Math 5110 Homework 2 Solutionsfogelson/5110_f18/HW2Solns_PM.pdf · Math 5110 Homework 2 Solutions...

Math 5110 Homework 2 Solutions

1. In 1973 Varley, Gradwell, and Hassell proposed the following model of a population witha density dependent reproduction rate ln 12

pn+1 =λ

αp1−bn (1)

which can be rewritten as

pn+1 =

(1

αp−bn

)(λpn) . (2)

(a) The second factor λpn is fairly straightforward to interpret. It is a growth facter basedon the current population level, where λ ≥ 0 is a unitless per-capita growth rate.The first factor 1αp

−bn is an inhibition factor based on the current population. The

parameter b ≥ 0 is dimensionless, while the parameter 1α > 0 has units of [population]b.

b acts as a measure of how the population inhibits itself, with b = 1 meaning that thepopulation is constant, b > 1 indicating strong levels of self-inhibition, and 0 < b < 1indicating mild levels of self-inhibition. b < 0 is no longer inhibition, and is notrelevant to the model. α can be thought of as how much this inhibition is magnified.

(b) The equilibrium equation p∗ = λα(p∗)1−b has two solutions: p∗ = 0, the trivial solution,

and the positive solution p∗ =(λα

) 1b , which is when the inhibition factor takes on a

value of 1.

Set

f(p) =λ

αp1−bn .

Investigating the stability of the equilibria, we have that |f ′(p)| =∣∣∣λ(1−b)α 1pb ∣∣∣ ap-

proaches ∞ as p approaches 0, while at the positive equilibrium p∗ =(λα

) 1b , we have

f ′(p∗) = 1 − b. This indicates that p∗ =(λα

) 1b is stable for 0 < b < 2 and unstable

otherwise, while p∗ = 0 is always unstable (see Figure 1 below).

Figure 1: Bifurcation diagram for λα = 3. Solid lines indicate stability, while dashed lines indicatean unstable equilibrium. Note the sigular behavior of the positive equilibrium as b→ 0.

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(c) When the stability condition for the positive equilibrium is violated, i.e. b > 2, anon-physical pair of period 2 points apppear at p = 0 and p = ∞, which solutionsequences will approach. This is due to the fact that for b > 2, the right-hand sideof equation (1) behaves like 1pcn

, c = b − 1 > 1, which has strong singular behavioraround pn = 0.

Figure 2: Plot of the solution sequence for b = 2.01, λα = 3, up to n = 50.

Figure 3: Cobweb plot for b = 2.01, λα = 3, up to n = 50.

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2. Consider the tumor growth model we discussed in class

Cn+1 = (m σ(Cn) + p)Cn. (3)

Remember that p represents the base-line survival rate of mature cancer cells, m representsthe per-capita number of new cancer cells produced in the tumor, and σ is the tumor sizedependent survival rate, which represents the body’s ability to fight a tumor. We decidedthat for this model to be reasonable σ(0) = 1, and that as C → ∞, σ(C) → 0. In classwe discussed σ(C) = 1/(1 + αC). Another function that behaves this way is

σ(C) = e−αC . (4)

Using this new model of immune response, we can say the following.

(a) After dividing both sides of the steady-state equation by C∗ the non-zero tumor sizeequilibrium can be found by solving

1 = me−αC∗

+ p, (5)

resulting in

C∗ =1

αln

(m

1− p

). (6)

This equilibrium is positive when 1− p < m, i.e. when the birth rate m exceeds thedeath rate 1− p. Note that 1− p is always positive since p is a proportion.

(b) Setting f(C) = (me−αC∗

+ p)C, the stability condition for the positive equilibrium is

− 1 < (p+me−αC(1− αC))∣∣∣C=C∗

< 1, (7)

which yields

− 1 < 1 + (1− p) ln(

m

1− p

)< 1. (8)

Rearranging these inequalities so that m in in the middle yields

1− p < m < (1− p)e2

1−p . (9)

The left inequality is automatically satisfied if we have a positive equilibrium. Theright inequality is a bit trickier, but to get a sense of how large the right-hand side

is, a quick calculation shows ddp(1− p)e2

1−p > 0 for 0 < p < 1, so the minimum value

of (1 − p)e2

1−p is e2 ≈ 7.39 at p = 0, while a series expansion of the exponential inpowers of 1/(1− p) or the use of L’Hopital’s rule shows (1− p)e

21−p →∞ as p→ 1.

On the other hand, looking at the stability of C∗ = 0, the stability condition becomes

1− p > m, (10)

which is only satisfied when there is no positive equilibrium solution.

(c) Set p = 1/2. Based on part (b), the positive equilibrium will be stable for 1/2 < m <12e

4 ≈ 27.3, and will only be present for m > 1/2. For values of m between 12e4 and

30, numerical simulations show a pair of stable period 2 points appear.

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Figure 4: Bifurcation diagram for 0 < m < 30 with α = 0.9, p = 1/2.

Figure 5: Cobweb plot for α = 0.9, p = 1/2, m = 0.4, up to n = 50.

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Figure 6: Cobweb plot for α = 0.9, p = 1/2, m = 3, up to n = 50.

Figure 7: Cobweb plot for α = 0.9, p = 1/2, m = 9.4, up to n = 50. Oscillatory behavior clearlypresent.

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Figure 8: Cobweb plot for α = 0.9, p = 1/2, m = 25, up to n = 50. Equilibrium is still stable,but a periodic solution is close to emerging.

Figure 9: Cobweb plot for α = 0.9, p = 1/2, m = 30, up to n = 500. A pair of period 2 point(marked on the cobweb plot in green) has emerged.

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Figure 10: Solution sequence for α = 0.9, p = 1/2, m = 30, up to n = 500.

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3. Suppose that a population of parasitic wasps (denoted Pn) and host insects (denoted Hn)follow the Nicholson-Bailey model, except that each year some of the hosts manage to findsafe refuge and therefore cannot be attacked by waps.

(a) Suppose that a set fraction θ of the hosts manage to find safe refuge. The originalNicholson-Bailey model is

Hn+1 = λHne−αPn (11)

Pn+1 = cHn(1− e−αPn), (12)

Where 1− e−αPn is the probability of a wasp encountering a host, e−αPn is the proba-bility of a host not encountering a wasp, α is a search efficiency for the wasps, λ is theper-capita growth rate for the hosts, and c is the number of eggs layed by the waspsper host. If we modify this so that a certain proportion θ of the hosts is exposed toattack, and thus only a certain proportion 1− θ of the hosts is exposed to the wasps,then the modified equations are

Hn+1 = λθHn + λ(1− θ)Hne−αPn (13)Pn+1 = c(1− θ)Hn(1− e−αPn). (14)

Looking at the equilibrium equations

H∗ = λθH∗ + λ(1− θ)H∗e−αP ∗ (15)P ∗ = c(1− θ)H∗(1− e−αP ∗) (16)

one equilibrium is given by H∗ = P ∗ = 0. Dividing the first equation by H∗ andsolving for P ∗ gives

P ∗ =1

αln

(λ(1− θ)1− λθ

). (17)

Substituting this back into the second equation results in

H∗ =λ

cα(λ− 1)ln

(λ(1− θ)1− λθ

). (18)

These equilibrium values are only positive simultaniously if 1 < λ < 1/θ (note that0 ≤ θ ≤ 1 is a proportion).Now to investigate the stability. The Jacobain for this system is

J =

[λθ + λ(1− θ)e−αP −αλ(1− θ)He−αPc(1− θ)(1− e−αP ) αc(1− θ)He−αP

]. (19)

Evaluating this at the zero equilibrium gives

J =

[λ 00 0

], (20)

and since this is a triangular matrix, we can read off the eigenvalues λ, 0 from thediagonal of the matrix. This equilibrium is only stable if λ < 1, in which case theother equilibirum is not in the positive quadrant.

Using the fact that e−αP∗

= 1−λθλ(1−θ) and 1−e−αP ∗ = k−1λ(1−θ) , the Jacobian at the second

equilibrium reduces to

J =

1 −λ(1−λθ)c(λ−1) ln(λ(1−θ)1−λθ )c(λ−1)λ

1−λθλ−1 ln

(λ(1−θ)1−λθ

) , (21)8


which has trace τ and determinant ∆ given by

τ = 1 +1− λθλ− 1

ln

(λ(1− θ)1− λθ

)(22)

∆ =λ(1− λθ)λ− 1

ln

(λ(1− θ)1− λθ

). (23)

To get a sense of the magnitude of the eigenvalues, we can look at 1 − τ + ∆ and1 + τ + ∆ , which are the values of the characteristic polynomial for J evaluated at−1 and 1. Substituting in the forms for τ and ∆ above, we find that

1− τ + ∆ = (1− λθ) ln(λ(1− θ)1− λθ

)> 0 (24)

1 + τ + ∆ = 1 +(λ+ 1)(1− λθ)

(λ− 1)ln

(λ(1− θ)1− λθ

)> 0, (25)

indicating that if the eigenvalues are not complex conjugates, they have magnitudeless than 1, which indicates stability. If the eigenvalues are complex conjugates, thento investigate stability, we need to look at when |∆| = 1 (note that since ∆ ≥ 0 whenthis equilibrium is positive, or more generally when λ ≥ 0, the absolute value is notneeded here). In particular, we would like look for solutions to ∆ = 1 for 1 < λ < 1/θ,0 ≤ θ ≤ 1. While this cannot be solved for in closed form, we can plot the roots usingMATLAB or other software (see Figure 11). The most notable result from this is thatif θ, the fraction of hosts that are safe, is greater than or equal to 1/2, there are novalues of λ in the physically relevant range such that ∆ = 1 has a solution. Since ∆is a continuous function of λ for 0 ≤ λ < 1/θ and ∆ = 0 at λ = 0, this indicates that∆ < 1 everywhere it is defined. In the case of complex conjugate eigenvalues, themagnitude of both eigenvalues is equal to ∆1/2 since the determinant is the productof the eigenvalues, so θ ≥ 1/2 implies that the eigenvalues have magnitude less than1 regardless if they are real or complex.

Figure 11: Solution to ∆ = 1 as a function of λ and θ (red curve). The blue curves λ = 1 andλ = 1/θ bound the region where the second equilibrium to the system is in the positive (first)quadrant.

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The physical interpretation behind this is that a positive stable equilibrium existsif over half of the hosts are safe from attack. This is notable since in the originalNicholson-Bailey model, the positive equilibrium is always unstable when it exists,as is the zero equilibrium, and the system exhibits oscillating or periodic dynamicinstead of converging to a fixed point.

Analysis: It is more difficult to show that there are no solutions to ∆ = 1 for θ ≥ 1/2analytically. To start, note that limλ→1/θ ∆ = 0, and that by using L’Hopital’s ruleonce, we can also show limλ→1 ∆ = 1. Taking the derivative of ∆ with respect to λ,we find that

d∆

dλ=

(2θλ− θλ2 − 1) ln(λ(1−θ)1−λθ

)+ λ− 1

(λ− 1)2, (26)

which is continuous for 0 < λ < 1/θ as well.

By applying L’Hopital’s rule twice, it is possible to show limλ→1d∆dλ = 1− 2θ. Com-

bining this with limλ→1 ∆ = 1, limλ→1/θ ∆ = 0, we can conclude that if θ < 1/2,there will be at least 1 value of λ where ∆ = 1 since then the derivative at λ = 1 willthen be positive, and hence there will be a range of values where there is at least oneeigenvalue with magnitude greater than 1.

On the other hand, if we can establish that d∆dλ ≤ 0 for 0 < λ < 1/θ, θ ≥ 1/2, then wecan conclude that ∆ starts at a value of 1 at λ = 1, and is decreasing until it reaches0 at λ = 1/θ, so that ∆ < 1 in the relevant range of values, and all our eigenvaluesare less than 1 in magnitude, resulting in a stable equilibirum.

To start, we will try and bound the derivative by a simpler function. Note that thedenominator (λ− 1)2 is always positive, so we can just consider the numerator of thederivative when determining the sign. Next, the quadratic in λ given by 2θλ−θλ2−1is always negative, and has a maximum value of θ − 1 at λ = 1, so to show d∆dλ ≤ 0for θ ≥ 1/2 and 1 < λ < 1/θ, it is sufficient to show

f(λ) := (θ − 1) ln(λ(1− θ)1− λθ

)+ λ− 1 < 0

for θ ≥ 1/2 and 1 < λ < 1/θ.To do this, we can first show that limλ→1 f(λ) = 0, limλ→1/θ f(λ) = −∞. If we canthen show that dfdλ < 0, we can conclude f(λ) < 0 over the desired range of λ, θ.

Now,df

dλ=−θλ2 + λ+ θ − 1

λ(1− λ).

Since the denominator is always positive for 1 < λ < 1/θ, we will just look at thenumerator −θλ2 + λ+ θ − 1. It take on the values 0 at λ = 1, θ − 1 ≤ 0 at λ = 1/θ,and has a maximum value at λ = 1/(2θ) of

1

4θ+ θ − 1.

This maximum value is always ≥ 0, however the maximum occurs in the interval0 < λ ≤ 1 for θ ≥ 1/2. Combining this with the values of dfdλ at λ = 1, 1/θ, we canconclude that dfdλ ≤ 0 for θ > 1/2, and following the chain of logic established, we havethat there are no values of λ in 0 < λ < 1/θ, 1/2 ≤ θ ≤ 1 such that the eigenvalues ofthe Jacobian J have magnitude greater than 1.

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(b) If we now assume instead that the “refuge” can only hold a certain number S of hostswhose per-capita reproductive rate is 1, the resulting equations are

Hn+1 = λ(Hn − S)e−αPn + S (27)Pn+1 = c(Hn − S)(1− e−αPn). (28)

There are two solutions to the steady-state equations

H∗ = λ(H∗ − S)e−αP ∗ + S (29)P ∗ = c(H∗ − S)(1− e−αP ∗). (30)

The first equilibrium is H∗ = S, P ∗ = 0. The second can be solved by dividing thefirst equation by H∗ − S and solving for P ∗, then using substitution in the secondequation to solve for H∗. The result is H∗ = λαc(λ−1) ln(λ) + S, P

∗ = 1α ln(λ), whichis a positive solution for λ > 1.

The Jacobian for this system is given by

J =

[λe−αP −αλ(H − S)e−αP

c(1− e−αP ) αc(H − S)e−αP]. (31)

At H∗ = S, P ∗ = 0, J has the same for as in part (a), with eigenvalues k, 0. It istherefore stable if the second equilibrium is not positive.

At H∗ = λαc(λ−1) ln(λ) + S, P∗ = 1α ln(λ), the Jacobian is

J =

[1 − λc(k−1) ln(λ)

c(λ−1)λ

1λ−1 ln(λ)

]. (32)

We will assume that λ > 1 so the equilibrium is physical. Proceeding along the samelines as in part (a), we find that the trace and determinant are given by

τ = 1 +1

λ− 1ln(λ) (33)

∆ =λ

λ− 1ln(λ). (34)

and that

1− τ + ∆ = ln(λ) > 0 (35)

1 + τ + ∆ = 1 +(λ+ 1)(1− λ)

(λ− 1)ln(λ) > 0. (36)

We then have that if the eigenvalues of J are real, they are less than 1 in magnitude.

For the complex conjugate case, the analysis is also similar, although simpler, to part(a). We have limλ→1 ∆ = 1, and

d∆dλ =

λ−1−ln(λ)(λ−1)2 > 0 for λ > 1, therefore ∆ > 1 for

λ > 1. This indicates that this second equilibrium is always unstable, and exhibitsoscillations, since ∆ > 1 rules out real eigenvalues based on (34) and (35).

This is different from the previous case in part because this second equilibrium isunstable, in contrast to part (a), where some parameter regions induced stability.The fundamental different, though, is that in part (a) the number of hosts that cantake refuge is dependent on the current host population, while in part (b) it is a fixednumber not dependent on the current population. Instead of stablizing the secondequilibrium for some parameter values, this shifts the zero equilibrium to a non-trivialequilibrium where the hosts can survive without any parasitoids present, unlike in par(a).

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4. Suppose that we have a host/parasitoid system where the host organism has an agestructure that we wish to model. Juveniles are not bulnerable to attack by parasitoids,but adults are. Suppose that the rate of attacking of vulnerable hosts and the growth ofparasitoids follows the Nicholson-Bailey model. The adults that escape attack produce(on average) λ offspring, λ ≥ 0. The fraction of juveniles that survive to become adultsis given by σ, 0 ≤ σ ≤ 1.

(a) Let the juvenile population after step n be denoted by hn. The equations describingthe evolution of the populations are

hn+1 = λHne−αPn (37)

Hn+1 = σhn (38)

Pn+1 = cHn(1− e−αPn). (39)

(Note: This model assumes a year delay in the time it takes for juveniles to replaceadults, so juveniles produced by adults in year n replace the adults in year n + 2. Ifthe juveniles produced by adults in year n replace the adults in the next year n + 1instead of having this additional one year delay, we would instead be able to reducethis model the original Nicholson-Bailey model with the parameter λ repaced by theproduct λσ.)

(b) To solve for the non-zero equilibrium, first substitute h∗ = λH∗e−αP∗

into the secondequation to get

H∗ = σλH∗e−αP∗. (40)

Assuming H∗ 6= 0, this equation gives

P ∗ =1

αln(λσ). (41)

Substituting this into the third equation in the system allows us to solve for H∗

H∗ =λσ

cα(λσ − 1)ln(λσ), (42)

an a third substitution back into the second equation in the system gives us the finalequilibrium value

h∗ =λ

cα(λσ − 1)ln(λσ). (43)

These three values are all positive if λσ > 1, which just means that the total fractionof new hosts based purely on host birth and survival rates is greater than 1.

The Jacobian of this system is then

J =

0 λe−αP −αλHe−αPσ 0 00 c(1− e−αP ) αcHe−αP

. (44)Evaluating this at the non-zero equailibrium gives us

J =

01σ −

λc(λσ−1) ln(λσ)

σ 0 0

0 c(λσ−1)λσ1

λσ−1 ln(λσ)

. (45)

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(c) Since the determinant ∆ is the product of the (in this case three) eigenvalues ofthe Jacobian, a determinate greater than 1 in magnitude corresponds to an unstableequilibirum. (Note, however, the converse is not necessarily true. A determinant lessthan 1 in magnitude does not mean all the eigenvalues are less than 1 in magnitudeas well, just their product.)

In this case, the determinant of J is mot conveniently calculated by expanding aboutthe first column (or second row), since two of the entries are 0. This gves

∆ = −σ(

1

σ(λσ − 1)ln(λσ) +

1

σln(λσ)

)= − λσ

λσ − 1ln(λσ). (46)

It helps to define a new parameter γ = λσ since only their product appears in thedeterminant. The condition for a positive equilibrium is then γ > 1. Since we have

∆ = − γγ − 1

ln(γ) (47)

limγ→1+

∆ = −1 (48)

d∆

dγ=

ln(γ)− (γ − 1)(γ − 1)2

< 0 for γ > 1, (49)

the determinant is always greater than 1 in magnitude when there is a positive equi-librium.

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