Ch 9.4: Competing Species

Ch 9.4: Competing Species

In this section we explore the application of phase plane analysis to some problems in population dynamics.

These problems involve two interacting populations and are extensions of those discussed in Section 2.5, which dealt with a single population.

Although the relationships discussed here are overly simple compared to the complex relationships in nature, it is still possible to acquire some insight into ecological principles from a study of these model problems.

Logistic Equations

Suppose that in some closed environment there are two similar species competing for a limited food supply.

For example, two species of fish in a pond that do not prey on each other but do compete for the available food.

Let x and y be the populations of the two species at time t.

As in Section 2.5, assume that the population of each species, in the absence of the other, is modeled by the logistic equation.

Thus

where 1 and 2 are the growth rates of the two populations, and 1/1 and 2 /2 are their saturation levels.

),(/),(/ 2211 yydtdyxxdtdx

Competing Species Equations

However, when both species are present, each will impinge on the available food supply for the other. In effect, they reduce each other’s growth rates and saturation populations.

The simplest expression for reducing growth rate of species x due to the presence of species y is to replace the growth factor 1 - 1x by 1 - 1x - 1y, where 1 is a measure of the degree to which species y interferes with species x.

Similarly, we model the reduced growth rate of species y due to presence of species x by replacing the growth factor 2 - 2y by 2 - 2y - 2x.

Thus we have the system of equations)(/),(/ 222111 xyydtdyyxxdtdx

Example 1: Population Equations (1 of 8)

Consider the system of equations

To find the critical points, we solve

obtaining (0,0), (0,0.75), (1,0), and (0.5,0.5). These critical points correspond to equilibrium solutions.

The first three points involve the extinction of one or both species. Only the fourth critical point corresponds to the long term survival of both species.

Other solutions are represented as trajectories in the xy-plane that describe the evolution of the populations over time.

)5.075.0(/),1(/ xyydtdyyxxdtdx

,0)5.075.0(,0)1( xyyyxx

Example 1: Direction Field (2 of 8)

A direction field for our system of equations is given below.

Only the first quadrant is depicted, as this corresponds to positive population size.

The heavy dots in the figure correspond to the critical points.

Based on the direction field, (0.5,0.5) attracts other solutions and therefore appears to be asymptotically stable.

The other three critical points appear to be unstable.

To confirm these observations,we can examine the linear approximations to each point.

Example 1: Linearization (3 of 8)

Our system of equations,

is almost linear, since F and G are twice differentiable.

To obtain the linear system near a critical point (x0, y0), we use the results of Section 9.3, given below.

Thus

0

0

0000

0000where,

),(),(

),(),(

yy

xx

v

u

v

u

yxGyxG

yxFyxF

v

u

dt

d

yx

yx


v

u

xyy

xyx

v

u

dt

d

000

000

5.0275.05.0

21

Example 1: Critical Point at (0,0) (4 of 8)

For the critical point (0,0), the approximating linear system is

The eigenvalues and eigenvectors are

and hence the general solution for this linear system is

Thus the origin is an unstable node of both the linear and nonlinear systems. The trajectories near origin are all tangent to y-axis, except for one trajectory that lies along the x-axis.

y

x

y

x

dt

d

75.00

01

1

0,75.0;

0

1,1 )2(

2)1(

1 ξξ rr

tt ececy

x 75.021 1

0

0

1





Thus (1,0) is an unstable saddle point of both the linear and nonlinear systems. One pair of trajectories approach (1,0) along the x-axis, while all other trajectories depart from (1,0).

v

u

v

u

dt

d

25.00

11

5

4,25.0;

0

1,1 )2(

2)1(

1 ξξ rr

tt ececv

u 25.021 5

4

0

1

Example 1: Critical Point at (0,0.75) (6 of 8)

For the critical point (0,0.75), the linear system is



Thus (0,0.75) is an unstable saddle point of both the linear and nonlinear systems. One pair of trajectories approach (0,0.75) along y-axis, while all other trajectories depart from (0,0.75).

v

u

v

u

dt

d

75.0375.0

025.0

1

0,75.0;

3

8,25.0 )2(

2)1(

1 ξξ rr

tt ececv

u 75.02

25.01 1

0

3

8

Example 1: Critical Point at (0.5,0.5) (7 of 8)

For the critical point (0.5,0.5), the linear system is



Thus (0.5,0.5) is an asymptotically stable node of both the linear and nonlinear systems.

v

u

v

u

dt

d

5.025.0

5.05.0

1

2,854.0

4

22;

1

2,146.0

4

22 )2(2

)1(1 ξξ rr

tt ececv

u 854.02

146.01

1

2

1

2

Example 1: Phase Portrait (8 of 8)

A phase portrait is given below, along with the direction field, for our nonlinear system

Note that the quadratic terms in these equations are all negative, and hence x' < 0 and y' < 0 for large x and y.

Thus the trajectories are directed inward towards (0.5,0.5).


Example 2: Population Equations (1 of 9)

Consider the system of equations

The critical points are (0,0), (1,0), (0,2), and (0.5,0.5). These critical points correspond to equilibrium solutions.

The first three points involve the extinction of one or both species. Only the fourth critical point corresponds to the long term survival of both species.

Other solutions are represented as trajectories in the xy-plane that describe the evolution of the populations over time.

)75.025.05.0(/),1(/ xyydtdyyxxdtdx

Example 2: Direction Field (2 of 9)

A direction field for our system of equations is given below, where the heavy dots correspond to the critical points.

Based on the direction field, (0.5,0.5) appears to be a saddle point, and hence unstable, while (1,0) and (0,2) appear to be asymptotically stable.

Thus only one species will eventually survive, and this species is determined by the initial conditions.





Thus the origin is an unstable node of both the linear and nonlinear systems. All trajectories leave the origin tangent to y-axis, except for one trajectory that lies along the x-axis.

y

x

y

x

dt

d

5.00

01

1

0,5.0;

0

1,1 )2(

2)1(

1 ξξ rr

tt ececy

x 5.021 1

0

0

1



The general solution for this linear system is

Thus (1,0) is an asymptotically stable node of both the linear and nonlinear systems.

One pair of trajectories approach (1,0) along the x-axis

All other trajectories approach (1,0) tangent to the line with slope –3/4, determined by the eigenvector (2).

v

u

v

u

dt

d

25.00

11

tt ececv

u 25.021 3

4

0

1


For the critical point (0,2), the linear system is

The general solution for this linear system is

Thus (0,2) is an asymptotically stable node of both the linear and nonlinear systems.

One trajectory approaches (0,2) along the line with slope 3, determined by the eigenvector (1).

All other trajectories approach (0,2) along the y-axis.

v

u

v

u

dt

d

5.05.1

01

tt ececv

u 5.021 1

0

3

1


For the critical point (0.5,0.5), the linear system is



v

u

v

u

dt

d

125.0375.0

5.05.0

5687.0

1

8/573

1,7844.0

16

575

;3187.1

1

8/573

1,1594.0

16

575

)2(2

)1(1

ξ

ξ

r

r

tt ececv

u 7844.02

1594.01 5687.0

1

3187.1

1


Thus for the critical point (0.5,0.5), we have

It follows that (0.5,0.5) is an unstable saddle point of both the linear and nonlinear systems.

One pair of trajectories approaches (0.5,0.5) as t , while the others depart from (0.5,0.5).

The entering trajectories approach (0.5,0.5) tangent to the line with slope of 0.5687, determined by the eigenvector (2).

tt ececv

u 7844.02

1594.01 5687.0

1

3187.1

1


A phase portrait is given below, along with the direction field.

Of particular interest is the pair of trajectories that enter the saddle point. These trajectories form a separatrix that divides the first quadrant into two basins of attraction.


Trajectories starting above the separatrix approach the node at (0,2), while those below approach the node at (1,0).

If initial state lies on separatrix, then the solution will approach the saddle point, but the slightest perturbation will send the trajectory to one of the nodes instead.

Thus in practice, one species will survive the competition and the other species will not.

Coexistence Analysis (1 of 7)

Examples 1 and 2 show that in some cases the competition between two species leads to an equilibrium state of coexistence, while in other cases the competition results in eventual extinction of one of the species.

We can predict which situation will occur by examining the governing equations

Note that this system is almost linear, since F and G are quadratic polynomials.

There are four cases to be considered, depending on the relative orientation of the lines

)(/),(/ 222111 xyydtdyyxxdtdx

0,0 222111 xyyx

Coexistence Analysis: Nullclines (2 of 7)

The graphs below show the relative orientation of the lines

The lines are called the x and y nullclines, respectively, because x' = 0 on the first and y' = 0 on the second.

The heavy dots represent equilibrium solutions.

Thus sustained coexistence is not possible in cases (a) and (b).

We show that sustained coexistence happens only in case (d).

0,0 222111 xyyx

Coexistence Analysis: Linear System (3 of 7)

Let (X,Y) be a critical point, with corresponding linear system

Since (X,Y) is a critical point, we solve

to obtain nonzero values of X and Y, as given below:

Further, with (X,Y) a critical point, the linear system reduces to

v

u

XYY

XYX

v

u

dt

d

2222

1111

2

2

0,0 222111 XYYX

2121

2112

2121

1221 ,

YX

v

u

YY

XX

v

u

dt

d

22

11

Coexistence Analysis: Eigenvalues (4 of 7)

The eigenvalues of the linear system are

If 12 – 12 < 0, then the eigenvalues are real and of opposite sign, and hence (X,Y) is an unstable saddle point, and thus sustained coexistence is not possible.

If 12 – 12 > 0, then the eigenvalues are real, negative and unequal, or complex. It can be shown that the eigenvalues are not complex, and hence (X,Y) is an asymptotically stable node, and thus sustained coexistence is possible.

In Example 1, 12 – 12 = (1)(1) - (1)(0. 5) = 0.5 > 0.

In Example 2, 12 – 12 = (1)(0.25) - (1)(0.75) = -0.5 < 0.

2

)(4)( 21212

21212,1

XYYXYXr

Coexistence Analysis: Case (c) (5 of 7)

In case (c), we have

These inequalities, together with

yield 12 – 12 < 0.

Therefore sustained coexistence is not possible in case (c).

21121

1

2

21221

2

2

1

1 or ,or

,0,02121

2112

2121

1221

YX

Coexistence Analysis: Case (d) (6 of 7)

In case (d), we have

These inequalities, together with

yield 12 – 12 > 0.

We can also show that the other criticalpoints are unstable, and thus the two populations approach the equilibriumstate of coexistences in case (d).

21121

1

2

21221

2

2

1

1 or ,or

,0,02121

2112

2121

1221

YX

Discussion of Coexistence Conditions (7 of 7)

For our competitive species equations,

sustained coexistence is possible when 12 – 12 > 0.

The ’s are a measure of the inhibitory effect that the growth of each population has on itself.

The ’s are a measure of the inhibitory effect that the growth of each population has on the other species.

Thus when 12 – 12 > 0, the competition is weak, and the species can coexist.

When 12 – 12 < 0, the competition is strong, and the species cannot coexist – one must die out.

)(/),(/ 222111 xyydtdyyxxdtdx

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Ch 9.4: Competing Species