Past exam of the short options course of the University of Oxford / 2021 / 9 / b / i

At a constant steady state $N_t=N$, the discrete equation reduces toWrite $s=\delta +1>1$. For $r<s$, no nonzero real root is possible, so zero is the only real steady state. For $r>s$ the real roots areThus there is exactly one strictly positive steady state, $N^{\ast }=\sqrt{(r-s)/(bs)}$. There are three real steady states in total, two of them nonnegative. At $r=s$ the only one is zero. This distinction keeps the positive-population statement separate from the count over all real numbers.