Question: Is the origin a nonlinear center for the system
$$x'=-y-x^2$$ $$y'=x$$
We have a theorem that states if the system is reversible orbits close to the origin are closed. It seems that this system is not reversible and so I was wondering if the converse holds. Does this imply that the center is not a nonlinear center? any help is appreciated, thanks!
On
This is a reversible system. The general definition of reversibility is first we define a type of mapping $R:X\to X$ for the phase space $X$ that satisfies $R(R(x))= x$ for all $x \in X$. Then any $C^1$ (not sure if $C^1$ is needed but its nice to ensure uniqueness) vector field $f:X\to X$ is reversible if there is some $R$ which satisfy the above and so that $\dot{x} = f(x)$ is invariant under the change of variable $t\to -t$ and $x\to R(x)$.
For your question, to prove reversibility you can define $R(x,y) = (-x,y)$ so that indeed $\dot{x} = (f(x,y),g(x,y))$ given in your example is invariant under the change of variables $(t,x,y) \to (-t,-x,y)$. And since its reversible the origin is a non-linear center.
Of course another fast way to check reversibility if you are using the notion of verifying $f(x,-y) = -f(x,y)$ and $g(x,-y) = g(x,y)$ is to simply swap the variables x and y.
(I took most of this from strogatz non-linear dynamics chapter 6.6)
Eliminate $x$ to get $$ y''+y'^2+y=0. $$ This has an integrating factor $e^y$ giving $$ (e^y)''+e^yy=0 $$ which can be integrated to $$ E=\frac12[(e^y)']^2+V(y),~~ V'(y)=(e^y)'e^yy=e^{2y}yy'=\frac14(e^{2y}(2y-1))' $$ or $$ E(x,y)=\frac12x^2e^{2y}+\frac14e^{2y}(2y-1) $$ All solutions follow the level curves of this first integral.