Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 1 / B2

For the squared-denominator population map, a fixed point satisfiesThere is always the extinction fixed point $\bar{x}=0$. For a positive fixed point, divide by $\bar{x}$ and use $b+\bar{x}>0$ to obtainAt equality the candidate merges with zero and is not positive.

Differentiate the squared-denominator population map:At the positive fixed point, $b+\bar{x}=\sqrt{r}$, so its fixed-point multiplier isSince $0<b/\sqrt{r}<1$, we have $-1<F^{\prime }(\bar{x})<1$. A small perturbation ${\epsilon }_t$ obeys ${\epsilon }_{t+1}=F^{\prime }(\bar{x}){\epsilon }_t+O({\epsilon }_t^2)$, and hence shrinks locally. Thus the positive fixed point is locally asymptotically stable by fixed point stability for an iteration.

This is a stage-structured population model with a census once per maturation interval. Only adults present at census $t$ contribute to the juvenile cohort at census $t+1$; each contributes $r$ new juveniles on average. Thus $r$ is an effective fecundity, including any survival of offspring needed to reach the next census. Juveniles present at $t$ do not contribute a separate birth term during this interval.

The term $c{y}_t$ consists of juveniles that survive and mature into adults by the next census, so $c$ is their survival-to-maturity fraction. The term $(1-d)z_t$ consists of surviving adults; $d$ is the adult mortality fraction per interval. All surviving juveniles leave the juvenile compartment in one step, adults remain adults, and there is no immigration or density dependence. The reproduction term is applied to adults counted at the start of the interval, rather than only to those surviving to its end.

Eliminate the juvenile variable from the stage-structured population model. Since $y_{t+1}=2z_t$,The characteristic polynomial of this linear recurrence isso $z_t=A(3/2{)}^t+B(-1{)}^t$. The initial conditions give $z_0=0$ and $z_1=(3/4)10=15/2$, hence $A+B=0$ and $(3/2)A-B=15/2$. ThereforeThe alternating term records the initial imbalance between life stages; the growing $(3/2{)}^t$ term dominates eventually.

At a nonzero fixed point of the nonlinear stage-structured population model, the adult balance gives $d\bar{z}=c\bar{y}$. Substitute $\bar{y}=d\bar{z}/c$ into the juvenile equation and cancel $\bar{z}>0$:ThusBoth coordinates are positive precisely when $\boxed{cr>d{b}^2}$.

The Jacobian matrix at the extinction fixed point iswith characteristic polynomialUnder $cr>d{b}^2$, $p(1)=d-cr/b^2<0$, whereas $p(q)\to \infty$ as $q\to \infty$. Hence its positive eigenvalueis greater than one. Its eigenvector can be taken with both components positive: choose $(r/(b^2{q}_{+}),1{)}^T$. Thus there are biologically admissible positive perturbations that grow. The extinction fixed point is unstable by linear stability analysis for a discrete dynamical system.