- #1
KFC
- 477
- 4
I have a set of ODE of the following form
<br /> \begin{cases}<br /> \displaystype{\frac{dx(t)}{dt}} = F(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm]<br /> \displaystype{\frac{dy(t)}{dt}} = G(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm]<br /> \displaystype{\frac{dz(t)}{dt}} = H(z, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})<br /> \end{cases}<br />
where \delta, \Delta, \omega 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 (\delta, \Delta are very small number. So when \delta \to 0 and \Delta \to 0, the steady solutions are x^{(0)}, y^{(0)}, z^{(0)}. Take x as example, the first order corrections of the steady solution is of the form
x = x^{(0)} + y^{(1)} \delta e^{i\omega t} + z^{(1)} \Delta e^{-i\omega t}
I wonder why the above steady solution is time dependent? In this sense, can I conclude that y^{(1)}, z^{(1)} are time independent?
<br /> \begin{cases}<br /> \displaystype{\frac{dx(t)}{dt}} = F(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm]<br /> \displaystype{\frac{dy(t)}{dt}} = G(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm]<br /> \displaystype{\frac{dz(t)}{dt}} = H(z, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})<br /> \end{cases}<br />
where \delta, \Delta, \omega 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 (\delta, \Delta are very small number. So when \delta \to 0 and \Delta \to 0, the steady solutions are x^{(0)}, y^{(0)}, z^{(0)}. Take x as example, the first order corrections of the steady solution is of the form
x = x^{(0)} + y^{(1)} \delta e^{i\omega t} + z^{(1)} \Delta e^{-i\omega t}
I wonder why the above steady solution is time dependent? In this sense, can I conclude that y^{(1)}, z^{(1)} are time independent?