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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.

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1 Answer 1

up vote 2 down vote accepted

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 $\textrm{d}f(0)=0$.

Moreover, Lemma 1 (page 20) tells you that all solutions converge exponentially fast to the center manifold. Therefore, all equilibria near the origin belongs to any center manifold. Recall that center manifolds are not unique in general.

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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 :) –  user58533 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

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