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I solved many cases for the following dynamical system $\dot{x} = x (1-x-ay)$ and $\dot{y} = c y (1- b x -y)$. However, I reached the case where $c>0$ and $a>1$, $b=1$ and I ended up with the Jacobian of the fixed point $(1,0)$, which is

$$A_{(1,0)} = \begin{pmatrix} -1 & -a\\ 0 & 0 \end{pmatrix}$$

It is clear that we have a zero real part eigenvalue, so we need to find the center manifold. Here is the problem, every time i assume $y=h(x)=c_1 x^2 + c_2 x^3 +...$ I get all the coefficients zero! What is the problem

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    $\begingroup$ You are computing the wrong invariant manifold. y=0 is invariant, but it is not the center manifold. $\endgroup$
    – Michael Renardy
    Commented Jan 31, 2022 at 14:19
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    $\begingroup$ Don't mean to nitpick, but in what sense is this a symbolic dynamics question? $\endgroup$
    – Ronnie Pavlov
    Commented Feb 4, 2022 at 14:08
  • $\begingroup$ @MichaelRenardy I had exam in this question and I have not solve it till now! How can I determine the stability of the equilibria? $\endgroup$
    – Mr. Proof
    Commented Feb 5, 2022 at 0:58
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    $\begingroup$ @SelfLearner: The center manifold is tangent to the eigenvector for the zero eigenvalue. The manifold you are computing it tangent to the other eigenvector! You may also want to read up on what symbolic dynamics is. $\endgroup$
    – Michael Renardy
    Commented Feb 5, 2022 at 2:41
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    $\begingroup$ I really don't mean it rudely, but perhaps you shouldn't use a tag for a topic that you're not familiar with? You could learn more here if curious: scholarpedia.org/article/Symbolic_dynamics $\endgroup$
    – Ronnie Pavlov
    Commented Feb 9, 2022 at 17:45

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The center manifold for the first problem is not $y = O(x^2)$. It is tangent to the center subspace of the linearized operator, which is not $x$-axis in this case. Please find an eigenvector associated with zero eigenvalue and change the coordinates so that the center subspace corresponds one of the coordinate axes. In the new coordinates, you will get the answer soon.

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  • $\begingroup$ Could you show me the full solution please. $\endgroup$
    – Mr. Proof
    Commented Feb 9, 2022 at 1:47

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