= Solution
Assume $\varepsilon>0$. At an <equilibrium>, $w=\gamma v=f(v)$, so
$$
v\bigl[(a-v)(v-1)-\gamma\bigr]=0.
$$
The nonzero roots are
$$
v_\pm=\frac{1+a\pm\sqrt{(1-a)^2-4\gamma}}2,\qquad w_\pm=\gamma v_\pm.
$$
Consequently, 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 $f'(0)=-a$, the <Jacobian matrix> at the origin is
$$
J_0=\begin{pmatrix}-a/\varepsilon&-1/\varepsilon\\\gamma&-1\end{pmatrix},\qquad
\operatorname{tr}J_0=-\frac a\varepsilon-1<0,\qquad
\det J_0=\frac{a+\gamma}{\varepsilon}>0.
$$
Thus the origin is always <asymptotically stable>. Its <eigenvalues> are
$$
\lambda_\pm=\frac{-(a+\varepsilon)\pm\sqrt{(a-\varepsilon)^2-4\varepsilon\gamma}}{2\varepsilon}.
$$
Writing $\gamma_N=(a-\varepsilon)^2/(4\varepsilon)$, 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 $a=\varepsilon$, 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>.
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