Question

In Exercises 3–6, assume that any initial vector \({x_0}\) has an eigenvector decomposition such that the coefficient \({c_1}\) in equation (1) of this section is positive.

4. Determine the evolution of the dynamical system in Example 1 when the predation parameter p is .125. (Give a formula for \({x_k}\) .) As time passes, what happens to the sizes of the owl and wood rat populations? The system tends toward what is sometimes called an unstable equilibrium. What do you think might happen to the system if some aspect of the model (such as birth rates or the predation rate) were to change slightly?

\(\)

Short answer

The general solution is\({{\rm{x}}_k} = {c_1}{\left( 1 \right)^k}\left( {\begin{aligned}{}{0.8}\\1\end{aligned}} \right)\).

As \(k \to \infty \), in this situation, the population finds equilibrium, with ten thousand rats for every eight owls. The population size is solely determined by the values of \({c_1}\).

Because slight changes in birth rates or predation rates might modify the condition, this equilibrium is not considered stable.

Step-by-step solution

Given term for the vector and a matrix 

The owl and wood rat populations at time k are described by\({{\rm{x}}_k} = \left( {\begin{aligned}{}{{O_k}}\\{{R_k}}\end{aligned}} \right)\).

Where k is the number of months in a year, and the number of owls in the study area is \({O_k}\), while the number of rats is \({R_k}\) (measured in thousands). Because owls consume rats, the population of one species should have an impact on the other.

The changes in these populations can be described by the equations:

\(\)\(\begin{aligned}{}{O_{k + 1}} &= \left( {0.5} \right){O_k} + \left( {0.4} \right){R_k}\\{R_{k + 1}} &= \left( { - p} \right){O_k} + \left( {1.1} \right){R_k}\end{aligned}\)

Where p is a positive parameter to be specified.

In the matrix form,

\({{\rm{x}}_{k + 1}} = \left( {\begin{aligned}{}{0.5}&{}&{0.4}\\{ - p}&{}&{1.1}\end{aligned}} \right){{\rm{x}}_k}\)

Put the value of p in the above matrix.

\(A = \left( {\begin{aligned}{}{0.5}&{}&{0.4}\\{ - 0.125}&{}&{1.1}\end{aligned}} \right)\)

Find the eigenvalue 

For finding eigenvalue,

\(\det \left( {A - \lambda I} \right) = \left( {\begin{aligned}{}{0.5 - \lambda }&{0.4}\\{ - 0.125}&{1.1 - \lambda }\end{aligned}} \right)\)

So, the characteristics equation is,

\(\)

\(\begin{aligned}{}0 &= \left( {0.5 - \lambda } \right)\left( {1.1 - \lambda } \right) + \left( {0.4} \right)\left( {0.125} \right)\\ &= 0.55{\rm{ }} - {\rm{ }}1.6\lambda {\rm{ }} + {\rm{ }}{\lambda ^2}{\rm{ }} + {\rm{ }}0.05\\ &= {\lambda ^2} - {\rm{ }}1.6\lambda {\rm{ }} + {\rm{ }}0.6{\rm{ }}\\ &= \left( {\lambda {\rm{ }} - {\rm{ }}1} \right)\left( {\lambda {\rm{ }} - {\rm{ }}0.6} \right)\end{aligned}\)

The eigenvalues are 1 and \(0.6\).

When\(\lambda = 1\),

\(\begin{aligned}{}{E_1} &= Nul\left( {\begin{aligned}{}{ - 0.5}&{}&{0.4}\\{ - 0.125}&{}&{1.1}\end{aligned}} \right)\\{E_1} &= \left( {\begin{aligned}{}1&{}&{ - \frac{4}{5}}\\0&{}&0\end{aligned}} \right)\end{aligned}\)

An eigenvector is \({{\rm{v}}_1} = \left( {\begin{aligned}{}{0.8}\\1\end{aligned}} \right)\).

When \(\lambda {\rm{ }} = {\rm{ }}0.6{\rm{ }}\),

\(\begin{aligned}{}{E_{0.6}} &= Nul\left( {\begin{aligned}{}{ - 0.1}&{}&{0.4}\\{ - 0.125}&{}&{0.5}\end{aligned}} \right)\\{E_{0.6}} &= Nul\left( {\begin{aligned}{}1&{}&{ - 4}\\0&{}&0\end{aligned}} \right)\end{aligned}\)

An eigenvector is \({{\rm{v}}_2} = \left( {\begin{aligned}{}4\\1\end{aligned}} \right)\).

The general equation is,

\(\begin{aligned}{}{{\rm{x}}_k} = {c_1}{\left( 1 \right)^k}\left( {\begin{aligned}{}{0.8}\\1\end{aligned}} \right) + {c_2}{\left( {0.6} \right)^k}\left( {\begin{aligned}{}4\\1\end{aligned}} \right)\\ \to {c_1}{\left( 1 \right)^k}\left( {\begin{aligned}{}{0.8}\\1\end{aligned}} \right)\end{aligned}\)

As \(k \to \infty \), in this situation, the population finds equilibrium, with ten thousand rats for every eight owls. The population size is solely determined by the values of \({c_1}\).

Because slight changes in birth rates or predation rates might modify the condition, this equilibrium is not considered stable.