Finding the central manifold of a dynamical system Take the dynamical system: $$x' = 0.5(1-x)xy$$ $$y' = -y(1-x)^3-y^2(2x^2-1.5x+0.5)-2x(1-x)^4+x(1-x)^3.$$
I want to find the central manifold and deduce its dynamics (stable or unstable). The above system is already in the following required form: $$ x' = Ax + f(x,y)$$ $$y' = By + g(x,y)$$
where necessarily $A=0$ and $B=-1$. Given this, we can parameterise the centre manifold by: $$h(x) = ax^2+bx^3+cx^4 +O(x^5).$$ First, we compute $y' = \frac{dh}{dx}x'$ which is: $$ y' = a^2x^4 + O(x^5)$$ and we compare it with the $y'$ from the above dynamical system, which is: $$y' = -x+(5-a)x^2+(3a-b-9)x^3+(-\frac{a^2}{2}-3a+3b-c+7)x^4 + O(x^5).$$ Comparing coefficients between the two $y'$'s gives $a=5$, $b=6$ and $c=-27.5$. This means that the centre manifold should be parameterised by: $$h(x) = 5x^2+6x^3-27.5x^4 +O(x^5).$$
Question: I do not believe the stated $h(x)$ to be the correct approximation to the manifold. You can see the correct centre manifold in the figure of the phase plane for the system I have attached. If you plot $h(x)$ on something like Desmos, you can clearly see that it is not a good approximation. Can you spot an error in my working or have I not included something I should have? Thanks

 A: Your linearized system is wrong. It should be
$$\left\{ \begin{array}{rcl} x' & = & 0x + 0y + f(x,y) \\
y' &=& -x-y+g(x,y) \end{array} \right.$$
where bith $f$ and $g$ are second order terms. In particular, the center manifold is tangent to $\{x+y=0\}$, i.e. to the vector $(1,-1)$, which is what you observe on the plot. The expression for $h$ has to be changed accordingly.
A: Considering the system
$$
\dot x = f(x,y)\\
\dot y = g(x,y)
$$
It has at the origin the jacobian
$$
J_{0,0}=\left(
\begin{array}{cc}
 0 & 0 \\
 -1 & -1 \\
\end{array}
\right)
$$
with eigenvalues $(-1,0)$ so we have an one-dimensional manifold.
Making $y = h(x) = \sum_{k=1}^n a_k x^k$ and substituting into the $y$ dynamics we have
$$
h'(x)f(x,h(x))=g(x, h(x))
$$
and with $n = 2$ we obtain after grouping powers of $x$
$$
\cases{a_1+1=0\\ \frac{a_1^2}{2}-3 a_1+a_2-5=0}
$$
and after solving
$$
\left\{a_1= -1,a_2=\frac{3}{2}\right\}
$$
and the approximation to the manifold
$$
h(x) = -x+\frac{3 x^2}{2}
$$
The flow along the central manifold is given by
$$
\dot x = f(x,h(x)) = -\frac{3 x^4}{4}+\frac{5 x^3}{4}-\frac{x^2}{2}
$$
which is unstable.

