Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 14A / a / Solution

The equilibrium points are $(0,0)$ and $(\pm 1,\pm (b-a))$, with correlated signs. The Jacobian matrix is $\left(\begin{array}{cc}a-3b{x}^2& 1\\ 3x^2-1& 0\end{array}\right)$. At the origin its eigenvalues satisfy ${\lambda }^2-a\lambda +1=0$. It is a focus for $0<|a|<2$, a repeated-eigenvalue node for $|a|=2$, and a node for $|a|>2$; stability is determined by the sign of $a$, with stability for $a<0$. Its Poincaré index is $+1$. At either other equilibrium the determinant is $-2$, so it is a saddle point with Poincaré index $-1$.