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.

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.

  • $\begingroup$ @Paul Why not? According to me it is, with period $2\pi$. I.e. $(x(t),y(t))=(x(t+2\pi),y(t+2\pi))$ $\endgroup$ – Richie Nov 25 '17 at 12:43

$$ x'=\ \ y+x(1-x^2-y^2)\\ y'= -x+y(1-x^2-y^2) $$ 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.

  • $\begingroup$ What are the vecotr fields $F_1,F_2$? Are those my $f_1,f_2$? $\endgroup$ – Richie Nov 25 '17 at 13:48
  • $\begingroup$ No, these are the full fields, $F_1=\binom{y}{-x}+(1-x^2-y^2)\binom{1}{1}$ your original one and $F_2=\binom{y}{-x}+(1-x^2-y^2)\binom{x}{y}$ the one I proposed. Your $f$ is the difference, $f=F_2-F_1=(1-x^2-y^2)\binom{x-1}{y-1}$ $\endgroup$ – Lutz Lehmann Nov 25 '17 at 14:08
  • $\begingroup$ Great solution, thanks. $\endgroup$ – Richie Nov 25 '17 at 14:18
  • $\begingroup$ Could you please take a look at the edit, I don't know how to show asymptotic stability $\endgroup$ – Richie Nov 25 '17 at 20:50
  • $\begingroup$ Just adding the terms and using $x^2+y^2=1$ I get $tr(J)=2(λ-1)(x+y)-2λ$. While indeed the first term integrates to $0$, the second constant term integrates to $-4πλ$. $\endgroup$ – Lutz Lehmann Nov 25 '17 at 22:19

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