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Analytically solve second-order coupled ODE (damping term coupled)

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    I need to (analytically) solve a system of coupled second-order ODEs:
    (A) [itex]\frac{du}{dt} - fv = \Omega^2x[/itex]
    (B) [itex]\frac{dv}{dt} + fu = \Omega^2y[/itex]

    where

    [itex]u = \frac{dx}{dt}[/itex]
    [itex]v = \frac{dy}{dt}[/itex]

    subject to the initial conditions [itex]u(t=0) = U[/itex] and [itex]v(t=0) = 0[/itex].

    2. Relevant equations
    ---

    3. The attempt at a solution
    (1) I first converted the ODEs to:
    [itex]\frac{d^2x}{dt^2} - f\frac{dy}{dt} = \Omega^2x[/itex]
    [itex]\frac{d^2y}{dt^2} + f\frac{dx}{dt} = \Omega^2y[/itex]

    and then added and subtracted them to get:
    [itex]\frac{d^2(x+y)}{dt^2} - f\frac{d(x-y)}{dt} = \Omega^2(x+y)[/itex]
    [itex]\frac{d^2(x-y)}{dt^2} - f\frac{d(x+y)}{dt} = \Omega^2(x-y)[/itex]

    then making the substitution
    [itex]\alpha = x + y[/itex]
    [itex]\beta = x - y[/itex]

    which just leads me to the exact problem I started with.

    Since I got stuck here, I tried it a different way...

    (2) Making note that, from (B), [itex]u=\frac{1}{f}(\Omega^2y - \frac{dv}{dt})[/itex]. Plugging into A, we get:
    [itex]\frac{1}{f}(\Omega^2\frac{dy}{dt} - \frac{d^2v}{dt^2})-fv=\Omega^2x[/itex]
    (...after rearranging...)
    [itex]\frac{d^2v}{dt^2} + v(f-\frac{\Omega^2}{f}) = -\Omega^2x[/itex]

    Since [itex]v=\frac{dy}{dt}[/itex], this is now a third order coupled ODE (after we do something similar to above for [itex]u[/itex]), and I don't know how to solve it.

    Does anyone know where to go from here?
     
  2. jcsd
  3. Oct 5, 2014 #2

    Ray Vickson

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    It looks like the easiest way to solve your system is to use Laplace transforms. Let ##X = X(s) = {\cal L}[x(t)](s)## and ##Y = Y(s) = {\cal L}[y(t)](s)##, then get two coupled linear equations for ##X(s),Y(s)##. You would need to use standard results for relating the Laplace transforms of ## x'(t)## and ##x''(t)## to that of ##x(t)##, etc.
     
  4. Oct 5, 2014 #3
    Hi Ray,

    Thank you for your suggestion! It's been quite awhile since I've seen Laplace transforms and had forgotten about them. Unfortunately I ran into a bit of trouble when trying to find a solution...

    ##{\cal L}\{x(t)\}=X(s)##
    ##{\cal L}\{\frac{dx}{dt}\} = sX(s) - x(0) = sX(s)## (with the initial condition ##x(t=0) = 0##)
    ##{\cal L}\{\frac{du}{dt}\} = s^2X(s) - u(0) = s^2X(s) - U##

    and

    ##{\cal L}\{y(t)\}=Y(s)##
    ##{\cal L}\{\frac{dy}{dt}\} = sY(s) - y(0) = sX(s)## (with the initial condition ##y(t=0) = 0##)
    ##{\cal L}\{\frac{dv}{dt}\} = s^2Y(s) - v(0) = s^2Y(s) ##

    Then, taking the Laplace transforms of the coupled ODEs in the problem statement, we get:
    ##s^2X(s) + U - fsY(s) = \Omega^2X(s)##
    ##s^2Y(s) + fsX(s) = \Omega^2Y(s)##

    Isolating ##X## from the second equation, we get ##X(s) = \frac{Y(s)(\Omega^2 - s^2)}{fs}##

    Then, plugging into the first equation,
    ##s\frac{Y(s)(\Omega^2 - s^2)}{fs} + U - fsY(s) = \Omega^2\frac{Y(s)(\Omega^2 - s^2)}{fs}##

    Leading to
    ##Y(s)=\frac{-U}{\frac{\Omega^2s - s^3}{f}-fs-\frac{\Omega^4 - s^2\Omega^2}{fs}} = \frac{fUs}{f^2s^2 + s^4 - 2s^2\Omega^2 + \Omega^4}##

    I don't see how I can decompose this into partial fractions so that I can get ##y(t) = {\cal L}^{-1}\{Y(s)\}##. Have I gone wrong somewhere or is this problem not solvable using Laplace transforms?
     
  5. Oct 5, 2014 #4

    pasmith

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    If you set [itex]z = x + iy[/itex] then your system reduces to the second-order autonomous linear ODE with constant coefficients[tex]
    \frac{d^2 z}{dt^2} + if\frac{dz}{dt} - \Omega^2 z = 0[/tex] subject to [itex]z'(0) = U[/itex]. The standard method of solving such ODEs will still work even though one of the coefficients is not real, so the general solution will be [tex]z(t) = Ae^{k_1 t} + Be^{k_2 t}[/tex] where [itex]A[/itex] and [itex]B[/itex] are complex constants and [itex]k_1[/itex] and [itex]k_2[/itex] are the roots of [tex]k^2 + ifk - \Omega^2 = 0.[/tex] It is possible to obtain four linearly-independent real-valued solutions from this general solution.
     
  6. Oct 5, 2014 #5
    Hi pasmith,

    Thank you for your suggestion! That did the trick, and I was able to come up with a solution for ##x(t)## and ##y(t)##. I forgot to mention in my problem statement that ##f=2\Omega##, so the characteristic equation ##k^2+ifk-\Omega^2## has a double root, and the solution is of the form ##z(t)=Ae^{rt}+Bte^{rt}##.

    My solutions were
    ##x(t) = Utcos(\Omega t)##
    ##y(t) = -Utsin(\Omega t)##
    which satisfy the initial coupled ODEs.

    Thank you again for your help!
     
  7. Oct 5, 2014 #6

    vela

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    Another possible approach: You have
    \begin{eqnarray*}
    \dot{x} &= u \\
    \dot{y} &= v \\
    \dot{u} &= fv + \Omega^2 x \\
    \dot{v} &= -fu + \Omega^2 y
    \end{eqnarray*} which you can solve using matrix methods.
     
  8. Oct 6, 2014 #7
    Hi vela, I only know how to solve the matrix problem numerically. Do you know how I can get an analytical solution with matrix methods?
     
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