 Solution

At any equilibrium, the first balance gives $u=\mu$ and the second gives $f(\mu ,v)=0$. For every fixed $u>0$,andThe intermediate value theorem and strict monotonicity give precisely one positive root $v=g(u)$ for each $u>0$. Since $f_v\ne 0$, the implicit function theorem makes this a smooth nullcline. For every $\mu >0$, therefore,is the unique positive equilibrium. Graphically, draw the vertical line $u=\mu$ through the given graph: it meets the graph once, even on its steep descending section. This vertical line is the condition for simultaneous stationarity; the full $u$ nullcline at finite $\epsilon$ is actually $f(u,v)=(\epsilon /\gamma )(\mu -u)$.