# Steady solution of ODE

1. May 17, 2009

### KFC

I have a set of ODE of the following form

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

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?

2. May 18, 2009

### HallsofIvy

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!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook