 Solution

Let $D=Q/V_0$ be the dilution rate. Population loss is $DN$, and nutrient loss by consumption is $NK(C)/V_0$, so mass conservation givesUse the dimensionless variables $\tau =Dt$, $c=C/K_0$, and $n=N{K}_{\max }/(Q{K}_0)$. Substitution into the Monod equation gives the stated system withFor a positive equilibrium, the first equation requires $\alpha c/(1+c)=1$. Hence $c_{\ast }=1/(\alpha -1)$ and $n_{\ast }=\alpha (\beta -c_{\ast })$, both positive exactly under the specified inequalities.

Writing $A=\alpha n_{\ast }/(1+c_{\ast }{)}^2>0$, the Jacobian matrix at this equilibrium isBoth eigenvalues are negative, so the state is locally asymptotically stable. The Monod chemostat equilibrium and relaxation also makes one decay rate transparent: $z=c+n/\alpha$ satisfies $z^{\prime }=\beta -z$, independently of the uptake kinetics.