Consider an equation of the form

,

where

is periodic with period 1, that is,

. The particular function

that we examine is taken from [1]:

.

Because the differential equation is non-autonomous in

, the solution trajectory, denoted by

, depends on both

and

.

By selecting the "time trajectory" tab and varying the sliders

and

, you can explore the solution's dependence on these two parameters. In all cases the solution approaches a 1-periodic solution

, that is:

as

.

Another interesting property of 1-periodic equations is

.

By selecting "amplitude shifted" and varying the sliders

,

, and

, you can examine how the trajectory

is affected by comparing it with

. As noted earlier, all trajectories ultimately approach a 1-periodic state.

By selecting "time-amplitude shifted" and taking

, you can see that the trajectories

and

are simply translates of each other in time.

By integrating the 1-periodic differential equation in time, the solution trajectory always approaches a periodic state, if that state is stable. It turns out for this 1-periodic equation that there is another 1-periodic state, but it is not stable and so cannot be found by time integration. One method for investigating the family of periodic states is to examine the Poincaré map drawn from the solution trajectories. By selecting "Poincaré map", you can examine the Poincaré map for the differential equation. You can vary the iterations on the map with the slider

The map is constructed by finding the

iterates of

.

The intersection of the Poincaré map with the line

gives the fixed points for the map. In this case there are two fixed points that are 1-periodic solutions to the differential equation. The intersection near

is an unstable fixed point. The stability of a fixed point can be deduced from the slope of the Poincaré map at the intersection point or by computing the Floquet exponents, which is done in this Demonstration. Negative exponents represent stable solutions.