Question

Find all equilibrium points for the predator-prey model [x = - xy/2 y = -3y/4 + y/4 and describe the trajectories near those points. Have WolframAlpha or some other program to draw some trajectories. (Here x is the population of the prey (rabbits) and y is the population of the predators (foxes).)

Answer 1

An application of the nonlinear system of differential equations inmathematical biology / ecology: to model the predator-prey relationship of a simple eco-system.

Suppose in a closed eco-system (i.e. no migration is allowed into or out of the system) there are only 2 types of animals: the predator and the prey. They form a simple food-chain where the predator species hunts the prey species, while the prey grazes vegetation. The size of the 2 populations can be described by a simple system of 2 nonlinear first order differential equations (a.k.a. the Lotka-Volterra equations, which originated in the study

of fish populations of the Mediterranean during and immediately after WW I). Let x(t) denotes the population of the prey species, and y(t) denotes the population of the predator species. Then

x) = a x − xy

y) = −c y + xy

Here a, c, , and are positive constants.

This system has two critical points. One is the origin, and the other is in the

first quadrant.

0 = x) = a x − xy = x(a − y) 1 x = 0 or a − y = 0

0 = y) = −c y + xy = y(−c + x) 1 y = 0 or −c + x = 0

Find next with picture. Fig -1

Kindly look on Fig - 1 ...

This system has two critical points. One is the origin, and the other is in the first quadrant. c a Therefore, the critical poiats are (0,0) and The Jacobian matrix is At (0, 0). the linearized system has coefficient matrix The eigenvalues are a and-c. Hence, it is an unstable saddle point. c a At the linearized system has coefficient matrix The eigenvalues are aci. It is a stable center. (Previously, we have learned that the purely imaginary eigenvalues case in a nonlinear system is ambiguous, with several possible behaviors. But in this example it really is a center. See its phase portrait on the next page.)

Variations of the basic Lotka-Volterra equations One obvious shortcoming of the basic predator-prey system is that the population of the prey species would grow unbounded, exponentially. in the absence of We'll just replace the exponential growth term in the first equation by the two-term logistic growth expression: e is an easy solution to this unrealistic behavior (ax- Therefore, in the absence of predators, the first equation becomes the logistic equation. The prey population would instead stabilize at the carrying capacity given by the logistic equation In this new system there will be 3 critical points: at the origin, in the first quadrant, and the third on the positivex-axis. This third critical point is at the environmental carying capacity of prey x, while y-0. That is, the additional critical point is at (k, 0). where kis the environmental carrying capacity for the prey species seen earlier in the course. It is the limiting population of the prey species in the absence of the predator species. The critical point in the first quadrant could be an asymptotically stable node or spiral point.

If the predator species has an alternate food source (omnivores such as bears, for example, could possibly subsist on plants alone), then it needs not to die out due to starvation even if the preys are totally absence. In this case we

could replace the −c y term in the second equation by logistic-growth terms as well:

x) = (a x − r x2) − xy

y) = (b y − c y2) + xy

More complex food-chains can be similarly constructed as systems of morethan 2 equations. For example, suppose there is an enclosed eco-syste containing 3 species. Species x is a grass-grazer whose population inisolation would obey the logistic equation, and that it is preyed upon by species y who, in turn, is the sole food source of species z. Then their respective population might be modeled by the 3-equation system:

x) = (a x − r x2) − xy

y) = −c y + xy − 　 yz

z) = −d z + - yz

Result :

What do the equations mean?

(1) At the bottom of the food chain, population x grows logistically in the absence of its natural predator (y, in this case); while it decreases due to hunting by y.

(2) Population y starves in the absence of its sole food source (x, in this example); it grows by hunting and eating x; it is, in turn, hunted by z and therefore it would decrease due to interaction with z.

(3) Population z sits atop the food chain. It starves in the absence of its sole food source (y, in this example); it grows by hunting and eating y. Populations x and z do not interact directly.