Suppose we have some ODE of the form $$\dot{x}=f(\alpha,x),$$ where $\alpha \in \mathbb{R}^k$ is some vector of parameters. And for $\alpha \in A$ we have $x=e_{\alpha}$ stable hyperbolic equilibriums for the corresponding equations. Now, lets change $\alpha$ for $\alpha (t)$ with $\alpha (t) \in A,\forall t$. Can we ensure (under some conditions on $f$ and $\alpha (t)$) that the equilibrium "will turn" into a trajectory $e(t)$ which attracts solutions (locally) of the non-autonomous equation $\dot{x}=f(\alpha(t),x)$ (forward or in a pull-back sense)?
In particular, I'm working with a system for which I know the equilibriums for $\alpha \in A$. And if I change $\alpha$ for a periodic function, I can prove (using degree theory) that a periodic solution exists, provided $\alpha (t) \in A$ and the amplitude of oscillation is small. Is this solution a "perturbation" of the equilibrium? does it preserve the attractiveness?
I hope the question is well posed, Thanks!
Edit: Thank you RPA for your answer. My problem is actually a little bit more complex because my system involves time delays. I could write the equations here if you are interested. But basically I have 2 equations which depend on the parameters $\gamma$, $K$, and $\delta$. If they satisfy certain inequalities its not hard to solve for the non trivial equilibrium. And for $\gamma(t)$, $K(t)$, and $\delta(t)$ T-periodic, if the same inequalities hold $\forall t$ then I can probe existence of a non trivial periodic solution. So I believe that if the amplitude of this functions gets small, then this periodic solution should also have small amplitude until it becomes a point. But I don't know how to see this.
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, \begin{equation} \mathbf{\dot x} = A(t) \mathbf{x} = \begin{bmatrix} -1 & e^{2 t} \\ 0 & -1 \end{bmatrix}\bf x \nonumber \end{equation} 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, $$\Phi[t,0] = \begin{bmatrix} e^{-t} & 0.5(e^t-e^{-t}) \\ 0 & e^{-t} \end{bmatrix}$$ So the solution is $x(t) = \Phi[t,0]x(0)$ which is not stable because of the $e^{t}$ term.
So I guess we would need a bit more info on your system.