6 Sex and age structure

Up to now all individuals in a population were treated as identical. But individuals differ: in sex, in age, in size, in where they live. Such differences matter because birth and death rates depend on them. This week we introduce the two most important kinds of structure, sex and age. Both lead to linear models whose long-term behaviour is exponential growth with a stable composition of the population.

Learning outcomes for this week

By the end of this week you should be able to:

formulate sex-structured models with different mating functions and explain their assumptions;
derive the long-term sex ratio from the ODE for \(M/F\);
construct a Leslie matrix from a life table of survival probabilities and birth numbers;
project an age-structured population forward in time by hand;
use eigenvalues and eigenvectors of the Leslie matrix to find the long-term growth factor and the stable age distribution.

Open the week 6 notebook in Colab

6.1 Sex-structured populations

We keep track of the numbers \(F(t)\) of females and \(M(t)\) of males: \[ \frac{dF}{dt}=-\mu_FF+b_F\,\varphi(F,M),\qquad \frac{dM}{dt}=-\mu_MM+b_M\,\varphi(F,M). \tag{6.1}\] The mating function \(\varphi(F,M)\) describes how the rate of births depends on the numbers of females and males. \(b_F\) and \(b_M\) describe how many of the offspring are daughters and sons, and \(\mu_F\) and \(\mu_M\) are the death rates. These sex-specific rates are examples of vital rates.

Group task: pooling the launch results

Compare what you found for the three mating functions in the launch investigation.

For \(\varphi=FM\): what happens for small and for large initial populations? Is this realistic?
For \(\varphi=F\) and \(\varphi=2FM/(F+M)\): how do the numbers behave in the long run? What about the sex ratio \(M/F\)? Does it depend on the initial numbers?
What is the ecological assumption behind each of the three mating functions? For which species might each be appropriate?

Explain: the mating function \(\varphi=FM\)

Take equal vital rates, \(\mu_F=\mu_M=\mu\) and \(b_F=b_M=b\).

Show that \(F-M\) decays exponentially. Hence if \(F(0)=M(0)\) then \(F=M\) for all time.
With \(F=M\) show that \(dF/dt=-\mu F+bF^2\). Find the steady states and their stability.
For large \(F\), neglect \(-\mu F\) and solve. Show that the population becomes infinite in finite time. When?

Explain: female-limited reproduction, \(\varphi=F\)

If there are always enough males, the birth rate depends only on the number of females.

Solve the equation for \(F(t)\).
Substitute your solution into the equation for \(M\) and solve it using an integrating factor.
Show that, if \(b_F>\mu_F\), the sex ratio tends to \[s=\lim_{t\to\infty}\frac{M}{F}=\frac{b_M}{b_F-\mu_F+\mu_M}.\]
In the US in 1992 the rates per 100 000 per year were \(\mu_F=807\), \(\mu_M=902\), \(b_F=1480\) and \(b_M=1550\). What sex ratio does the model predict? (The actual ratio was \(0.953\).) Why is the model not appropriate for humans?

Explain: the harmonic mean, \(\varphi=2FM/(F+M)\)

This mating function is limited by females when males are abundant and by males when females are abundant.

Show that \(\varphi(\lambda F,\lambda M)=\lambda\varphi(F,M)\): doubling both sexes doubles the birth rate.
Show that \(s=M/F\) satisfies \(\dfrac{ds}{dt}=\dfrac1{F^2}\left(F\dfrac{dM}{dt}-M\dfrac{dF}{dt}\right)\) and, using item 1, that this is an ODE involving \(s\) only.
Find its steady state, and hence show that the long-term sex ratio is \[s=\frac{2b_M+\mu_F-\mu_M}{2b_F+\mu_M-\mu_F}.\]
Once \(M=sF\), show that \(F\) grows exponentially, and find the growth rate.

Check: feedback quiz 6.1

You will be asked to find long-term sex ratios and growth rates for sex-structured models with given mating functions and parameter values.

Consolidation: sex-structured models

\(\varphi=FM\) (with symmetric rates): \(dF/dt=-\mu F+bF^2\) has an unstable steady state \(\mu/b\). Smaller populations die out, larger ones blow up in finite time (roughly at \(t=1/(bF_0)\)). The model is not ecologically sensible because the per-capita birth rate grows without bound.
\(\varphi=F\) (female-limited): \(F=F_0e^{(b_F-\mu_F)t}\) and \[M(t)=M_0e^{-\mu_Mt}+\frac{b_MF_0}{b_F-\mu_F+\mu_M}\left(e^{(b_F-\mu_F)t}-e^{-\mu_Mt}\right),\] so \(M/F\to b_M/(b_F-\mu_F+\mu_M)\). For the US data this gives \(s\approx0.984\).
\(\varphi=2FM/(F+M)\) (harmonic mean): the ratio \(s=M/F\) obeys its own ODE, with stable steady state \(s=\dfrac{2b_M+\mu_F-\mu_M}{2b_F+\mu_M-\mu_F}\), and then both sexes grow exponentially at rate \[r=\frac{2b_Mb_F-\mu_Mb_F-\mu_Fb_M}{b_M+b_F}.\]

The trick used for the harmonic mean works for every mating function that is homogeneous of degree one, \(\varphi(\lambda F,\lambda M)=\lambda\varphi(F,M)\): the ratio \(M/F\) approaches a constant and the population grows or decays exponentially. To get a stable population size, density dependence (a carrying capacity) must be added.

6.2 Age structure: the Leslie matrix

For species that breed once a year it is natural to use discrete time and discrete age classes. Let \(N_{a,t}\) be the number of individuals of age \(a\) in year \(t\). The life table of a species gives:

\(S_a\): the probability that an individual of age \(a\) survives to age \(a+1\);
\(b_a\): the number of offspring an individual produces when it reaches age \(a\).

Explain: deriving the model

We count the population just after the breeding season each year.

Explain why \(N_{a,t+1}=S_{a-1}N_{a-1,t}\) for \(a\geq1\).
The newborns in year \(t+1\) are produced by individuals who have just reached age \(a+1\). Explain why \[N_{0,t+1}=\sum_{a\geq0}F_aN_{a,t}, \qquad F_a=b_{a+1}S_a.\] \(F_a\) is called the fecundity of age class \(a\).
Suppose no individual reproduces after age \(m\). Explain why we only need to follow the age classes \(0,\ldots,m-1\), and write the model as \(\mathbf N_{t+1}=L\mathbf N_t\) with a matrix \(L\). What does \(L\) look like?

Group task: a first life table

A population has the following life table.

\(a\)	\(b_a\)	\(S_a\)
0	0	1
1	0	1/2
2	1	1/2
3	2	1/2
4	0	⋮

Compute the fecundities \(F_0,F_1,F_2\) and write down the \(3\times3\) Leslie matrix.
Starting with 10 newborns, \(\mathbf N_0=(10,0,0)\), calculate by hand \(\mathbf N_1\), \(\mathbf N_2\), \(\mathbf N_3\) and \(\mathbf N_4\). Share the work in your group and check each other.
Continue with the notebook (part C). What happens in the long run? What happens for other initial conditions?

Check: feedback quiz 6.2

You will be asked to compute fecundities and Leslie matrices from life tables and to project populations forward a few years by hand.

Consolidation: the Leslie model

With a maximal reproductive age \(m\) the model is \[ \begin{pmatrix}N_0\\N_1\\N_2\\\vdots\\N_{m-1}\end{pmatrix}_{t+1}= \begin{pmatrix} F_0&F_1&\cdots&F_{m-2}&F_{m-1}\\ S_0&0&\cdots&0&0\\ 0&S_1&\cdots&0&0\\ \vdots&&\ddots&&\vdots\\ 0&0&\cdots&S_{m-2}&0 \end{pmatrix} \begin{pmatrix}N_0\\N_1\\N_2\\\vdots\\N_{m-1}\end{pmatrix}_t, \qquad F_a=b_{a+1}S_a. \tag{6.2}\] The matrix \(L\) is the Leslie matrix. Older, non-reproducing age classes can be computed afterwards from \(N_{a,t+1}=S_{a-1}N_{a-1,t}\). If instead all rates are constant from some age \(m\) on, all individuals of age \(\geq m\) can be lumped into a “plus group”, which adds the entry \(S_m\) in the bottom-right corner.

In the example, \(L=\begin{pmatrix}0&1/2&1\\1&0&0\\0&1/2&0\end{pmatrix}\) and \(\mathbf N_1=(0,10,0)\), \(\mathbf N_2=(5,0,5)\), \(\mathbf N_3=(5,5,0)\), \(\mathbf N_4=(2.5,5,2.5)\). The population converges to \((4,4,2)\): a stable age distribution that no longer changes.

6.3 Stable age distribution and eigenvalues

In the example the total population eventually stopped changing. In general the yearly growth factor \(N_{t+1}/N_t\) of the total population and the proportions in each age class settle down, but the total population keeps growing or shrinking.

Explore: notebook part D

Part D projects a population with a Leslie matrix of your choice and also prints the eigenvalues and eigenvectors of \(L\).

For several Leslie matrices, compare the long-term growth factor \(N_{t+1}/N_t\) with the eigenvalues of \(L\). What do you notice?
Compare the long-term proportions in each age class with the eigenvectors.
How quickly does the population approach its stable age distribution? Can you relate this to the other eigenvalues?

Explain

Try the Ansatz \(\mathbf N_t=\lambda^t\mathbf v\) in \(\mathbf N_{t+1}=L\mathbf N_t\). What conditions must \(\lambda\) and \(\mathbf v\) satisfy?
Assume \(L\) has distinct eigenvalues \(\lambda_1,\ldots,\lambda_n\) with eigenvectors \(\mathbf v_1,\ldots,\mathbf v_n\). Write down the general solution. How are the coefficients determined by \(\mathbf N_0\)?
Suppose \(|\lambda_1|>|\lambda_i|\) for all \(i\neq1\). Show that \(\mathbf N_t/\lambda_1^t\to c_1\mathbf v_1\).
For the example, show that \(\det(L-\lambda I)=(1-\lambda)(\lambda^2+\lambda+\tfrac12)\). Find all eigenvalues and check that \((4,4,2)\) is an eigenvector.

Check: feedback quiz 6.3

You will be asked to find characteristic polynomials, dominant eigenvalues and stable age distributions of small Leslie matrices.

Consolidation: long-term behaviour from eigenvalues

The Ansatz \(\mathbf N_t=\lambda^t\mathbf v\) works if \(L\mathbf v=\lambda\mathbf v\). With distinct eigenvalues the general solution is \[ \mathbf N_t=c_1\lambda_1^t\mathbf v_1+\cdots+c_n\lambda_n^t\mathbf v_n, \qquad \mathbf N_0=c_1\mathbf v_1+\cdots+c_n\mathbf v_n. \] If \(\lambda_1\) is dominant (\(|\lambda_1|>|\lambda_i|\) for \(i\neq1\)), then \(\mathbf N_t\approx c_1\lambda_1^t\mathbf v_1\) for large \(t\):

the population grows (or shrinks) by the factor \(\lambda_1\) each year;
the age distribution is proportional to \(\mathbf v_1\), the stable age distribution, whatever the initial condition (if \(c_1\neq0\));
the convergence is faster the smaller \(|\lambda_2|/|\lambda_1|\) is.

In the example the eigenvalues are \(1\) and \(-\frac12\pm\frac12i\) (with modulus \(1/\sqrt2<1\)), so the population converges to a multiple of \((4,4,2)\) with damped oscillations. Next week we find out when a dominant eigenvalue is guaranteed to exist.

6.4 Exam-style questions

Exam-style question 6.1: geometric mean mating function

In the sex-structured model Eq. 6.1 take \(\varphi(F,M)=\sqrt{FM}\).

Determine the asymptotic sex ratio \(s=\lim_{t\to\infty}M/F\). [5 marks]
Evaluate it for \(\mu_F=2\), \(\mu_M=1\), \(b_F=1/2\), \(b_M=3/2\) per year. [2 marks]

Solution

At the steady state of the ratio, \(F\,dM/dt-M\,dF/dt=0\). Dividing by \(FM\): \(-\mu_M+b_M\sqrt{F/M}+\mu_F-b_F\sqrt{M/F}=0\). With \(r=\sqrt{M/F}\) and multiplying by \(-r\): \(b_Fr^2-(\mu_F-\mu_M)r-b_M=0\), whose positive root is \(r=\frac{\mu_F-\mu_M+\sqrt{(\mu_F-\mu_M)^2+4b_Mb_F}}{2b_F}\). So \[s=\left(\frac{\mu_F-\mu_M+\sqrt{(\mu_F-\mu_M)^2+4b_Mb_F}}{2b_F}\right)^2.\]
\(s=\left(\frac{1+\sqrt{1+3}}{1}\right)^2=9\).

Exam-style question 6.2: dominance structure

A primate population consists of alpha males \(M\), alpha females \(F\) and subordinates \(Q\). Only alpha individuals breed. Alpha females rely on food gathered by the subordinates. The model is \[ \frac{dF}{dt}=b_FF-\frac{\mu_FF}{Q}F,\qquad \frac{dM}{dt}=b_MF-\mu_MM,\qquad \frac{dQ}{dt}=b_QF-\mu_QQ, \] with positive constants, \(b_F>\mu_F\) and \(b_Q>\mu_Q\).

Interpret the term \(\mu_FF^2/Q\). [2 marks]
Derive an ODE for \(F/Q\) and recognise it as a logistic equation. Hence show that \(F/Q\to\dfrac{b_F+\mu_Q}{b_Q+\mu_F}\). [5 marks]
Find the long-term ratio \(R=\lim F/M\) and show that it is positive. [4 marks]
Describe a limitation of the birth terms and suggest an improvement. [2 marks]

Solution

The per-capita death rate of alpha females is \(\mu_FF/Q\): it increases when there are few subordinates per alpha female to gather food.
\(\frac{d}{dt}\frac FQ=\frac{F'Q-FQ'}{Q^2}=(b_F+\mu_Q)\frac FQ\left(1-\frac{b_Q+\mu_F}{b_F+\mu_Q}\frac FQ\right)\), a logistic equation with rate \(b_F+\mu_Q\) and carrying capacity \(K=\frac{b_F+\mu_Q}{b_Q+\mu_F}\), so \(F/Q\to K\).
\(\frac{d}{dt}\frac FM=\frac FM\left(b_F+\mu_M-\mu_F\frac FQ-b_M\frac FM\right)\). In the limit \(F/Q\to K\), so the non-zero steady state is \(R=\frac{b_F+\mu_M-\mu_FK}{b_M}=\frac{b_Fb_Q+\mu_Mb_Q+\mu_M\mu_F-\mu_F\mu_Q}{b_M(b_Q+\mu_F)}\), which is positive because \(b_Fb_Q>\mu_F\mu_Q\).
Births are independent of \(M\), which is realistic only if alpha males are plentiful. A harmonic-mean-type term such as \(b_iFM/(F/n+M)\) would be better.

Exam-style question 6.3: a Leslie matrix

In a population every individual produces 1 offspring when it turns 1 and 4 offspring when it turns 2, after which it dies. Half the newborns survive to age 1, and a quarter of the one-year-olds survive to age 2.

Give \(b_a\) and \(S_a\) and compute the fecundities. [2 marks]
Write down the Leslie matrix. [2 marks]
Starting with 12 newborns only, compute the numbers in each age class for \(t=1,2,3\). [3 marks]
Find the long-term growth factor and the stable age distribution. [4 marks]

Solution

\(b_1=1\), \(b_2=4\), \(b_a=0\) otherwise. \(S_0=1/2\), \(S_1=1/4\), \(S_2=0\). \(F_0=b_1S_0=1/2\), \(F_1=b_2S_1=1\), \(F_2=0\).
Only ages 0 and 1 are needed: \(L=\begin{pmatrix}1/2&1\\1/2&0\end{pmatrix}\).
\(\mathbf N_0=(12,0)\), \(\mathbf N_1=(6,6)\), \(\mathbf N_2=(9,3)\), \(\mathbf N_3=(7.5,4.5)\).
\(\det(L-\lambda I)=\lambda^2-\lambda/2-1/2=(\lambda-1)(\lambda+1/2)\). The dominant eigenvalue is \(\lambda_1=1\), so the population eventually stays constant. Its eigenvector satisfies \(N_1=S_0N_0/\lambda=N_0/2\), so the stable age distribution is proportional to \((2,1)\): two thirds newborns, one third one-year-olds. (Indeed \(\mathbf N_t\to(8,4)\).)

Exam-style question 6.4: Fibonacci population

Individuals produce one offspring when they turn 1 and another when they turn 2, and then die. There is no mortality before age 2.

Give \(b_a\), \(S_a\) and the Leslie matrix. [3 marks]
Starting with one newborn, compute \(\mathbf N_t\) for \(t=1,\ldots,5\). [2 marks]
Find the long-term growth factor and the stable age distribution. [4 marks]

Solution

\(b_1=b_2=1\), \(S_0=S_1=1\) (and \(S_2=0\)). \(F_0=b_1S_0=1\), \(F_1=b_2S_1=1\), so \(L=\begin{pmatrix}1&1\\1&0\end{pmatrix}\).
\((1,0)\to(1,1)\to(2,1)\to(3,2)\to(5,3)\to(8,5)\): Fibonacci numbers.
\(\lambda^2-\lambda-1=0\), so \(\lambda_1=\frac{1+\sqrt5}2\approx1.618\) (the golden ratio). The eigenvector has \(N_1=N_0/\lambda_1\), so the stable age distribution is proportional to \((\lambda_1,1)\), i.e. about 61.8% newborns.

Summary

In sex-structured models the mating function \(\varphi(F,M)\) encodes how births depend on both sexes. \(FM\) gives blow-up, female-limited \(F\) and the harmonic mean give exponential growth with a stable sex ratio.
For mating functions that are homogeneous of degree one, the ODE for \(M/F\) gives the long-term sex ratio.
Leslie model: \(\mathbf N_{t+1}=L\mathbf N_t\) with fecundities \(F_a=b_{a+1}S_a\) in the first row and survival probabilities \(S_a\) on the subdiagonal.
The dominant eigenvalue of \(L\) is the long-term growth factor and its eigenvector is the stable age distribution.

6.5 Launch: when does the age distribution settle down?

Launch investigation for week 7

Open the week 7 notebook in Colab

Consider a population in which individuals reproduce only once, when they turn 3, producing 6 offspring each, and then die. All newborns survive to age 1, half of those survive to age 2, and a third of those survive to age 3. Predict: starting with 10 newborns, what will the population look like in the long run?
Use part A of the notebook to check. Does the population approach a stable age distribution?
Now try the life histories in part A in which individuals reproduce (a) at ages 2 and 3, (b) at ages 2 and 4, (c) at ages 3 and 6, (d) at ages 1, 2 and 3. In which cases does the age distribution settle down?
Formulate a conjecture: what property of the set of reproductive ages decides whether the age distribution settles down?