 Solution

Assume $\epsilon >0$. At an equilibrium, $w=\gamma v=f(v)$, soThe nonzero roots areConsequently, with ${\gamma }_c=(1-a{)}^2/4$, there are three distinct equilibria for $0<\gamma <{\gamma }_c$, two distinct equilibria for $\gamma ={\gamma }_c$, and only the origin for $\gamma >{\gamma }_c$. At equality the two nonzero equilibria coincide; for $0<\gamma <{\gamma }_c$ they both have $a<v<1$. Geometrically, ${\gamma }_c$ is the maximum of the concave quadratic $(a-v)(v-1)$.

Since $f^{\prime }(0)=-a$, the Jacobian matrix at the origin isThus the origin is always asymptotically stable. Its eigenvalues areWriting ${\gamma }_N=(a-\epsilon {)}^2/(4\epsilon )$, the origin is a stable node for $0<\gamma <{\gamma }_N$ and a stable focus for $\gamma >{\gamma }_N$. At $\gamma ={\gamma }_N>0$ it is a degenerate node: the repeated negative eigenvalue has only one independent eigenvector, since the nonzero off-diagonal entry prevents $J_0$ being a scalar matrix. If $a=\epsilon$, then ${\gamma }_N=0$ and every permitted $\gamma >0$ gives a stable focus. This change from node to focus does not change stability and is not a Hopf bifurcation.