 Solution

A simple zero eigenvalue occurs on $c=d$, except at $(1,1)$. The odd symmetry of the vector field gives a pitchfork bifurcation. Near this curve the critical eigenvalue isFor $0<d<1$ the other two eigenvalues are stable: this is a supercritical pitchfork bifurcation as $c$ decreases through $d$. For $d>1$ it is a subcritical pitchfork bifurcation in the center direction, with an already unstable transverse direction.

For a Hopf bifurcation, factor the characteristic polynomial as $(m+\tau )(m^2+{\omega }^2)$. Comparing coefficients givesThus the trivial branch has the local bifurcation curvesThe second curve is a Hopf bifurcation with a stable third eigenvalue. The Routh-Hurwitz stability criterion gives the stable trivial region $0<d<1$, $d<c<(d-2{)}^2$. Its Hopf bifurcation is subcritical. To check the sign, normalize the critical right eigenvector by $q_x=1$ and putThe second component of the normalized left projection is $G$. There is no quadratic term at the origin, and the cubic derivative in the second component is $C_2(q,q,\bar{q})=2$. The cubic Hopf coefficient is therefore $l_1=\mathrm{Re}G/\omega =4/[\omega ((4-d{)}^2+4{\omega }^2)]>0$. A small repelling periodic orbit lies on the stable-origin side. The diagram also includes the nonzero-branch information derived next.