Suppose we study the population of two species in a region - hawks and mice. Hawks are the predator and mice are the prey. Let [math]h_k[/math] and [math]m_k[/math] be the population of the hawks and mice in the region after [math]k[/math] months respectively. We assume the following formulas for the transition of populations:[br][br][math]\left\{\begin{eqnarray}h_k & = & \frac12 h_{k-1}+\frac1{100}m_{k-1}\\m_k & = & -\frac{50}4 h_{k-1}+\frac54 m_{k-1}\end{eqnarray}\right.[/math][br][br][u]Interpretation[/u]: [br][br][list][*]The "[math]\frac12 h_{k-1}[/math]" term: Without any mice to eat, the hawk population will decrease by half in a month.[/*][*]The "[math]\frac54 m_{k-1}[/math]" term: Without any hawk, the mice population will increase by a factor of [math]\frac{5}{4}[/math] in a month.[/*][*]The "[math]\frac{1}{100}m_{k-1}[/math]" term: On average, 100 mice can support an additional hawk in a month.[/*][*]The "[math]-\frac{50}{4}h_{k-1}[/math]" term: On average, 4 hawks can kill 50 mice in a month.[/*][/list][br][u]Question[/u]: Suppose there are 50 hawks and 1600 mice in the region initially. Will the populations eventually become stabilized?[br][br]First of all, it is given that [math]h_0=50,m_0=1600[/math]. And the transition of populations can be written as follows:[br][br][math]\begin{pmatrix}h_k\\m_k\end{pmatrix}=A\begin{pmatrix}h_{k-1}\\m_{k-1}\end{pmatrix}[/math], where [math]A=\begin{pmatrix}\frac12&\frac1{100}\\-\frac{50}4&\frac54\end{pmatrix}[/math].[br][br]Then [math]\begin{pmatrix}h_k\\m_k\end{pmatrix}=A\begin{pmatrix}h_{k-1}\\m_{k-1}\end{pmatrix}=A^2\begin{pmatrix}h_{k-2}\\m_{k-2}\end{pmatrix}=\cdots=A^k\begin{pmatrix}h_0\\m_0\end{pmatrix}[/math][br][br]If we can write down the entries of [math]A^k[/math] explicitly in terms of [math]k[/math], then we can study the behaviour of [math]h_k[/math] and [math]m_k[/math] when [math]k[/math] is large.[br][br]We can diagonalize [math]A[/math]:[br][br][math]A=\begin{pmatrix}\frac1{50}&\frac1{25}\\1&1\end{pmatrix}\begin{pmatrix}1&0\\0&\frac34\end{pmatrix}\begin{pmatrix}-50&2\\50&-1\end{pmatrix}[/math][br][br]Hence [math]A^k=\begin{pmatrix}\frac1{50}&\frac1{25}\\1&1\end{pmatrix}\begin{pmatrix}1^k&0\\0&\left(\frac34\right)^k\end{pmatrix}\begin{pmatrix}-50&2\\50&-1\end{pmatrix}=\begin{pmatrix}-1+2\left(\frac34\right)^k&\frac1{25}\left[1-\left(\frac34\right)^k\right]\\50\left[-1+\left(\frac34\right)^k\right]&2-\left(\frac34\right)^k\end{pmatrix}[/math][br][br]When [math]k\to \infty[/math],[math]\left(\frac34\right)^k\to 0[/math]. Hence [math]A^k\to \begin{pmatrix}-1&\frac1{25}\\-50&2\end{pmatrix}[/math].[br][br]Therefore, [math]\begin{pmatrix}h_k\\m_k\end{pmatrix}\to \begin{pmatrix}-1&\frac1{25}\\-50&2\end{pmatrix}\begin{pmatrix}50\\1600\end{pmatrix}=\begin{pmatrix}14\\700\end{pmatrix}[/math].[br]