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Maple: Variation of Predator-Prey Model

Saturday, 17 of November , 2007 at 9:06 pm

> restart; a := 20; alpha := 2; sigma := .3; c := 30; g := 2

20
2
0.3
30
2

> e1 := diff(x(t), t) = ax(t)-alphax(t)y(t)-sigmax(t)^2;
> e2 := diff(y(t), t) = -cy(t)+gx(t)*y(t);

$\frac{d}{dt}x(t) = 20x(t) - 2x(t)y(t) - .3x(t)^{2}$
$\frac{d}{dt} y(t) = -30y(t) + 2 x(t)y(t)$

> solve({rhs(e1) = 0, rhs(e2) = 0}, {x(t), y(t)}):
> sol(1):
> with(DEtools): with(plots):
> DEplot({e1, e2}, {x(t), y(t)}, t = 0 .. 5, [[x(0) = 15, y(0) = 7]], stepsize = 3)

> plotX := DEplot({e1, e2}, {x(t), y(t)}, t = 0 .. 4, [[x(0) = 1, y(0) = 1]], stepsize = 3, scene = [t, x], arrows = NONE, linecolor = RED);
> plotY := DEplot({e1, e2}, {x(t), y(t)}, t = 0 .. 4, [[x(0) = 1, y(0) = 1]], stepsize = 3, scene = [t, y], arrows = NONE, linecolor = BLUE);
> plots[display]([plotX, plotY])

Comments (1)

Category: Differential Equations, Maple

Answer: Variation of Predator-Prey Model

Saturday, 17 of November , 2007 at 1:49 am

Answer: The critical points for the following equations

$\frac{\partial x}{\partial t} = x (a - \alpha y - \sigma x)$
$\frac{\partial y}{\partial t} = y (-c + \gamma x)$
will occur when
$\frac{\partial}{\partial t}\begin{pmatrix}x \\ y\end{pmatrix} = 0$
which will occur when
$\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}, \begin{pmatrix}\frac{a}{\sigma} \\ 0\end{pmatrix}, \begin{pmatrix}\frac{c}{\gamma} \\ \frac{a}{\alpha} - \frac{c \sigma}{\alpha \gamma}\end{pmatrix}$

The first critical point at can be shown to be an unstable saddle point. The entrance to this critical point is along the y axis while all other paths will depart from it. This means that the only way for both the predator and prey to become extinct is to force the prey’s population down to zero.

The second critical point at $< \frac{a}{\sigma} , 0 >$ can be shown to be an unstable saddle point when $\frac{c}{\gamma} < \frac{a}{\sigma}$ and an asymptotically stable sink when $\frac{c}{\gamma} >= \frac{a}{\sigma}$ .

In this case, the predator will become extinct and the prey will grow until it is saturated.

The third critical point at $<\frac{c}{\gamma} , \frac{a}{\alpha} - \frac{c \sigma}{\alpha \gamma} >$ has the most interesting behavior. Since the model is meaningless when , we only need to analyze the characters of the critical point when $\frac{a}{\sigma}\geq\frac{c}{\gamma}$ . For the conditions of interest, the critical point is an asymptotically stable spiral sink when $\frac{a}{\sigma} > \frac{c}{\gamma}(\frac{\sigma}{16 \gamma}+1)$ . Otherwise, the point is an asymptotically stable sink.

In either case, the predator and prey will reach an equilibrium at this critical point as long as $\frac{a}{\sigma}\geq\frac{c}{\gamma}$ .

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Category: Answers, Differential Equations

Question: Variation of Predator-Prey Model

Sunday, 11 of November , 2007 at 6:25 pm

The Lotka-Volterra Predator-Prey Model attempts to describe the dynamics of a biological system where two species interact, one a predator and one its prey. The model is described by the following first order, non-linear, differential equations:

$\frac{\partial x}{\partial t} = x (a - \alpha y)$
$\frac{\partial y}{\partial t} = y (-c + \gamma x)$
where

x = number of prey

y = number of predators

a = growth rate of the prey

c = death rate of the predator

$\gamma$ = predation rate coefficient

$\alpha$ = reproduction rate of predators per 1 prey eaten

Question: A major flaw in this model is that in the absence of a predator, the prey would grow without bound. If we assume that the population growth of the prey reduces to a logistic equation in the absence of a predator, we can get the following variation of Lotka-Volterra’s differential equation.

$\frac{\partial x}{\partial t} = x (a - \alpha y - \sigma x)$
$\frac{\partial y}{\partial t} = y (-c + \gamma x)$
where $\sigma$ is the rate of saturation of the prey.

Find the critical points and discuss their nature and stability characteristics.

Comments (4)

Category: Differential Equations, Questions

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