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Use of Mathematical Modeling in Biology and Ecology

Prepared by:

Catherine KiuSandra William

Sharon YeoYong Bing Sing

Use of Mathematical

Modeling in Biology and Ecology

Predator-Prey Model

SIR (Susceptible–Infected–

Recovered) Model

Stella Model

Predator-Prey Model

Lotka-Volterra Model

• Vito Volterra (1860-1940)– famous Italian mathematician– Retired from pure mathematics

in 1920– His son-in-law, Humberto

D'Ancona, was a biologist who studied the populations of various species of fish in the Adriatic Sea.

• Alfred J. Lotka (1880-1949)– American mathematical

biologist– primary example: plant

population/herbivorous animal dependent on that plant for food

Rules and Regulation

1. This game involves 1 pupil as a guard, 4 pupils who act as predators (snakes) and 14 pupils who act as preys (chickens).

2. A paper will be stick on the back of the preys. 3. Predator has to ‘eat’ the preys by taking paper which was

sticked on the back of preys and draw 20 stars on the paper. 4. Guard will examine the 20 stars drawn.

(20 stars drawn = prey is dead)5. Other predator is not allowed to ‘eat’ the preys that are

‘dead’. 6. Predators are given 1 minute to ‘eat’ their preys.7. Preys that are “dead”, have to stay with the guard. 8. Amount of preys that eaten by predators are recorded.

Predators Number of Preys

A 3

B 6

C 2

D 2

Guard

Number of Predators Number of Preys

1 A= 3, B=6, C=2, D=2

2 A+B=9, A+C= 5,

A+D = 5, B+C=8,

B+D = 8, C+D=4

3 A+B+C = 11, A+B+D=11, B+C+D = 10,

C+D+A = 7

4 A+B+C+D = 13

predators

preysPhase portrait when p = 1.5

+

Lotka-Volterra Model

• The Lotka-Volterra equations are a pair of first order, non-linear, differential equations that describe the dynamics of biological systems in which two species interact.

• Forms the basis of many models used today in the analysis of population dynamics

Predator-Prey Differential Equations

Assumptions• The predator species is totally dependent on

the prey species as its only food supply. • The prey species has an unlimited food supply

and • no threat to its growth other than the specific

predator.

• If there were no predators (y), the second assumption would imply that the prey (x) species grows exponentially.

• If is the size of the prey (x) population at time t, x = x(t),then we would have

= ax.

• But there are predators (y), which must account for a negative component in the prey (x) growth rate. Suppose we write y = y(t) for the size of the predator (y) population at time (t). Here are the crucial assumptions for completing the model:– The rate at which predators encounter prey is

jointly proportional to the sizes of the two populations.

– A fixed proportion of encounters leads to the death of the prey.

• These assumptions lead to the conclusion that the negative component of the prey growth rate is proportional to the product xy of the population sizes.

= ax – bxy

• Now we consider the predator population. If there were no food supply, the population would die out at a rate proportional to its size, we would find

• (Keep in mind that the "natural growth rate" is a composite of birth and death rates, both presumably proportional to population size. In the absence of food, there is no energy supply to support the birth rate.)

= -my

• But there is a food supply: the prey (x). And what's bad for prey (x) is good for predator (y). That is, the energy to support growth of the predator (y) population is proportional to deaths of prey (x), so = -my + nxy

• This discussion leads to the Lotka-Volterra Predator-Prey Model:

where a, b, m, and n are positive constants.• This equations are called Lotka-Volterra

equations.• The equations are autonomous because the

rates do not explicitly depend on time (t).

= ax – bxy

= -my + nxy

• The Lotka-Volterra model consists of a system of linked differential equations that cannot be separated from each other and that cannot be solved in closed form. Nevertheless, there are a few things we can learn from their symbolic form.

• Explain why

Note: This is a general result in calculus.

dydy dt

dxdxdt

= x

= (shown)

Thank You

Question• Given a=0.04, b=0.002, m=0.08, n=0.0004.

x = ___ , ____y= ___ , ____

So, (x,y) = (0, 0) or (200, 20)For the general Predator-Prey equations,

What is equilibrium point?• Consider the system of 2 autonomous differential

equations

The first step is find the equations of the zero isoclines, which are defined as the set of points that satisfy

),(

),(

yxgdt

dy

yxfdt

dx

),(0

),(0

yxg

yxf

• Each equation results in a curve in the x-y space.

• Point equilibrium occur where the two isoclines intersect (Figure 1).

• A point equilibrium therefore simultaneously satisfies the two equations f(x, y) = 0 and g(x, y) = 0

• We will call point equilibrium.

What is equilibrium point?

x

y

0),( yxg

0),( yxf

Equilibrium

Figure 1: Zero isoclines corresponding to the two differential equations. Equilibrium occur where the isoclines intersect.

What is equilibrium point?

Analyzing Predator-Prey equations graphically

• With initial conditions of x=200 and y=10, what do you observe?

t (month)

Analyzing Predator-Prey equations graphically

Fill in the blanks.1. The chicken and snakes populations oscillate

periodically between their _________ maximum and minimum.

2. For chicken, the population ranges from about ____________.

3. For snakes, the population ranges from about ____________.

4. About _________ months after the chicken population peaks, the snakes population peaks.

• With initial conditions of x=200 and y=10, what do you observe? Plot it.

Slope-fields of Predator-Prey equations

Slope-fields of Predator-Prey equations

• Starting from (200, 10), the solution curve can be drawn such that the tangent follows the slope fields. This gives a close curve in the xy-plane as follows.

Slope-fields of Predator-Prey equations

Slope-fields of Predator-Prey equations

Slope-fields of Predator-Prey Equations

Based on the graph, at (200,10),

dx/dt = ?dy/dt = ?Please give the answers.

Slope-fields of Predator-Prey equations

So, the chicken population is increasing at (200, 10). This means that we should trace the curve counterclockwise as t increases.

Slope-fields of Predator-Prey equations

Slope-fields of Predator-Prey equations

• Chicken • snak

e

• Chicken • sna

ke

ExercisesSystemr’(t) = 2 r(t) – 0.01 r(t) f(t)f’(t) = - f(t) + 0.01 r(t) f(t)Exercise 1: Exercise 2:Initial Conditions Initial Conditionsr(0) = 120 r(0) = 100f(0) = 200 f(0) = 200

Exercise 3: Exercise 4:Initial Conditions Initial Conditionsr(0) = 400 r(0) = 800f(0) = 100 f(0) = 20

SOFTWARE

FORMULArabfox2 = NDSolve [{r’[t] == 2r[t]-0.01r[t] f[t],

f’[t] == -f[t]+0.01r[t] f[t], r[0] ==200, f[0] ==10}, {r[t], f[t]}, {t, 0, 10}]

rabfoxplot2 = ParametricPlot [{rabfox2[[1, 1, 2]], rabfox2[[1, 2, 2]]}, {t, 0, 10}]

MATHEMATICA 4.0

Slope-fields of Predator-Prey equations

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