mathematics//differential equations//ODE//autonomous system

A system is autonomous when the rule that generates the field does not contain time: \(\dot x=f(x)\). The velocity at a state is the same whenever you pass through it.

A system is autonomous when the rule that generates the field does not contain time: x˙=f(x)\dot x=f(x)x˙=f(x). The velocity at a state is the same whenever you pass through it.

r˙=ar−brf,f˙=crf−df.\dot r=ar-brf,\qquad \dot f=crf-df.r˙=ar−brf,f˙​=crf−df.

Rabbits rrr and foxes fff: rabbits breed on their own and get eaten when they meet foxes; foxes grow by eating and die on their own. Nothing in the rule knows what time it is. The two interaction terms carry the encounters, and the whole picture is one fixed sea of arrows in the (r,f)(r,f)(r,f) plane. That is a three-dimensional autonomous field once you add a third species, and still a fixed field.

Trajectories are still parametrized by ttt, in the autonomous case as much as in any other. The difference is only whether ttt appears in the rule: the clock tells you where you are along the path, it does not change the rules of the terrain (time).

In a non-autonomous system, x˙=−x+sin⁡t\dot x=-x+\sin tx˙=−x+sint for instance, the sea of arrows itself moves: the same state has a different velocity at a different instant, and the field has effectively one more dimension (external forcing).

Two consequences of autonomy: trajectories never cross, because a state has one velocity (initial condition), and any non-autonomous system can be made autonomous by adding ttt as a state with t˙=1\dot t=1t˙=1, at the price of that extra dimension.

The same split exists in discrete probabilistic form: a time-homogeneous Markov chain, whose transition matrix does not change from step to step, is autonomous (the rule is fixed and only the state moves), and a chain with a time-varying matrix is the non-autonomous case.

equationẋ = 1.0 x (1 − x) startx(0) = 0.08 One equation, a family of solutions: every curve follows the slopes of the field, and the starting point picks which curve is the solution.

Drag the start of the logistic equation and the copper curve, the same start two units later, stays the green one moved right; switch to the forced equation and the two part.