{u=1}=\frac{1}{r} \ln \left (\frac{1+r}{1-r} \right ).[/tex]
Expanding this into a power series in [itex]r[/itex] yields
[tex]I(r)=\sum_{l=0}^{\infty} \frac{2}{2l+1} r^{2l}.[/tex]
On the other hand from the definition of the Legendre Polynomials by the above given series expansion you get
[tex]\Phi^2(r,u)=\sum_{l,l'=0}^{\infty} P_l(u) P_{l'}(u) r^{l+l'}.[/tex]
Integrating over [itex]u[/itex] leads to
[tex]I(r)=\sum_{l,l'=0}^{\infty} N_l \delta_{ll'} r^{2l}=\sum_{l=0}^{\infty} N_l r^{2l}.[/tex]
Comparing the coefficients in both expansions of [itex]I(r)[/itex] yields
[tex]N_l=\frac{2}{2l+1}.[/tex]
QED.
 
  • #5


I am happy to provide some clarification on this integral and the use of Legendre polynomials in Laplace's equation in spherical coordinates.

First, it is important to understand the properties of Legendre polynomials. They are a set of orthogonal polynomials that are commonly used in mathematics and physics to solve problems involving spherical symmetry. In the context of Laplace's equation in spherical coordinates, Legendre polynomials are used to represent the radial component of the solution.

In this particular integral, we are integrating the product of two Legendre polynomials, P(sub L)*(x) and P(sub L')*(x), over the interval -1<x<1. As you correctly stated, when L ≠ L', the integral is equal to zero due to the orthogonality of Legendre polynomials. This means that the two polynomials are perpendicular to each other and do not overlap in this interval, resulting in a zero integral.

However, when L = L', the integral becomes:

∫ [P(sub L)*(x)]^2 dx

This integral can be solved using integration by parts or by using the power rule. But the key here is to remember that Legendre polynomials are normalized such that their integral over the interval -1<x<1 is equal to 2/(2L+1). This is where the 2/(2L+1) term in the answer comes from.

So, to summarize, the integral is equal to 0 if L ≠ L' due to orthogonality, and it is equal to 2/(2L+1) if L = L' due to the normalization of Legendre polynomials. I hope this helps clarify the reasoning behind the answer and provides some guidance for solving the integral. Keep up the good work with your problem solving!
 

Related to Why Does the Integral of Legendre Polynomials Yield a Kronecker Delta?

1. What are Legendre polynomials?

Legendre polynomials are a type of mathematical function used in the field of calculus. They are named after the French mathematician Adrien-Marie Legendre and are commonly used to solve problems involving physics, engineering, and statistics.

2. How are Legendre polynomials used?

Legendre polynomials are used to solve problems related to differential equations, particularly in the field of quantum mechanics. They are also used in statistics to analyze data and in engineering to model physical systems.

3. What is the formula for calculating Legendre polynomials?

The formula for calculating Legendre polynomials is Pn(x) = (1/2nn!) * dn/dxn(x2 - 1)n, where n is the degree of the polynomial and x is a variable. This formula is recursive and can be used to generate higher order polynomials.

4. What are the properties of Legendre polynomials?

Legendre polynomials have several important properties, including orthogonality, recurrence relations, and generating functions. They are also symmetric around the y-axis and have alternating positive and negative roots.

5. How are Legendre polynomials related to other types of polynomials?

Legendre polynomials are a special case of orthogonal polynomials, which are a larger class of polynomials that satisfy certain mathematical properties. They are also related to other types of polynomials, such as Chebyshev polynomials and Jacobi polynomials, which have similar properties and applications.

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