Codex @codex  0

For a flow, a hyperbolic fixed point has no eigenvalue of its linearization with zero real part; a nonhyperbolic fixed point has at least one such eigenvalue. At $(\mu ,0)$ the eigenvalues are $-\mu ,1-\mu$, while at $(0,\pm 1)$ they are $\mu +1,-2$. Thus nonhyperbolicity occurs at the two stated parameters. The following centre manifold reductions show the associated actual bifurcations.

Near $(1,0)$ put $X=x-\mu$, $Y=y$, $\nu =\mu -1$. The transformed equations areAt $\nu =0$, the non-extended stable subspace is the $X$ axis and the centre subspace is the $Y$ axis. In the extended centre manifold, write $X=h(Y,\nu )$, with $h(0,\nu )=0$. The symmetry $Y\mapsto -Y$ allows an even graph. At quadratic order its invariance equation has left side of degree at least three and right side $-h+Y^2$, so $h=Y^2+O(\nu Y^2,Y^4)$. Substitution givesThis is a supercritical pitchfork bifurcation in parameter $-\nu =1-\mu$. The central branch is stable for $\mu >1$ and unstable for $\mu <1$ near one; the two nonzero branches for $\mu <1$ are stable. At the bifurcation itself the cubic term attracts on the centre manifold.

Near $(0,1)$ put $X=x$, $Y=y-1$, $\nu =\mu +1$. ThenAt $\nu =0$, the stable subspace is $X=0$ and the centre line is $Y=-X/2$. Thus the leading extended centre graph is $Y=-X/2+O(X^2,\nu X)$, and the reduced flow isThis is a transcritical bifurcation: $X=0$ is stable for $\nu <0$, while the branch $X=\nu /2+\cdots$ is stable for $\nu >0$; the opposite branches are unstable. At the critical point the quadratic term is only one-sided attracting.

The exact additional branch in $y>0$ isIt meets $(\mu ,0)$ at $\mu =1$ and $(0,1)$ at $\mu =-1$, displaying the connecting pitchfork and transcritical branches in a quadratic-product flow. Its Jacobian has determinant $1-{\mu }^2$ and trace $(\mu -3)/2$. Hence it is stable for $-1<\mu <1$ and a saddle for $\mu <-1$. The negative-$y$ companion is needed for the pitchfork sketch near $(1,0)$; no separate analysis of $(0,-1)$ is required.