Recently, I dealt with determining stable/ unstable/ center manifold. Here is one task.
Determine a stable and a center manifold at the rest point of the system $$ \dot{x}=x^2,\qquad\dot{y}=-y. $$
I think this is not that difficult here. The linearization matrix is $$ A=\begin{pmatrix}0 & 0\\0 & -1\end{pmatrix} $$ so the eigenvalues are $\lambda_1=0,\lambda_2=-1$.
I think a center manifold is given by $$ W^c: y=h(x)\text{ with }h(0)=h'(0)=0\text{ and } \dot{y}=h'(x)\dot{x} $$ So, formally, one can make the start $h(x)=ax^2+bx^3+...$ and determine the coeffcients. Here, I simply get $h(x)\equiv 0$ so that the center manifold is simply $$ W^c: y=0 $$ Similarly, I get for the stable manifold that $$ W^s: x=g(y)=0. $$
Am I right?
I know that the stable manifold is unique but the center manifold is not. What would be another center manifold?
Perhaps it is appropriate to note explicitly the following: in some cases the center manifold is unique.
This comment (promoted to answer, since there is nothing more to add) applies both to local center manifolds and global center manifolds, even irrespectively of the space in which we look for the functions of which the manifold is a graph.