 Solution

A steady-state bifurcation is a change in the local equilibrium branches or their stability as a parameter varies. At such a bifurcation the Jacobian matrix at the equilibrium has a zero eigenvalue, so the implicit function argument for a unique persistent local equilibrium fails.

For this system,At $(0,\mu )$ its eigenvalues are $1-{\mu }^2,-\mu$; at $(1,0)$ they are $-2,\mu -1$. Both have one simple zero eigenvalue at $\mu =1$, with the other eigenvalue negative. Interior equilibrium branches, obtained below, meet these boundary equilibria there, establishing the actual bifurcations rather than merely a necessary eigenvalue test.