Past exam of the differential equations 1 course of the University of Oxford / 2021 / 3 / b / iii / Solution

One suitable autonomous planar system isIts equilibrium points are exactly the four combinations $(x,y)=(\pm 1,\pm 1)$. Its Jacobian matrix isThus $(-1,-1)$ and $(1,1)$ have eigenvalues $\pm 2$ and are saddle equilibria. At the other two points the eigenvalues are $\pm 2i$. To establish the nonlinear centre equilibrium classifications, use the first integralIts Hessian matrix is $\mathrm{diag}(2x,-2y)$, a negative-definite matrix at $(-1,1)$ and a positive-definite matrix at $(1,-1)$. Thus each is a strict local extremum, and nearby regular level sets are closed curves supporting periodic orbits, by the planar first integral criterion for a centre.

Immediately to the right of $(-1,1)$, the vector field points down because $x^2-1<0$, so that centre is clockwise. Immediately to the right of $(1,-1)$ it points up because $x^2-1>0$, so that centre is counterclockwise. This verifies both the nonlinear classifications and the prescribed orientations, with exactly four equilibrium points.