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Steady solution of ODE

  1. May 17, 2009 #1

    KFC

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    I have a set of ODE of the following form

    [tex]
    \begin{cases}
    \displaystype{\frac{dx(t)}{dt}} = F(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm]
    \displaystype{\frac{dy(t)}{dt}} = G(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm]
    \displaystype{\frac{dz(t)}{dt}} = H(z, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})
    \end{cases}
    [/tex]

    where [tex]\delta, \Delta, \omega[/tex] are constants.

    If only concern about the steady solution, can I conclue that the solution must be time-independent?

    The equations is quite complicate so one must consider the small pertubration ([tex]\delta, \Delta[/tex] are very small number. So when [tex]\delta \to 0[/tex] and [tex]\Delta \to 0[/tex], the steady solutions are [tex]x^{(0)}, y^{(0)}, z^{(0)}[/tex]. Take x as example, the first order corrections of the steady solution is of the form

    [tex]x = x^{(0)} + y^{(1)} \delta e^{i\omega t} + z^{(1)} \Delta e^{-i\omega t}[/tex]

    I wonder why the above steady solution is time dependent? In this sense, can I conclude that [tex]y^{(1)}, z^{(1)}[/tex] are time independent?
     
  2. jcsd
  3. May 18, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A "steady state solution" is by definition a solution constant in time. Yes, it is independent of time.

    Again, the definition of "steady state solution" is that it is time independent!
     
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