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Eigenfunctions and ODE.

  1. Jan 19, 2014 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I have the following ODE:
    y′′−2xy′+2αy=0
    I'd like to determine the first three eigenfunctions.

    2. Relevant equations

    3. The attempt at a solution
    The solution y(x) may be recursively represented as:
    an+2=an(2n−2α)/[(n+2)(n+1)]
    I have found the eigenvalues to be −2α, however I find the manner whereby the eigenfunctions are determined to be rather perplexing. I'd sincerely appreciate an explanation. For instance, I know that for α=0, a2=a0(0−0)/2, but why would that entail y0(x)=a0? I mean, how was that derived?
     
  2. jcsd
  3. Jan 19, 2014 #2

    pasmith

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    Whenever you have a recurrence of the form
    [tex]
    a_{n+2} = F(n)a_n
    [/tex]
    then you know immediately that if [itex]a_n = 0[/itex] or [itex]F(n) = 0[/itex] then [itex]a_{n + 2m} = 0[/itex] for all [itex]m \geq 0[/itex].

    Starting from [itex]a_0[/itex] will give you the even terms [itex]a_{2n}[/itex], but those terms don't affect the odd terms [itex]a_{2n+1}[/itex], which are obtained by starting from [itex]a_1[/itex]. Here we have a second-order linear ODE, so we expect that for each [itex]\alpha[/itex] there will be two linearly independent solutions. We can define these to be [itex]E_\alpha(x)[/itex] which is obtained by taking [itex]a_0 = 1[/itex] and [itex]a_1 = 0[/itex] and [itex]O_\alpha(x)[/itex] which is obtained by taking [itex]a_0 = 0[/itex] and [itex]a_1 = 1[/itex].

    Thus for [itex]\alpha = 0[/itex] we have [itex]E_0(x) = 1[/itex] and [itex]O_0(x) = x + \frac13 x^3 + \dots[/itex], and the general solution will be [itex]cE_0(x) + dO_0(x)[/itex]. In fact we can solve the ODE analytically when [itex]\alpha = 0[/itex] to find that
    [tex]
    O_0(x) = \int_0^x \exp(u^2)\,\mathrm{d}u.
    [/tex]

    You can see that if [itex]\alpha[/itex] is an even positive integer then the even series terminates, and if [itex]\alpha[/itex] is an odd positive integer then the odd series terminates, and if [itex]\alpha[/itex] is not a positive integer then neither series terminates.
     
  4. Jan 19, 2014 #3
    I am sorry, but I am not really following. I pretty much lost you at "thus...". In any case, I happen to know that the first three eigenfunctions are: a0, a1x and a0(1-2x2). If you could explain to me how these were obtained I'd be grateful.
     
  5. Jan 19, 2014 #4

    vela

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    When ##\alpha=0##, if you write out the series solution, you get
    $$y = a_0(1) + a_1\left(x + \frac{x^3}{3} + \cdots\right).$$ This is of the form ##y = a_0 y_1 + a_1 y_2## where ##a_0## and ##a_1## are arbitrary constants and ##y_1=1## and ##y_2=x+\cdots## are the two linearly independent solutions.

    For this problem, you're apparently looking for polynomial solutions. For ##\alpha=0##, you can do this be setting ##a_1=0## to get rid of the infinite series.

    It turns out, you can obtain polynomial solutions for only certain values of ##\alpha##. You should be able to deduce what these values are from the recurrence relation.
     
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