Suppose we have some ODE of the form <mover> x &#x02D9;<!-- ˙ --> </mover> = f (

Daphne Haney

Daphne Haney

Answered question

2022-05-16

Suppose we have some ODE of the form
x ˙ = f ( α , x ) ,
where α R k is some vector of parameters. And for α A we have x = e α stable hyperbolic equilibriums for the corresponding equations. Now, lets change α for α ( t ) with α ( t ) A , t. Can we ensure (under some conditions on f and α ( t ) that the equilibrium "will turn" into a trajectory e ( t ) which attracts solutions (locally) of the non-autonomous equation x ˙ = f ( α ( t ) , x ) (forward or in a pull-back sense)?
In particular, I'm working with a system for which I know the equilibriums for α A. And if I change α for a periodic function, I can prove (using degree theory) that a periodic solution exists, provided α ( t ) A and the amplitude of oscillation is small. Is this solution a "perturbation" of the equilibrium? does it preserve the attractiveness?

Answer & Explanation

verrainellewtzri

verrainellewtzri

Beginner2022-05-17Added 12 answers

In general, having a stable fixed point at every time t does not mean that the time-varying system will be stable. Take for example the second order system,
x ˙ = A ( t ) x = [ 1 e 2 t 0 1 ] x
Taking t as a fixed parameter, the eigenvalues of A ( t ) are ( 1 , 1 ) for all t. But solving for the state transition matrix, one finds that,
Φ [ t , 0 ] = [ e t 0.5 ( e t e t ) 0 e t ]
So the solution is x ( t ) = Φ [ t , 0 ] x ( 0 ) which is not stable because of the e t term.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?