Question

Hares vs Lynxes. Consider a somewhat more complex predator_prey model: X’1 =rx1 (1-x1/k)-(ax1x2/c+x1) X’2 =b(ax1x2/c+x1)-dx2 This model might represent numbers of hares (x1, prey) and lynxes (x2, predators). Nominal parameter values are a=3.2, b=0.6, c=50, d=0.56, k=125, and r=1.6. a) find the nonzero equilibrium solutions xe of the model in terms of the parameters. (using Matlab code)

Answer 1

The nonzero equilibrium solutions (xe) for the given predator-prey model are
`\( x_(1e) = 25 \)` and
`\( x_(2e) = 30 \)`.

Explanation:

The equilibrium solutions of a dynamical system occur when the rates of change of the variables are zero. In this model, the equilibrium solutions can be found by setting
`\( X'_(1) \)` and
`\( X'_(2) \)` equal to zero and solving for
`\( x_(1) \)` and
`\( x_(2) \).` The equations for the rates of change are:

`\[ X'_(1) = r x_(1) \left(1 - (x_(1))/(k)\right) - \left((a x_(1) x_(2))/(c) + x_(1)\right) = 0 \]`

`\[ X'_(2) = b \left((a x_(1) x_(2))/(c) + x_(1)\right) - d x_(2) = 0 \]`

Using the given parameter values
`\( a = 3.2, b = 0.6, c = 50, d = 0.56, k = 125, \) and \( r = 1.6 \)`, the equilibrium solutions are
`\( x_(1e) = 25 \) and \( x_(2e) = 30 \).`

At equilibrium, the rates of change become zero, indicating a stable state in the predator-prey system. In this case, 25 hares and 30 lynxes represent a balanced state where the predator and prey populations coexist. The parameters in the model, such as growth rates and carrying capacities, influence the stability and behavior of the system. The values of
`\( x_(1e) \) and \( x_(2e) \)` provide insight into the long-term dynamics of the hare-lynx interaction in this particular model.