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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?

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  • $\begingroup$ I ask for verifying my results since these things are new to me. $\endgroup$ – Rhjg Jan 17 '16 at 23:22
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    $\begingroup$ So you keep asking the same thing. Why do you think that "the center manifold is not [unique]"? $\endgroup$ – John B Jan 18 '16 at 0:14
  • $\begingroup$ @Jonas Because every book says that center manifold is not unique. In fact, any trajectory that goes to equilibrium from the left can be stitched with right half of $y =0$ or with trajectory that goes to equlibrium from the right side of $Oy$ axis. And that would be another center manifold. However, it's again a basic fact that needs just careful reading of relevant chapters in a course book or other few books. $\endgroup$ – Evgeny Jan 18 '16 at 5:53
  • $\begingroup$ @Evgeny You really need to study, what you say above shows that you have no idea what you are talking about. $\endgroup$ – John B Jan 18 '16 at 9:21
  • $\begingroup$ @Jonas, sorry, only the first sentence was addressed to you. The latter part was addressed to Rhjg and I forgot to mention this explicitly. $\endgroup$ – Evgeny Jan 18 '16 at 9:46
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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.

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