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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, π, in patch one (and the other proportion, (1 β π), 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:
ππ₯
ππ‘= π₯ (π (π1 (1 β
ππ₯
πΎ1) β ππ§1) + (1 β π) (π2 (1 β
(1βπ)π₯
πΎ2) β ππ§2))
ππ§1
ππ‘= π§1(βπ1 + π1πππ₯ β π1π§1)
ππ§2
ππ‘= π§2(βπ2 + π2π(1 β π)π₯ β π2π§2)
ππ
ππ‘= ππ(1 β π) ((π1 (1 β
ππ₯
πΎ1
) β ππ§1) β (π2 (1 β(1 β π)π₯
πΎ2
) β ππ§2))
Where π₯ is the population of prey at time π‘, π§1 and π§2 are the populations of predators in patch 1 and 2
respectively, and π is the proportion of time prey spend in patch 1. In patch i, the growth rate of prey is
ππ and the carrying capacity is πΎπ . Predator i has an intrinsic death rate of ππ, coefficient of intraspecific
competition ππ, and conversion of prey to predator fitness ππ. The interaction coefficient between prey
and predator π§1 is π and between prey and predator π§2 is π. Lastly, π is the time-scale separation
coefficient. The expression for ππ
ππ‘ 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 π in patch 1
and (1 β π) 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) π 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 π 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 π,
denoted by (Ξ»,Β΅,Ο,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 π₯, we know that the fitness in both patches should
be zero so we have the linear equations (in terms of p)
π1 (1 βππ
πΎ1
) β πΒ΅ = 0 (1)
πππ π2 (1 β(1 β π)π
πΎ2
) β πΟ = 0 (2)
Additionally, the fitness of both predators π§1 and π§2 must also be zero, and again we have a set of linear equations (in terms of p)
βπ1 + π1πππ β π1Β΅ = 0 (3)
πππ β π2 + π2π(1 β π)π β π2Ο = 0 (4)
And lastly for the strategy p, ππ
ππ‘= 0 if and only if
(π1 (1 βππ
πΎ1
) β πΒ΅ β π2 (1 β(1 β π)π
πΎ2
) + πΟ) = 0
Where if (1) and (2) are satisfied so is this equation for ππ
ππ‘.
Solving for Β΅ in both (1) and (3) then set the equations equal to each other (similarly for Ο using (2) and
(4)) . This generates two equations with π equal to a function of p. Setting these equations equal to each
other yields an expression for the unique value of π at the equilibrium, given that π β (0,1),
π = [(
π1ππ1
+π1
πΎ1 π)
(π2ππ2
+π2
πΎ2π)
(π2
π2+
π2
π)
(π1
π1+
π1
π)
+ 1]
β1
(5)
Thus there is a unique value of p described by equation (5) that admits an equilibrium (Ξ»,Β΅,Ο,p) for some set of parameter values.
Computing the Jacobian matrix at this equilibrium yields
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π½ =
βΞ»(π2π1
πΎ1+
(1 β π)2π2
πΎ2) βππΞ» βπ(1 β π)Ξ» Ξ» ((1 β π)
π2
πΎ2β π
π1
πΎ1)
π1ππΒ΅ βπ1Β΅ 0 π1πλ¡
π2π(1 β π)Ο 0 βπ2Ο βπ2πΞ»Ο
βπ (ππ1
πΎ1+ (1 β π)
π2
πΎ2) βππ ππ βπΞ» (
π2
πΎ2+
π1
πΎ1)
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 π1 = 0.8, πΎ1 = 3, πππ π2 = 0.7, πΎ2 = 2.5.
For the effect parameters of predator on prey set π = 1 πππ π = 1.
The death rates of predators in absence of prey as π1 = 0.5 πππ π2 = 0.5.
For predator conversion rates, set π1 = 0.5 πππ π2 = 0.75.
For intraspecific competition between predators, set π1 = 0.1 πππ π2 = 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 π = 1. The equilibrium is
(1.801,0.506,0.504,0.6113).The Jacobian at this equilibrium is
π½|(1.8,0.5,0.5,0.6) =
β0.255 β1.1 β0.7 β0.0970.154 β0.05 0 0.4550.147 0 β0.025 β0.681
β0.271 β1 1 β0.984
Which has eigenvalues
πΏ1 = β0.557766 + 0.7609806π πΏ2 = β0.557766 β 0.7609806π πΏ3 = β0.99934 + 0.6075781π
πΏ4 = β0.99934 β 0.6075781π
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
(Ξ»,Β΅,0,p))? Assume that for any population of prey, the fitness function for predator π§2 is negative, i.e.
βπ2 + π2π(1 β π)π < 0 (6)
In this scenario, the fitness of prey in both patches must be zero (equations (1) and (2)), but since π§ = 0, from (2) we have that
π2 (1 β(1 β π)π
πΎ2
) = 0
Which can be simplified to π = 1 βπΎ2
π which can be substituted into inequality (6) which eliminates π
and π to give the inequality
πΎ2 <π2
ππ2 (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 π =πΎ1
π and πΎ1 <
π1
ππ1.
The Jacobian for when z2=0 is
π½ =
βΞ»(π2π1
πΎ1+
(1 β π)2π2
πΎ2) βππΞ» βπ(1 β π)Ξ» Ξ» ((1 β π)
π2
πΎ2β π
π1
πΎ1)
π1ππΒ΅ βπ1Β΅ 0 π1πλ¡0 0 βπ2 + π2π(1 β π)Ξ» 0
βπ (ππ1
πΎ1+ (1 β π)
π2
πΎ2) βππ ππ βπΞ» (
π2
πΎ2+
π1
πΎ1)
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Let us choose parameters so that inequality (7) is satisfied (i.e. take πΎ2 = 0.15 <0.5
1β0.75= 0.66).
For the patches we will set π1 = 0.8, πΎ1 = 3, πππ π2 = 0.7, πΎ2 = 0.15.
For the effect parameters of predator on prey set π = 1 πππ π = 1.
The death rates of predators in absence of prey as π1 = 0.5 πππ π2 = 0.5.
For predator conversion rates, set π1 = 0.5 πππ π2 = 0.75.
For intraspecific competition between predators, set π1 = 0.1 πππ π2 = 0.05.
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
6.16ππ3 + (0.108 + 7.02π)π2 + (0.032 + 3.13π)π + 0.47π = 0
If we solve the Routh-Hurwitz Criteria for the third order equation from in terms of π, we get the quadratic equation
19.075π2 + 0.562π + 0.0034 > 0
Which suggests that for these parameters, the system is stable for all π β₯ 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 π, let π2 = 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) π = 1 (black solid line), the system evolves to (1.25,0.506,0,0.88), and b) π = 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
π >π2π(1 β π) β
π1ππ
β (π2π1
πΎ1+
(1 β π)2π2
πΎ2)
π1
πΎ1+
π2
πΎ2
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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 π2 > ππ2πΎ2, or if
2) the conversion rate of the second predator satisfies π2 <π2
ππΎ2.
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:
βπ1 + π1πππ < 0 (8)
πππ β π2 + π2π(1 β π)π < 0 (9)
While the prey population at this equilibrium can be determined using equations (1) and (2). Since π§1 = 0 and π§2 = 0, we can solve (1) as
πΎ1 = ππ
And (2) gives us
πΎ2 = (1 β π)π
Substituting the first equation into the second shows that the prey equilibrium population is simply
π = πΎ1 + πΎ2
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
πΎ1 <π1
ππ1
πππ πΎ2 <π2
ππ2
The Jacobian is
π½ =
βΞ»(π2π1
πΎ1+
(1 β π)2π2
πΎ2) βππΞ» βπ(1 β π)Ξ» Ξ» ((1 β π)
π2
πΎ2β π
π1
πΎ1)
0 βπ1 + π1ππΞ» 0 00 0 βπ2 + π2π(1 β π)Ξ» 0
βπ (ππ1
πΎ1+ (1 β π)
π2
πΎ2) βππ ππ βπΞ» (
π2
πΎ2+
π1
πΎ1)
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 Ο on the behavior of the system.
For this example, we will define the parameters as follows:
For the patches we will set π1 = 0.4, πΎ1 = 1.5, πππ π2 = 0.9, πΎ2 = 4.
For the effect parameters of predator on prey set π = 0.6 πππ π = 0.7.
The death rates of predators in absence of prey as π1 = 0.4 πππ π2 = 0.65.
For predator conversion rates, set π1 = 0.8 πππ π2 = 0.3.
For intraspecific competition between predators, set π1 = 0.03 πππ π2 = 0.08.
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 Ο (Ο=1 is the black line, Ο=0.25 is the red-dotted line, and Ο=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 Ο: when Ο=0.25,
even after 150 time units the population is not at the exact equilibrium (πΟ=0.25,t=150 = 0.2103249).
Additionally, the solutions of p vary in period and amplitude.
For p:
1) When Ο=5 there is an increase in the period and amplitude of p in the first 25 time units
when compared to Ο=1. The increased initial amplitude could be interpreted biologically
through Ο 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 Ο=0.25, there is a decreased amplitude and increased period when compared to Ο=1.
The decreased amplitude can be explained biologically through Ο as prey taking longer to
learn to move to the more favourable patch. The increased period can be explained through Ο as prey responding slowly to changes in which patch is more favourable.
For y:
1) When Ο=5, the population of y decreases faster than when Ο=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 Ο=0.25, the population of y decreases slower than when Ο=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.