2.4: Rabbits are an Age-Structured Population
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Fibonacci’s rabbits form an age-structured population and we can use this simple case to illustrate the more general approach. Fibonacci’s rabbits can be categorized into two meaningful age classes: juveniles and adults. Here, juveniles are the newborn rabbits that can not yet mate; adults are those rabbits at least one month old. Beginning with a newborn pair at the beginning of the first month, we census the population at the beginning of each subsequent month after mated females have given birth. At the start of the \(n\)th month, let \(u_{1, n}\) be the number of newborn rabbit pairs, and let \(u_{2, n}\) be the number of rabbit pairs at least one month old. Since each adult pair gives birth to a juvenile pair, the number of juvenile pairs at the start of the \((n+1)\)-st month is equal to the number of adult pairs at the start of the \(n\)-th month. And since the number of adult pairs at the start of the \((n+1)\)-st month is equal to the sum of adult and juvenile pairs at the start of the \(n\)-th month, we have
\[\begin{aligned} &u_{1, n+1}=u_{2, n} \\[4pt] &u_{2, n+1}=u_{1, n}+u_{2, n} \end{aligned} \nonumber \]
or written in matrix form
\[\left(\begin{array}{l} u_{1, n+1} \\[4pt] u_{2, n+1} \end{array}\right)=\left(\begin{array}{ll} 0 & 1 \\[4pt] 1 & 1 \end{array}\right)\left(\begin{array}{l} u_{1, n} \\[4pt] u_{2, n} \end{array}\right) . \nonumber \]
Rewritten in vector form, we have
\[\mathbf{u}_{n+1}=L \mathbf{u}_{n \prime} \nonumber \]
where the definitions of the vector \(\mathbf{u}_{n}\) and the matrix \(L\) are obvious. The initial conditions, with one juvenile pair and no adults, are given by
\[\left(\begin{array}{l} u_{1,1} \\[4pt] u_{2,1} \end{array}\right)=\left(\begin{array}{l} 1 \\[4pt] 0 \end{array}\right) \text {. } \nonumber \]
The solution of this system of coupled, first-order, linear, difference equations, (2.4.2), proceeds similarly to that of coupled, first-order, linear, differential equations. With the ansatz, \(\mathbf{u}_{n}=\lambda^{n} \mathbf{v}\), we obtain upon substitution into (2.4.2) the eigenvalue problem
\[\mathbf{L} \mathbf{v}=\lambda \mathbf{v} \nonumber \]
whose solution yields two eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\), with corresponding eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\). The general solution to (2.4.2) is then
\[\mathbf{u}_{n}=c_{1} \lambda_{1}^{n} \mathbf{v}_{1}+c_{2} \lambda_{2}^{n} \mathbf{v}_{2} \nonumber \]
with \(c_{1}\) and \(c_{2}\) determined from the initial conditions. Now suppose that \(\left|\lambda_{1}\right|>\left|\lambda_{2}\right|\). If we rewrite (2.4.5) in the form
\[\mathbf{u}_{n}=\lambda_{1}^{n}\left(c_{1} \mathbf{v}_{1}+c_{2}\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{n} \mathbf{v}_{2}\right) \nonumber \]
then because \(\left|\lambda_{2} / \lambda_{1}\right|<1, \mathbf{u}_{n} \rightarrow c_{1} \lambda_{1}^{n} \mathbf{v}_{1}\) as \(n \rightarrow \infty\). The long-time asymptotics of the population, therefore, depends only on \(\lambda_{1}\) and the corresponding eigenvector \(\mathbf{v}_{1} .\) For our Fibonacci’s rabbits, the eigenvalues are obtained by solving \(\operatorname{det}(\mathrm{L}-\lambda \mathrm{I})=\) 0 , and we find
\[\begin{aligned} \operatorname{det}\left(\begin{array}{cc} -\lambda & 1 \\[4pt] 1 & 1-\lambda \end{array}\right) &=-\lambda(1-\lambda)-1 \\[4pt] &=0 \end{aligned} \nonumber \]
or \(\lambda^{2}-\lambda-1=0\), with solutions \(\Phi\) and \(-\phi\). Since \(\Phi>\phi\), the eigenvalue \(\Phi\) and its corresponding eigenvector determine the long-time asymptotic population age-structure. The eigenvector may be found by solving
\[(\mathrm{L}-\Phi \mathrm{I}) \mathbf{v}_{1}=\mathbf{0} \nonumber \]
or
\[\left(\begin{array}{cc} -\Phi & 1 \\[4pt] 1 & 1-\Phi \end{array}\right)\left(\begin{array}{l} v_{11} \\[4pt] v_{12} \end{array}\right)=\left(\begin{array}{l} 0 \\[4pt] 0 \end{array}\right) \nonumber \]
The first equation is just \(-\Phi\) times the second equation (use \(\Phi^{2}-\Phi-1=0\) ), so that \(v_{12}=\Phi v_{11}\). Taking \(v_{11}=1\), we have
\[\mathbf{v}_{1}=\left(\begin{array}{c} 1 \\[4pt] \Phi \end{array}\right) \nonumber \]
The asymptotic age-structure obtained from \(\mathbf{v}_{1}\) shows that the ratio of adults to juveniles approaches the golden mean; that is,
\[\begin{aligned} \lim _{n \rightarrow \infty} \frac{u_{2, n}}{u_{1, n}} &=v_{12} / v_{11} \\[4pt] &=\Phi . \end{aligned} \nonumber \]