Computing center manifolds Consider the following system $x' = y, y'=-y+3x+xy$.
find the center manifold and by converting to coordinates. Is the origin stable, semi-stable, or unstable?
I am confused if I am doing it correctly or if I answered the question.
For linearization, $\frac{\partial x'}{\partial x} = 0$, $\frac{\partial x'}{\partial y} = 1$
$\frac{\partial y'}{\partial x} = 6x + y$, $\frac{\partial y'}{\partial y} = -1 + x$.
So, $J(x, y)= \begin{bmatrix} 0 & 1\\ 6x+y & -1+x \end{bmatrix}$
$J(0, 0) = \begin{bmatrix} 0 & 1\\ 0 & -1 \end{bmatrix}$
$\begin{bmatrix} x'\\y'\end{bmatrix} = \begin{bmatrix} 0 & 1\\ 0 & -1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$
$x' = y$ and $y'=-y$
Eigenvalues of matrix are $\lambda (\lambda + 1) = 0 \Longrightarrow \lambda = 0, 1$.
 A: Assuming that $\dot y = -y + 3x^2 + xy$, the work is correct, but the 0 eigenvalue begs for a center manifold reduction. I will leave some of the details for you to fill in.
First, we can show that the eigenvector of $\lambda = 0$ is $(1,0)$, which implies that we should seek an approximation of the center manifold given by $y = y(x)$. Differentiation gives $\dot y = y'(x) \dot x $ and so
$$ -y + 3 x^2 + x y = y'(x)y. $$
Let's look for a power series representation $y = \sum_{n=2}^\infty a_n x^n$ (why can I skip the first two terms?). Plugging this into the above expression leads to
$$ x \sum_{n=2}^\infty a_n x^n + 3 x^2 - \sum_{n=2}^\infty a_n x^n = \left(\sum_{n=2}^\infty a_n n x^{n-1} \right) \left(\sum_{m=2}^\infty a_m x^m \right). $$
We balance terms to find the first few coefficients:
\begin{align*}
& \mathcal O(x^2): \quad 3 - a_2 =0 \\
& \mathcal O(x^3): \quad a_2 - a_3= 2a_2^2
\end{align*}
This implies that $a_2 = 3$ and $a_3 = -15$ and so the center manifold is approximated by $y(x) = 3x^2 - 15 x^3$ for $x \ll 1$. Plugging this into the equation for $\dot x$ we see that
$$\dot x = 3x^2 + \mathcal O(x^3)$$
thus the point is actually a saddle.
