Question 7.4.1: Finding Equilibrium Solutions of a System of Equations Find ...
Finding Equilibrium Solutions of a System of Equations
Find and interpret all equilibrium solutions of the predator-prey model
\left\{\begin{array}{l}x^{\prime}(t)=0.2 x(t)-0.1[x(t)]^2-0.4 x(t) y(t) \\y^{\prime}(t)=-0.1 y(t)+0.1 x(t) y(t),\end{array}\right.where x and y represent the populations (in hundreds of animals) of a prey and a predator, respectively.
If (x, y) is an equilibrium solution, then the constant functions x(t) = x and y(t) = y satisfy the system of equations with x′(t) = 0 and y′(t) = 0. Substituting into the equations, we have
0=0.2 x-0.1 x^2-0.4 x y0=-0.1 y+0.1 x y
This is now a system of two (nonlinear) equations for the two unknowns x and y. There is no general method for solving systems of nonlinear algebraic equations exactly. In this case, you should solve the simpler equation carefully and then substitute solutions back into the more complicated equation. Notice that both equations factor, to give
0=0.1 x(2-x-4 y)0=0.1 y(-1+x).
The second equation has solutions y = 0 and x = 1. We now substitute these solutions one at a time into the first equation.
Taking y = 0, the first equation becomes 0=0.1 x(2-x), which has the solutions x = 0 and x = 2. This says that (0, 0) and (2, 0) are equilibrium solutions of the system. Note that the equilibrium point (0, 0) corresponds to the case where there are no predators or prey, while (2, 0) corresponds to the case where there are prey but no predators. Taking x = 1, the first equation becomes 0=0.2-0.1-0.4 y, which has the solution y=\frac{0.1}{0.4}=0.25. A third equilibrium solution is then (1, 0.25), corresponding to having both populations constant, with four times as many prey as predators. Since we have now considered both solutions from the second equation, we have found all equilibrium solutions of the system: (0, 0), (2, 0) and (1, 0.25).