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?

Question

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 &quot;will turn&quot; 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&#039;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 &quot;perturbation&quot; of the equilibrium? does it preserve the attractiveness?

verrainellewtzri · Accepted Answer

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                      ]    xTaking   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.

Suppose we have some ODE of the form <mover> x ˙ </mover> = f (

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