Lyapunov Stability Let $\dot{x}=v(x)$ with $v(x)=Ax+O(\left \| x \right \|^2)$, $v\in C^k(U)$, $U\subset \mathbb{R}^n$, $  n\geq 2$, is true or not that if the origin is a singular point Lyapunov stable for $\dot{x}=Ax$? then is a point Lyapunov stable for $\dot{x}=v(x)$
I think this is false, let 
${x}'=-y+x^3 \\ {y}'=x+y^3$
so we have the origin is a singular point Lyapunov stable for $\dot{x}=Ax$ because if I take $g(x,y)=\frac{x^2}{2}+\frac{y^2}{2}+1>0$ then I have $\bigtriangledown g(x,y)(-y,x)=0$ then the origin is Lyapunov stable, then I would find another $g´$ such that $g´(x,y)({x}',{y}')>0$  bye Lyapunov´s Stability theorem I will have that the origin is not stable, well that´s my idea, I just want to know if I'm on the right way
Thanks for any comments!
 A: If $\dot x = Ax$ is stable at the origin in the sense of Lyapunov, there exists a positive definite $P$ matrix such that $AP + A^TP \leq 0$.
For the case where the Lyapunov equation is $AP + A^TP < 0$, If $v(x)$ can be written as $v(x) = Ax + O(||x||^2||)$, where $O(||x||^2) \to 0$ when $||x|| \to 0$ (Taylor series residue), then the Jacobian of the system $\dot x = v(x)$ about the origin is $A$, which is Hurwitz and therefore the system is stable about the origin. Note that there is not any information about the region of convergence.
On the other hand, if you can not find such $P$ for $AP + A^TP < 0$, but the system is stable in the sense of Lyapunov, you have that $AP + A^TP = 0$. Therefore, $A$ is not Hurwitz, but its eigenvalues have negative real part and zero real part (they are on the imaginary axis). Therefore, checking the stability of the non-linear system $\dot x = v(x)$ via the Jacobian fails, and you have to go for other analysis tool such as Centre Manifold theory.
For the first case, the system is stable about the origin. In the second case you can not assert that the system is stable or unstable only checking the linearization of the system about the origin.
