Let $$x'=1+y-x^2-y^2+af_1(x,y)$$ $$y'=1-x-x^2-y^2+af_2(x,y)$$ where $a>0$ is very small.
What is a perturbation $af(x,y)$ that preserves periodic orbits but makes it asymptotically stable?
What I've done:
So we perturb the system
$$x'=1+y-x^2-y^2$$
$$y'=1-x-x^2-y^2$$
A periodic orbit of this is given by $(x,y)=(\cos t,-\sin t)$ (found this earlier), so we must have
$$-\sin t=-\sin t+af_1(\cos t,-\sin t)$$
$$-\cos t=-\cos t+af_2(\cos t,-\sin t)$$
So we want
$$f_1(\cos t,-\sin t)=f_2(\cos t,-\sin t)=0$$
However, how do I find such $f_1,f_2$ s.t. we obtain asymptotic stability? I am supposed to determine asymptotic stability by finding the Floquet exponents.
Edit:
Following LutzL's answer below I want to compute the Floquet multipliers. For that we have to compute the Jacobian of $F_1+\lambda(F_2-F_1)$, then plugging in our periodic solution $(\cos t,-\sin t)$ in the Jacobian $J$, we find the multiplier to be
$$m=\exp\left(\int_0^{2\pi}\text{tr }J dt\right)$$
I found
$$J_{11}=\frac{\partial x'}{\partial x}=-2x+\lambda(1-x^2-y^2)-2\lambda x(x-1)$$
$$J_{22}=\frac{\partial y'}{\partial y}=-2y+\lambda(1-x^2-y^2)-2\lambda y(y-1)$$
But plugging in our periodic solution $(\cos t,-\sin t)$ and integrating from $0$ to $2\pi$ gives multiplier 1, wherea I need $<1$ for asymptotic stability.
$$ x'=\ \ y+x(1-x^2-y^2)\\ y'= -x+y(1-x^2-y^2) $$ has the unit circle as stable solution. A convex combination of both vector fields should shift your system towards stability. As $$(1-\lambda)F_1+λF_2=F_1+λ(F_2-F_1)$$ you can write this convex combination also as perturbation.