# Predator-Prey Problems Goal: Investigate the interaction of species via

Question

LAB #9

Predator-Prey Problems

Goal: Investigate the interaction of species via a particular predator-prey problem.

Required tools: Matlab routines pplane , dﬁeld and fplot.

Discussion

You will examine a predator-prey problem that has historical roots as noted in the

following excerpt of an article.

Predator-Prey Problem:

Why the Percentage of Sharks Caught in the Mediterranean Sea

Rose Dramatically During World War I

In the mid 1920’s the Italian biologist Umberto D’Ancona was studying the population variations of various species of ﬁsh that interact with each other. During his

research he came across some data on percentages-of-total-catch of several species of

ﬁsh that were brought into diﬀerent Mediterranean ports in the years that spanned

WWI. The data gave the percentage-of-total-catch of selachians (sharks, skates, rays,

etc) which are not very desirable as food ﬁsh. The data for the port of Fiume, Italy,

during 1914-1923 was as follows:

1914 1915

11.9% 21.4%

1916 1917 1918

22.1% 21.2% 36.4%

1919 1920 1921

27.3% 16.0% 15.9%

1922 1923

14.8% 10.7%

D’Ancona was puzzled by the very large increase in the percentage of selachians

during the period of the war. Obviously, he reasoned, the increase in the percentage of

selachians was due to the greatly reduced level of ﬁshing during this period. But how

does the intensity of ﬁshing aﬀect the ﬁsh population ? The answer to this question

was of great concern to D’Ancona in his research on the struggle for existence between

competing species. It was also of concern to the ﬁshing industry, since it would have

obvious implications for the way ﬁshing should be done.

What distinguishes the selachians from the food ﬁsh is that the selachians are predators, while the food ﬁsh are their prey; the selachians depend on the food ﬁsh for their

survival. At ﬁrst, D’Ancona thought that this accounted for the large increase of selachians during the war. Since the level of ﬁshing was greatly reduced during this period,

there were more prey available to the selachians, who therefore thrived and multiplied

rapidly. However, this explanation does not hold any water since there were also more

food ﬁsh during this period. D’Ancona’s theory only shows that there are more selachians when the level of ﬁshing is reduced; it does not explain why a reduced level of ﬁshing

is more beneﬁcial to the predators than their prey.

After exhausting all possible biological explanations of this phenomenon, D’Ancona

turned to his colleage, the famous Italian mathematician Vito Volterra. Hopefully,

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Volterra would formulate a mathematical model of the growth of the predator (selachians) and their prey (food ﬁsh), and this model would provide the answer to D’Ancona’s

question. Volterra began his analysis of this problem by separating all the ﬁsh into

the prey population x(t) and the predator population y(t). He then reasoned that the

food ﬁsh do not compete very intensively among themselves since their food is generally

plentiful. Hence, in the absence of the selachians, the food ﬁsh would grow according to

the Malthusian growth model x = ax, for some positive constant a. Next, he reasoned

that the number of contacts per unit time between predators and prey is bxy, for some

positive constant b. Hence x = ax − bxy. Similarly, Volterra concluded that the predators have a natural rate of decrease −cy, proportional to their present number, and that

they also increase at a rate dxy, proportional to their present number y and their food

supply x. Thus,

dx

dt = ax − bxy = x(a − by)

(∗)

dy

= −cy + dxy = y(−c + dx)

dt

This system of equations governs the interaction of the selachians and food ﬁsh in the

absence of ﬁshing.

Assignment

For the system (∗), we shall take a = 1, b = 0.5, c = 0.75, d = 0.25. Hence the

system becomes

x = x(1 − 0.5y)

y = y(−0.75 + 0.25x)

(∗∗)

You will investigate the behavior of solutions to this system and, at the end, examine

what happens when ﬁshing is permitted.

(1) Find the equilibrium points for this system. These are the points (x, y) obtained

by setting x = y = 0 (see page 350).

(2) Enter the above system into pplane . Choose your scale so that the origin is at the

center of the picture and the other equilibrium point is centered in the 1st quadrant.

Plot several orbits in each quadrant. It appears that all of the trajectories in the

1st quadrant are closed curves. (You will see why shortly.) Notice that they all

contain the equilibrium point. This is due to a general theorem which says that for

a “nice” autonomous system, any closed orbit must contain an equilibrium point.

What is the signiﬁcance for the populations of food ﬁsh and selachians of the fact

that the orbits are closed loops ?

(3) The pplane plot gives a plot of y against x. Explain why, as a function of x, the

system (∗∗) yields

dy

y(−0.75 + 0.25x)

=

.

dx

x(1 − 0.5y)

Use dﬁeld (not pplane ) to plot some solutions to this equation for various initial

data. Use the same ranges on x and y as part (2). You should ﬁnd that the solution

2

curves here are the same as the orbits in part (2) except that the top half and the

bottom half of each closed orbit must be plotted separately. Can you think of a

mathematical reason for this ? (Think about the fact that we are representing y

as a function of x.)

(4) The equation in part (3) is a separable equation. Solve it to show that the general

solution is given implicitly by

C=

(5) Let C =

1

2

yx0.75

.

e0.5y e0.25x

in the above equation. This should describe one orbit.

(a) Approximate the value of y which corresponds to x = 1. To do this, note

that when C = 1 the equation from part (4) is equivalent to

2

e0.25x

y

= 0.5y .

2×0.75

e

If x = 1, the value of the left side is approximately 0.6420. You can ﬁnd the

corresponding values of y by plotting (using fplot) the function on the right

side of this equality and reading oﬀ the appropriate values.

(b) Find the values of y corresponding to x = 2, 3, 4, · · · , 10. Explain why you

typically get 2 such values (if you get any). You will eventually ﬁnd that

there is no y value for the given x. Approximate the largest value x0 of x for

which there is a corresponding y. (For this, a graph of the left side of the

equation in (a) might be useful.) Explain why there is only one value of y

corresponding to x0 .

(c) Approximate the smallest value x1 of x for which there is a corresponding y.

Explain why there is only one value of y corresponding to x1 .

The fact that y exists only for x1 ≤ x ≤ x0 says that the orbit in question lies

over the interval [x1 , x0 ] on the x-axis. The fact that we get two y values for

each x strictly between x1 and x0 says that the orbit has a top and a bottom.

The fact that we get only one y value corresponding to x1 and x0 says that

the top and bottom curves meet at the end. This is almost a proof that the

orbit in question is a closed curve. In fact, it is not hard to complete this

analysis to show that the orbit is indeed a closed curve.

(6) In this part, we continue part (5) to show why all of the orbits are closed. Deﬁne

the functions F (x) and G(y) as follows

F (x) =

x0.75

e0.25x

and G(y) =

y

e0.5y

.

Notice that the orbits all are described by an equation of the form

G(y) =

3

C

.

F (x)

Furthermore, if the orbit begins at a point in the 1st quadrant, then C > 0. Why?

Using fplot, plot the graph of G(y) over the range 0 ≤ y ≤ 6. Using your graph,

explain why (typically) if there is a positive value of y for which the above equation

is valid, then there will be two. These values correspond to the top and bottom

of the closed orbit. Explain why lim F (x) = 0. Use this together with the graph

x→∞

of G(y) to explain why if x is suﬃciently large, then there are no values of y for

which the above equation is true. Similarly, show that for x suﬃciently near 0,

there are no such y. This shows that the orbit lies over a closed interval on the

x-axis. Thus, each orbit has a top and a bottom and lies over a closed interval in

the x-axis.

(7) Let us now return to our system (∗∗). Using pplane plot x and y as functions of

t for several diﬀerent initial conditions from the 1st quadrant. Do they appear to

be periodic ? If so, does the period appear to depend upon the initial condition?

(8) It can be proved that x(t) and y(t) are indeed periodic. (This follows from the

existence and uniqueness theorem together with the fact that the orbits are closed.

Ask your classroom instructor for details.) Let T be the period. We can compute

the average values of x(t) and y(t) by the formulas:

x=

1

T

T

x(t) dt and y =

0

1

T

T

y(t) dt .

0

Remarkably, we can evaluate these integrals exactly without knowing either x(t)

or y(t). Give reasons for each of the steps below for evaluating x and use a similar

argument to compute y for yourself:

y

= −0.75 + 0.25x

y

1

T

T

0

1

y (t)

dt =

y(t)

T

T

(−0.75 + 0.25x(t)) dt

0

1

1 T

(ln |y(T )| − ln |y(0)|) = −0.75 +

0.25x(t) dt

T

T 0

0 = −0.75 + 0.25x

3= x

Notice that the average value of x(t) does not depend upon how big the orbit is.

Where, in your phase plane picture, does the point (x, y) appear ?

(9) Explain how the system

x = x(1 − 0.5y) − αx

,

y = y(−0.75 + 0.25x) − αy

where α is a small constant, describes the eﬀect of ﬁshing on the population. Plot

some orbits if α = 0.05. Are there still closed orbits ? Compute the average

4

selachian and food ﬁsh populations if α = 0.05 and compare with the average

computed in part (8). How does this result relate to the original question asked

by D’ Ancona to Volterra ?

5

Lab Expectations – Lab 9: A “PredatorPrey” Problem

1. Find the equilibrium points for the system.

2. Plot in pplane with correct scale (scale w/ origin at center and other equilibrium at

center of first quadrant)

3. What is the significance for the populations of food fish and selachians of the fact

that the orbits are closed loops?

4. Explain why, as a function of x, that equation is true.

5. Plot this equation in dfield on the same range as above.

6. Why must the top and bottom half each closed orbit be plotted separately?

7. Solve the diff eq dy/dx.

8. a. With x=1 and C= 1∕2 , find y by plotting.

9. b. Find the values of y corresponding to x = 2, 3, … 10.

10. Why do you get two values, if any? Approximate the largest value xto 3 decimals)

0

(

for which there is no y value. Why is there only one yvalue corresponding to x

0

?

11. c. Approximate the smallest value xto three decimals) for which there is a

1

(

corresponding y. Why is there only one y value corresponding to x

1

?

12. PlotG(y)over0<y<6.

13. Explain why – if there is a positive value for y, then there will be two. Why is lim

xinf

F(x) = 0? Why is it that for large x, there are no y values and for small x, there is no

such y.

14. Plot x and y as functions of t.

15. Do they appear to be periodic? Does the period depend on the initial condition?

16. Give reasons for each step of the x integral evaluation. Evaluate the y integral

yourself. Where does the point (xbar, ybar) appear?

17. Explain how the system describes the effect of fishing on the population. Plot orbits

for alpha = .05. Are they still closed? Compute average selachian and food fish

populations. Compare these to averages from #8. How does this result relate to the

question of the last page?

Graphs – 2, 3, 5, 6, 7, 9

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