ssg6.5-6 stability and bifurcations

6.5. PHASE LINES AND CLASSIFYING EQUILIBRIA 805

6.5 Phase Lines and Classifying Equilibria

In this section and the next section, we focus on autonomous (independent of time) differential equations

dy

dt= f(y)

In the previous section we noted that the slope field for an autonomous differential equation is time-independent.

Since each vertical line in the slope field contains all the information about the slopes, the slope fields contains an

infinite amount of redundancy. In this section we trim off this redundancy using phase lines and discuss classifying

equilibria , the y-values for which f(y) = 0.

Phase lines

In the last section we sketched slope fields by determining where the slope is zero (nullcline), and where it is positive

and where it is negative. In this section, we consider a phase-line diagram that collapses the two-dimensional slope

field to the y-axis without losing any information regarding the qualitative behavior of solutions to the differential

equation dydt = f(y) (e.g. see Figure 6.26). The following procedure creates a phase line.

Phase Lines

To draw a phase line for dydt = f(y),

Step 1. Draw a vertical line corresponding to the y-axis.

Step 2. Draw solid circles on this line corresponding to the equilibria of dydt = f(y).

That is, y-values where f(y) = 0.

Step 3. Draw an upward arrow on intervals where f(y) > 0. On these intervals,

solutions of the differential equation are increasing.

Step 4. Draw a downward arrow on intervals where f(y) < 0. On these intervals,

the solutions of the differential equation are decreasing.

Example 1. Phase lines for clonal genotypes

Consider two clonally reproducing lines of the same species (i.e. individual replicate themselves rather than

reproducing sexually) exhibiting two genotypes a and A and whose per-capita growth rates are ra and rA, respectively.

Suppose these two clonal lines are growing together in the same population and let y denote the proportion of genotype

©2010 Schreiber, Smith & Getz

806 6.5. PHASE LINES AND CLASSIFYING EQUILIBRIA

Figure 6.26: An illustration of how the three qualitatively different solution zones y < 0, 0 < y < 1, and y > 1,separated by the two equilibrium solutions y = 0 and y = 1 associated with the logistic equation dy

dt = y(1 − y), canbe collapsed on the y axis by removing (or projecting down) the time axis t.

a in this population. It it left as an exercise(see Problem 39) to show that the variable y satisfies the equation

dy

dt= (ra − rA)y(1 − y).

a. Draw the phase line for this equation when ra > rA.

b. Draw the phase line for this equation when ra < rA.

c. Discuss why this makes sense.

Solution.

a. Begin by drawing the y-axis. The equilibria are determined by the solutions of

0 = (ra − rA)y(1 − y)

Since the equilibria are y = 0 and y = 1, we draw solid circles on the y axis at these y-values. Since

ra > rA, we have dydt > 0 for 0 < y < 1 and we draw an upward arrow on this interval. Since dy

dt < 0 for

y > 1 and y < 0, we draw downward arrows on these intervals. This results in the phase line illustrated

in Figure 6.27a.



0

1

0

1

a. ra > rA b. ra < rA

Figure 6.27: Phase lines for dydt = (ra − rA)y(1 − y)

b. Again begin by drawing the y-axis. The equilibria are determined, as before, by the solutions of

0 = (ra − rA)y(1 − y)

Since the equilibria are y = 0 and y = 1, we draw solid circles on the y-axis at these y-values. Since

ra < rA, we have dydt < 0 for 0 < y < 1 and we draw a downward arrow on this interval. Since dy

dt > 0 for

y > 1 and y < 0, we draw upward arrows on these intervals. This results in the phase line illustrated in

Figure 6.27b.

c. If the per-capita growth rate of genotype a is greater than the per-capita growth rate of genotype A, then

we would expect genotype a to become more and more prevalent in the population. Hence, provided that

y > 0 initially, y approaches 1 as seen in the phase line for part a. Conversely, if the per-capita growth

of genotype a is less than the per-capita growth rate of genotype A, then we would expect a to become

less and less prevalent in population. Hence y should approach 0 as seen in the phase line for part b.

2

In the last example, we found the phase lines from an equation, but sometimes we have a graph (or data leading

to a graph) and not an equation. The next example shows us how to find the phase lines in such a case.

Example 2. From graphs to phase lines to solutions

Let the graph of f(y) be as shown in Figure 6.28.

a. Draw a phase line for dydt = f(y).

b. Sketch solutions for this differential equation that satisfy y(0) = −1.1, y(0) = 1.1, and y(0) = 0.9.

Solution.



Figure 6.28: Graph of dydt = f(y)

a. Since the graph of f(y) intersects the y-axis at the points −2, −1, 1, and 2, these y-values are the

equilibria of y′ = f(y). We draw solid circles at these points of the phase line. Since f(y) > 0 on the

intervals (−∞,−2) and (1, 2), we draw upward arrows on these intervals, as shown in Figure 6.29a. For

all the other intervals, (−2,−1), (−1, 1), and (2,∞), we draw downward arrows.

-1

1

-2

2

0.2 0.4 0.6 0.8 1 1.2 1.4t

-2

-1

0

1

2

y

a. Phase line b. Solutions to differential equation

Figure 6.29: Phase line and solutions for given graph

b. According to the phase line, a solution initiated at y = −1.1 initially decreases slowly (as it is near

the equilibrium y = −1), decreases more rapidly, and numerical/analytical methods can be use to show

that this solution asymptotes at the equilibrium y = −2. A solution initiated at y = 1.1, initially

increases slowly, increases more rapidly, and and numerical/analytical methods can be use to show that

this solution at y = 2. A solution initiated at y = 0.9, initially decreases slowly, decreases more rapidly,

and and numerical/analytical methods can be use to show that this solution at y = −1. These solutions

are shown in Figure 6.29b.

2

The equation y = (ra − rA)y(1 − y) in Example 1 has a special name in the context of evolutionary game



theory. It is called the replicator equation. Evolutionary game theory was developed in the late 1970s, by the

eminent theoretical evolutionary biologist, John Maynard-Smith (1920-2004). (For more information about one of

the world’s greatest evolutionary biologists see the HISTORICAL QUEST in the problem set). Perhaps the best known

of his games is the Hawk–Dove game which describes under what conditions non-aggressive behaviors can persist in

a population.

In general, for any two inherited contrasting strategies, the growth rates ra and rA for genotype a (e.g. hawks) and

genotype A (e.g. doves) respectively in the replicator equation are constructed from a two-by-two table. This table

is known as the pay-off matrix and it tells us how much payoff (benefit if positive, cost if negative) an individual gets

following a pairwise interaction with another individual. The payoffs when a meets a is denoted by Paa. Similarly,

we use PaA, PAa, and PAA to denote the payoffs when a meets A, A meets a, and A meets A, respectively. We

summarize this information in Table 6.3

Table 6.3: Payoff MatrixType a Type A

Type a (proportion y) Paa PaA

Type A (proportion (1 − y)) PAa PAA

To determine the per-capita (i.e. proportional) growth rate of a genotype, we find the expected payoff by cal-

culating the product of the chance of meeting an individual playing a particular strategy and the corresponding

payoff.

For genotype a, this expected payoff is

ra = yPaa + (1 − y)PaA

and for genotype A the expected payoff is

rA = yPAa + (1 − y)PAA

Substituting these expressions for ra and rA into dydt = y(1−y)(ra−rA), we get the two-strategy replicator equations.



Two-Strategy

Replicator Equations

The replicator equation describing the proportion y(t) of the population of geno-

types a and A, each playing different strategies with payoff interactions Pij , (i, j = a

and A) is

dy

dt= y(1 − y)(ra − rA)

= y(1 − y) [PaA − PAA + y(Paa + PAA − PaA − PAa)]

The following assumptions are made regarding the Hawk-Dove game:

A population of individuals competes for a limiting “resource” such as mates, food, or shelter. To win this

resource, individuals engage in pair-wise contests and play one of two strategies, hawk or dove. Individuals playing

the hawk strategy constantly escalate the intensity of the contest until they either they get the resource or they get

injured. Individuals playing the dove strategy leave the contest whenever their opponent escalates the conflict.

We consider this game in the next two examples.

Example 3. The Hawk-Dove replicator equation

Suppose a hawk gets a payoff of V > 0 every time it meets a dove and the dove gets 0. Further every time two

doves meet they share the payoff V , while if two hawks meet they escalate the contest until one gets the net payoff

V and the other pays a cost C > 0. What are the payoff matrix entries for this contest and the replicator equation

that describes the frequency of doves in the population?

Solution. Let a denote doves and A denote hawks. In this game, the payoffs are:

Paa =V

2, PaA = 0, PAa = V, and PAA =

V − C

2

This last value represents what the average hawk obtains in a hawk-hawk encounter. If we now substitute these

values in the two-strategy replicator equation we obtain

dy

dt= y(1 − y)(ra − rA)

= y(1 − y)

[

0 − V − C

2+ y

(V

2+

V − C

2− 0 − V

)]

= y(1 − y)

[C − V

2− Cy

2

]



2

Example 4. Dynamics of a Hawk-Dove game

Consider a Hawk-Dove game with a payoff of 2 and a cost of 3. Sketch the phase line and then discuss the

evolutionary implications.

Solution. When V = 2 and C = 3, we obtain from the previous example the specific replicator equation

dy

dt= y(1 − y)

(1

2− 3y

2

)

The equilibria solutions are values for which dy/dt = 0:

y = 0, y = 1, y = 1/3

For 0 < y < 1/3, dydt > 0. To see this, choose a value in the interval, say y = 1

6 and calculate

dy

dt=

1

6

(

1 − 1

6

) (1

2− 1/2

2

)

> 0

For 1/3 < y < 1, dydt < 0. To see this, choose a representative value, say y = 1

2 and calculate

dy

dt=

1

2

(

1 − 1

2

) (1

2− 3/2

2

)

< 0

Finally, for y > 1, dydt > 0. To see this, choose some representative value, say y = 2. Then,

dy

dt= 2(1 − 2)

(1

2− 6

2

)

> 0

The phase line is shown in Figure 6.30a.

The phase line implies that if initially hawks and doves are present, then the population approaches an equilibrium

consisting of 13 doves and 2

3 hawks. This approach to this equilibrium is illustrated in Figure 6.30b. 2

The equilibrium in Example 3 support multiple strategies in the population. Such an equilibrium is called a

polymorphic equilibria. We can understand the growth rates of hawks and doves at low frequencies as follows.

Imagine the population consists mainly of doves and only a few hawks. Individuals are most likely to have a contest

with a dove. For, a dove this means that they get on average a payoff of V/2 = 2 × 12 = 1. For a hawk this means

they get a payoff of V = 2. Therefore, the hawk numbers would grow at twice the rate of doves. Alternatively,



1/3

1

0

5 10 15 20t

0.2

0.4

0.6

0.8

1y

a. Phase line b.Solution to the differential equation

Figure 6.30: Phase line and solutions to a Hawk-Dove game

consider a population consisting mostly of hawks and only a few doves. An individual engaging in a contest is most

likely to encounter a hawk. A hawk, on average, gets a payoff of (V − C)/2 = (2 − 3)/2 = −1/2. A dove gets a

payoff of 0. So hawk frequency will decline.

Classifying Equilibria

When a system starts at an equilibrium, it remains there for all time. However, in the real world, biological systems

are constantly subject to environmental perturbations (small changes). Thus, if a system starting at equilibrium

is slightly perturbed from equilibrium, we need to ask does it tend to return to the equilibrium or not? When

the system tends to return to the equilibrium, we call the equilibrium stable. Otherwise, we call it unstable. More

precisely, we make the following definitions.

Classification of

Equilibria

An equilibrium y∗ for dydt = f(y) is classified as follows:

Stable: f(y) > 0 for all y < y∗ near y∗ and f(y) < 0 for all y > y∗ near y∗.

Solutions initiated near the equilibrium tend toward the equilibrium in forward

time (i.e. as t → ∞).

Unstable: f(y) < 0 for all y < y∗ near y∗ and f(y) > 0 for all y > y∗ near

y∗. Solutions initiated near the equilibrium tend toward the equilibrium in

backward time (i.e. as t → −∞).

Semi-stable: Either f(y) < 0 for all y 6= y∗ near y∗ or f(y) > 0 for all y 6= y∗ near

y∗ Solutions initiated near one side (resp. other side) of the equilibrium tend

toward the equilibrium in backward (resp. forward) time.

Graphical depictions of these definitions are provided in Figure 6.31.



y* y* y*

Stable Unstable Semistable

Figure 6.31: Graphical characterization of classifying equilibria

Example 5. Classifying equilibria

Classify the equilibria for dydt = f(y) where the graph of f(y) is the graph given in Figure 6.27 in Example 2 which

we repeat here for convenience.

-2 -1 1 2y

-10

-5

5

Solution. Previously, we sketched the phase lines for dydt = f(y) and found four equilibria: y = −2, y = +2, y =

−1, y = +1. From the phase line sketch in Example 2, Fig. 6.29, we classify the equilibria as follows: y = −2 and

y = 2 are stable, y = 1 is unstable, and y = −1 is semi-stable. 2

Example 6. Membrane potential

The voltage V across the membrane of a neuron is maintained by voltage-gated (i.e. controlled) protein channels

embedded in the cell membrane. These channels regulate the flow of positively charged potassium ions and negatively

charged organic molecules out of the cell, and negatively charged chlorine ions and positively charged sodium ions

into the cell. If the membrane is perturbed from its resting potential V0 by a small input current (e.g. coming from

another neuron), it will return to its resting potential. If this perturbing current, however, is sufficiently large to

cause V (t) to drop below a critical threshold level Vc, then the sodium ions flow across the membrane until the

voltage stabilizes at a new depolarized equilibrium level Vd. Show that model

dV

dt= −k(V − V0)(V − Vc)(V − Vd)



exhibits these characteristics by finding and classifying its equilibria for the values V0 = −70 mV, Vc = −30 mV,

Vd = 55 mV and k = 1.

Solution. For the constants in question, the right-hand-side of the equation is

f(V ) = −(V + 70)(V + 30)(V − 55)

This function is a cubic in the variable V with roots at V = −70, −30, and 55. The graph of this cubic is given by

−80 −60 −40 −20 0 20 40 60−1

−0.5

0

0.5

1

1.5

2x 10

5

V

From the graph, we see that V0 = −70 mV and Vd = 55 mV are stable as required, and that Vc = −30 mV is

unstable, also as required.

2

Linearization

An analytical approach to classifying equilibria involves linearizing about the equilibria. Suppose y∗ is an equilibrium

for dydt = f(y). Consider a = f ′(y∗). Since f(y∗) = 0, a linear approximation to f(y) for y near y∗ is given by

f(y) ≈ f(y∗) + f ′(y∗)(y − y∗) linear approximation (6.1)

= a(y − y∗) since f(y∗) = 0 and f ′(y∗) = a (6.2)

Hence,

dy

dt≈ a(y − y∗)

for y values near y∗. As you are asked to show in Problem 37, the solution to

dy

dt= a(y − y∗)



satisfying y(0) = y0 is

y(t) = (y0 − y∗)eat + y∗

We can use this solution as a first-order approximation for the solution to dydt = f(y) satisfying y(0) = y0. This

approximation y(t) = (y0 − y∗)eat + y∗ to the solution remains reasonable provide y(t) remains near y∗. Using this

approximation, one can prove the following theorem.

Theorem 6.1. Linearization

Let dydt = f(y) have an equilibrium at y = y∗ is

Stable if f ′(y∗) < 0 then y∗

Unstable if f ′(y∗) > 0 then y∗ is unstable.

Indeterminable if f ′(y∗) = 0: no conclusion is possible without looking at higher order derivatives.

Informally the result follows from the fact that

y(t) ≈ (y0 − y∗)eat + y∗ where a = f ′(y∗)

implies the rate at which solutions move towards or away from y∗ is given approximately by eat. Thus when

a = f ′(y∗) < 0 solutions starting near y∗ move toward y∗ and when a = f ′(y∗) > 0 solutions starting near y∗ move

away from y∗.

Example 7. Population resilience

Consider two populations whose dynamics are described by

dN

dt= N

(

1 − N

10, 000

)

anddP

dt= 0.5 P

(

1 − P

10, 000

)

a. Find the equilibria and use linearization to classify.

b. Describe in what ways the populations are similar and dissimilar.

Solution.



a. For both populations, the equilibria are given by 0 and 10, 000. For the first model set f(N) = N(1 −

N/10, 000), we find

f ′(N) =

(

1 − N

10, 000

)

− N

10, 000

Checking the equilibria:

Equilibria Evaluate Classification

N = 0 f ′(0) = 1 unstable

N = 10, 000 f ′(10, 000) = −1 stable

For the second model, set g(P ) = dP/dt, we find

Equilibria Evaluate Classification

P = 0 g′(0) = 12 unstable

P = 10, 000 g′(10, 000) = − 12 stable

b. The populations are similar in that both populations have equilibria at 0 and 10, 000 which are unstable

and stable, at 0 and 10,000, respectively. Hence, populations tend to approach the equilibrium value of

10, 000.

The populations differ in that P (the second model) tends to grow less rapidly at low densities; i.e.

g′(0) = 12 < f ′(0) = 1. Moreover, if the populations are at the equilibrium of 10,000, the P population

recovers less rapidly from a perturbation; i.e. g′(10, 000) = − 12 > f ′(10, 0000) = −1.

2

When one population recovers more rapidly from environmental perturbations than another population (as with

P versus the N population in Example 7) it is said to be more resilient.

Example 8. Hawk-Dove game revisited

Consider the Hawk-Dove game

dy

dt= y(1 − y)

(1

2− 3y

2

)

where y is the frequency of doves in the population.

a. Use linearization to classify each of the equilibria.

b. Use your work from part a to determine whether the hawks increase more rapidly at low frequencies or

the doves increase more rapidly at low frequencies.



Solution.

a. Let

f(y) = y(1 − y)

(1

2− 3y

2

)

As we have seen, the equilibria are y = 0, y = 1, and y = 1/3. To linearize, we need the derivative:

f ′(y) = (1 − y)

(1

2− 3y

2

)

− y

(1

2− 3y

2

)

− y(1 − y)3

2

Evaluated at y = 0, we obtain

f ′(0) =1

2> 0

Hence, the equilibrium y = 0 is unstable. Since

f ′(1) = 1 > 0,

the equilibrium y = 1 is unstable. Since

f ′(1

3) = −1

3< 0,

the equilibrium y = 13 is stable.

b. Since f ′(1) = 1 > f ′(0) = 12 we see that hawks at low frequency increase more rapidly than doves at low

frequency.

2

Example 9. Linearization of membrane voltage model

By considering the linearization of the model

dV

dt= −k(V − V0)(V − Vc)(V − Vd)

classify the equilibria for the case V0 < Vc < Vd.

Solution. Define

f(V ) = −k(V − V0)(V − Vc)(V − Vd)

The roots of the function f(V ) are the equilibria V = V0, Vc, and Vd. which by two applications of the product rule



implies

f ′(V ) = −k[(V − Vc)(V − Vd) + (V − V0)(V − Vd) + (V − V0)(V − Vc)]

At V = V0, we have

f ′(V0) = −k[(V0 − Vc)(V0 − Vd) + 0 + 0] = −k(V0 − Vc)(V0 − Vd).

Since V0 < Vc < Vd, f ′(V0) < 0 and V = V0 is stable. At V = Vc, we have

f ′(Vc) = −k(Vc − V0)(Vc − Vd)

Since V0 < Vc < Vd, f ′(Vc) > 0 and V = Vc is unstable. At V = Vd, we have

f ′(Vd) = −k(Vd − V0)(Vd − Vc)

Hence, f ′(Vd) < 0 and V = Vd is stable. 2

Problem Set 6.5

LEVEL 1 – DRILL PROBLEMS

Draw phase lines, classify the equilibria, and sketch a solution satisfying the specified initial value for the equations

in Problems 1 to 10.

1. dydt = 1 − y2, y(0) = 0

2. dydt = 2 − 3y, y(0) = 2

3. dydt = −7, y(0) = −2

4. dydt = 10, y(0) = 5

5. dydt = y(y − 10)(20 − y), y(0) = 9

6. dydt = y(y − 5)(25 − y), y(0) = 7

7. dydt = sin y, y(0) = 0.1

8. dydt = 1 − sin y, y(0) = −0.6

9. dydt = y2 − 2y + 1, y(0) = 0



10. dydt = y3 − 4y, y(0) = 0.1

Draw a phase line for dydt = f(y) for the graphs shown in Problems 11 to 14. Sketch the requested solutions.

11. y(0) = −1.1, y(0) = 1.1, y(0) = 0.9

12. y(0) = −0.1, y(0) = 0.9, y(0) = 1.1

13. y(0) = −2, y(0) = 1, y(0) = 2

14. y(0) = −0.1, y(0) = 1.9, y(0) = 3.

Linearize about the equilibrium in Problems 15 to 20 and classify it.

15. dydt = 4 − y2, y∗ = 2



16. dydt = cos y, y∗ = π

2

17. dydt = 1√

2− cos y, y∗ = π/4

18. dydt = 2y − y2 − y10, y∗ = 0

19. dydt = 3 − y, y∗ = 3

20. dydt = y(10 − y)(100 − y), y∗ = 100

Sketch the phase line and classify the equilibria for the Hawk-Dove game with the values V and C given in Problems

21 to 24.

21. V = 2, C = 2

22. V = 4, C = 2

23. V = 3, C = 2

24. V = 2, C = 4

Sketch the phase line and classify the equilibria for the replication equations with the indicated payoffs in Problems

25 to 28.

25. Paa = 2, PaA = 1, PAa = 1, and PAA = 2

26. Paa = 1, PaA = 2, PAa = 3, and PAA = 4

27. Paa = −1, PaA = 2, PAa = 1, and PAA = −1

28. Paa = 2, PaA = −1, PAa = −1, and PAA = 3

LEVEL 2 – APPLIED PROBLEMS AND THEORY

29. “The Stag Hunt” is a story told by Rousseau in A Discourse on Inequality that became a game. In the story

we read

If it was a matter of hunting a deer, everyone well realized that he must remain faithful to his post;

but if a hare happened to pass within reach of one of them we cannot doubt that he would have gone

off in pursuit of it without scruple...



To turn this into an evolutionary game, consider a population of individuals that can engage in group hunting

for larger game (e.g. packs of wolves etc.) Each individual in this population can play one of two strategies,

hunt stag (i.e. remain loyal to the pack even if an alternative prey comes along) or hunt hare (i.e. run after

hares whenever he see them). In his writing, Thomas Hobbes present informal arguments about this game that

suggest the following payoff matrix

Hunt Stag Hunt Hare

Hunt Stag 7.5 4

Hunt Hare 7 5

a. Find the replicator equation.

b. Sketch the phase line, and classify the equilibria.

c. Discuss how the outcome of the evolutionary game depends on the initial composition of the popu-

lation.

30. Consider two scenarios based on Problem 29:

i. In a population of stag hunters, a few individuals decide to hunt hares.

ii. In a population of hare hunters, a few individuals decide to hunt stag.

Use linearization to determine in which of these scenarios, the “defecting” individuals are more rapidly excluded.

31. (Evolution of Cooperation, Part I) Consider a population with two strategies, cooperate and defect. Individuals

that cooperate provide a benefit B to their opponent and pay a cost C for providing this benefit. Defectors

provide no benefits to their opponents and pay no cost. Under these assumptions, we get the following payoff

matrix.

Cooperate Defect

Cooperate B − C −C

Defect C 0

a. Write down a replicator equation for this payoff matrix.

b. Assuming B > 0 and C > 0, sketch the phase line for the replicator equation.

c. Discuss the implications of your phase line.

32. (Evolution of Cooperation, part II) In Problem 31, cooperation could not evolve. However, cooperation is seen

in natural populations. In this problem, we investigate how individuals that interact frequently and respond to



the strategy of their opponents can promote the evolution of cooperation. Let us imagine that each time two

opponents meet they interact on average n times. Individuals can play one of two strategies: defect always or

tit-for-tat in which case an individual initially cooperates but switches to defecting if their opponent defected.

a. If each time individuals interact the individuals payoffs are as in Problem 31, then discuss why the

payoff matrix should be

Tit for Tat Defect

Tit for Tat n(B − C) −C

Defect B 0

b. Write down a replicator equation for this game.

c. Assume B = 3 and C = 2. Sketch phase lines for n = 2, 3, 4.

d. Discuss the implications for the evolution of cooperation.

33. To account for the effect of a generalist predator (with a type II functional response) on a population, ecologist

often write differential equations of the form

dN

dt= 0.1N

(

1 − N

1, 000

)

− 10N

1 + N

where N is the population abundance and t is time (in years). The first term of the equation corresponds to

logistic growth and the second term corresponds to saturating predation.

a. Sketch the phase line for this system.

b. Discuss how the fate of the population depends on its initial abundance.

34. Construct the phase line for the model

dV

dt= −2V 3 − 20V 2 + 3000V

and hence demonstrate that this equation belongs to the class of membrane voltage models presented in

Example 6.

35. Use a phase line diagram to discuss the behavior of the membrane voltage models presented in Example 6 with

constants k = 3, V0 = −65 mV, Vc = 40 mV and Vd = 40 mV. Does this membrane have the property that it

is able to switch between two states when perturbed by a current?

36. Historical Quest John Maynard Smith, or JMS as he was almost always known, was professor emeritus at

the University of Sussex, and one of the world’s great evolutionary biologists.



John Maynard Smith (1920-2004)

JMS introduced mathematical modeling from game theory into the study of mathematical biology, and com-

pletely revolutionized the way that biologists think about behavioral evolution. Jonathan Weiner wrote “A

Conversation With John Maynard Smith” which was published in the September 2000 issue of Natural History,

before JMS died. Here is what he said:

A classical geneticist and leading theorist in evolutionary biology, John Maynard Smith started out

as an engineer and worked as a “stress man” during World War II, calculating the stresses in airplane

wings. Since then, he has applied his knowledge of mathematics to some of the greatest problems in

evolution–exploring the stress points, the places where the theory threatens to pop its rivets.

Maynard Smith is best known for using game theory to explain the jousting matches that one sees

among the males of many species, from sticklebacks to sea lions, from stag beetles to stags. “You’d

simply expect them to sort of hit the other chap in the groin as quickly as possible,” he says, “and

yet there’s rather little escalated fighting and a great deal of display in settling contests.” It’s almost

as if the combatants are cooperating–a paradox the biologist explains by invoking the mathematics

of nonzero-sum contests and win-win situations.

At the University of Sussex in England, where he works, Maynard Smith is closely involved with a

group of colleagues he calls “The Institute for the Study of Tiny Minds”: neurobiologists working

on the behavior of ants, bees, worms, and snails. He also talks daily with colleagues across disci-

plines who, like him, are trying to apply the theory of natural selection to the design of robots and

computers.



Because of his stature, he received numerous prestigious awards, and for this Historical Quest, you should

research and say a few words about each of these awards achieved by JMS, or in the case of the last one,

established in his honor.

a. Balzan Prize

b. Crafoord Prize

c. Kyoto Prize

d. John Maynard Smith Prize

37. Verify that the solution to

dy

dt= a(y − y∗)

satisfying y(0) = y0 is given by

y(t) = (y0 − y∗)eat + y∗

38. Show that the linearization theorem is inconclusive when the derivative equals zero at the equilibrium by

considering the stability of the equilibria to the equations

dydt = y3 dy

dt = −y3 dydt = y2

39. Consider a population of clonally reproducing individuals consisting of two genotypes a and A with per-capita

growth rates, ra and rA, respectively. If Na and NA denote the densities of genotypes a and A, then

dNa

dt= ra Na

dNA

dt= rA NA

Also, let y = Na

Na+NAbe the fraction of individuals in the population that are genotype a. Show that y satisfies

dy

dt= (ra − rA)y(1 − y)


6.6. BIFURCATIONS 825

6.6 Bifurcations

Biological systems can exhibit a multitude of dynamical behaviors which can change abruptly or gradually in

response to external perturbations. The term bifurcation is used in the context of differential equation models to

denote a change in the stability of equilibria or the types of solutions that occur as a parameter in the model is

varied. In this section, we provide an introduction to bifurcation theory. This theory provides a systematic approach

to studying qualitative changes in the dynamical behavior of a differential equation. We will use the notation

dy

dt= f(y, a)

to represent an expression in y and a where y is the variable and a is a parameter. Our goal is to understand how

the qualitative behavior of this equation depends on a. More precisely, we will study how the phase line varies with

the parameter a.

In this section, we illustrate bifurcation theory with populations subjected to harvesting and the firing rates of

neural populations.

Sudden population disappearances

Example 1. Harvesting queen conch

Consider a population of queen conch in the Bahamas whose dynamics are given by

dy

dt= 10 y

(

1 − y

10, 000

)

− a

where t is time in years, y is number of conch, and a is the constant annual harvesting rate.

a. Draw phase lines for a = 0, a = 21, 000, and a = 30, 000.

b. Discuss the biological implications of these phase lines.

c. Determine how the number of equilibria depends on a.

Solution.


826 6.6. BIFURCATIONS

a. Consider a = 0. The equilibria are given by the solutions of

0 = 10y

(

1 − y

10, 000

)

− 0

Solving this equation yields the equilibria y = 0 and y = 10, 000. Since dydt > 0 for 0 < y < 10, 000 and

dydt < 0 for the other intervals, we obtain the phase line as shown in Figure 6.32a.

0

10,000

3,000

7,000

a = 0 a = 21, 000 a = 30, 000

Figure 6.32: Phase lines for the density (y) of conch for the three harvesting levels a, as labeled, inserted into theconch harvesting equation.

Consider a = 21, 000. The equilibria are given by solutions to

0 = 10 y

(

1 − y

10, 000

)

− 21, 000

which yields y = 3, 000 and y = 7, 000. Since dydt > 0 for 3, 000 < y < 7, 000 and dy

dt < 0 elsewhere, we get

a phase line as shown in Figure 6.32b.

Finally, consider a = 30, 000. In this case there are no equilibria because

0 = 10y

(

1 − y

10, 000

)

− 30, 000

has no real roots. Since dydt < 0 for all y, we get a phase line as shown in Figure 6.32c.

b. The phase lines in Figure 6.32 show that as a increases, the number of equilibria goes from two to zero.

In particular, at sufficiently high harvesting rates, the population is unable to persist at an equilibrium.

c. To determine how the equilibria depend on the harvesting rate a, we need to solve

0 = 10 y

(

1 − y

10, 000

)

− a

for y. Using the quadratic formula,

y = 5, 000± 100

√

2, 500 − a

10



Hence, we obtain two equilibria provided that

2, 500 − a/10 > 0

which occurs if and only if a < 25, 000. If a = 25, 000, then we get only one equilibrium given by

y = 5, 000. Finally, if a > 25, 000, then(2, 500− a

10

)is negative and there are no equilibria. Therefore,

a change in the number of equilibria occurs at a = 25, 000.

2

Example 1 illustrates that the phase line of dydt = f(y, a) can vary substantially as you vary the parameter a.

Moreover, it shows that at certain parameter values (i.e. a = 25, 000 in Example 1) there is a qualitative change in the

phase line. These values are important enough to have their own name: bifurcation values. We define bifurcation

values as the value of a parameter in an equation where either the number of equilibrium solutions changes or the

stability properties of these solutions undergo a transition from stable to unstable. A simple way to graphically

summarize how the behavior of the system depends on a is to graph something known as a bifurcation diagram.

The procedure for constructing such a diagram is summarized as follows.

Bifurcation diagram

A bifurcation diagram summarizes the behavior of a system in the a–y plane and

can be created as follows

Step 1. Draw that a-axis (horizontal) and the y-axis (vertical).

Step 2. Sketch the set of equilibria in the ay-plane. That is, the set of points (a, y)

that satisfy 0 = dydt = f(y, a).

Step 3. Determine in which regions of the ay-plane, dydt is positive or negative.

Step 4. For a collection of a values, draw a phase line. In particular, draw phase

lines at bifurcation values of a and at values of a that lie between bifurcation

values.

Example 2. Sudden queen conch disappearances

Sketch a bifurcation diagram for Example 1.

dy

dt= 10 y

(

1 − y

10, 000

)

− a



with a ≥ 0 and y ≥ 0. Discuss the implications for population harvesting.

Solution. We begin by solving

0 = 10 y

(

1 − y

10, 000

)

− a

for a and graphing a = 10y(1 − y/10, 000) in the ay-plane. The graph is a parabola as shown in Figure 6.33a.

0 5000 10000 15000 20000 25000 30000a

0

2000

4000

6000

8000

10000

y

a. Graph of dydt = 0 b. Bifurcation diagram

Figure 6.33: The curve of equilibria and bifurcation diagram for dydt = 10y(1 − y/10000)− a

Choosing a point inside the parabola, say (0, 5000), we obtain dy/dt = 10 · 5, 000(1− 1/2) = 25, 000 > 0. Hence,

dy/dt > 0 inside of the parabola. Choosing a point outside of the parabola, say (10, 0), we obtain dy/dt = −10 < 0.

Hence dy/dt < 0 outside of the parabola.

Next, we can sketch phase lines for several a values, say a = 0, a = 20, 000, a = 25, 000, and a = 30, 000. For

each of these values of a, we draw a vertical line. Where the line intersects the parabola we draw a solid circle (in

red in Fig. 6.33b.) as this corresponds to points where dy/dt = 0. Where the line lies inside the parabola, we draw

an upward arrow. Where the line lies outside the parabola, we draw downward arrows. The resulting bifurcation

diagram is illustrated in Figure 6.33b. Notice that for a = 0, a = 20, 000, and a = 30, 000, we get the same phase

lines as in Example 1.

This bifurcation diagram indicates that for

0 < a < 25, 000

there are two equilibria. The lower equilibrium is unstable equilibrium and the upper equilibrium is stable. When

the two equilibria coalesce at a = 25, 000, the resulting equilibrium is semi-stable—that is, solutions starting above

the density y = 5, 000 decrease to asymptotically approach the equilibrium y = 5, 000, while solutions that start

below 5, 000 also decrease to asymptotically approach 0.



Noting that this critical semi-stable equilibrium value y = 5, 000 is half the carrying capacity K = 10, 000, it

follows that for harvesting rates over the range 0 < a < 25, 000, the population can persist provided that its initial

population abundance is sufficiently large. Moreover, the stable population equilibrium is always greater than 5, 000.

On the other hand, if the population is harvested at a rate a > 25, 000, it will eventually be driven to 0, at which

point the harvesting must necessarily be set to 0 since a population that has 0 individuals can no longer be harvested.

2

An important implication of the bifurcation diagram in Example 2 is that gradual changes in harvesting can

bring about discontinuous changes in the population abundance. More specifically, when the harvesting rate is ever

so slightly increased beyond the bifurcation value (a = 25, 000 in Example 1) the population begins slowly at first

but then more rapidly to decline from abundance to extinction. Such population disappearances have been observed

in natural populations. Dramatic examples include the precipitous drop of blue pike (stizostedion vitreum glaucum)

from annual catches of 10 million pounds to less than one thousand pounds in the mid 1950s, or the unexpected

collapse of the Peruvian anchovy population in 1973, as illustrated in Figure 6.34, and the sudden reduction of Great

Britain’s grey partridge (perdix perdix ) population in 1952.

1955 1960 1965 1970year

2·106

4·106

6·106

8·106

1·107

1.2·107

metric tons caught

Figure 6.34: Catch data for Peruvian anchovies in the 20th century

The bifurcation occurring at a = 25, 000 in the queen conch example is a saddle node bifurcation because the

transition from two equilibria to no equilibria is preceded by the appearance of a semi-stable (or saddle) equilibrium.

A more colorful name for this bifurcation is a blue sky catastrophe as two equilibria vanish into the blue sky as a

increases past the value 25, 000. Other types of bifurcations are possible, such as the pitchfork bifurcation illustrated

by the next example. A look ahead at Figure 6.35 indicates the source of this name: one equilibrium bifurcates into

three as the value of the bifurcation parameter increases to create a pitchfork looking object.

Example 3. Pitchfork bifurcation



Sketch a bifurcation diagram for

dy

dt= ay − y3

Solution. The equilibria are given by

0 = y(a − y2)

Hence, either y = 0 or y2 = a so that for a ≥ 0 the right-hand-side of the differential equation is f(y) = y(y −√

a)(y +√

a). The sketches of the curves y = 0 for all a and y = ±√a for a ≥ 0 in the ay-plane yields Figure 6.35a.

These curves determine four regions in the ay-plane: the regions above and below the pitchfork and the upper and

-1 -0.5 0 0.5 1 1.5 2a

-2

-1

0

1

2

y

a. Graph of dydt = 0 b. Bifurcation diagram

Figure 6.35: Pitchfork bifurcation

lower parabolic wedges of the pitchfork. Using the point (a, y) = (0, 1), we obtain dydt = −1 < 0. Hence, dy

dt < 0

in the region above the pitchfork. Using the point (a, y) = (0,−1), we obtain dydt = 1 > 0. Hence, dy

dt > 0 in the

region below the pitchfork. Using the point (a, y) = (2, 1), we obtain dydt = 1 > 0. Hence, dy

dt > 0 in the upper

parabolic wedge of the pitchfork. Using the point (a, y) = (2,−1), we obtain dydt = −1 < 0. Hence, dy

dt < 0 in the

lower parabolic wedge of the pitchfork.

To complete the bifurcation diagram, it suffices to sketch phase lines for a negative a-value (i.e. only one

equilibrium), a positive a-value (i.e. three equilibria), and the bifurcation value a = 0. Drawing vertical lines at

these a values, solid circles at the equilibria, upward arrows where dydt > 0, and downward arrows where dy

dt < 0,

results in the bifurcation diagram illustrated in Figure 6.35b. 2

Modelling memory formation

Behind the motions and thoughts of every animal lies a vast network of cells, the nervous system. The network

comprises billions of cells called neurons. A typical “network” neuron is illustrated in Figure 6.36, although neurons

with various types of morphologies make up the total neural system of any animal.



Figure 6.36: A neuron

Neurons specialize in carrying “messages” from one part of the body to another through an electrochemical

process that typically causes a voltage spike to travel along the membrane of the neural cell. The message is received

by dendrites, which look like tentacles attached to the cell body. The chemical messages pass down these tentacles

into the cell body and then out through one main long axon. The end of this axon then communicates with dendrites

of neurons further down the neural chain, thereby passing the message along from one neuron to the next. Messages

between two neurons are usually passed in the form of a chemical flux of so-called neurotransmitters. Excitatory

neurotransmitters trigger “go” signals that allow the message to be passed to the next neuron in the communication

line and inhibitory neurotransmitters produce “stop” signals that prevent the message from being forwarded. A

single neuron “integrates” the incoming signals to determines whether or not to pass the information along to other

cells. The activity within a single neuron is typically measured by the rate which it “fires” voltage spikes.

The simplest model of a population of neurons is the Wilson-Cowan model. It assumes that the entire

population of neurons fire at the same rate y (units are number of spikes/msec) and are of the same type (i.e. release

the same type of neurotransmitters).



Wilson-Cowan Neural

Population Model

Let a be the rate at which an external source produces neurotransmitters that

stimulate the dendrites of a population of neurons. If a is positive or negative,

then the external neurotransmitters are respectively excitatory or inhibitory.

Let b be the rate at which each individual neuron release neurotransmitters when

it fires. If b is positive or negative, then the internal neurotransmitters re-

spectively are excitatory or inhibitory.

Let c be the rate at which the firing of an active neuron decays exponentially in

the absence of external stimulation.

Then the firing rate y (measured in spikes per unit time) of each neuron in the

network is modeled by the equation

dy

dt= −cy +

1

1 + e−a−by

Example 4. Modeling memory formation

Consider an application of the Wilson-Cowan model in which b = 6 (that is, the neurons are excitatory) and

c = 1 (that is, in one unit of time the firing rates has dropped by a factor e−1 = 1/e).

a. Sketch the bifurcation diagram with respect to parameter a.

b. Create a plot of y(t) that corresponds to a population of neurons that is initially quiescent—that is

y(0) = 0—and is then subject to an external stimulus that has the following “switching” characteristics:

a = −3 for 0 ≤ t < 20 (units of t are ms), a = −1 on 20 ≤ t ≤ 40 and a = −3 on 40 < t ≤ 100.

c. Discuss the implications of what you have found.

Solution.

a. This equation is too complicated to plot by hand, so we will graph it using technology. Some computer

programs will graph this equation as shown, but most will require that we solve for one of the variables.

Solving for a in terms of y under equilibrium conditions yields

1 + e−ae−6y =1

y



e−a = e6y

(1

y− 1

)

ea = e−6y

(y

y − 1

)

a = −6y − ln

(1 − y

y

)

Using technology, we find that the graph of this curve is shown in Figure 6.37. Using the point (a, y) =

(−5, 1), we obtain dydt < 0. Hence, dy

dt < 0 in the left region. Using the point (a, y) = (0, 0), we obtain

dydt > 0. Hence, dy

dt > 0 in the right region. To complete the bifurcation diagram, we draw five phase lines.

One at each bifurcation value (i.e. a ≈ −2.5 and a ≈ −3.5) and one to either side of the bifurcation

values. Doing so, we obtain the bifurcation diagram illustrated in Figure 6.37b.

-5 -4 -3 -2 -1 0a

0

0.2

0.4

0.6

0.8

1

y

a. Graphs of dydt = 0 b. Bifurcation diagram

Figure 6.37: The curve of equilibria and bifurcation diagram for a Wilson-Cowan model

b. We will use technology to solve the differential equation (a = −3):

dy

dt= −y +

1

1 + e3−6y

for 0 ≤ t < 20 with y(0) = 0. This solution is shown below:

5 10 15 20t

0.02

0.04

0.06

0.08

0.1y

Then, as the domain shifts to 20 ≤ t ≤ 40, we are given that a = −1, so we again use technology to graph



a solution of

dy

dt= −y +

1

1 + e1−6yy(20) ≈ 0.07

25 30 35 40t

0.2

0.4

0.6

0.8

1y

Finally, a returns to −3 for the domain 40 < t ≤ 100, and we use from the above graph an initial value

of y(40) = 1, to give the following graph:

50 60 70 80 90 100t

0.2

0.4

0.6

0.8

1y

We use technology to put together these parts into a single graph as shown in Figure 6.38.

20 40 60 80 100t

0.2

0.4

0.6

0.8

1y

Figure 6.38: Graph of how a population of neurons records that it has been subject to a change in background firingrate a on the interval t ∈ [20, 40] (ms).

c. We see from Figure 6.38 that the population of neurons initially rises from 0 to asymptote at a low firing

rate around y = 0.06. In terms of the bifurcation the bifurcation diagram (Figure 6.37b). The activity of

the population has risen from 0 to the lower arm of the S-shaped null-cline in ya-space. When a rises from

−3 to −1, the lower arm ceases to exist and the population rises, as we see in Figure 6.38, to the upper



arm which has a value close to 1. After “switching back” at t = 40 ms to the value a = −1, the neural

firing rate starts to decline, but now it is only able to drop to the upper arm of the S-shaped null-cline in

the bifurcation diagram (Figure 6.37b). By remaining at the high firing rate, the population of neurons is

effectively “remembering” that the background stimulus was in one state (a = −3), switched to a second

state (a = −1) for some period of time and then switched back to the first state again (a = −3). In

this way the neuron has recorded an ”off-on-off” event and is said to now remember that it was once

”switched on.” To clear the memory, the background stimulus a would need to drop below approximately

−3.5 (see Figure 6.37b)

2



Problem Set 6.6

LEVEL 1 – DRILL PROBLEMS

Draw the phase lines requested in Problems 1 to 6.

1. dydt = y

(1 − y

100

)− a; a = 0, a = 9, a = 25

2. dydt = 2y

(

1 − y1,000

)

− a; a = 0, a = 180, a = 600

3. dydt = 450 − ay; a = −10, a = 0, a = 10

4. dydt = (100 − y)(y − 250)− a; a = 0, a = 2000, a = 5000

5. dydt = y2 − ay + 1; a = 0, a = 2, a = 4

6. dydt = 2y2 − ay + 240; a = 0, a = 50, a = 200

Sketch bifurcation diagrams for the equations in Problems 7 to 12.

7. dydt = ay − y2

8. dydt = y2 − a

9. dydt = 1 + ay

10. dydt = 1 − ay2

11. dydt = sin y − a

12. dydt = y2 − ay + 2

Consider an application of the Wilson-Cowan model for the values of b and c in Problems 13 to 18. Sketch the

bifurcation diagram with respect to parameter a.

13. b = 5, c = 1

14. b = 4, c = 1

15. b = 8, c = 1

16. b = 4, c = 2

17. b = 8, c = 2



18. b = 12, c = 2

LEVEL 2 – APPLIED PROBLEMS AND THEORY

SIS model in Epidemiology

Mathematical epidemiologists often use the symbol S to denote the number of individuals in a population that

are susceptible to a disease, I the number of people infected with the disease, and R the number of individuals that

have recovered and are now immune. If no individuals die from the disease then the total number of individuals in

the population is N = S + I + R. This kind of model is called and SIR model. In the case that all individuals that

recover are immediately susceptible, then R = 0 for all time and the model is called an SIS model. Many sexually

transmitted infections, for example gonorrhea, do not confer immunity, and are best described by SIS models. This

is one reason why a single round of antibiotics, even if applied widely on a population basis will not have a long-term

effect in lowering incidence of STIs.

Let us assume that a susceptible individual encounters and gets infected by infected individuals at a rate pro-

portional to the density of infected in the population. Call this proportionality constant b ≥ 0. The constant b is

known as the transmission rate in the epidemiological literature. Let us also assume that individuals infected with

the disease recover from the disease at a constant rate r ≥ 0. Under these assumptions we obtain the SIS model:

dI

dt= bIS − rI

Since I + S = N , we know S = N − I so

dI

dt= bI(N − I) − rI

Rearranging terms yields

dI

dt= I (bN − r − bI)

In Problems 19 to 22 sketch a bifurcation diagram with respect to b if r = 1 and with respect to r if b = 1. Discuss

under what conditions the disease persists in a population confined to living in a group of indicated size.

19. N = 1, 000 (boarding school)

20. N = 10, 000 (army camp)

21. N = 100, 000 (isolated town in Alaska)

22. N = 1, 000, 000 (isolated city in remote region of Asia)

Habitat destruction



Consider a population living in a patchy environment. Let y be the fraction of patches occupied by the species of

interest. Let c ≥ 0 denote the colonization rate (i.e. the rate at which individuals from one patch colonize an empty

patch), d ≥ 0 the rate at which individuals clear out of a patch, and 0 ≤ D ≤ 1 is the fraction of patches destroyed

by mankind. Then we get the following model of Harvard biologist, Richard Levins,

dy

dt= cy(1 − D − y) − dy

Sketch the bifurcation diagram for this differential equation for the information given in Problems 23 to 28.

23. Assume that D = 0. Sketch bifurcation diagrams for d when c = 1 and for c when d = 0. Under what

conditions does the population persist?

24. Assume that D = 0. Sketch bifurcation diagrams for d when c = 2 and for c when d = 1. Under what


25. Assume that D = 0.5. Sketch bifurcation diagrams for d when c = 1 and for c when d = 0.5. Under what


26. Assume that D = 0.5. Sketch bifurcation diagrams for d when c = 2 and for c when d = 2. Under what


27. Assume that c = 3/2. Sketch bifurcation diagrams for D when d = 1/2 and for d when D = 0. Under what


28. Assume that d = 2. Sketch bifurcation diagrams for D when c = 4 and for c when D = 1/2. Under what


Lotka-Volterra predation

In the 1920’s two mathematicians, Vito Volterra (1860-1940) and Alfred Lotka (1880-1949), considered models

of the density of a prey species, denoted here by the variable x, predated by a species at a density denoted here by

the variable y. They then wrote down two differential equations, one for the prey and one for the predator, in which

the prey equation included the predator density and the predator equation included the prey density. We have not

developed the theory on how to analyze a system of two interdependent differential equations, which rightly belongs

to a course on multivariate calculus, but at least we can analyze the behavior of the prey equation or the predator

equation where the density of the other species appears as a parameter. The general form of the prey equation is:

Prey Equation:dx

dt= xg(x) − yh(x),



where g(y) is a per capita growth rate of the prey species and h(x) is the rate at which each unit of predator is able

to extract prey. Note it is assumed that both g(0) = 0 and h(0) = 0. The general form of the predator equation is:

Predator Equation:dy

dt= ybh(x) − yf(y),

where h(x) is the prey extraction rate per predator appearing in the above prey equation, 0 < b < 1 is the efficiency

with which predators can convert a unit of consumed prey into their own biomass (ingestion, digestion, metabolism,

etc.), and f(y) is the rate at which predators die when they have no prey species to feed upon.

In Problems 29 to 31 it sketch the bifurcation diagram for the specified growth and extraction functions in the prey

equation in which the density y of the predators is regarded as a parameter in the prey equation and in Problems

32 to 33 sketch the bifurcation diagram for the specified extraction and mortality functions in the predator equation

in which the density x of the prey species is regarded as a parameter.

29. In the classic Lotka-Volterra model g(x) is constant or or is a decreasing linear function and h(x) is homogeneous

(i.e. h(0) = 0) linear, so assume g(x) = 0.5(1−x/3) and h(x) = x. Under what conditions does the population

persist?

30. In the modified Lotka-Volterra model g(x) is constant or is a decreasing linear function and h(x) is a saturating

function of prey density x, so assume g(x) = 0.5 and h(x) = xx+2 . Under what conditions is the population

reigned in by predation?

31. Assume g(x) = 0.5(1−x/3) and h(x) = xx+2 . Under what conditions is the prey population driven to extinction

by predation?

32. In the classic Lotka-Volterra model f(y) is constant. Thus, assuming, b = 0.2 and f(y) = 1 under what

conditions does the predator population persist when h(x) = x?

33. Assume b = 0.2 and f(y) = y under what conditions does the predator population persist when h(x) = x?

34. A self regulatory genetic network Smolen et al. (1998, 1999) investigated a model of a single transcription

factor, TF-A, that activates its own transcription TF-A forms a homodimer that activates transcription by

binding to enhancers (TF-REs). A rapid equilibrium is assumed between monomeric and dimeric TF-A. The

transcription rate saturates with TF-A dimer concentration to a maximal rate a, which is proportional to TF-A

phosphorylation. Responses to stimuli are modeled by varying the degree of TF-A phosphorylation. A basal

synthesis rate d is present, as well as a first-order process for degradation, −cy. If y denotes the concentration

of TF-A then the model is given by

dy

dt=

a y2

b + y2− c y + d



Assume that b = 1, c = 1, and d = 0.1. Sketch a bifurcation diagram over the region 1 ≤ a ≤ 3 and 0 ≤ y ≤ 3.

Discuss when you expect to see two stable equilibria.

35. (Evolution of Cooperation, part III) Problem 31 in Section 6.5 investigated how individuals that interact fre-

quently and respond to the strategy of their opponents can promote the evolution of cooperation. If opponents

interact on average n times and cooperation gives a benefit B to the opponent and a cost C to the cooperator,

then the pay off matrix for the strategies tit-for-tat and defect are given by

Tit for Tat Defect

Tit for Tat n(B − C) −C

Defect B 0

a. Write down a replicator equation for this game.

b. Assume B = 4 and C = 3, and sketch a bifurcation diagram with respect to the parameter n.

c. Discuss the implications for the evolution of cooperation.


ssg6.5-6 stability and bifurcations

Documents