= Solution
At any <equilibrium>, the first balance gives $u=\mu$ and the second gives $f(\mu,v)=0$. For every fixed $u>0$,
$$
f(u,0)=\frac{u^2}{1+u^2}>0,\qquad
\lim_{v\to\infty}f(u,v)=-\infty,
$$
and
$$
\boxed{f_v(u,v)=-\frac{u^4}{1+u^4}\frac{2v}{(1+v^2)^2}-\delta<0
\quad(v\ge0).}
$$
The <intermediate value theorem> and strict <monotonicity> give precisely one positive root $v=g(u)$ for each $u>0$. Since $f_v\ne0$, the <implicit function theorem> makes this a smooth <nullcline>. For every $\mu>0$, therefore,
$$
\boxed{(u_*,v_*)=(\mu,g(\mu))}
$$
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 $\varepsilon$ is actually $f(u,v)=(\varepsilon/\gamma)(\mu-u)$.
Back to article page
