Finding the central manifold of a dynamical system

Question

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

Answer 1

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.

Answer 2

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.