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Homework Help: Bessel Functions - Eigenvalues + Eigenfunctions

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm given a standard form of Bessel's equation, namely
    [tex] x^2y\prime\prime + xy\prime + (\lambda x^2-\nu^2)y = 0 [/tex]
    with [itex] \nu = \frac{1}{3} [/itex] and [itex] \lambda [/itex] some unknown constant, and asked to find its eigenvalues and eigenfunctions.

    The initial conditions are [itex] y(0)=0 [/itex] and [itex] y\prime (\pi)=0 [/itex].

    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 [itex] \sqrt{\lambda} [/itex] 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 [tex] y(x) = C_1 e^{ikx} + C_2 e^{-ikx} [/tex] then the eigenfunctions are [itex] e^{\pm ikx} [/itex] and the eigenvalues are [itex] C_1, C_2 [/itex], 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. jcsd
  3. Mar 27, 2010 #2


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    Homework Helper
    Gold Member

    Since when are Bessel functions just sines and cosines?

    Not quite, the eigenfunctions would be [itex] e^{\pm ikx} [/itex], but [itex]C_1[/itex] and [itex]C_2[/itex] 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 [itex]\pm ik[/itex].

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