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Mathematical methods of Physics, ODE

  1. Dec 8, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must find the constant K such that [itex]y''-\left ( \frac{1}{4}+\frac{K}{x} \right )y=0[/itex] for x>0 has a non trivial solution that is worth 0 when x tends to 0 and when x tends to infinity.


    2. Relevant equations
    Frobenius method.


    3. The attempt at a solution
    I proposed a solution of the form [itex]y(x)=\sum_{n=0}^{\infty}a_nx^{n+s}[/itex].
    This gave me [itex]y''(x)=\sum _{n=2}^\infty (n+s)(n+s-1)a_nx^{n+s-2}[/itex].
    Plugging y and y'' into the DE I get that [itex]\sum _{n=2}^\infty (n+s)(n+s-1)a_nx^{n+s-2} - \frac{1}{4}\sum_{n=0}^{\infty}a_nx^{n+s}+K\sum_{n=0}^{\infty}a_nx^{n+s-1}=0[/itex].
    Now I must equate the powers of x inside each series in order to put all the arguments of the series under a single series.
    So let's say I want to have as common power of x, n+s.
    This gives me [itex]\sum_{n=0}^{\infty}a_{n+2}x^{n+s} (n+s+2)(n+s+1)-\frac{1}{4}\sum_{n=0}^{\infty}a_nx^{n+s}+K\sum_{n=1}^{\infty}a_{n+1}x^{n+s}=0[/itex].
    I separate the first term of the series in order to get the indicial equation.
    Hence [itex]Ka_0 x^{s-1}+\sum_{n=0}^{\infty}x^{n+s} [a_{n+2}(n+s+2)(n+s+1)-\frac{1}{4}a_n+Ka_{n+1}]=0[/itex].
    So that the indicial equation is [itex]Ka_0=0[/itex]. I think this implies that [itex]a_0=0[/itex].
    Setting the argument of the infinite series equal to 0 gives me the recurrence relation [itex]a_{n+2}(n+s+2)(n+s+1)-\frac{a_n}{4}+Ka_{n+1}=0[/itex].
    Now I know I should get all terms in function of [itex]a_1[/itex], but I don't know how to mathematically show this.
    For example if I calculate [itex]a_2[/itex], I get that it's worth [itex]-\frac{Ka_1}{(s+1)(s+2)}[/itex]. I'd get [itex]a_3[/itex] in terms of [itex]a_2[/itex] and [itex]a_1[/itex] and thus of [itex]a_1[/itex] alone, etc.
    So how can I proceed further?

    EDIT: Nevermind, I've made lots of errors, I'm going to redo all.
     
    Last edited: Dec 8, 2011
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  3. Dec 8, 2011 #2

    fluidistic

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    Yes I had made some errors.
    The indicial equation is [itex]x^{s-2}a_0 s(s-1)=0[/itex]. Now I know that [itex]a_0[/itex] can't be 0 so that s(s+1)=0 (indicial equation) which gives [itex]s=0[/itex] or [itex]s=1[/itex].
    I also get [itex]a_1=-\frac{Ka_0}{s(s+1)}[/itex].
    And again a non beautiful recurrence relation: [itex]a_{n+2}= \left ( \frac{a_n}{4}-Ka_{n+1} \right ) \frac{1}{(n+s+1)(n+s+2)}[/itex].
    I'm still confronted to the same problem as before, how to express [itex]a_{n+2}[/itex] in terms of [itex]a_{n}[/itex] only.
    Hmm now that I think about it, s can't be 0 otherwise [itex]a_1[/itex] isn't definied. So that s=1.

    I guess I have to find a patern by calculating a few first coefficient [itex]a_n[/itex]. I'm trying this, but this doesn't come easily to me.
     
    Last edited: Dec 8, 2011
  4. Dec 8, 2011 #3

    vela

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    When you calculate y'', the index should still start at n=0. Keep in mind that s isn't necessarily an integer, so you can't assume differentiation will cause certain terms to vanish.

    Try using the approach used on

    http://quantummechanics.ucsd.edu/ph130a/130_notes/node236.html

    where you factor out the asymptotic behavior first, and then look for a series solution.
     
  5. Dec 8, 2011 #4

    fluidistic

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    Hi vela, yes I'm aware of my mistake for n=0, n=2. I've redone everything and fixed it for post 2.
    Also I'm not really understanding Frobenius's method. Why a solution of the form [itex]\sum _{n=0}^{\infty}a_n x^{n+s}[/itex] rather than just [itex]\sum _{n=0}^{\infty}a_n x^{n}[/itex]?
    Also I wonder why in all my problems I never get [itex]a_{n+1}[/itex] in function of [itex]a_n[/itex] but instead I get [itex]a_{n+2}[/itex] in function of both [itex]a_{n+1}[/itex] and [itex]a_n[/itex] which doesn't simplificate things out. In all the examples I've read on the internet, they always get [itex]a_{n+1}[/itex] in function of [itex]a_n[/itex].
     
  6. Dec 9, 2011 #5

    vela

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    It's because the coefficient of y in the differential equation is singular at x=0, and you're expanding your series about this point. If the coefficients weren't singular at the point you're expanding about, you could use the plain old Taylor series-type expansion.

    It's because you're trying to solve hard equations.

    One trick you can try is to factor out certain behaviors of the solution. In this problem, when x gets large, the differential equation becomes approximately
    [tex]y''-\frac{1}{4}y = 0,[/tex] which has solutions [itex]e^{\pm x/2}[/itex]. Because you want the solution to vanish at infinity, you eliminate the positive-exponent solution and posit that the solution can be written in the form [itex]y=f(x)e^{-x/2}[/itex]. Substitute this back into the original differential equation to find the differential equation for f, and then try solving for f using the Frobenius method. This time you will find a recursion relation that's easier to deal with.
     
  7. Dec 9, 2011 #6

    fluidistic

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    Ah ok thank you very much, very clear explanation.
    The DE then becomes [itex]ce^{-x^2/2} \left [ \frac{f(x)}{4}-f'(x)+f''(x) \right ]- \left [ \frac{1}{4}+\frac{K}{x} \right ] f(x)ce^{-x^2/2}=0[/itex].
    After simplificating things out, this is equivalent to the DE [itex]f''(x)-f'(x)-\frac{Kf(x)}{x}=0[/itex]. I'm going to try Frobenius's method on this DE. I hope I didn't make any mistake.
     
  8. Dec 9, 2011 #7

    fluidistic

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    After a lot of algebra I reach that the indicial equation gives [itex]s=0[/itex] or [itex]s=1[/itex] but I can discard s=0 because otherwise [itex]a_1[/itex] cannot be definied.
    I get that [itex]a_{n+1}= \frac{a_n (n+K)}{n(n+1)}[/itex].
    I've calculated the first 3 terms, setting [itex]a_0=1[/itex] for convenience.
    I noticed a patern and I think that in general, [itex]a_n=\frac{(K+n)!}{n!(n+1)!(K+1)}[/itex].
    So that [itex]f(x)=\sum_{n=0}^\infty \frac{x^n}{(K+1)} \frac{(K+n)!}{n!(n+1!)}[/itex].
    But instead of considering [itex]y(x)=f(x)e^{-x/2}[/itex], I considered [itex]y(x)=f(x)ce^{-x/2}[/itex].
    Let me know if you think I made an error, so that I post all my work.
     
  9. Dec 9, 2011 #8

    vela

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    This should be
    [tex]a_{n+1} = \frac{n+K+1}{(n+1)(n+2)}a_n[/tex]but otherwise your results look fine. Now in the large-n limit, you have
    [tex]\frac{a_{n+1}}{a_n} \sim \frac{1}{n}[/tex]which is the same behavior as the series for ex. This means if your solution for f(x) is an infinite series, f(x) will behave like ex when x is large, which is exactly what you don't want because that means y(x) ~ ex/2, which diverges as x→∞. So you don't want f(x) to be an infinite series. How can you achieve this?
     
  10. Dec 9, 2011 #9

    fluidistic

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    Whoops for what I wrote. I copied from my draft the wrong relation (I get yours for s=1), which would be valid in case s=0 was allowed.
    Abour your question, maybe truncate the series... but I don't know up to which n. I could keep I think in principles any finite power of x. But what's the difference if I truncate after n=3 and n=120, I don't really know. Do I get a better "approximation" of the solution? If so, I don't really understand why taking an infinite series doesn't lead to the "real" solution. I know it will diverge and that if I truncate it it won't diverge but no more than this.
     
  11. Dec 9, 2011 #10

    vela

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    You're not finding the solution. You're finding a whole set of solutions. If you truncate at n=3, that's one solution, for a certain value of K=K3. If you truncate at n=120, that's a different solution, for a different value of K=K120. The point is that you can find these truncated solutions for only certain values of K. These are exactly the Ks you're looking for. It's only for these values of K that you can find solutions which satisfy the boundary conditions.
     
  12. Dec 10, 2011 #11

    fluidistic

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    Ok I see.
    Let's say I truncate at n=0 in order to look for [itex]K_0[/itex]. This means [itex]f(x)=\frac{K_0!}{K_0+1}[/itex]. So [itex]y_0 (x)=ce^{-x/2}\frac{K_0!}{K_0+1}[/itex].
    So [itex]y''_0(x)=\frac{c}{4}e^{-x/2}\frac{K_0!}{K_0+1}[/itex].
    Plugging this into the original DE, I reach [itex]ce^{-x/2}\frac{K_0!K_0}{x(K_0+1)}=0[/itex]. This means [itex]K_0=0[/itex]. Is this possible?

    I asked a friend about this problem yesterday, he said I should get 2 functions as solution because it's a DE of second order. He said I should truncate the infinite series when n is such that [itex]a_{n+1}=0[/itex] but [itex]a_n \neq 0[/itex]. So in this case it would be [itex]n=-s-K[/itex]. He also said we use Frobenius method when we want to know how the solution function behaves in the singular points. Something you can't get with a simple Taylor series (this comments makes sense to me).
    I'd appreciate if you could comment on what he told me... he's just a student like me.
     
  13. Dec 10, 2011 #12

    vela

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    If the series truncates at the first term, you have f(x) = a0x1. (Remember your solution is for s=1.) Therefore, [itex]y_0 = xe^{-x/2}[/itex]. If you plug this into the original DE, you get (1+K)e-x/2=0, which is satisfied when K=-1. This is consistent with what your friend said below, namely that K=-(s+n), which is equal to -1 when n=0.

    You don't get a second solution because you eliminated it by requiring the wave function remain finite at the origin and at infinity.

    Yup, that's right.
     
  14. Dec 10, 2011 #13

    fluidistic

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    Ok thank you vela for these explanations.
    I've some troubles however to calculate [itex]K_0[/itex].
    [itex]a_{n+1}=a_n \frac{(n+s+K)}{(n+s)(n+s+1)}[/itex]. I take s=1 so [itex]a_{n+1}=a_n \frac{(n+1+K)}{(n+1)(n+2)}[/itex] in agreement with you.
    Now if [itex]a_0=1[/itex], [itex]a_1=\frac{K+1}{2}[/itex], [itex]a_2=a_1 \frac{K+2}{2\cdot 3}[/itex], [itex]a_3=\frac{(K+1)(K+2)(K+3)}{2\cdot 1 \cdot 2 \cdot 3 \cdot 3 \cdot 4}[/itex].
    In general [itex]a_n = \frac{(K+n)!}{n!(n+1)!K}[/itex] which sligtly differ from my previous post (this current one looks more correct to me).
    Thus [itex]f(x)= \sum_{n=0}^{\infty } \frac{x^n (K+n)!}{n!(n+1)!K}[/itex]. If I truncate this series at the first term ([itex]n=0[/itex]), I get [itex]f(x)=\frac{K_0!}{K_0}=(K_0-1)![/itex].
    So [itex]y_0(x)=ce^{-x/2}(K_0-1)![/itex]. I don't understanding why I don't get your value "x" for f(x).
     
  15. Dec 10, 2011 #14

    vela

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    The numerator would be (K+n)!/K! if K were a non-negative integer, but K actually turns out to be negative.

    It's easier if you simply solve
    [tex]a_{n+1} = \frac{K+n+1}{(n+1)(n+2)} a_n = 0[/tex]for k, assuming an≠0.
     
  16. Dec 11, 2011 #15

    fluidistic

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    Ah ok thank you, I think I finally understand now.
     
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