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Frobenius Method When Initial Value of A Sum is not 1

  1. Dec 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve

    \begin{equation*}
    36x^2y''+(5-9x^2)y=0
    \end{equation*}

    using the Frobenius method

    2. Relevant equations

    Assume a solution of the form

    \begin{equation*}
    y=\sum_{n=0}^{\infty}{a_nx^{n+s}}
    \end{equation*}

    then

    \begin{equation*}
    y''=\sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s-2}}
    \end{equation*}

    3. The attempt at a solution

    I substitute the expanded forms of ##y## and ##y''## into the D.E

    \begin{equation*}
    36x^2 \sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s-2}}+5 \sum_{n=0}^{\infty}{a_nx^{n+s}} -9x^2 \sum_{n=0}^{\infty}{a_nx^{n+s}} =0
    \end{equation*}

    and put the terms of ##x## into the sums

    \begin{equation*}
    36 \sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s}}+5 \sum_{n=0}^{\infty}{a_nx^{n+s}} -9\sum_{n=0}^{\infty}{a_nx^{n+s+2}} =0
    \end{equation*}

    I then modify the last term so that the ##x## factor is ##x^{n+s}##

    \begin{equation*}
    36 \sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s}}+5 \sum_{n=0}^{\infty}{a_nx^{n+s}} -9\sum_{n=2}^{\infty}{a_{n-2}x^{n+s}} =0
    \end{equation*}

    This is where I get confused, the last term starts at ##n=2##, all the examples I have seen have their highest starting value for a sum equal to ##n=1##, in those cases I would comine the terms with the sum starting at ##n=0## and take out the first term of ##n=0## resulting in the indicial equation multiplied by ##a_o## which I would use to solve the D.E. Now however I have an extra term, just to be clear below is how I have found the indicial equation:

    \begin{equation*}
    \sum_{n=0}^{\infty}{[36(n+s)(n+s-1)+5]a_nx^{n+s}}-9\sum_{n=2}^{\infty}{a_{n-2}x^{n+s}}=0 \\
    [36(s)(s-1)+5]a_0x^s+\sum_{n=1}^{\infty}{[36(n+s)(n+s-1)+5]a_nx^{n+s}}-9\sum_{n=2}^{\infty}{a_{n-2}x^{n+s}}=0
    \end{equation*}

    So I have my indicial equation but I can't combine the second and third terms together because their initial values don't match. I'm not sure if I should pull out the ##n=1## term on the second term so I can create a recursion relation to use however I am not sure what to do with that term I pull out.

    Also a somewhat unrelated question, in this course we are also working with the Legendre D.Es and Bessel's equations, would I be able to solve both of these using the forbenius method? Or do I need some other techniques?

    Any help would be appreciated!
     
    Last edited: Dec 15, 2015
  2. jcsd
  3. Dec 16, 2015 #2

    haruspex

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    Yes.
    What would you do with the n=0 term?
     
  4. Dec 17, 2015 #3
    Thanks for the reply. For the n=0 term since we can't assume ##a_0=0## then we have to find a value for ##s## that eliminates the ##a_0## term. I get ##s=\frac{5}{6}## and ##s=\frac{1}{6}##. I assume then I would do the same for the ##n=1## term. Below I have my equation with the ##n=1## taken out

    \begin{equation*}
    (36s(s-1)+5)a_0x^s+(36s(s+1)+5)a_1x^{s+1}+\sum_{n=2}^{\infty}{\big[(36(n+s)(n+s-1)+5)a_nx^{n+s}-9a_{n-2}x^{n+s}\big]}
    \end{equation*}

    Thinking about this wouldn't this mean that the only possible values for ##s## would be common roots of the ##n=0## term and the ##n=1## term? This is the only way we could garuntee that the terms including ##a_0## and ##a_1## would equal zero. The issue now is I don't have a common ##s## value for the ##n=0## and ##n=1## terms. For ##n=0## I have ##s=\frac{5}{6}## and ##s=\frac{1}{6}## and for ##n=1## I have ##s=-\frac{5}{6}## and ##s=-\frac{1}{6}##. Does this mean I have made an error somewhere in my calculation or is there something I don't understand?
     
  5. Dec 17, 2015 #4

    haruspex

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    Right, because of the way s is defined.
    Why would the same restriction apply to that term?
     
  6. Dec 17, 2015 #5
    I didn't write it out in the equation above but that entire equation equals zero. From my understanding all the ##a_n## terms are unknown constants so the only way we can garuntee that this equation equals zero is if the ##s## portion of the terms equals zero.

    Or do we define ##a_0## as zero and for any other ##a_n## where ##n>0## we can choose if it equals zero? Is there somewhere else in the equations that indicates that ##a_1## is zero? Would finding the recursion relations using ##s=\frac{5}{6}## and ##s={1}{6}## give an ##a_1=0## maybe? I'll try this out in the morning if it would help.

    Reading my textbook it mentions that since ##a_0x^s## is the first term in the series we assume it is not zero but it doesn't say why, is there somewhere I could read up on why?

    Sorry for all the questions, seems like once I figure something out five more questions arise.

    Thanks again for the help, I really appreciate it.
     
  7. Dec 17, 2015 #6

    haruspex

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    In the assumed form of the equation, s is defined to be the lowest power with a nonzero coefficient. Hence a0 is not zero. But any other coefficient can be zero.
     
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