Question 17:Construct a stage-matrix model for an animal species that has two life stages: juvenile (up to 1 year old) and adult. Suppose the female adults give birth each year to an average of 1.6 female juveniles. Each year, 30% of the juveniles survive to become adults and 80% of the adults survive. For \(k \ge 0\), let \({x_k} = \left( {{j_k},{a_k}} \right)\), where the entries in \({x_k}\) are the numbers of female juveniles and female adults in year \(k\).

1. Construct the stage-matrix \(A\) such that \({x_{k + 1}} = A{x_k}\) for \(k \ge 0\).
2. Show that the Population is growing, compute the eventual growth rate of the Population, and give the eventual ratio of juveniles to adults.
3. (M) Suppose that initially there are 15 juveniles and 10 adults in the Population. Produce four graphs that show how the Population changes over eight years: (a) the number of juveniles, (b) the number of adults, (c) the total Population, and (d) the ratio of juveniles to adults (each year). When does the ratio in (d) seem to stabilize? Include a listing of the program or keystrokes used to produce the graphs for (c) and (d).

Here, as per the data given, each adult gave birth to 1.6 juveniles on average. So, for the particular year, we have:

\({j_{k + 1}} = 1.6{a_k}\)

Also,30% of the juveniles survive to become adults, and 80% of the adults survive. So:

\({a_{k + 1}} = 0.3{j_k} + 0.8{a_k}\)

So, the stage matrix can be given as:

\(A = \left( {\begin{array}{*{20}{c}}0&{1.6}\\{0.3}&{0.8}\end{array}} \right)\)

Hence, this is the required matrix.

Enter this matrix in MATLAB as:

>> \(A = \left( {\begin{array}{*{20}{c}}{0\,\,1.6;}&{0.3\,\,0.8}\end{array}} \right)\);

For eigenvalues, enter instruction as:

>> \(E = {\rm{eigs}}\left( A \right)\);

We get:

\(E = \left( {\begin{array}{*{20}{c}}{1.2}\\{ - 0.4}\end{array}} \right)\)

Now, for eigenvectors, enter instruction as:

>> \(\left( {\begin{array}{*{20}{c}}A&B\end{array}} \right) = {\rm{eigs}}\left( A \right)\);

So, we have:

\(\begin{array}{l}{v_1} = \left( {\begin{array}{*{20}{c}}{ - 0.8}\\{ - 0.6}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right)\\{v_2} = \left( {\begin{array}{*{20}{c}}{ - 0.970143}\\{0.242536}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\end{array}\)

Thus, these are the required eigenvectors.

Now, the general solution can be given as:

\(x\left( t \right) = {c_1}{\left( {\left| {1.2} \right|} \right)^k}\left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right) + {c_2}{\left( {\left| { - 0.4} \right|} \right)^k}\left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\)

Hence, this is the required solution.

Now, from the question, we have the initial vector as:

\(\left( {\begin{array}{*{20}{c}}{{j_0}}\\{{a_0}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{15}\\{10}\end{array}} \right)\)

So, for \(k = 0\), the general solution is:

\(\left( {\begin{array}{*{20}{c}}{15}\\{10}\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\)

The argument matrix reduced to row echelon form using MATLAB as:

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}4&{ - 4}&{15}\\3&1&{10}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&{\frac{{55}}{{16}}}\\0&1&{ - \frac{5}{{16}}}\end{array}} \right)\end{array}\)

Thus, the final general solution is:

\({x_k} = \left( {\frac{{55}}{{16}}} \right){\left( {\left| {1.2} \right|} \right)^k}\left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right) + \left( { - \frac{5}{{16}}} \right){\left( {\left| { - 0.4} \right|} \right)^k}\left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\)

Hence, the Population is increasing in the ratio: 4 juveniles for every 3 adults.

Since initially, we have 15 juveniles and 10 adults. Then, the graphs that show rate variation of Population in 8 years are:

c. The total Population.

Hence, these are the required graphs, and the ratio is stabilized from the very beginning.