Find the stability of $O(0,0)$ for the different values of the parameters a,b.
The system of differential equations is:
$$x'=-x+y+x^2+ax^3$$ $$y'=x-y+ax^2+bxy$$
ATTEMPT:
$F=(F_1,F_2):R^2\rightarrow R^2$
$F(0)=0$, $DF(0)=0$ Consider that:$$x'=-x+y+F_1(x,y)$$ $$y'=x-y+F_2(x,y)$$ with $$F_1(x,y)=x^2+ax^3$$ and $$F_2(x,y)=ax^2+bxy$$
$$\iff$$ $$\begin{bmatrix}x' \\ y' \end{bmatrix}=\begin{bmatrix} -1 &1 \\ 1&-1 \end{bmatrix}+F(x,y)$$
Then we have that :$y(0)=y'(0)=0$ and $x-y+F_2(x,y)$ From these 2 equations we have that $y(x)=\dfrac{-x-ax^2}{bx-1}$ with $b\ne 0$ and $x \ne \dfrac{1}{b}$
For the stationary point i have that $y(0)=0$
So,$$F_1(x,y(x)=\dfrac{-x-ax^2}{bx-1})=-x+\dfrac{-x-ax^2}{bx-1}+x^2+ax^3$$ From this point i don't know how to proceed so to find the kind of stability for the point $O(0,0)$ for the different values of a and b. Am i right until now or no?
I would really appreciate a thorough solution and explanation, since I've just started working on these problems and I have to clear my mind on them.
Thanks in advance !
As in the linked posts, you will need to change the variables so that the "fast" and "slow" coordinates are separated, on in other words, you need coordinates along the center manifold and perpendicular to it. As before, these are $$ u=x+y,\quad v=x-y $$ so that \begin{align} u'&=(1+a)\frac{(u+v)^2}4+b\frac{u^2-v^2}4+O((u,v)^3)\\ v'&=-2v+(1-a)\frac{(u+v)^2}4-b\frac{u^2-v^2}4+O((u,v)^3) \end{align} The zero set of the second equation is in first order given by $2v=O((u,v)^2)$ so that $v=O(u^2)$. Thus the next-lowest order terms give $$v=\frac18(1-a-b)u^2.$$ By the same reasoning the lowest order terms for the first equation give $$u'=\frac14(1+a+b)u^2+O(u^3)$$ which only gives interesting results if the coefficient is zero so that the quadratic terms in $u$ cancel.