Let be

$$ \left\{ \begin{array}{lcc} \dot{x_1}=-ax_1^2+bx_1x_2+cx_2^2 \\ \\ \dot{x_2}=-x_2+dx_1^2+mx_1x_2+nx_2^2 \end{array} \right.$$ a system; $a,b,c,d,m,n\in \mathbb{R}$ constants.

$1.$ If $a\neq0$ then $\bar{x}=0$ is inestable.

$2.$ If $a=0$ and $bd<0$ then $\bar{x}=0$ is asymptotically stable.

$3.$ If $a=0$ and $bd>0$ then $\bar{x}=0$ is unestable.

$4.$ If $a=b=0$ and $cd\ne0$ then $\bar{x}=0$ is unestable.

$5.$ If $a=b=c=0$ then $\bar{x}=0$ is stable but not asymptotically stable.

$6.$ If $a=d=0$ then $\bar{x}=0$ is stable but not asymptotically stable.

Let be $f_1(x_1, x_2)=-ax_1^2+bx_1x_2+cx_2^2$ and $f_2(x_1, x_2)=dx_1^2+mx_1x_2+nx_2^2.$ The idea is to find a function $\phi(x_1)$ such that $-\phi(x_1)+f_2(x_1, \phi(x_1))=0$ and $\phi(0)=0,\phi'(0)=0$ and apply the following result:

Theorem: Let's suppose that $f_1(x_1, \phi(x_1))=px_1^k+\Theta(|x_1|^{k+1})$ when $x_1\to 0$,$p\ne 0$ and $k\in \mathbb{N}.$

Then the equilibrium $\bar{x}=0$ of system is asymptotically stable if $p<0$ and $k$ odd and unstable otherwise.

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    $\begingroup$ What is your question exactly ? $\endgroup$ – Rebellos Nov 27 '17 at 19:31
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    $\begingroup$ @MrYouMath I wouldn't say that all of these questions are totally different. All of them would be answered after finding center manifold of this system — it seems that this is the reason for introducing $\phi(x_1)$ although conditions for it are a bit strange to me. Also you can't linearize in "easy case", but it's easy to get analytical solution in that case. $\endgroup$ – Evgeny Nov 28 '17 at 5:32
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    $\begingroup$ @Evgeny: You are right that linearization does not work in the easy case. $\endgroup$ – MrYouMath Nov 28 '17 at 6:36
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    $\begingroup$ Yeah, but you can solve this system of linear equations easily. $\endgroup$ – Evgeny Nov 29 '17 at 20:38
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    $\begingroup$ There is a theorem: if you compute Jacobi matrix at equilibrium and all eigenvalues satisfy ${\rm Re} \; \lambda \leqslant 0$, then the stability of equilibrium only depends on center manifold. Center manifold is tangent to the subspace spanned by eigenspaces with ${\rm Re} \; \lambda = 0$. $\endgroup$ – Evgeny Nov 30 '17 at 4:33

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