# Homework Help: Bessel Functions - Eigenvalues + Eigenfunctions

1. Mar 27, 2010

### heilEuler

1. The problem statement, all variables and given/known data
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 $y\prime (\pi)=0$.

3. 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.

Last edited: Mar 27, 2010
2. Mar 27, 2010

### gabbagabbahey

Since when are Bessel functions just sines and cosines?

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 $\pm ik$