I must prove that the differential equation below does not have an analytic centre manifold:
$$ \dot{x}=x^3, \dot{y}=2y-2x^2 $$
I try:
The linearisation of the system at the origin is: $$\dot{x}=DX(0)x= \left( \begin{matrix} 0 & 0 \\ 0 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) $$
and find the eigenvalues with the corresponding eigenvectors:
$\lambda_1=0$: $(1,0)$
$\lambda_2=2$: $(0,1)$
and from the Centre Manifold Theorem, I find that there exist locally a centre manifold containing the origin, and tangent to the vector $(1,0)$ at the origin.
My question is, then... How do I prove that I can not find it analytically?
The orbits of my system are: $$ \frac{\dot{y}}{\dot{x}}=\frac{2y-2x^2}{x^3} $$
Have I understood correctly the problem? What should I do?
Thank you
Looking for analytic functions $y(x)=\sum\limits_{n\geqslant0}a_nx^n$ such that $$x^3y'(x)=2(y(x)-x^2),$$ and $y(0)=y'(0)=0$, one gets the conditions $a_0=a_{2n+1}=0$ for every $n\geqslant0$, $a_2=1$, and $a_{2n+2}=na_{2n}$ for every $n\geqslant1$, hence $$y(x)=\sum_{n\geqslant0}n!\,x^{2n+2}.$$ The radius of convergence of this series is zero hence the center manifold is not analytic at $(0,0)$.
streamplot[{x^3,2y−2x^2},{x,-2,2},{y,-1,2}]