Calculate the center manifold of a 2-dimensional system I need to calculate the center manifold $W^c((0,0))$ up to and including terms of order 3, and determine the flow near (0,0), for the following system:
$$
\left\{
\begin{array}{c}
\dot{x} = \beta xy-x^3+y^2 \\ 
\dot{y} = -y+x^2+xy
\end{array}
\right. 
$$
where $\beta \in \mathbb{R}$. I'm struggling to find a solid example of how to go about finding the center manifold. My textbook lacks worked examples, and the few examples found here either don't go into detail or appear to be special cases. If anyone could please step me through the process or direct me to a well explained worked example so I can learn to utilise the method myself.
Thanks in anticipation of your responses!
 A: The linearized system has $\dot x=0$ and $\dot y=-y$, so that the "fast" part of the dynamical system converges towards the $x$ axis, which is the null-space of the kernel of the Jacobian. 
Sufficiently far away from the $x$ axis, this remains the dominant behavior, for $|x|\ll |y|$ the leading terms of the equations are
$$
\dot x=y^2,~~~\dot y=-y\\
\implies y(t)=y_0e^{-t},~~ x(t)=x_0-\frac12y_0^2(e^{-2t}-1)
$$
giving a fast movement along the quadratic curve $x=x_0-\frac12(y^2-y_0^2)$ towards the point $(x_0+\frac12y_0^2,0)$ on the $x$-axis.
For the slow motion close to the $x$-axis one expects a negligible dynamic in $y$ direction, so that in a first approximation $0=\dot y=-y+x^2+xy$ or $y=\frac{x^2}{1-x}\approx x^2+x^3$ for the slow or center manifold. Insert $y=\phi(x)=x^2+a_3x^3+a_4x^4+O(x^5)$ into the differential equations to find 
$$
\left\{\begin{aligned}
\dot{x} &= \beta (x^3+a_3x^4+a_4x^4+..)-x^3+(x^4+2a_3x^5+(a_3^2+2a_4)x^6+..) \\ 
(2x+3a_3x^2+4a_4x^3+..)\dot x &= -(a_3x^3+a_4x^4+..)+(x^3+a_3x^4+a_4x^5+..)\\ 
(2+3a_3x+4a_4x^2+..)\dot x &= -(a_3x^2+a_4x^3+..)+(x^2+a_3x^3+a_4x^4+..)
\end{aligned}\right.
$$
so that comparing terms in both equations gives that there can not be second order terms on the right side of the last equation, so $a_3=1$. Inserting this coefficient and expanding one gets
$$
\left\{\begin{aligned}
\dot{x} &= (β-1)x^3 +(β+1)x^4+(βa_4+2)x^5+.. \\ 
(2+3x+4a_4x^2+..)\dot x &= (1-a_4)x^3+(a_4-a_5)x^4+..
\end{aligned}\right.
$$
Equality in the lowest degree terms then demands $(1-a_4)=2(β-1)$, $(a_4-a_5)=2(β+1)+3(β-1)$, so that $a_4=3-2β$, $a_5=4-7β$.

The direction of the slow flow close to the origin depends on the sign of $(β-1)$, for $β>1$ it is away from the origin, for $β<1$ towards the origin, for $β=1$ it is increasing outside a smaller neighborhood of $x=0$. 
