Bessel Functions - Eigenvalues + Eigenfunctions

[heilEuler]

Homework Statement

I'm given a standard form of Bessel's equation, namely
x^2y\prime\prime + xy\prime + (\lambda x^2-\nu^2)y = 0
with \nu = \frac{1}{3} and \lambda some unknown constant, and asked to find its eigenvalues and eigenfunctions.

The initial conditions are y(0)=0 and y\prime (\pi)=0.

The Attempt at a Solution

This is a single question assignment, so it's supposed to be reasonably extensive. What troubles me is that as far as I know, with this being THE typical Bessel equation, aren't the eigenfunctions pretty much a given? (i.e. they will be one sine and one cosine function of \sqrt{\lambda} and x, right?)

***Check my understanding please; the eigenfunctions are the functions in the general solution, and the eigenvalues are their respective coefficients? For example if some simple equation has the solution y(x) = C_1 e^{ikx} + C_2 e^{-ikx} then the eigenfunctions are e^{\pm ikx} and the eigenvalues are C_1, C_2, right? ***

I mean I can derive them, but it's not particularly difficult, especially since we did it in class, which makes me wonder if it's really what the professor wants.

The eigenvalues I should be able to determine with the two initial conditions I'm given, I think. So mainly, I'm curious if I'm doing the right thing, or if I'm completely off base with my interpretation of the question.

 

[gabbagabbahey]

What troubles me is that as far as I know, with this being THE typical Bessel equation, aren't the eigenfunctions pretty much a given? (i.e. they will be one sine and one cosine function of \sqrt{\lambda} and x, right?)

Since when are Bessel functions just sines and cosines?

***Check my understanding please; the eigenfunctions are the functions in the general solution, and the eigenvalues are their respective coefficients? For example if some simple equation has the solution y(x) = C_1 e^{ikx} + C_2 e^{-ikx} then the eigenfunctions are e^{\pm ikx} and the eigenvalues are C_1, C_2, right? ***

Not quite, the eigenfunctions would be e^{\pm ikx}, but C_1 and C_2 are not the eigenvalues (they are just constant used to express the general solution as a linear combination of the two eigenfunctions). The eigenvalues are actually \pm ik.

Why? Well, the example general solution you give satisfies the differential equation y''(x)=-k^2 y(x). Compare that to the eigenvalue equation Av=\lambda v (Where, as usual, A is any linear operator, v one of its eigenfunctions, and \lambda the corresponding eigenvalue). The two eigenfunctions, v_{\pm}= e^{\pm ikx}, satisfy the equation \frac{d}{dx}v_{\pm}=\pm ik v_{\pm}. With \frac{d}{dx} being a linear operator, this is clearly an eigenvalue equation with eigenfunctions v_{\pm} and corresponding eigenvalues \pm ik