Question on Solving Laplace Equation Potentials.

[Fjolvar]

Hello, I'm learning how to solve Laplace's equation to find Potentials in Cartesian, Cylindrical, and Spherical Coordinates and let's just say it's not going as smoothly as I'd like. In particular, I'm having difficulty with the Spherical case which involves Legendre Polynomials, Method of Frobenius, Orthogonality, etc.

In the Spherical case, how do you determine Pl(X) in the Angular Equation of V(r,\vartheta) where \Theta(\vartheta) = Pl(cos(\vartheta))..

What determines l (lower case L) in the Legendre Polynomials when solving for Pl(X)..

I know that when l=0, Pl(X) = 1. When l=1, Pl(X) = X, etc. So what does l depend on and how does it relate to the order of the equation and the physics of a problem?

 

[diazona]

You're familiar with separation of variables, right? Remember that when you're solving a differential equation that can be rearranged into the form
f(r) + h(\phi) = 0
each of those parts individually must be equal to a constant. In the case of the Laplace equation, you find that the constant value of f(r) has to take the form l(l+1), where l can be any integer. If you set h(\phi) equal to that particular constant, you get the Laplace differential equation, whose solutions are P_l^m(\cos\phi). So the value of l is determined by your choice of the separation constant. In practice, it can be any positive integer. (Or negative integer, but you get the same values so it doesn't matter; by convention we limit l to positive values.)

A detailed explanation is available at http://mathworld.wolfram.com/LaplacesEquationSphericalCoordinates.html