Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 7E / Solution

The simultaneous zeros are the origin and the four corners $(\pm 1,\pm 1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, with eigenvectors $(1,1),(1,-1)$, so the origin is a saddle equilibrium of Poincaré index $-1$. At a corner $J=-2xyI$. Thus $(1,1)$ and $(-1,-1)$ are stable nodes; $(1,-1)$ and $(-1,1)$ are unstable nodes, all of index $+1$. Although the node eigenvalues coincide, the equilibria are hyperbolic and their stability follows from their strictly signed real eigenvalues.

The lines $x=\pm 1$ and $y=\pm 1$ are invariant and divide the plane into nine convex regions. A nonconstant periodic trajectory cannot cross any of these lines by uniqueness, and cannot be contained in one of them because one-dimensional autonomous flows have no nonconstant periodic solutions. It must therefore lie inside one region. The central region contains only the origin, whose index is $-1$; all other regions contain no equilibria. But a simple periodic orbit has index $+1$, equal to the sum of the enclosed equilibrium indices. Its interior stays in the same convex region, giving a contradiction in either case. There are no periodic orbits.

Away from the invariant lines, trajectory curves have the first integralsince both logarithmic derivatives equal $-2xy$. The diagonal $y=x$ is the unstable separatrix of the origin, leading toward the two sinks; $y=-x$ is its stable separatrix, coming from the two sources. The diagram shows these separatrices, the invariant lines and the direction field.