Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (c)/Solution 1

Let $0<R<1$. Solving for the steady state $0=R-\frac{S}{2+S}$, we obtain $S=\frac{2R}{1-R}$. For $S$ very close to $0$, we would have $R-\frac{S}{2+S}$ close to $R$, which is positive, so $\frac{dS}{dt}>0$ for $S<\frac{2R}{1-R}$. On the other hand, for $S$ very large, we would have $\frac{S}{2+S}$ close to $1$, so $R-\frac{S}{2+S}$ is negative, and therefore $\frac{dS}{dt}<0$.

The situation is described by the following phase diagram: