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Method of Frobenius

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Use method of Frobenius to solve this equation:

    ##y''(x)-y'(x)=x##
    2. Relevant equations
    ------

    3. The attempt at a solution
    Seek an answer of the form

    ##y=\sum _{n=0}^{\infty } a_n x^{n+r}##

    Plug into the equation to get...

    ##\sum _{n=0}^{\infty } a_{n+1} (n+r) (n+r-1) x^{n+r-2}-\sum _{n=0}^{\infty } a_n
    (n+r) x^{n+r-1}=x##

    Now, for every previous attempt I have had at Frobenius method, I am used to having a homogeneous equation. I'm likely having a brain fart because I do not see how I would put this equation into a homogeneous form so I can get the correct indicial equation and solve for my roots. Does anyone have a suggestion for the next proper step to take?
     
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  3. Dec 15, 2013 #2

    vanhees71

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    You can first solve for the homogeneous equation. Try to find two linearly independent solutions. Then the general solution of the homogeneous equation is given by the general linear combination of these two solutions. Since you are told to use the Frobenius method, use it. Of course, the equation is easier solved in closed form with the standard ansatz [itex]y(x)=\exp(\lambda x)[/itex], but of course you can also use the Frobenius method.

    Then you only need one particular solution of the inhomogeneous equation. It's clear that for this, the ansatz [itex]y(x)=a x^2+bx[/itex] must lead to one.
     
  4. Dec 15, 2013 #3
    So, by solving the homogeneous equation and equating powers of x, I end up with the indicial equation
    ##r\left(r-1\right)a_0##=0

    This gives me an integer difference for my powers of ##r##. 0 and 1. If I set ##r=0##, then I get the following relationships for the constants.
    No equation is yielded for ##a_0## since 0=0 for the case of ##n=0##
    For higher powers of ##n##, I get a series generated, which I can add to the a_0.

    ##y=a_0+a_1 x+\frac{a_1 x^2}{2}+\left(\frac{1}{3!}+\frac{a_1}{3!}\right)
    x^3+\left(\frac{1}{4!}+\frac{a_1}{4!}\right) x^4+\ldots##

    This is basically the solution. What I do not understand is what happened with the case of r=1? I am apparently fully finished with the solution after only touching the case of r=0. Isn't there a natural log that term that I need if the powers differ by an integer? Is it because I actually ended up with two constants instead of all of the constants relating to each other?
     
  5. Dec 16, 2013 #4

    vanhees71

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    First of all, setting [itex]r=r_1=0[/itex], for the [itex]a_k[/itex] with [itex]k \geq 1[/itex] you get a recursion relation that determines them in terms of [itex]a_0 \neq 0[/itex] which is arbitrary. To find one of the linearly independent solutions, [itex]y_1(x)[/itex] you can set [itex]a_0=1[/itex].

    Since the indical equation has [itex]r=r_2=1[/itex] as a second solution, we have [itex]r_1-r_2=1 \in \mathbb{N}_0[/itex]. Then we know that we can make the ansatz
    [tex]y(x)=\gamma \ln x y_1(x)+\sum_{k=0}^{\infty} b_k x^{r_2+k}.[/tex]
    Plugging this into the differential equation, you'll see you find a solution with [itex]b_0=1[/itex] and and arbitrary [itex]b_1[/itex]. You can set [itex]b_1=0[/itex]. The constant [itex]\gamma[/itex] is uniquely defined.
     
  6. Dec 16, 2013 #5

    vela

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    When the roots differ by an integer, the larger of the roots will yield a solution. You essentially considered the r=1 case when you moved onto considering the coefficients ##a_1## and beyond. You obtained the solution you found that was proportional to ##a_1## and the particular solution. The smaller root may or may not result in a second solution depending on how the coefficients are related. In this problem, it did yield a solution, namely ##y_2(x) = a_0##. Since you found two independent solutions, you're done.

    Fuch's theorem guarantees one solution. Only if the series method doesn't yield a second independent solution do you need to use the method vanhees described to find it.
     
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