Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 5A / b / Solution

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.