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Second order diff. eq. Frobenius

  1. Apr 1, 2012 #1
    Hi there. I have this exercise, which says:

    Demonstrate that:

    [tex]xy''+(1-x)y'+\lambda y=0[/tex]

    has a polynomial solution for some λ values.
    Indicate the orthogonality relation between polynomials, the fundamental interval, and the weight function.

    So I thought I should solve this using Frobenius method. I have one singular point at x=0, which is regular. I assumed a solution of the form:
    [tex]y(x)=\sum_0^{\infty}a_n x^{n+r}[/tex]

    And then replacing in the diff. eq. I get:
    [tex]\sum_0^{\infty}a_n (n+r)(n+r-1) x^{n+r-1}+\sum_0^{\infty}a_n (n+r)x^{n+r-1}-\sum_0^{\infty}(n+r)a_n x^{n+r}+\lambda \sum_0^{\infty}a_n x^{n+r}=0[/tex]
    [tex]\sum_0^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0[/tex]
    [tex]a_0r^2x^{r-1}+\sum_1^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0[/tex]

    Therefore r=0.

    Then replacing r=0, and changing the index for the first summation, with m=n-1, n=m+1:
    [tex]\sum_0^{\infty}a_{m+1} (m+1)^2 x^{m}-\sum_0^{\infty}a_n (n-\lambda) x^{n}=0[/tex]
    And now calling m=n
    [tex]\sum_0^{\infty}x^m \left ( a_{m+1} (m+1)^2 x^{m}-a_m (m-\lambda) \right )=0[/tex]
    So I have the recurrence relation:

    Trying some terms:

    I'm not sure what this gives, I tried this:
    This is wrong, because the factorial in the numerator is only defined for positive values of (n-1-λ), and if n=1 I get (-\lambda)!, which wouldn't work for a_1, unless λ=0, which gives the trivial solution. But I think it works for n>1.
    Last edited: Apr 1, 2012
  2. jcsd
  3. Apr 1, 2012 #2


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    Science Advisor

    Hey Telemachus.

    You can define the factorial for negative values, but the values can not be integers: if this holds then the factorial function does extend to the negative real line (minus the integers). Just in case you need more details:

  4. Apr 1, 2012 #3
    Thank you chiro. Do you think that what I did is ok?

    I should take the diff. eq. into the self-adjoint form to get the weight function. About the fundamental interval, I think I should look at the convergence radius for the solution, right?
    Last edited: Apr 1, 2012
  5. Apr 2, 2012 #4
    Ok. I worked this in a different fashion:


    And now I called:

    Then λ-n can't be a negative integer, and the polynomials would be given by:
    Anyway, I think the an are wrong again, because if I take n=1 I get [tex]a_1=-a_0 \Gamma(\lambda-1)[/tex] which doesn't fit.

    There is another solution, it is given by using the Frobenius theorem, and it involves a logarithm, but I think it isn't needed.

    I actually think that I didn't have to get this explicit solution. To demonstrate what the problem asks I think I should take the equation to the self adjoint form.
    [tex]xy''+(1-x)y'+\lambda y=0\rightarrow y''+(\frac{1}{x}-1)y'+\frac{\lambda}{x}y=0[/tex]

    Multiplying by [tex]r(x)=e^{\ln (x) -x}[/tex]
    I get:
    [tex]\frac{d}{dx}\left ( e^{\ln (x) -x}\frac{dy}{dx} \right) +\lambda\frac{e^{\ln (x) -x}}{x}y=0[/tex]
    This is the self adjoint form for my differential equation. Then the weight function is given by: [tex]p(x)=\frac{e^{\ln (x) -x}}{x}[/tex]

    I don't know how to get the fundamental interval.

    By the way, should I post this in homework and coursework questions? if it is so, please move it, and I'm sorry.
    Last edited: Apr 2, 2012
  6. Apr 2, 2012 #5
    Ok. It's solved.
  7. Apr 2, 2012 #6


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    Science Advisor

    The original problem was show that the equation "has a polynomial solution for some λ values." So you really just need to show that for some [itex]\lambda[/itex], The coefficients are eventually 0.
  8. Apr 2, 2012 #7
    Yes, but for which λ? besides, the coefficients doesn't seem that easy to get. I actually couldn't. I used some theorems on the sturm liouville theory to solve this, I didn't get the coefficients explicitly. I've tried, but I couldn't find the coefficients. I would like to find the right expression for the a_n in the recurrence relation, but it doesn't seem to be that easy.
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