1) The document describes a two-patch, two-predator, one-prey system with differential equations modeling the population dynamics. 2) There exists a unique stable equilibrium where all three species can coexist if prey adopt an optimal adaptive strategy (p) of dividing time between patches. 3) The system can reduce to a one-predator, one-prey refuge system if the parameters satisfy an inequality causing one predator population to go extinct.

1. 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

2. 2
Considera2-predator1-preysystemthathas two patches.Preyare free tomove betweenthe
patcheswhile eachpredatorisrestrictedtoone patch. Additionally,preygrowthineitherpatchis
logisticandtheyspendaproportionof time, 𝑝,in patch one (andthe otherproportion, (1 − 𝑝), inpatch
two).Predatorfunctional responsesare bothof Holling-type Iandintraspecificcompetitionispresentin
bothpredatorspecies. Thiscanbe describedusingthe followingsystem,[1],of fourdifferential
equations:
𝑑𝑥
𝑑𝑡
= 𝑥 (𝑝 (𝑟1 (1 −
𝑝𝑥
𝐾1
) − 𝑎𝑧1) + (1 − 𝑝)(𝑟2 (1 −
(1−𝑝) 𝑥
𝐾2
) − 𝑏𝑧2))
𝑑𝑧1
𝑑𝑡
= 𝑧1(−𝑚1 + 𝑘1 𝑎𝑝𝑥 − 𝑐1 𝑧1)
𝑑𝑧2
𝑑𝑡
= 𝑧2(−𝑚2 + 𝑘2 𝑏(1 − 𝑝) 𝑥 − 𝑐2 𝑧2)
𝑑𝑝
𝑑𝑡
= 𝜏𝑝(1 − 𝑝) (( 𝑟1 (1 −
𝑝𝑥
𝐾1
) − 𝑎𝑧1) − ( 𝑟2 (1 −
(1 − 𝑝) 𝑥
𝐾2
) − 𝑏𝑧2))
Where 𝑥 is the populationof preyattime 𝑡, 𝑧1 and 𝑧2 are the populationsof predatorsinpatch1 and 2
respectively, and 𝑝 isthe proportionof time preyspendinpatch1. In patch i, the growth rate of preyis
𝑟𝑖 and the carrying capacityis 𝐾𝑖. Predatori has an intrinsicdeathrate of 𝑚 𝑖, coefficientof intraspecific
competition 𝑐𝑖,andconversionof preytopredatorfitness 𝑘 𝑖.The interactioncoefficientbetweenprey
and predator 𝑧1 is 𝑎 and betweenpreyandpredator 𝑧2 is 𝑏.Lastly, 𝜏 is the time-scale separation
coefficient. The expressionfor
𝑑𝑝
𝑑𝑡
isthe derivativeof the fitnessfunctionof the prey.
Let usassume that
1) preyare free to move betweenpatch1 and 2 andspenda proportionof time 𝑝 inpatch 1
and (1 − 𝑝) inpatch 2. Thisproportionshoulddependonthe observedfitnessof individuals
ineitherpatch(i.e.if a preyinpatch 1 seesthata preyin patch 2 hashigherfitness,it
shouldmigrate topatch 2),
2) 𝑝 is the strategythat the whole prey populationplays,and
3) preyin eitherpatchhave some wayto evaluate the fitnessof preyinthe otherpatch so
that theycan maximize theirownfitness.
If there existsavalue of 𝑝 such that the fitnessof preyinbothpatchesis zero,thenthere will be
no netmovementof preybetweenpatchesandastable equilibriumexists.Thismaybe acoexistence
equilibriumof all three species,orthe two-predatorone-preysystemmayreduce toa one-predator
one-preyrefugesystem,orbothpredatorsgo extinctandpreygrowthisonlylimitedbytheircarrying
capacity.

3. 3
Existence of Equilibria
Three-species coexistence equilibrium
First,we will considerathree-speciescoexistenceequilibriumwithpreyplaying adaptive strategy 𝑝,
denotedby(λ,µ,σ,p) usingsystem[1].Isthere a unique strategythatallowsall three speciestocoexist?
At a stable internal equilibriumfor the prey species 𝑥,we know thatthe fitnessinbothpatchesshould
be zero sowe have the linear equations (intermsof p)
𝑟1 (1 −
𝑝𝜆
𝐾1
) − 𝑎µ = 0 (1)
𝑎𝑛𝑑 𝑟2 (1−
(1 − 𝑝) 𝜆
𝐾2
) − 𝑏σ = 0 (2)
Additionally,the fitnessof bothpredators 𝑧1 and 𝑧2 must alsobe zero,and againwe have a setof linear
equations (intermsof p)
−𝑚1 + 𝑘1 𝑎𝑝𝜆 − 𝑐1µ = 0 (3)
𝑎𝑛𝑑 − 𝑚2 + 𝑘2 𝑏(1 − 𝑝) 𝜆 − 𝑐2σ = 0 (4)
Andlastlyforthe strategy p,
𝑑𝑝
𝑑𝑡
= 0 if and onlyif
( 𝑟1 (1 −
𝑝𝜆
𝐾1
) − 𝑎µ − 𝑟2 (1 −
(1 − 𝑝) 𝜆
𝐾2
) + 𝑏σ) = 0
Where if (1) and(2) are satisfiedsoisthisequationfor
𝑑𝑝
𝑑𝑡
.
Solvingforµ inboth (1) and (3) thensetthe equationsequal toeachother(similarly forσ using(2) and
(4)) . Thisgeneratestwoequationswith 𝜆 equal toafunctionof p.Settingthese equationsequal toeach
otheryieldsanexpressionforthe unique value of 𝑝 atthe equilibrium, giventhat 𝑝 ∈ (0,1),
𝑝 = [
(
𝑘1 𝑎
𝑐1
+
𝑟1
𝐾1 𝑎
)
(
𝑘2 𝑏
𝑐2
+
𝑟2
𝐾2 𝑏
)
(
𝑚2
𝑐2
+
𝑟2
𝑏
)
(
𝑚1
𝑐1
+
𝑟1
𝑎
)
+ 1]
−1
(5)
Thus there isa unique value of p describedby equation(5) thatadmitsanequilibrium(λ,µ,σ,p)forsome
setof parametervalues.
Computingthe Jacobianmatrix atthisequilibriumyields

4. 4
𝐽 =
−λ(
𝑝2
𝑟1
𝐾1
+
(1 − 𝑝)2
𝑟2
𝐾2
) −𝑎𝑝λ −𝑏(1 − 𝑝)λ λ ((1 − 𝑝)
𝑟2
𝐾2
− 𝑝
𝑟1
𝐾1
)
𝑘1 𝑎𝑝µ −𝑐1µ 0 𝑘1 𝑎λµ
𝑘2 𝑏(1 − 𝑝)σ 0 −𝑐2σ −𝑘2 𝑏λσ
−𝜏(𝑝
𝑟1
𝐾1
+ (1 − 𝑝)
𝑟2
𝐾2
) −𝜏𝑎 𝜏𝑏 −𝜏λ (
𝑟2
𝐾2
+
𝑟1
𝐾1
)
If the eigenvaluesof the characteristicequationforthismatrix all have negative real part,the
equilibriumisasymptoticallystable.
For a numerical example,we will define the parameters(arbitrarily) asfollows:
For the patcheswe will set 𝑟1 = 0.8, 𝐾1 = 3, 𝑎𝑛𝑑 𝑟2 = 0.7, 𝐾2 = 2.5.
For the effectparametersof predatoronpreyset 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 1.
The deathrates of predatorsinabsence of preyas 𝑚1 = 0.5 𝑎𝑛𝑑 𝑚2 = 0.5.
For predatorconversionrates,set 𝑘1 = 0.5 𝑎𝑛𝑑 𝑘2 = 0.75.
For intraspecificcompetitionbetweenpredators,set 𝑐1 = 0.1 𝑎𝑛𝑑 𝑐2 = 0.05.
If the eigenvaluesof the Jacobianevaluatedatthe equilibriumforthisparametersetall have negative
real part, the systemisasymptoticallystable.Set 𝜏 = 1.The equilibriumis
(1.801,0.506,0.504,0.6113).The Jacobianat thisequilibriumis
𝐽|(1.8,0.5,0.5,0.6) =
−0.255 −1.1 −0.7 −0.097
0.154 −0.05 0 0.455
0.147 0 −0.025 −0.681
−0.271 −1 1 −0.984
Whichhas eigenvalues
𝛿1 = −0.557766 + 0.7609806𝑖
𝛿2 = −0.557766 − 0.7609806𝑖
𝛿3 = −0.99934 + 0.6075781𝑖
𝛿4 = −0.99934 − 0.6075781𝑖
Whichall have negative real partso the systemisasymptoticallystable.
For these valuesof parameters,we canpredict(usingthe above equationforp) the value thatpwill
take at equilibriumusingequation(5).Inthiscase,the predictedvalueispp=0.6112956. From the
model,after150 time steps,the observedvalue of pat the equilibriumis pobs=0.6112956, thuspp=pobs
(note thatif the initial conditionsare notnearthe equilibriumpopulationsizes,the preystrategymay
not exactlymatchthe predictedvalue after150 time steps).Additionally,evenwheninitial conditions
are varied,the equilibriumpopulationsizesandstrategyremainthe same.

5. 5
One-predator one-prey refuge system
Next, whenwillsystem[1]reduce toa predator-preyrefugesystem (e.g.(x,z1,z2,p) evolvesto
(λ,µ,0,p))?Assumethatforany populationof prey,the fitnessfunctionforpredator 𝑧2 isnegative,i.e.
−𝑚2 + 𝑘2 𝑏(1 − 𝑝) 𝜆 < 0 (6)
In thisscenario,the fitnessof preyinbothpatchesmustbe zero(equations(1) and(2)),butsince 𝑧 = 0,
from(2) we have that
𝑟2 (1 −
(1 − 𝑝) 𝜆
𝐾2
) = 0
Whichcan be simplifiedto 𝑝 = 1 −
𝐾2
𝜆
whichcan be substitutedinto inequality(6) whicheliminates 𝜆
and 𝑝 to give the inequality
𝐾2 <
𝑚2
𝑏𝑘2
(7)
Therefore,if thisinequality (7) issatisfied,thensystem [1]reducestoa predator-preyrefugesystem. As
the intrinsicdeathrate of the predatorincreases,the carryingcapacityof the patch that yieldsarefuge
systemincreases (i.e.the predatorsdie outquicklysotheyneedmore preypresentto save themfrom
extinction). Asthe abilityof predatorstoconvertpreytofitnessincreases,the carryingcapacityof the
patch to cause extinctionof the predatordecreases(i.e.since predatorsare betterutilizingeachprey,
they can tolerate lowerpreypopulations).
The solutionissimilarforz1 to go extinct,where 𝑝 =
𝐾1
𝜆
and 𝐾1 <
𝑚1
𝑎𝑘1
.
The Jacobianfor whenz2=0 is
𝐽 =
−λ(
𝑝2
𝑟1
𝐾1
+
(1 − 𝑝)2
𝑟2
𝐾2
) −𝑎𝑝λ −𝑏(1 − 𝑝)λ λ ((1 − 𝑝)
𝑟2
𝐾2
− 𝑝
𝑟1
𝐾1
)
𝑘1 𝑎𝑝µ −𝑐1µ 0 𝑘1 𝑎λµ
0 0 −𝑚2 + 𝑘2 𝑏(1 − 𝑝)λ 0
−𝜏(𝑝
𝑟1
𝐾1
+ (1 − 𝑝)
𝑟2
𝐾2
) −𝜏𝑎 𝜏𝑏 −𝜏λ (
𝑟2
𝐾2
+
𝑟1
𝐾1
)

6. 6
Let uschoose parameterssothat inequality(7) issatisfied(i.e.take 𝐾2 = 0.15 <
0.5
1∗0.75
= 0.66).
For the patcheswe will set 𝑟1 = 0.8, 𝐾1 = 3, 𝑎𝑛𝑑 𝑟2 = 0.7, 𝐾2 = 0.15.
For the effectparametersof predatoronpreyset 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 1.
The deathrates of predatorsinabsence of preyas 𝑚1 = 0.5 𝑎𝑛𝑑 𝑚2 = 0.5.
For predatorconversionrates,set 𝑘1 = 0.5 𝑎𝑛𝑑 𝑘2 = 0.75.
For intraspecificcompetitionbetweenpredators,set 𝑐1 = 0.1 𝑎𝑛𝑑 𝑐2 = 0.05.
Thissystemevolvesfrom(x,z1,z2,p))=(1,0.5,0.5,0.4) to(1.25,0.506,0,0.880). The characteristic
polynomial of thissystemis
6.16𝜏𝜆3 + (0.108 + 7.02𝜏) 𝜆2 + (0.032 + 3.13𝜏) 𝜆 + 0.47𝜏 = 0
If we solve the Routh-HurwitzCriteriaforthe thirdorderequationfrom intermsof 𝜏, we get the
quadraticequation
19.075𝜏2 + 0.562𝜏 + 0.0034 > 0
Whichsuggeststhat forthese parameters,the systemisstable forall 𝜏 ≥ 0.Thusevenif preydonot
behave adaptively(inthiscase),the systemcanstill persist.

7. 7
In orderto geta restrictionon 𝜏, let 𝑟2 = 0.1, and the initial conditionforpwouldhave tobe
verysmall.Thisway,preyare startinginthe lessfavourable patchbutnotmovingto the firstpatch in
time to allowforthe predator-preyrefugesystemtopersist(i.e.predatorz1 goesextinctbeforeenough
preymove intopatch 1).
Thisis the behaviourof the systemwithinitial conditions(X,Z1,Z2,P)=(1,0.5,0.5,0.1) where
a) 𝜏 = 1 (blacksolidline), the systemevolvesto(1.25,0.506,0,0.88), and
b) 𝜏 = 0.08 < 0.099 (dottedline),the systemevolvesto (3.21,0,0,0.93) after150 time steps.
Thisrestrictionisderivedusingtr(J),whichgive the inequality
𝜏 >
𝑘2 𝑏(1 − 𝑝) −
𝑐1 𝜇
𝜆
− (
𝑝2
𝑟1
𝐾1
+
(1 − 𝑝)2
𝑟2
𝐾2
)
𝑟1
𝐾1
+
𝑟2
𝐾2

8. 8
Due to inequality(7),we wouldexpecttosee thiskindof behaviourif we chose parameterssuchthat:
1) the mortalityrate of the secondpredatorsatisfies 𝑚2 > 𝑏𝑘2 𝐾2,or if
2) the conversionrate of the secondpredatorsatisfies 𝑘2 <
𝑚2
𝑏𝐾2
.
One-prey system
The conditionsforbothpredatorsto go extinctinsystem [1]are simply derivedfromwhenbothof the
followinginequalitiesare satisfied:
−𝑚1 + 𝑘1 𝑎𝑝𝜆 < 0 (8)
𝑎𝑛𝑑 − 𝑚2 + 𝑘2 𝑏(1− 𝑝) 𝜆 < 0 (9)
While the preypopulationatthisequilibriumcanbe determinedusingequations(1) and(2).
Since 𝑧1 = 0 and 𝑧2 = 0, we can solve (1) as
𝐾1 = 𝑝𝜆
And(2) givesus
𝐾2 = (1 − 𝑝)𝜆
Substitutingthe firstequationintothe secondshowsthatthe preyequilibriumpopulationissimply
𝜆 = 𝐾1 + 𝐾2
Usingthese identitiesforpand(1-p) and substitutingthemintoequations(8) and(9) we getthat both
are satisfied forall valuesof p if and onlyif
𝐾1 <
𝑚1
𝑎𝑘1
𝑎𝑛𝑑 𝐾2 <
𝑚2
𝑏𝑘2
The Jacobianis
𝐽 =
−λ(
𝑝2
𝑟1
𝐾1
+
(1 − 𝑝)2
𝑟2
𝐾2
) −𝑎𝑝λ −𝑏(1 − 𝑝)λ λ ((1 − 𝑝)
𝑟2
𝐾2
− 𝑝
𝑟1
𝐾1
)
0 −𝑚1 + 𝑘1 𝑎𝑝λ 0 0
0 0 −𝑚2 + 𝑘2 𝑏(1 − 𝑝)λ 0
−𝜏 (𝑝
𝑟1
𝐾1
+ (1 − 𝑝)
𝑟2
𝐾2
) −𝜏𝑎 𝜏𝑏 −𝜏λ (
𝑟2
𝐾2
+
𝑟1
𝐾1
)
Thus system[1]can evolve toone of three differentoutcomes dependingonparametervalues
(assumingpreygrowthratesare both non-zero):

9. 9
i) A three-speciestwo-predatorone-preysystem,
ii) A two-speciespredator-preyrefugesystem, or
iii) A one-speciessystemwhereonlythe prey survives.
Invasion by prey playing a different strategy
The above showsthat a stable three-speciesequilibriumcanindeedbe establishedandthe preyevolve
to playthe strategyp=0.6113 for the set of parameters outlined. Considerinvasionof thissystembyan
alternate preyspeciesWthatplaysa fixedstrategyq=0.5.Additionally,fix the strategyof preyXto
p=0.6113. At t=200, an invadingpopulationof W=0.3 entersthe system.
It isclear fromthese graphs that W cannot invade the systemandthatthe system eventually returnsto
itsoriginal equilibriumforp=0.6113. Att=400, the populationsare (X,Z1,Z2,W)=(1.801,0.506,0.503,0)
whichisthe original equilibrium. CouldaninvadingpreyWplayingq=0.6113 invade the systemwhere
preyX isplayingp=0.5?

10. 10
If residentpreyXisplayingp=0.5, thenit hasan equilibrium(X,Z1,Z2)=(1.47,1.09,0).
If invadingpreyW isplayingq=0.6113, thenit can invade the system(asseenabove).Att=400, the
populationsizesare (X,Z1,Z2,W)=(0,0,1.25,1.93).However,anyinvadingpreyplayingq>p=0.5could
invade the system.

11. 11
Can an invaderplayingp=0.6112996 invade aresidentpopulationplayingasimilarvalue?Letustestfor
residentpreyplayingp=0.59 (dottedline) andp=0.59 (solidline). The vertical lineindicateswhenthe
invasionoccurs.
From thisgraph,we can see thatinboth casesW can successfullyinvadeevenwhenXisplayinga
strategyclose to q=0.6112996.

12. 12
Time-Scale Separation
Nextwe will investigate the effectof the time scale coefficientτonthe behaviorof the system.
For thisexample, we will define the parametersasfollows:
For the patcheswe will set 𝑟1 = 0.4, 𝐾1 = 1.5, 𝑎𝑛𝑑 𝑟2 = 0.9, 𝐾2 = 4.
For the effectparametersof predatoronpreyset 𝑎 = 0.6 𝑎𝑛𝑑 𝑏 = 0.7.
The deathrates of predatorsinabsence of preyas 𝑚1 = 0.4 𝑎𝑛𝑑 𝑚2 = 0.65.
For predatorconversionrates,set 𝑘1 = 0.8 𝑎𝑛𝑑 𝑘2 = 0.3.
For intraspecificcompetitionbetweenpredators,set 𝑐1 = 0.03 𝑎𝑛𝑑 𝑐2 = 0.08.
We can calculate pp using(5) to getthat pp=0.2104549. From the simulationsafter100 time steps,the
systemevolvesto(4.045307,0.2882813,0.2590909,0.2104581), andafter 150 time stepsthe p value
becomes0.2104549, whichconfirmsourpredictionforp.
Considerthe parametersfromabove,wherep=0.210 at equilibrium.The behaviorof solutions
for differentvaluesof τ(τ=1 is the blackline,τ=0.25 is the red-dottedline,andτ=5is the green-dotted
line) isshownbelow.

13. 13
From thiscomparison,we cansee that forthisexample,there isamore pronounceddifference inthe
behaviorof yand p, thoughthe equilibriaare all almostequal fordifferentvaluesof τ:when τ=0.25,
evenafter150 time unitsthe populationisnotatthe exactequilibrium( 𝑝τ=0.25,t=150 = 0.2103249).
Additionally,the solutionsof pvaryin periodandamplitude.
For p:
1) Whenτ=5 there isan increase in the periodandamplitude of pinthe first25 time units
whencomparedtoτ=1. The increased initial amplitude couldbe interpretedbiologically
throughτ in thatsince preychange theirbehaviorquickly,if apatchismore favourable than
another, initiallypreywill move intothispatchinlarge numberswhichultimately decreases
the fitnessof all preyinthat patch.The decreasedperiodcanbe explainedasprey
respondingquickertochangesinpatch fitnesssothey move betweenpatchesmore
frequently.
2) Whenτ=0.25, there isa decreasedamplitudeandincreasedperiodwhencomparedtoτ=1.
The decreasedamplitudecanbe explainedbiologicallythroughτasprey taking longerto
learnto move tothe more favourable patch.The increasedperiodcanbe explainedthrough
τ as preyrespondingslowlytochangesinwhichpatchis more favourable.
For y:
1) Whenτ=5, the populationof ydecreasesfasterthanwhen τ=1.Thisis due to more prey
movingintopatch2 initially,sothere islesspreyavailable inpatch1 and predatorycannot
sustaina highpopulation.
2) Whenτ=0.25, the populationof ydecreasesslowerthanwhen τ=1.Thisisbecause lessprey
are movingtothe secondpatch, sothe populationof yhasmore prey available inthatinitial
time interval.
If the initial conditionforp=0,thenall the preywill be inpatch 2 andthe populationof ywill decreaseto
0. Let the initial conditionsbe (2,1,1,0),thenthe systemevolvesto(0.7,0,0.503,0). Note thatthe
populationof zat equilibriumisthe same forwhenthere isathree-speciescoexistenceequilibriumand
that the populationof x isjust(1-0.6113)*1.8=0.7. Whenp starts at 1, the systemevolvesto
(1.1,0.506,0,1). The population of yis the same as at the three-speciesequilibriumandthe populationof
x is0.6113*1.8=1.1.