Solution of 2nd order ODE

Discussion in 'Math' started by boks, Nov 10, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
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    Can someone explain to me why the solution of \frac{d^{2}\Phi (\phi)}{d\phi^{2}} = -m_{l}^{2} is \Phi = e^{im_{l}\phi}?
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    You probably know that the derivative of exp(c*x) with respect to x is c*exp(c*x).

    You also need to know that i=sqrt(-1). You may know this, or perhaps you usually use j=sqrt(-1). They tend to use i in physics, and j in electrical engineering.

    Anyway, take the derivative of exp(i*ml*phi) with respect to phi twice and remember that i squared is minus one.
     
  3. Ratch

    New Member

    Mar 20, 2007
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    boks,

    Actually the solution is Phi = -(1/12)*m^4 + C1*m + C2 , where C1 and C2 are arbitary constants. How did you get a complex term for a solution?

    Ratch
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    Ratch, you misread the question. The derivatives are with respect to  \phi , not with respect to  m_l . You may have been thrown off by the typo in the equation.

    That must have been a good vacation you took! :)
     
    Last edited: Nov 17, 2008
  5. Ratch

    New Member

    Mar 20, 2007
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    steveb,

    So what is ml, a constant? And what is the correct equation? I still don't see a sinusoid in the solution.

    Yes, my wife and I went to Hot Springs, Arkansas.

    Ratch
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    This equation shows up in the quantum mechanical solution for the hydrogen atom, and  m_l is the magnetic quantum number. There is a typo in the way boks wrote it, but it was clear from the context what he meant. One issue is the proper amplitude for the solution. There is a customary way to normalize the solution, but I assumed that was not the sticking point.
     
    Last edited: Nov 18, 2008
  7. Mark44

    Well-Known Member

    Nov 26, 2007
    626
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    Based on the problem that was posted, Ratch's solution was a lot closer than the one the OP provided.
    For the given problem, I get
    \Phi = \frac{-m^2\phi ^2}{2}  + C\phi + D

    SteveB, you mentioned that boks had a typo in the equation he posted. My guess is that the correct DE was missing a factor of \Phi on the right side, like so:
    \frac{d^2 \Phi}{d\phi ^2 } = -m^2 \Phi
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    Well, I would say your answer is the correct answer based on the exact equation posted. It's a matter of judgement whether ratch or the OP is closer. Technically they are both wrong.

    Your corrected equation with the  \Phi put in on the right hand side is correct, of course. I guess it's good to point this out for the benefit of anyone that is not familiar with the context of the quantum mechanical equation. When I first answered the question I really didn't even notice. It's like one of those sentences that has has a repeated word, and your mind just doesn't even notice it.
     
  9. Ratch

    New Member

    Mar 20, 2007
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    Mark44,

    If d2(Phi)/phi = -m^2*Phi , then the solution is Phi = A*cos(m*phi) + (B/m)*sin(m*phi)

    Ratch
     
  10. Mark44

    Well-Known Member

    Nov 26, 2007
    626
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    Nope, it's not. Two can play at this pedantic game.:)
    The solution is
    \Phi = A cos(m\phi) + B sin(m\phi)
    To show that this function is the general solution, take the first and second aking derivatives.
    \frac{d}{d\phi}\Phi = -Am sin(m\phi) + Bm cos(m\phi)
    \frac{d^2}{d\phi ^2}\Phi = -Am^2 cos(m\phi) - Bm^2 cos(m\phi)
    = -m^2(A cos(m\phi) + B sin(m\phi)) = -m^2\Phi

    In addition, the linear combination of cosine and sine of the solution can also be written as a linear combination of exponential functions (of im\phi), similar to what boks originally posted.
     
  11. Ratch

    New Member

    Mar 20, 2007
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    Mark44,

    Oh yes it is. Our solutions are equivalent. Your "A" is what I designate as "B/m" and your "B" is what I designate as A. They are all constants, albeit different constants. Each solution's double derivative equates to -m^2*Phi . I will admit that your expression of the solution looks nicer.

    Would you be so kind as to give me a quick example. Make A and B any nontrivial value, and show the equivalency of the exponential sums.

    Ratch
     
  12. steveb

    Senior Member

    Jul 3, 2008
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  13. Mark44

    Well-Known Member

    Nov 26, 2007
    626
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    I'll take your word for the equivalence of our solutions. For an example,
    here's what I was talking about, which was that sine and cosine of the same whatever could be expressed in terms of exponentials. Acos(x) + B sin(x) would be just A times the first thing on the right + B times the second thing on the right. See the wikipedia article on Euler's formula at http://en.wikipedia.org/wiki/Euler's_formula.
    cos x = \frac{e^{ix} + e^{-ix} }{2}
    sin x = \frac{e^{ix} - e^{-ix} }{2i}
     
  14. Ratch

    New Member

    Mar 20, 2007
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    Mark44,

    OK, thanks.

    Ratch
     
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