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Can a basic knowledge of perturbation theory solve this?

  1. Oct 19, 2013 #1
    Hello all,

    I have boiled a very long physics problem down to the point that I need to solve the coupled equations

    [tex] \frac{\partial^2 x}{\partial u^2} + xf(u) + yg(u) = 0 [/tex]

    [tex] \frac{\partial^2 y}{\partial u^2} + yf(u) - xg(u) = 0 [/tex]

    We may assume that[tex] |f| ,|g| << 1.[/tex] and that both f and g are periodic on the same interval T, i.e. [tex]f(u) = f(u+T),g(u) = g(u+T)[/tex]
    I was wondering if this is something that could be solved with little knowledge of perturbation theory or if this was a very advanced problem, and I was wondering if anyone had any useful resources. So far I have found that this would be relatively easy to solve if they weren't coupled, but this is a level of complexity which is beyond me

    Thank you.
     
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 20, 2013 #2

    mfb

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    Set a=x+y, b=x-y
    This should decouple your system.
     
  4. Oct 20, 2013 #3
    I tried this substitution but I'm very certain it will not actually uncouple these ode's, unless I am missing a very interesting algebra trick
     
  5. Oct 20, 2013 #4

    mfb

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    Oh sorry, I missed the minus sign in the second equation.
    Hmm... it is impossible to decouple them with linear transformations.

    Edit: Oh, complex numbers are great.
     
    Last edited: Oct 20, 2013
  6. Oct 20, 2013 #5

    pasmith

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    If you set [itex]z = x + iy[/itex] and (EDIT:) [itex]w = f - ig[/itex] then your system becomes
    [tex]
    \frac{d^2 z}{d u^2} + w(u)z = 0
    [/tex]
    which may be easier to analyze.
     
    Last edited: Oct 20, 2013
  7. Oct 20, 2013 #6
    Depending on w(u), it may still be difficult to find the general solution.
    If this second order ODE admits Liouvillian solutions, they can be found using Kovacic' algorithm, you can do this with e.g. Maple.
     
  8. Oct 21, 2013 #7
    Yes you can probably use perturbation theory, but you have to first define your ordering. Its not enough that f and g are small. They have to be small compared to something.

    As an example lets assume [itex] xg /\frac{\partial^2 x}{\partial u^2} [/itex] is order [itex]\epsilon[/itex] small. We also assume that x and y are of comparable magnitude, as are f and g. The validity of these assumptions depend on the problem, and you might have to alter the following approach depending on the problem. And its important to check that your final answer is consistent with this ordering.

    With these assumptions we next expand x (and y) in terms of order [itex]\epsilon[/itex]: [itex] x=\epsilon^0 x_0 + \epsilon^1 x_1 + \epsilon^2 x_2 + \dots [/itex]. Here [itex] \epsilon [/itex] is just an marker of smallness. We also note that [itex] \frac{\partial^2 }{\partial u^2} [/itex] is order [itex]1/\epsilon [/itex] large and and it is helpful to write it as [itex] \frac{1}{\epsilon}\frac{\partial^2 }{\partial u^2} [/itex]

    Now we plug the expansions for x and y into the equation, and equate terms of the same order in [itex] \epsilon [/itex]. This creates a hierarchy of equation that we can solve.

    To lowest order we have:
    [tex] \frac{\partial^2 x_0}{\partial u^2} = 0 [/tex]

    [tex] \frac{\partial^2 y_0}{\partial u^2} = 0 [/tex]

    The next order equation is
    [tex] \frac{\partial^2 x_1}{\partial u^2} + x_0f(u) + y_0 g(u) = 0 [/tex]

    [tex] \frac{\partial^2 y_1}{\partial u^2} + y_0f(u) - x_0 g(u) = 0 [/tex]

    And all higher order equations are of this form:
    [tex] \frac{\partial^2 x_j}{\partial u^2} + x_{j-1}f(u) + y_{j-1} g(u) = 0 [/tex]

    [tex] \frac{\partial^2 y_j}{\partial u^2} + y_{j-1}f(u) - x_{j-1} g(u) = 0 [/tex]

    Note that in each equation for [itex] x_j,y_j [/itex] the terms [itex] x_{j-1},y_{j-1} [/itex] are known from the previous equation.
     
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