Adaptive Dynamics Draft

1

Existence and behaviour of a two-patch two-predator

one-prey system

By:

James Duncan

Undergraduate Student Research Award: Mathematics

Supervisors: Dr. Ross Cressman and Dr. Yuming Chen

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Consider a 2-predator 1-prey system that has two patches. Prey are free to move between the

patches while each predator is restricted to one patch. Additionally, prey growth in either patch is

logistic and they spend a proportion of time, 𝑝, in patch one (and the other proportion, (1 − 𝑝), in patch

two). Predator functional responses are both of Holling-type I and intraspecific competition is present in

both predator species. This can be described using the following system, [1], of four differential

equations:

𝑑𝑥

𝑑𝑡= 𝑥 (𝑝 (𝑟1 (1 −

𝑝𝑥

𝐾1) − 𝑎𝑧1) + (1 − 𝑝) (𝑟2 (1 −

(1−𝑝)𝑥

𝐾2) − 𝑏𝑧2))

𝑑𝑧1

𝑑𝑡= 𝑧1(−𝑚1 + 𝑘1𝑎𝑝𝑥 − 𝑐1𝑧1)

𝑑𝑧2

𝑑𝑡= 𝑧2(−𝑚2 + 𝑘2𝑏(1 − 𝑝)𝑥 − 𝑐2𝑧2)

𝑑𝑝

𝑑𝑡= 𝜏𝑝(1 − 𝑝) ((𝑟1 (1 −

𝑝𝑥

𝐾1

) − 𝑎𝑧1) − (𝑟2 (1 −(1 − 𝑝)𝑥

𝐾2

) − 𝑏𝑧2))

Where 𝑥 is the population of prey at time 𝑡, 𝑧1 and 𝑧2 are the populations of predators in patch 1 and 2

respectively, and 𝑝 is the proportion of time prey spend in patch 1. In patch i, the growth rate of prey is

𝑟𝑖 and the carrying capacity is 𝐾𝑖 . Predator i has an intrinsic death rate of 𝑚𝑖, coefficient of intraspecific

competition 𝑐𝑖, and conversion of prey to predator fitness 𝑘𝑖. The interaction coefficient between prey

and predator 𝑧1 is 𝑎 and between prey and predator 𝑧2 is 𝑏. Lastly, 𝜏 is the time-scale separation

coefficient. The expression for 𝑑𝑝

𝑑𝑡 is the derivative of the fitness function of the prey.

Let us assume that

1) prey are free to move between patch 1 and 2 and spend a proportion of time 𝑝 in patch 1

and (1 − 𝑝) in patch 2. This proportion should depend on the observed fitness of individuals

in either patch (i.e. if a prey in patch 1 sees that a prey in patch 2 has higher fitness , it

should migrate to patch 2),

2) 𝑝 is the strategy that the whole prey population plays, and

3) prey in either patch have some way to evaluate the fitness of prey in the other patch so that they can maximize their own fitness.

If there exists a value of 𝑝 such that the fitness of prey in both patches is zero, then there will be

no net movement of prey between patches and a stable equilibrium exists. This may be a coexistence

equilibrium of all three species, or the two-predator one-prey system may reduce to a one-predator

one-prey refuge system, or both predators go extinct and prey growth is only limited by their carrying capacity.

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Existence of Equilibria

Three-species coexistence equilibrium

First, we will consider a three-species coexistence equilibrium with prey playing adaptive strategy 𝑝,

denoted by (λ,µ,σ,p) using system [1]. Is there a unique strategy that allows all three species to coexist?

At a stable internal equilibrium for the prey species 𝑥, we know that the fitness in both patches should

be zero so we have the linear equations (in terms of p)

𝑟1 (1 −𝑝𝜆

𝐾1

) − 𝑎µ = 0 (1)

𝑎𝑛𝑑 𝑟2 (1 −(1 − 𝑝)𝜆

𝐾2

) − 𝑏σ = 0 (2)

Additionally, the fitness of both predators 𝑧1 and 𝑧2 must also be zero, and again we have a set of linear equations (in terms of p)

−𝑚1 + 𝑘1𝑎𝑝𝜆 − 𝑐1µ = 0 (3)

𝑎𝑛𝑑 − 𝑚2 + 𝑘2𝑏(1 − 𝑝)𝜆 − 𝑐2σ = 0 (4)

And lastly for the strategy p, 𝑑𝑝

𝑑𝑡= 0 if and only if

(𝑟1 (1 −𝑝𝜆

𝐾1

) − 𝑎µ − 𝑟2 (1 −(1 − 𝑝)𝜆

𝐾2

) + 𝑏σ) = 0

Where if (1) and (2) are satisfied so is this equation for 𝑑𝑝

𝑑𝑡.

Solving for µ in both (1) and (3) then set the equations equal to each other (similarly for σ using (2) and

(4)) . This generates two equations with 𝜆 equal to a function of p. Setting these equations equal to each

other yields an expression for the unique value of 𝑝 at the equilibrium, given that 𝑝 ∈ (0,1),

𝑝 = [(

𝑘1𝑎𝑐1

+𝑟1

𝐾1 𝑎)

(𝑘2𝑏𝑐2

+𝑟2

𝐾2𝑏)

(𝑚2

𝑐2+

𝑟2

𝑏)

(𝑚1

𝑐1+

𝑟1

𝑎)

+ 1]

−1

(5)

Thus there is a unique value of p described by equation (5) that admits an equilibrium (λ,µ,σ,p) for some set of parameter values.

Computing the Jacobian matrix at this equilibrium yields

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𝐽 =

−λ(𝑝2𝑟1

𝐾1+

(1 − 𝑝)2𝑟2

𝐾2) −𝑎𝑝λ −𝑏(1 − 𝑝)λ λ ((1 − 𝑝)

𝑟2

𝐾2− 𝑝

𝑟1

𝐾1)

𝑘1𝑎𝑝µ −𝑐1µ 0 𝑘1𝑎λµ

𝑘2𝑏(1 − 𝑝)σ 0 −𝑐2σ −𝑘2𝑏λσ

−𝜏 (𝑝𝑟1

𝐾1+ (1 − 𝑝)

𝑟2

𝐾2) −𝜏𝑎 𝜏𝑏 −𝜏λ (

𝑟2

𝐾2+

𝑟1

𝐾1)

If the eigenvalues of the characteristic equation for this matrix all have negative real part, the equilibrium is asymptotically stable.

For a numerical example, we will define the parameters (arbitrarily) as follows:

For the patches we will set 𝑟1 = 0.8, 𝐾1 = 3, 𝑎𝑛𝑑 𝑟2 = 0.7, 𝐾2 = 2.5.

For the effect parameters of predator on prey set 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 1.

The death rates of predators in absence of prey as 𝑚1 = 0.5 𝑎𝑛𝑑 𝑚2 = 0.5.

For predator conversion rates, set 𝑘1 = 0.5 𝑎𝑛𝑑 𝑘2 = 0.75.

For intraspecific competition between predators, set 𝑐1 = 0.1 𝑎𝑛𝑑 𝑐2 = 0.05.

If the eigenvalues of the Jacobian evaluated at the equilibrium for this parameter set all have negative

real part, the system is asymptotically stable. Set 𝜏 = 1. The equilibrium is

(1.801,0.506,0.504,0.6113).The Jacobian at this equilibrium is

𝐽|(1.8,0.5,0.5,0.6) =

−0.255 −1.1 −0.7 −0.0970.154 −0.05 0 0.4550.147 0 −0.025 −0.681

−0.271 −1 1 −0.984

Which has eigenvalues

𝛿1 = −0.557766 + 0.7609806𝑖 𝛿2 = −0.557766 − 0.7609806𝑖 𝛿3 = −0.99934 + 0.6075781𝑖

𝛿4 = −0.99934 − 0.6075781𝑖

Which all have negative real part so the system is asymptotically stable.

For these values of parameters, we can predict (using the above equation for p) the value that p will

take at equilibrium using equation (5). In this case, the predicted value is pp=0.6112956. From the

model, after 150 time steps, the observed value of p at the equilibrium is pobs=0.6112956, thus pp=pobs

(note that if the initial conditions are not near the equilibrium population sizes, the prey strategy may

not exactly match the predicted value after 150 time steps). Additionally, even when initial conditions are varied, the equilibrium population sizes and strategy remain the same.

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One-predator one-prey refuge system

Next, when will system [1] reduce to a predator-prey refuge system (e.g. (x,z1,z2,p) evolves to

(λ,µ,0,p))? Assume that for any population of prey, the fitness function for predator 𝑧2 is negative, i.e.

−𝑚2 + 𝑘2𝑏(1 − 𝑝)𝜆 < 0 (6)

In this scenario, the fitness of prey in both patches must be zero (equations (1) and (2)), but since 𝑧 = 0, from (2) we have that

𝑟2 (1 −(1 − 𝑝)𝜆

𝐾2

) = 0

Which can be simplified to 𝑝 = 1 −𝐾2

𝜆 which can be substituted into inequality (6) which eliminates 𝜆

and 𝑝 to give the inequality

𝐾2 <𝑚2

𝑏𝑘2 (7)

Therefore, if this inequality (7) is satisfied, then system [1] reduces to a predator-prey refuge system. As

the intrinsic death rate of the predator increases, the carrying capacity of the patch that yields a refuge

system increases (i.e. the predators die out quickly so they need more prey present to save them from

extinction). As the ability of predators to convert prey to fitness increases, the carrying capacity of the

patch to cause extinction of the predator decreases (i.e. since predators are better utilizing each prey, they can tolerate lower prey populations).

The solution is similar for z1 to go extinct, where 𝑝 =𝐾1

𝜆 and 𝐾1 <

𝑚1

𝑎𝑘1.

The Jacobian for when z2=0 is

𝐽 =

−λ(𝑝2𝑟1

𝐾1+

(1 − 𝑝)2𝑟2

𝐾2) −𝑎𝑝λ −𝑏(1 − 𝑝)λ λ ((1 − 𝑝)

𝑟2

𝐾2− 𝑝

𝑟1

𝐾1)

𝑘1𝑎𝑝µ −𝑐1µ 0 𝑘1𝑎λµ0 0 −𝑚2 + 𝑘2𝑏(1 − 𝑝)λ 0

−𝜏 (𝑝𝑟1

𝐾1+ (1 − 𝑝)

𝑟2


𝑟2

𝐾2+

𝑟1

𝐾1)

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Let us choose parameters so that inequality (7) is satisfied (i.e. take 𝐾2 = 0.15 <0.5

1∗0.75= 0.66).

For the patches we will set 𝑟1 = 0.8, 𝐾1 = 3, 𝑎𝑛𝑑 𝑟2 = 0.7, 𝐾2 = 0.15.

For the effect parameters of predator on prey set 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 1.




This system evolves from (x,z1,z2,p))=(1,0.5,0.5,0.4) to (1.25,0.506,0,0.880). The characteristic polynomial of this system is

6.16𝜏𝜆3 + (0.108 + 7.02𝜏)𝜆2 + (0.032 + 3.13𝜏)𝜆 + 0.47𝜏 = 0

If we solve the Routh-Hurwitz Criteria for the third order equation from in terms of 𝜏, we get the quadratic equation

19.075𝜏2 + 0.562𝜏 + 0.0034 > 0

Which suggests that for these parameters, the system is stable for all 𝜏 ≥ 0. Thus even if prey do not behave adaptively (in this case), the system can still persist.

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In order to get a restriction on 𝜏, let 𝑟2 = 0.1, and the initial condition for p would have to be

very small. This way, prey are starting in the less favourable patch but not moving to the first patch in

time to allow for the predator-prey refuge system to persist (i.e. predator z1 goes extinct before enough prey move into patch 1).

This is the behaviour of the system with initial conditions (X,Z1,Z2,P)=(1,0.5,0.5,0.1) where

a) 𝜏 = 1 (black solid line), the system evolves to (1.25,0.506,0,0.88), and b) 𝜏 = 0.08 < 0.099 (dotted line), the system evolves to (3.21,0,0,0.93) after 150 time steps.

This restriction is derived using tr(J), which give the inequality

𝜏 >𝑘2𝑏(1 − 𝑝) −

𝑐1𝜇𝜆

− (𝑝2𝑟1

𝐾1+

(1 − 𝑝)2𝑟2

𝐾2)

𝑟1

𝐾1+

𝑟2

𝐾2

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Due to inequality (7), we would expect to see this kind of behaviour if we chose parameters such that:

1) the mortality rate of the second predator satisfies 𝑚2 > 𝑏𝑘2𝐾2, or if

2) the conversion rate of the second predator satisfies 𝑘2 <𝑚2

𝑏𝐾2.

One-prey system

The conditions for both predators to go extinct in system [1] are simply derived from when both of the following inequalities are satisfied:

−𝑚1 + 𝑘1𝑎𝑝𝜆 < 0 (8)

𝑎𝑛𝑑 − 𝑚2 + 𝑘2𝑏(1 − 𝑝)𝜆 < 0 (9)

While the prey population at this equilibrium can be determined using equations (1) and (2). Since 𝑧1 = 0 and 𝑧2 = 0, we can solve (1) as

𝐾1 = 𝑝𝜆

And (2) gives us

𝐾2 = (1 − 𝑝)𝜆

Substituting the first equation into the second shows that the prey equilibrium population is simply

𝜆 = 𝐾1 + 𝐾2

Using these identities for p and (1-p) and substituting them into equations (8) and (9) we get that both are satisfied for all values of p if and only if

𝐾1 <𝑚1

𝑎𝑘1

𝑎𝑛𝑑 𝐾2 <𝑚2

𝑏𝑘2

The Jacobian is

𝐽 =

−λ(𝑝2𝑟1

𝐾1+

(1 − 𝑝)2𝑟2

𝐾2) −𝑎𝑝λ −𝑏(1 − 𝑝)λ λ ((1 − 𝑝)

𝑟2

𝐾2− 𝑝

𝑟1

𝐾1)

0 −𝑚1 + 𝑘1𝑎𝑝λ 0 00 0 −𝑚2 + 𝑘2𝑏(1 − 𝑝)λ 0

−𝜏 (𝑝𝑟1

𝐾1+ (1 − 𝑝)

𝑟2


𝑟2

𝐾2+

𝑟1

𝐾1)

Thus system [1] can evolve to one of three different outcomes depending on parameter values (assuming prey growth rates are both non-zero):

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i) A three-species two-predator one-prey system,

ii) A two-species predator-prey refuge system, or

iii) A one-species system where only the prey survives.

Invasion by prey playing a different strategy

The above shows that a stable three-species equilibrium can indeed be established and the prey evolve

to play the strategy p=0.6113 for the set of parameters outlined. Consider invasion of this system by an

alternate prey species W that plays a fixed strategy q=0.5. Additionally, fix the strategy of prey X to

p=0.6113. At t=200, an invading population of W=0.3 enters the system.

It is clear from these graphs that W cannot invade the system and that the system eventually returns to

its original equilibrium for p=0.6113. At t=400, the populations are (X,Z1,Z2,W)=(1.801,0.506,0.503,0)

which is the original equilibrium. Could an invading prey W playing q=0.6113 invade the system where prey X is playing p=0.5?

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If resident prey X is playing p=0.5, then it has an equilibrium (X,Z1,Z2)=(1.47,1.09,0).

If invading prey W is playing q=0.6113, then it can invade the system (as seen above). At t=400, the

population sizes are (X,Z1,Z2,W)=(0,0,1.25,1.93). However, any invading prey playing q>p=0.5 could invade the system.

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Can an invader playing p=0.6112996 invade a resident population playing a similar value? Let us test for

resident prey playing p=0.59 (dotted line) and p=0.59 (solid line). The vertical line indicates when the

invasion occurs.

From this graph, we can see that in both cases W can successfully invade even when X is playing a strategy close to q=0.6112996.

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Time-Scale Separation

Next we will investigate the effect of the time scale coefficient τ on the behavior of the system.

For this example, we will define the parameters as follows:

For the patches we will set 𝑟1 = 0.4, 𝐾1 = 1.5, 𝑎𝑛𝑑 𝑟2 = 0.9, 𝐾2 = 4.

For the effect parameters of predator on prey set 𝑎 = 0.6 𝑎𝑛𝑑 𝑏 = 0.7.




We can calculate pp using (5) to get that pp=0.2104549. From the simulations after 100 time steps, the

system evolves to (4.045307,0.2882813,0.2590909,0.2104581), and after 150 time steps the p value

becomes 0.2104549, which confirms our prediction for p.

Consider the parameters from above, where p=0.210 at equilibrium. The behavior of solutions

for different values of τ (τ=1 is the black line, τ=0.25 is the red-dotted line, and τ=5 is the green-dotted

line) is shown below.

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From this comparison, we can see that for this example, there is a more pronounced difference in the

behavior of y and p, though the equilibria are all almost equal for different values of τ: when τ=0.25,

even after 150 time units the population is not at the exact equilibrium (𝑝τ=0.25,t=150 = 0.2103249).

Additionally, the solutions of p vary in period and amplitude.

For p:

1) When τ=5 there is an increase in the period and amplitude of p in the first 25 time units

when compared to τ=1. The increased initial amplitude could be interpreted biologically

through τ in that since prey change their behavior quickly, if a patch is more favourable than

another, initially prey will move into this patch in large numbers which ultimately decreases

the fitness of all prey in that patch. The decreased period can be explained as prey

responding quicker to changes in patch fitness so they move between patches more

frequently.

2) When τ=0.25, there is a decreased amplitude and increased period when compared to τ=1.

The decreased amplitude can be explained biologically through τ as prey taking longer to

learn to move to the more favourable patch. The increased period can be explained through τ as prey responding slowly to changes in which patch is more favourable.

For y:

1) When τ=5, the population of y decreases faster than when τ=1. This is due to more prey

moving into patch 2 initially, so there is less prey available in patch 1 and predator y cannot

sustain a high population.

2) When τ=0.25, the population of y decreases slower than when τ=1. This is because less prey

are moving to the second patch, so the population of y has more prey available in that initial time interval.

If the initial condition for p=0, then all the prey will be in patch 2 and the population of y will decrease to

0. Let the initial conditions be (2,1,1,0), then the system evolves to (0.7,0,0.503,0). Note that the

population of z at equilibrium is the same for when there is a three -species coexistence equilibrium and

that the population of x is just (1-0.6113)*1.8=0.7. When p starts at 1, the system evolves to

(1.1,0.506,0,1). The population of y is the same as at the three-species equilibrium and the population of x is 0.6113*1.8=1.1.

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