# Determine a stable and a center manifold for the rest point

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?

• I ask for verifying my results since these things are new to me. – Rhjg Jan 17 '16 at 23:22
• So you keep asking the same thing. Why do you think that "the center manifold is not [unique]"? – John B Jan 18 '16 at 0:14
• @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. – Evgeny Jan 18 '16 at 5:53
• @Evgeny You really need to study, what you say above shows that you have no idea what you are talking about. – John B Jan 18 '16 at 9:21
• @Jonas, sorry, only the first sentence was addressed to you. The latter part was addressed to Rhjg and I forgot to mention this explicitly. – Evgeny Jan 18 '16 at 9:46