# Existence of periodic orbit of a nonautonomous system

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

• 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 $\epsilon$. – 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.