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Probing series solution of DE: irregular singular point

  1. Mar 11, 2012 #1

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    1. The problem statement, all variables and given/known data
    Consider
    [itex]y''+\frac{\alpha}{x^{r}}y'+\frac{\beta}{x^{s}}y=0[/itex]
    Suppose we try find a solution of the form<br />
    [itex]y=\sum_{n=0}^{\infty}a_{n}x^{n+\lambda}[/itex]
    Show that if r=2 and s=1 then there is only one possible value of [itex]\lambda[/itex] for which there is a formal solution in the form shown above.

    3. The attempt at a solution
    Substituting the general solution into the DE gives us:

    [itex]\sum_{n=0}^{\infty}(n+ \lambda )(n+ \lambda -1)a_{n}x^{n+ \lambda -2}+\frac{\alpha}{x^{2}}\sum_{n=0}^{\infty}(n+ \lambda )a_{n}x^{n+ \lambda -1}+\frac{\beta}{x}\sum_{n=0}^{\infty}a_{n}x^{n+ \lambda }=0[/itex]

    [itex]\sum_{n=0}^{\infty}(n+\lambda)(n+\lambda-1)a_{n}x^{n+ \lambda -2}+\alpha \sum_{n=0}^{\infty}(n+ \lambda )a_{n}x^{n+ \lambda -3}+\beta \sum_{n=0}^{\infty}a_{n}x^{n+ \lambda -1}=0[/itex]

    Pulling out some terms to make the starting terms in the summation have the same order of x:
    [itex]\lambda(\lambda-1)a_{0}x^{\lambda-2}+\sum_{n=1}^{\infty}(n+\lambda)(n+\lambda-1)a_{n}x^{n+\lambda-2}+\alpha \lambda a_{0}x^{\lambda-3}+\alpha(1+\lambda)a_{1}x^{\lambda-2 }+\alpha \sum_{n=2}^{\infty}(n+\lambda)a_{n}x^{n+\lambda-3}+\beta \sum_{n=0}^{\infty}a_{n}x^{n+\lambda-1}=0[/itex]


    Short of shifting the summations and doing some grouping(I don't feel like that is necessary to complete the answer to the question), at this point I'm not sure what I should do. Do I look at the coefficients of the lowest power of x as I would do if there was a quadratic indical equation? If that were the case I would simply say[itex]\alpha \lambda a_{0}=0[/itex] therefore the only possible value for [itex]\lambda[/itex] is zero(since alpha and [itex]a_{0}[/itex] are arbitrary). But then what does that mean for all the other co-efficients? Why would I be interested in that term anyway, since it is not actually an indical equation? I'm pretty sure that the solution lies in TRYING to get a quadratic indical equation and consequently failing

    I feel like I'm not far from deducing the answer, but I'm uneasy since I think I'm essentially dealing with a question that is telling me to 'show that the Frobenius method won't work for an irregular singular point'. I have no rules for trying to do this.

    Note: There is a similar question in Boyce & DiPrima Elementary Differential Equations and Boundary Value Problems(2nd edition)
     
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