Consider the system

$$\left\{ \begin{array} & x' & = & y\\ y' & = & -2y-x^3+1/\sqrt{27} + \epsilon \cos t\\ \end{array}\right.$$

Prove that, if $0<\epsilon \ll 1$, there exists a $2\pi$-periodic orbit of the system.

I don't know how to proceed. The professor solved another similar problem, and he used this argument:

Let $\phi (t;0,x_0,y_0,\epsilon)$ be the solution of the system (with initial conditions $x_0,y_0$), then there exists a $2\pi$-periodic orbit with initial condition $(x_\epsilon, y_\epsilon ) = (0,0) $ if the function

$$G(x_0,y_0,\epsilon)=\phi (2\pi;0,x_0,y_0,\epsilon)-(x_0,y_0)^t$$

has a zero (because the ODE has no fixed points). And then he uses the implicit function theorem.

I don't understand why this works, and I don't know how to use that in this exercise, because I don't have the initial condition $x_\epsilon, y_\epsilon$.

Any help is welcome

  • 4
    $\begingroup$ Note that at $\epsilon=0$ your system has a fixed point $(x,y)=(1/\sqrt3,0)$. Now analyse what happens to the Poincare return map near this point for nonzero but small $\epsilon$. $\endgroup$ – user10354138 Nov 14 '18 at 5:04

Thanks you. So, if $P:\mathbb{R}^3 \to \mathbb{R}^2$, $P(x,y,\epsilon ) =\phi(2\pi,x,y,\epsilon )$ is the Poincare return map and I define $G(x,y,\epsilon )=P(x,y,\epsilon) - (x,y)^t $, then $G(1/\sqrt 3 , 0 , 0 ) = P(1/\sqrt 3,0,0) -(1/\sqrt 3,0)^t=0$ because $(1/\sqrt 3,0)$ is a fixed point for $\epsilon =0$. With that, (using my professor's method) I check that G verify the Implicit-FT, so there are neighbourhoods of $(1/\sqrt 3 , 0 , \epsilon ) $ such that $G(1/\sqrt 3 , 0 , \epsilon )=0, \forall \epsilon $. Finally, I deduce that there exists a periodic orbit with initial condition $(1/\sqrt 3,0)$.

But actually I don't understand why the existence of zeros of $G$ in a neighbourhood of the fixed point guarantees the existence of a periodic orbit.

Thanks again for your help


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