TITLE: Periodic orbits of "even" perturbations of the differential system $x'=-y$, $y'=x$
QUESTION [24 upvotes]: Fix some even functions $f$ and $g$, differentiable, such that $f(0)=g(0)=0$ and $f'(0)=g'(0)=0$, and consider the autonomous differential system $$\left\{\ \begin{array}{lcr}x'&=&-y+f(x)\\ y'&=&x+g(y)\end{array}\right.$$

My question is whether every solution $t\mapsto(x(t),y(t))$ of this differential system which passes close enough to its fixed point $(0,0)$, is periodic.

If this helps, one can assume that the functions $f$ and $g$ are smooth, or polynomials, and/or that their sign is constant in a neighbourhood of $0$.
By hypothesis, $f(u)$ and $g(u)$ are negligible with respect to $u$ when $u\to0$. Thus, near the origin $(0,0)$, the differential system above is a perturbation of the linear differential system
$$\left\{\ \begin{array}{lcr}x'&=&-y\\ y'&=&x\end{array}\right.$$
Obviously, the solutions of this linear differential system are the circles centered at $(0,0)$, oriented positively.
Simulations based on the cases $f(u)=au^{2n}$ and $g(u)=bu^{2m}$, for various real constants $(a,b)$ and various (small) positive integers $n$ and $m$, seem to support the conjecture (but counterexamples would be welcome, naturally).
A recent question on the site is related to the case $f(u)\propto u^4$ and $g(u)\propto u^6$. Below is a simulation of the phase diagram when $f(u)=u^4$ and $g(u)=3u^6$, which seems to support the conjecture.

REPLY [7 votes]: To calculate the Lyapunov values, one must initially transform the system into the polar coordinates, which for this example takes the form
$$
\frac{dr}{d\theta}=\frac{f(r\cos \theta)\cos \theta+g(r\sin \theta)\sin \theta}{1+\frac{f(r\cos\theta)\sin\theta-g(r\sin\theta)\cos\theta}{r}}=R(r,\theta)=r R_1(\theta)+r^2R_2(\theta)+\ldots.\tag{1}
$$
This equation makes sense since for sufficiently small $r$ the denominator is strictly positive for $0<\theta<2\pi$.  
Let
$$
r=f(\theta,\theta_0,r_0)
$$
be the solution to this equation with the initial conditions $r(\theta_0)=r_0$. Consider the special case $\theta_0=0$ and get (assuming that the right hand sides are analytic)
$$
r=u_1(\theta)r_0+u_2(\theta)r_0^2+\ldots\tag{2}
$$
This function must solve $(1)$. After plugging $(2)$ into $(1)$ we find
$$
\dot u_1=R_1(\theta)u_1,\\
\dot u_2=R_1(\theta)u_2+R_2(\theta)u_1^2,\\
\ldots
$$
Clearly we must have $u_1(0)=1,u_i(0)=0$. Hence we can find all $u_j$ if given $R_j(\theta)$. In particular, for the example in the question $R_1(\theta)=0$, and hence $u_1(\theta)=1$.
By taking $\theta=2\pi$ we find $\alpha_j=u_j(2\pi)$.  So to find $\alpha_5$ one needs to find and consecutively solve 5 simple (but cumbersome) ODE. 
By the way, in the same book there is an expression for $L_2=\alpha_5$ for the system 
$$
\dot x=-y+\sum_{ij}a_{ij}x^iy^j\\
\dot y=x+\sum_{ij}b_{ij}x^iy^j
$$


but it seems to be easier to calculate $L_2$ with the help of diff equations (plus, while having fun working with your problem I found several typos in the book I referenced, so I am not sure whether this expression is 100% is correct).