# Center manifold of sets of equilibria

My question is regarding Center Manifolds containing a continuous set of equilibrium points. The theory I have studied talks about the existence of a center manifold for equilibrium points, but what happens if we do not have an isolated equilibrium point but a continuous set of equilibrium points? Let me give a toy example:

$\dot x=-y^2\\ \dot y=-y^2x\\ \dot z=-z+y^2$

The $x$-axis is a continuous set of equilibrium points, the linear part is (all along this line) $A=\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{bmatrix}$

So my interpretation is that there exists a 2-dimensional center manifold which contains the $x-$axis. Is this correct? Do you know a source for a proof?

Next if I would like to compute this center manifold? Does power series still apply? This is, can I propose an expression of the form:

$z=\displaystyle\sum_{i,j}a_{ij}x^iy^j$?

I guess it can´t be that "easy" since we would like to capture that the center manifold, at least in this case, is tangent to the 2-dimensional center space all along the $x-$axis. Anyway I am stuck here...

Thanks for any help in understanding this. Literature for reference is also appreciated.

The book of Jack Carr "Applications of Centre Manifold Theory" seems to be a good reference for problems related to center manifolds. In the example you provide, a center manifold does exist and it is of dimension 2. It is given locally by the graph of some function $h:\mathbb{R}^2 \longrightarrow \mathbb{R}$ satisfying $h(0)=0$ and $\textrm{d}f(0)=0$.
• But, as I understand, the function you write is tangent to the centre space only at the origin. Shouldn't it be then $h:\mathbb{R}^2\to\mathbb{R}$ of the form $h(x,0)=Dh(x,0)=0$??? Maybe I am complicating myself too much :) – PepeToro Dec 6 '13 at 9:21
• The system is written in $\mathbb{R}^3$ which writes $\mathbb{R}^2\times\mathbb{R}$ where $\mathbb{R}^2\times\{0\}$ is the center subspace. – Saïd Naciri Dec 21 '13 at 11:05