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Mathematical methods of Physics, ODE

  1. Dec 11, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    For what values of K does the DE [itex]xy''-2xy'+(K-3x)y=0[/itex] (1) has a bounded solution in [itex](0, \infty)[/itex]?


    2. Relevant equations
    Not so sure, Frobenius method maybe.


    3. The attempt at a solution
    First, I check what happens when x tends to infinity. I see that the DE behaves like [itex]\phi ''-2 \phi '-3 \phi =0[/itex]. Where [itex]\phi (x)=c_1e^{3x}+c_2e^{-x}[/itex]. I checked out and this indeed satisfies the DE (2).
    I propose a solution to (1) of the form [itex]y(x)=f(x) \phi (x)[/itex]. So I calculated y' and y'' and replaced all this into the DE (1) and after a lot of algebra I reached that the DE (1) is equivalent to the DE [itex]f''(x)+4f'(x)+f(x) \left ( \frac{Kc_1e^{3x}}{x} \right ) \cdot \left ( \frac{1}{c_1e^{3x}+c_2e^{-x}} \right )=0[/itex] (3) where these [itex]c_1[/itex] and [itex]c_2[/itex] differs from the ones given in the expression for [itex]\phi (x)[/itex] but this doesn't matter, they are constants. Now I must say I'm not really eager to use Frobenius's method on the DE (3). Mainly due to the denominator. I don't really know how to proceed... any help will be appreciated as usual.
     
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  3. Dec 11, 2011 #2

    LCKurtz

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    Not sure if it will help, but surely you wouldn't want to include the e3x in your calculations because it isn't bounded. I would try just the e-x part, but no guarantees; I haven't tried it myself.
     
  4. Dec 11, 2011 #3

    fluidistic

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    Whoops, you're absolutely right! I forgot about this. :redface:
     
  5. Dec 12, 2011 #4

    fluidistic

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    I progressed on the problem.
    I skip the algebra, and I must solve [itex]f''(x)-4f'(x)+\frac{K}{x}f(x)=0[/itex]. I use Frobenius's method on it and this gave me the indicial equation s=0 (that I discard later) and s=1.
    I reach the recurrence relation [itex]a_{n+1}=\frac{-a_n [4(n+s)+K]}{(n+2)(n+1)}[/itex].
    The values of K they ask for is when [itex]a_{n+1}=0[/itex] and [itex]a_n \neq 0[/itex]. So [itex]K=-4n-4[/itex] with n going from 0 to infinity. (This is, I hope, the answer to the problem).
    However say I want to find the general terms [itex]a_n[/itex] in [itex]\sum _{n=0}^{\infty} a_nx^{n+1}[/itex].
    I'm currently trying to find a pattern, by setting [itex]a_0=1[/itex] between the [itex]a_n[/itex]'s but I'm not getting sucessful in it.
    [itex]a_0=1[/itex], [itex]a_1=-\frac{4+K}{2}[/itex], [itex]a_2=\left ( \frac{4+K}{2} \right )\left ( \frac{8+K}{3\cdot 2} \right )[/itex], [itex]a_3=-\frac{(4+K)}{2} \left ( \frac{8+K}{3\cdot 2} \right ) \left ( \frac{12+K}{4\cdot 3} \right )[/itex].

    So far I realize that [itex]a_n=\frac{(-1)^n}{n!(n+1)!}[/itex] multiplied by something that I'm not sure (and don't find yet). Maybe something similar to [itex]\frac{(4n+K)!}{(4n+K-1)(4n+K-2)(4n+K-3)}[/itex]. Can someone help me on this task?
     
  6. Dec 12, 2011 #5

    vela

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    Did you perhaps make a sign error? I find the recursion relation to be
    [tex]a_{n+1} = -\frac{K-4(n+1)}{(n+1)(n+2)} a_n.[/tex]
     
  7. Dec 13, 2011 #6

    fluidistic

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    Yes I think I did, though now I get a different answer from ours.
    From [itex]f''(x)-4f'(x)+\frac{K}{x}f(x)=0[/itex] and [itex]f(x)=\sum_{n=0}^{\infty}a_n x^{n+s}[/itex], I reach that it's equivalent to [itex]\sum_{n=-1}^{\infty}a_{n+1}(n+s)(n+s+1) x^{n+s-2}-4\sum_{n=0}^{\infty}a_n(n+s) x^{n+s-1}-K \sum_{n=0}^{\infty}a_n x^{n+s-1}=0[/itex]
    [itex] \Rightarrow a_0s(s-1)x^{s-2}+\sum_{n=0}^{\infty}a_n x^{n+s-1} [-4(n+s)-K]+a_{n+1}(n+s+1)(n+s)x^{n+s-1}=0[/itex].
    So [itex]-a_n [4(n+s)+K]+a_{n+1}(n+s+1)(n+s)=0\Rightarrow a_{n+1}=\frac{a_n[4(n+s)+K]}{(n+s+1)(n+s)}[/itex].

    Mod note: fixed LaTeX
     
    Last edited by a moderator: Dec 13, 2011
  8. Dec 13, 2011 #7

    vela

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    You changed the sign on the K term, so you should get
    [tex]a_0s(s-1)x^{s-2}+\sum_{n=0}^{\infty}a_n x^{n+s-1} [K-4(n+s)]+a_{n+1}(n+s+1)(n+s)x^{n+s-1}=0.[/tex]Then you made one more sign error which got rid of the alternating sign.
     
  9. Dec 13, 2011 #8

    fluidistic

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    You are right, I made a typo when I copied the DE when I turned the page of my draft for the sign of K. I now reach your result.
    I'm trying to find a pattern by calculating the first terms, assuming [itex]a_0=1[/itex].
    [itex]a_1=-\frac{(K-4)}{2}, a_2=\frac{(K-4)(K-8)}{2\cdot 3\cdot 2}[/itex], etc.
    My attempt is [itex]a_n=\frac{(-1)^n(K-4)(K-8)...(K-4n)}{n!(n+1)!}[/itex] but this doesn't work for [itex]a_0[/itex] since I get K instead of 1.
     
  10. Dec 13, 2011 #9

    vela

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    If you require that the series terminates, you know K has the form K=4m where m=1, 2, 3, ....
     
  11. Dec 13, 2011 #10

    fluidistic

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    Yes indeed. [itex]K-4(n+1)=0 \Rightarrow k=4(n+1)[/itex] with n=0, 1, etc. which is equivalent to K=4m for m=1, 2, etc.
    I was just trying to do more than what they asked me. I wanted to express [itex]a_n[/itex] in terms of n only rather than [itex]a_{n+1}[/itex] in terms of [itex]a_n[/itex]. I don't think it really matters anyway, but since I've a weakness in doing it I wanted to eradicate it. :smile:
     
  12. Dec 13, 2011 #11

    vela

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    I didn't do this calculation out carefully yesterday so I may have made a mistake, but I think you can use the fact that K=4m to express an in terms of n and m.
     
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