Hi I'm stuck with this problem at first I didn't know how to begin so I copy an argument from [Carr, Application of Centre Manifold Theory]. But I don't know how can I find the coefficient from a, b and c? and also I don't understand the theory very well, it's very complicated to me. I'd appreciate if someone can help me with a detailed answer. Thank you
For the following system give an explicit formula for the centre manifold, what happening in the origin?
\begin{align*} \dot{x}&=-x+y^2-2x^2\\ \dot{y}&= \epsilon y-xy\end{align*}
The linearized problem has eigenvalues $-1, \epsilon$. This means that the results in the centre manifold theory does not apply directly. Now if we write the above equation as \begin{align*} \dot{x}&=-x+y^2-2x^2\\ \dot{y}&= \epsilon y-xy\\ \dot{\epsilon} &=0\end{align*}
We considered as an equation in $\mathbb{R}^3$ and the term $\epsilon x$ in the equation is nonlinear. Thus the linearized problem associated has eigenvalues $-1, 0,0$. Now the theory applies and by the centre manifold theorem for some $\delta_i>0$ for $i=1,2$ there is a map $h(y,\epsilon)=x$ for $|x|<\delta_1$ and $|\epsilon|<\delta_2$.
So we have $\dot{x}=h_y(y,\epsilon) \dot{y}$, where $h_y$ is the partial derivative with respect to y. From this, it follows that $h$ satisfies the functional equation
$$h_y(y,\epsilon)[\epsilon y-h(y,\epsilon)y]+h(y,\epsilon)-y^2+2h^2(y,\epsilon)=0$$
We have an expansion $h(y,\epsilon)=ay^2+by\epsilon+c\epsilon^2+\ldots+$ from here we have
\begin{align*} 0&=h_{y}(y, \epsilon)\dot{y}-\dot{x}\\ &=h_{y}(y, \epsilon)(\epsilon y -xy)+h(y, \epsilon)-y^{2}+2h^{2}\\ &= h_{y}\epsilon y-h_{y}xy+h-y^{2}+2h^{2}\\ &= (2ay+b\epsilon)\epsilon y-(2ay+b\epsilon)hy+h-y^{2}+2h^{2}\\ &= 2a\epsilon y^{2}+b\epsilon ^{2}y-(2ay+b\epsilon)(ay^{2}+by\epsilon +c\epsilon ^{2})y+ay^{2}+by\epsilon +c\epsilon ^{2}-y^{2}+2h^{2}\\ &= 2a\epsilon y^{2}+b\epsilon ^{2}y-(2ay^{2}+b\epsilon y)(ay^{2}+by\epsilon +c\epsilon ^{2})+ay^{2}+by\epsilon +c\epsilon ^{2}-y^{2}+2(ay^{2}+by\epsilon +c\epsilon ^{2})(ay^{2}+by\epsilon +c\epsilon ^{2})\\ &= 2a\epsilon y^{2}+b\epsilon ^{2}y-(2a^{2}y^{4}+2ab\epsilon y^{3}+2ac\epsilon ^{2}y^{2}+ab\epsilon y^{3}+b^{2}\epsilon ^{2}y^{2}+bc\epsilon ^{3}y)+ay^{2}+by\epsilon +c\epsilon ^{2}-y^{2}+2(a^{2}y^{4}+aby^{3}\epsilon +ac\epsilon ^{2}y^{2}+aby^{3}\epsilon +b^{2}y^{2}\epsilon ^{2}+bc\epsilon ^{3}y+ac\epsilon ^{2}y^{2}+bcy\epsilon ^{3}+c^{2}\epsilon ^{4}) \end{align*}
But from here we can find $b=c=0$ and $a=1$. Thus $h(y,\epsilon)=y^2 + \ldots$ and plugging this in the $y$ equation we have $\dot{y} =\epsilon y- y^3 + \ldots $ and when $\epsilon= 0$ the system is stable. Am I right?
Thanks