It is asked to analyze the singularities of the system $$\dot{x} = y e^y$$ $$\dot{y} = 1-x^2$$
I've found that the singularities are (1,0) and (-1,0)
The linearization of the sysyem give the matrix \begin{bmatrix} 0 & e^y + ye^y \\ -2x & 0 \end{bmatrix} We can only conclude something for the equilibris (-1,0), which is a saddle, since the eigenvalues are real. For the other point, we can conclude anything using this.
What should I do? A lyapunov function doesnt seems obvious, and, plotting the nullclines didnt told me muh( I couldnt interpretate well). Can someone tell me tne better way to analyze this?
Thanks!
The energy $$H(x,y)=3(y-1)e^y+x^3-3x$$ is constant on the solutions of the differential system. Furthermore, $H(1,0)=-5$ and $H\geqslant-4.21$ on the unit circle centered at $(1,0)$.
Thus, every solution starting from some initial condition $(x_0,y_0)$ in the unit disk centered at $(1,0)$ and such that $H(x_0,y_0)=h$ with $h$ in $(-5,-4.21)$, stays in this disk and on the level line $H(x,y)=h$. Thus, every such solution has a closed orbit. This proves that $(1,0)$ is a center of the non linearized system.