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Frobenius method for Hermite Equation

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm asked to solve the Hermite Differential Equation
    [tex] y''(x) - 2 x y'(x) + \lambda y(x) = 0 [/tex]

    using the Frobenius method

    2. Relevant equations

    I am to assume the solution is in the form
    [tex] y(x) = \sum a_n x^{n+r} [/tex]
    where r are the roots of the indicial equation that in this case reads
    [tex]r(r-1)=0 [/tex]

    3. The attempt at a solution
    Solving the indicial equation, I get r=0 and r=1. If r=0, then the Frobenius method is just the ordinary power series method (expanding around an ordinary point) and I every worked example of the same problem I'm attempting to solve do use the ordinary power series method without mentioning the indicial equation or the name of Frobenius.

    Indeed going with r = 0 I get
    [tex] y(x) = a_0 \left( 1 - \frac{\lambda}{2!} x^2 - \frac{(4-\lambda) \lambda}{4!} x^4 - \frac{(8-\lambda)(4- \lambda)\lambda}{6!} x^6 - ...\right) + a_1 \left( x + \frac{2-\lambda}{3!} x^3 + \frac{(6-\lambda)(2-\lambda)}{5!} x^5 + ...\right) [/tex]

    but going with r =1 I get
    [tex]
    \sum (n+1) n a_n x^{n-1} - 2 \sum (n+1) a_n x^{n+1} + \sum \lambda x^{n+1} = 0
    [/tex]
    [tex]
    a_{n+2} = \frac{2(n+1)- \lambda}{(n+2)(n+1)}
    [/tex]
    [tex]
    y(x) = a_0 \left(x + \frac{2 - \lambda}{2!} x^3 + \frac{(6-\lambda)(2-\lambda)}{4!} x^5 + ...\right) + a_1 \left( x^2 + \frac{4-\lambda}{3!} x^4 + \frac{(6-\lambda)(4-\lambda)}{5!} x^6 + ... \right)
    [/tex]

    Sou now I'm confused
    1. Because I get four different linearly independent solutions to a second order equation, even though they're very similar which makes me think they might be related
    2. I'm not sure how the method of Frobenius is supposed to work around ordinary points. I thought it was a more general power series method good for both ordinary and regular singular points, but now I'm not so sure if it's supposed to be used around ordinary points.
    3. Arfken&Weber says I should have an even solution for r=0, with all powers of x even in the power expansion, and for r=1 I should take only odd powers of x for the power series expansion. That might make sense, but I can't understand where that comes from, it seems just pulled out of hat.Why can't we have both even and odd powers for r=0 and r=1?
     
  2. jcsd
  3. Nov 14, 2014 #2

    haruspex

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    Are you sure?
     
  4. Apr 13, 2015 #3
    Well now I'm not ..
     
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