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Second order DE (hydrogen atom)

  1. Oct 28, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    The DE [itex]y''+\frac{2}{x}y'+ \left [ K+\frac{2}{x} - \frac{l(l+1)}{x^2} \right ]y=0[/itex], [itex]0<x< + \infty[/itex]. appears when working on the hydrogen atom. Find all the values of K (the eigenvalues) that generates solutions of the form [itex]\phi (x)[/itex] such that [itex]\phi (x)[/itex] remains finite when x tends to 0.

    2. Relevant equations

    I don't know.

    3. The attempt at a solution
    I don't know how to start. There's nothing said about the l's but I guess that they are natural numbers. Also why the solutions are phi's rather than y's, I don't know.
     
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  3. Oct 30, 2011 #2

    fluidistic

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    Can someone help me on this problem? It interests me a lot.
    It seems I can't really use the method of undetermined coefficients because the terms in front of y' and y aren't constant. I can't use either the method of variation of parameters since I don't know any solution of the -homogeneous DE-, in fact I want to solve the homogeneous DE.
    I see that I have an equation of the form [itex]y''+a(x)y'+b(x)y=0[/itex]. Where b is a somehow special function in the sense that it also depends on K and l.
     
  4. Oct 31, 2011 #3

    fluidistic

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    Ok I asked a friend about this problem, he told me to use power series solutions.
    I've never done this before and I'm stuck near the end. I'd appreciate some guidance.
    I proprose a solution of the form [itex]y(x)=\sum _{n=0}^{\infty} a_n x^n[/itex]. I derivated this expression twice. Plugging back into the orignal DE and setting the 3 power series with the same starting "n" (in my case n=0), I reach [itex]\sum _{n=0}^{\infty} x^n \{ (n+2)(n+1)a_{n+2}+\frac{2}{x}(n+1)a_{n+1}+ \left [ K+\frac{2}{x}-\frac{l(l+1)}{x^2} \right ] a_n \} =0[/itex].
    Since this is worth 0 for all x, I can conclude that [itex](n+2)(n+1)a_{n+2}+\frac{2}{x}(n+1)a_{n+1}+ \left [ K+\frac{2}{x}-\frac{l(l+1)}{x^2} \right ] a_n[/itex] must equal 0.

    Doing this I reach the recurrence relation [itex]a_{n+2}= \frac{\left [\frac{l(l+1)}{x^2}-\frac{2}{x}-K \right ]}{(n+2)(n+1)} a_n-\frac{2}{x(n+2)} a_{n+1}[/itex].

    I set [itex]n=0[/itex] to get [itex]a_2[/itex] in function of [itex]a_1[/itex] and [itex]a_0[/itex] and also [itex]n=1[/itex] to get [itex]a_3[/itex] in terms of [itex]a_1[/itex] and [itex]a_2[/itex].
    But I don't have an idea on how to determine [itex]a_0[/itex] and [itex]a_1[/itex] so that I can really "run" my recurrence relation.
    Any help is appreciated!
     
  5. Oct 31, 2011 #4
    I'm assuming the paramater "l" is an integer so the quantity l(l+1) is either zero or a positive integer. For now, just let [itex]l(l+1)=\nu[/itex]. Let's first write yours as:

    [tex]y''+\frac{2}{x}y'+\left(-\frac{\nu}{x^2}+\frac{2}{x}+K\right)y=0[/tex]

    The guiding principle for solving these is the "indicial" equation which I'll write as:

    [tex]c^2+c-\nu=0[/tex]

    for which we solve for the roots:

    [tex]\{c_1,c_2\}=\frac{-1\pm\sqrt{1+4\nu}}{2}[/tex]

    We then seek a solution of the form:

    [tex]y(x)=\sum_{n=0}^{\infty}a_nx^{n+c},\quad c=\{c_1,c_2\}[/tex]

    Ok, now, can you substitute that sum with just "c" for now, into the DE, adjust all the sum indices so that the powers on x are the same, and then letting a_0 be arbitrary, come up with a recursive relation for a_n for [itex]n\geq 1[/itex]?
     
    Last edited: Oct 31, 2011
  6. Oct 31, 2011 #5

    vela

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    Look up the method of Frobenius if you need more background on solving the differential equation using a series.
     
  7. Oct 31, 2011 #6

    fluidistic

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    Hey thanks for helping me jackmell. However I'm having troubles figuring out how you obtained your indicial or characteristic equation. Could you explain how you obtained it?
    Edit: I just saw your post vela, I'm definitely heading up to Frobenius.
     
  8. Nov 1, 2011 #7
    I'm afraid I caused some confusion above and can't edit. I should have written the DE as:

    [tex]x^2y''+2xy'+\left(Kx^2+2x-\nu\right)y=0[/tex]

    and now, substitute into that form, the expression:

    [tex]y(x)=\sum_{n=0}^{\infty} a_n x^{n+c}[/tex]

    Now, the indicial equation is contrived so that we can set [itex]a_0[/itex] arbitrary. You can see how it's derived in any DE text book. Tell you what, try getting these expressions:

    [tex]a_0: \text{arbitrary}[/tex]

    [tex]a_1=-\frac{2a_0}{c^2+3c-\nu}[/tex]

    [tex]n\geq 2:\quad a_n=-\frac{Ka_{n-2}+2a_{n-1}}{(n+c)(n+c-1)+2(n+c)-\nu}[/tex]
     
    Last edited: Nov 1, 2011
  9. Dec 6, 2011 #8

    fluidistic

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    Ok guys thank you for your help, I'm finally back to this problem (I had to pass a final exam of another course, was too busy on it. Now that I'm done with it, I can concentrate on Mathematical methods and classical mechanics).
    I've just downloaded a free online book on mathematical methods called "Mathematical Tools for Physics" by James Nearing. I've checked out Frobenius's method and indeed it looks like the way to go for my equation.
    Using his notation (replace c by s from jackmell's notation), I get:
    [itex]\sum _{l=0}^{\infty } a_l x^{l+s}[(l+s)(l+s+1)-v]+K\sum _{l=2}^{\infty } a_{l-2} x^{l+s}+2 \sum _{l=1}^{\infty } a_{l-1} x^{l+s}=0[/itex].
    I assume that [itex]a_0 \neq 0[/itex] so that the coefficient in front of the lower power of x (it appear in the first series only, when [itex]l=0[/itex]) must equal 0.
    This gives me the inditial equation [itex][s(s+1)-v]=0[/itex]. Thus [itex]s=\frac{-1 \pm \sqrt {1+4v} }{2}[/itex].
    Now I set the coefficients in front of [itex]x^{l+s}[/itex] equal to 0.
    I get: [itex]a_l [(l+s)(l+s+1)-v]+Ka_{l-2}+2a_{l-1}=0[/itex]. And I'm stuck here. I can get [itex]a_{l+2}[/itex] in terms of [itex]a_{l-1}[/itex] and [itex]a_{l}[/itex].
    So Jackmell, I don't know how to obtain your answer. :frown:
     
  10. Dec 6, 2011 #9

    vela

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    Looks good so far. Separating out the first and second term of the series, you have
    [tex]a_0[s(s+1)-v]x^s + [a_1((s+1)(s+2)-v)+2a_0]x^{s+1} +\sum_{l=2}^\infty [a_l((l+s)(l+s+1)-v)+Ka_{l-2}+2a_{l-1}]x^{s+l} = 0[/tex]Setting the coefficient of the first term equal to 0 yields the indicial equation, which you both agree on, and the general recursion relation, which again you both agree on, comes from setting the coefficient in the summation equal to 0. Jackmell's second equation comes from setting the coefficient of xs+1 equal to 0, though it looks like there's a term missing in the denominator.
     
  11. Dec 6, 2011 #10
    Ok, I made a typo in the equations for the coefficients. They should have been:

    [tex]a_0: \text{arbitrary}[/tex]

    [tex]a_1=-\frac{2a_0}{c^2+3c+2-\nu}[/tex]

    [tex]n\geq 2:\quad a_n=-\frac{Ka_{n-2}+2a_{n-1}}{(n+c)(n+c-1)+2(n+c)-\nu}[/tex]

    and it's not hard to (visually) check by first solving it numerically for some values of the parameters, then comparing that to the series solution. However, using numerical methods, we cannot find a solution valid at the origin because of the singularity of the DE. That doesn't mean the solution there is necessarilly singular though. So I just solved it for y(1)=y1 and y'(1)=y2 and the parameter values below. And using the above expressions, the solutions agree. Here's the code I used.
    Code (Text):

    k = 1;
    v = 1;
    y0 = 1;
    y1 = 1;

    mysol = NDSolve[{x^2*Derivative[2][y][x] + 2*x*Derivative[1][y][x] + (k*x^2 + 2*x - v)*y[x] == 0, y[1] == 1,
        Derivative[1][y][1] == 1}, y, {x, 1, 10}]

    ndsolveplot = Plot[y[x] /. mysol, {x, 1, 10}, PlotRange -> {{1, 10}, {-2, 2}}]

    c1 = (-1 + Sqrt[1 + 4*v])/2.;
    Subscript[a, 0] = 1;
    Subscript[a, 1] = -((2*Subscript[a, 0])/(c1^2 + 3*c1 + 2 - v));
    mya = Table[Subscript[a, n] = -((k*Subscript[a, n - 2] + 2*Subscript[a, n - 1])/((n + c1)*(n + c1 - 1) + 2*(n + c1) -
            v)), {n, 2, 25}];

    c2 = (-1 - Sqrt[1 + 4*v])/2.;
    Subscript[b, 0] = 1;
    Subscript[b, 1] = -((2*Subscript[b, 0])/(c2^2 + 3*c2 + 2 - v));
    myb = Table[Subscript[b, n] = -((k*Subscript[b, n - 2] + 2*Subscript[b, n - 1])/((n + c2)*(n + c2 - 1) + 2*(n + c2) -
            v)), {n, 2, 25}];

    my1[x_] := Sum[Subscript[a, n]*x^(n + c1), {n, 0, 25}]
    my2[x_] := Sum[Subscript[b, n]*x^(n + c2), {n, 0, 25}]

    thek = Solve[{k1*my1[1] + k2*my2[1] == y0, k1*(D[my1[x], x] /. x -> 1) + k2*(D[my2[x], x] /. x -> 1) == y1}, {k1, k2}]

    seriesplot = Plot[k1*my1[x] + k2*my2[x] /. thek, {x, 1, 10}, PlotRange -> {{1, 10}, {-2, 2}}]

    Show[{ndsolveplot, seriesplot}]
     
    Ok, so we got at least the series representation probably ok and to answer the OP's question about how to get the above expressions, well, you just have to substitute the series:

    [tex]y(x)=\sum_{n=0}^{\infty} a_n x^{n+c}[/tex]

    into the DE and solve for them. Work a few simple ones first to get the hang of it. But I do not see how this series solution is going to help determine for what values of K the solution is bounded at the origin except to just run it for a range of k and the parameters and just try to determine this numerically. However, sometimes the series representation can be converted to a managagble function. Not sure about this one though.
     
  12. Dec 6, 2011 #11

    fluidistic

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    Ah ok, that's a very clear explanation, thanks. Though if I set the coefficient of [itex]x^{s+1}[/itex] equal to 0, I get [itex]a_1 [(s+1)(s+2)-v+2a_0][/itex]. Since I assume [itex]a_1 \neq 0[/itex], I can divide by it. This gives me [itex]a_0 = -\left ( \frac{1}{2} \right ) [(s+1)(s+2)+v][/itex]. I already have s in function of v so that I have [itex]a_0[/itex] in function of v. Now I don't see why would [itex]a_0[/itex] be arbitrary, am I missing something?

    Edit: I'm reading your post jackmell, I didn't see it when I posted this one.
     
  13. Dec 6, 2011 #12
    The ability to set a_0 arbitrary comes from a consideration of the indical equation. If you take the first and second derivatives of that power series and substitute it into the DE you'll get after a change of index:

    [tex]\sum_{n=0} a_n(n+c)(n+c-1)x^{n+c}+2\sum_{n=0}a_n(n+c)x^{n+c}+K\sum_{n=2}a_{n-2}x^{n+c}+2\sum_{n=1} a_{n+1}x^{n+c}-v\sum_{n=0} a_n x^{n+c}=0[/tex]

    Now consider just the a_0 term out of all of that. It's

    [tex]a_0(c^2+c-v)=0[/tex]

    Well there you go, either a_0 has to be zero or the other one or both and we don't want a_0 to be zero so we let c^2+c-v (the indical equation) equal to zero. But if that's zero, then a_0 can be anything and the simplest, the one I used above, is to simply let it equal one.
     
  14. Dec 6, 2011 #13

    vela

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    fluidistic, are you sure the differential equation you started with is correct? I checked my quantum textbook, and the equation for the radial function is slightly different.
     
  15. Dec 6, 2011 #14

    fluidistic

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    I don't know if it's correct (doesn't look like correct if you say it isn't) but I'm 100% sure it's the one of an exercise of my course.
    You can see it in the attached file (exercise 1).
     

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  16. Dec 6, 2011 #15

    vela

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    You can't get rid of a1 altogether. You miswrote the coefficient. Only the first part is multiplied by a1, so you can't factor it out of the entire expression.

    Also, you can't assume a1≠0. You can take a0≠0 because by definition it's the coefficient of the lowest-order term of the series. If it were equal to 0, another term would be the lowest-order term of the series. You have no such requirement of a1.

    When you obtain the final answer, you'll find that all of the coefficients are proportional to a0, so a0 is essentially the arbitrary constant that appears when you solve a differential equation.
     
  17. Dec 7, 2011 #16

    vela

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    The indicial equation is [itex]c^2+c - l(l+1)=0[/itex]. You can solve that to find the two values of c, one of which you can exclude on physical grounds.
     
  18. Dec 7, 2011 #17

    fluidistic

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    Ok I finally reach [itex]a_1=-\frac{2 a_0}{(s+1)(s+2)-v}[/itex] and [itex]a_l = \frac{-2 a_{l-1}-Ka_{l-2}}{(l+s)(l+s+1)-v}[/itex] for [itex]l \geq 2[/itex].
    The inditial equation gives me [itex]s=c= \frac{-1 \pm \sqrt {1+4v}}{2}[/itex]. You are saying I can get rid of [itex]\frac{-1 - \sqrt {1+4v}}{2}[/itex] and keep [itex]\frac{-1 + \sqrt {1+4v}}{2}[/itex]? So I discard the negative value of s and take the positive... though I don't know why. I didn't relationate K and v to physical quantities yet. I'm guessing they are related to the energy of the H atom, but I'm not sure.

    Edit: c or s must be positive indeed, I mean I'm not taking a Laurent series or something like that.
     
    Last edited: Dec 7, 2011
  19. Dec 7, 2011 #18

    vela

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    Think about what happens as x approaches 0 if c is negative.
     
  20. Dec 7, 2011 #19

    fluidistic

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    Ok the whole function blows up to infinities and its first and second derivatives too.

    Edit: Am I right if I say that [itex]|K| \leq 2[/itex], to answer the question of the problem?
    This is my guess after examinating the behavior of [itex]a_l[/itex]. But I'm not sure at all of this answer.
     
    Last edited: Dec 7, 2011
  21. Dec 7, 2011 #20

    vela

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    No, that's not correct. The problem is essentially asking you to solve the radial equation and find the allowed energies.
     
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