Maini, P. K.

English

Oxford University Research Archive

Discrete mathematics and its application to ecology

Mathematical Institute Eprints Archive

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Mathematics Masterclasses
Stretchingtheimagination
Editedby
MICHAEL SEWELL
ProfessorofAppliedMathematics,Universityof Reading
OXFORD NEW YORK TOKYO
OXFORD UNIVERSITY PRESS
1997
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MathematicsMasterclasses-Stretchingtheimagination/ editedbyMichael Sewell.
1.Mathematics 1.Sewell,M. J. (Michael1.), 1934-
QA3.M3631997 51O-dc20 96-27450
ISBN 0198514948 (Hbk) ISBN 019851493 X (Pbk)
TypesetbyTechnicalTypesettingIreland
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MidsomerNorton,Avon
6 DiscreteMathematicsand its
Application to Ecology
byPHILIP MAINI
Centrefor MathematicalBiology,MathematicalInstitute,Universityof Oxford
1. Introduction
In thischapterwewill illustratehowsimplemathematicalideascanbeusedto
understandhowa populationgrows.We firstconsiderthewell-knownproblem
which generatesthe Fibonaccisequence.Considera populationof rabbits.
Supposethateverypairof rabbitscanreproduceonlytwice,whentheyareone-
andtwo-monthsold, andthateachtimetheyproduceexactlyonenewpair of
rabbits.Assumethatall newbornrabbitssurvive.The problemhereis to seeif
wecanpredicthowmanynewbornrabbitstherewill be in a certaingeneration
givenaninitialnewbornrabbitpopulation.
Beforebeginning,we needto usea convenientnotation.Let us definethe
quantityNmtobe thenumberof newbornpairsin generationm.Supposethat
webeginwithonenewbornpair.ThenwesaythatNo,thenumberof newborn
pairsin thezerogeneration(theAdamandEve of rabbits),is one.Now, after
onemonth,thispairof rabbitsgivesbirthto onepairof rabbits.Therefore,we
havethatNI, thenumberof newbornpairsin thefirstgeneration,is one.At the
endof thesecondmonth,AdamandEvegivebirthto theirsecondnewbornpair
andthenwill no longergivebirthtoanymorerabbits.But,thepairof rabbitsin
the first generationalsogivebirth to a pair of newborns(Adamand Eve's
'grandbunnies').Therefore,N2, the numberof newbornpairs in the second
generation,will be 1+1=2. In thethirdgeneration,newbornpairswill arise
fromthosepairsthatwerenewbornsin thesecondgeneration(N2),and from
thosenewbornsin thefirstgeneration(NI), Thereforethenumberof newborns
in the third generation,N3, will be N2+NI =2+ 1=3. The numberin the
fourthgeneration,N4,will be N3+N2=3+2=5 (seeFig. 1).We can now
provethefollowingtheorem.
Theorem1 (TheRabbitProblem)
If
Nm=numberof newbornpairsingenerationm
then
Nm+l =Nm +Nm-l' (1)
uu Ulscreteivlaltlt::1ilal1csandHsApplicationtoEcology DiscreteMathemal1csaudItsApphcal10ntoEcology
In
Proof
The numberof newbornrabbitsin the (m+ nth generationwill comeeither
from one-monthold parentswhowereborn in the mth generationor from
two-montholdparentswhowerebornin the(m - nth generation(seeFig. 1).
Henceresult. 0
To see howthistiesin withtheabovediscussion,letussupposethatNo= 1
(thismeansthatwebeginwithonepair of newbornsin thezero generation).
Then,we knowthat NI = 1.Now,if weput m=1 into equation(1),we have
thatN2 =NI +No, thatis, Nz= 1+ 1=2,whichis thenumberof newbornsin
the secondgeneration.Now puttingm=2 into equation(1), we havethat
N3=N2 +NI =2+1=3,thenumberof newbornsin thethirdgeneration.For
m=3, we haveN4=3+2=5, thenumberof newbornsin the fourthgenera-
tion.Continuinglike thiswe havethat N5=5+3=8, N6=8+5= 13,N7=
13+8= 21,etc.
The sequenceof numbers1,1,2,3,5,8,13,21,... is known as the Fibonacci
sequence.Equation(1) is an exampleof a recu"encerelationand this typeof
relationis verycommonin ecologicalstudies.A mathematicalformulationof
suchaproblemis knownasa mathematicalmodelof theproblem.The numbers
generatedby theFibonaccisequenceareknownastheFibonaccinumbersand
occurwidelyin nature.For example,thenumbersof spiralsrunningin opposite
directionson a ripeningsunfloweror a pine cone are often consecutive
Fibonaccinumbers.
We now moveon to whatappears,to beginwith, to be a very different
problem.We call it theBeetleProblem.Supposethatbeetlesof onegeneration
producethe larvae(immatureformof an insectwhicheventuallybecomesthe
insect)of the nextgenerationand then die so that only the larvaesurvive.
Assumethatall the larvaesurvive.The problemhereis, canwe predictthe
numberof larvaein a certaingeneration?In thiscase,wedefineNmto bethe
number(in someunits)of larvaein generationm. Supposethatthe average
numberof larvaeproducedperbeetleis 2. Let usassumethatin generation0,
No is 1. It is importantto note here that settingNI to be one does not
necessarilymeanthatin thefirstgenerationthereis onebeetle.This is because
we havenot specifiedour units.For example,if theunit wasthousands,then
No=1wouldmeanthattherewereonethousandbeetlesin generationO.Note
that No=! wouldnotmeanhalf a beetle,butwouldmean500(onehalfof a
thousand)beetles.
Assumingthat No=1,we havethat NI, the numberof beetlesin thefirst
generation,is 2X 1=2. The beetlesin the secondgenerationarisefrom the
larvaeproducedby thefirstgeneration,so Nz is 2X NI =4. In thesameway,
N3=2 X 4=8 etc.In fact,wecannowprovethefollowingtheorem.
Theorem2 (TheBeetleProblem)
If
Nm=numberofbeetles(insomeunits)ingenerationm
r = averagenumberof larvaeproducedper beetle
then
Nm=Norm (2)
w~
W~
W~
Proof
The generationm beetlescomefrom the (m - nth generation.Now each
beetlein the(m-nth generationproducedr larvae,soNm=rNm_I'Applying
thesameargumento the(m - nth generation,wehaveNm-I =rNm_z.Sowe
canwriteNm=rZNm_z'If wecarryonthisreasoningwededucethatNm=Norm
~~n 0
For ourexample,r =2 andNo=1.The formulathensaysthatNI =NoX 2=
2, Nz=NoX 4 =4, N3=NoX 8=8,etc.
Notice that if we want to calculatethe numberof beetlesin a certain
generation,thenallweneedto knowis Noand r andwehaveimmediatelythat
Nm=Norm.However,for the rabbitproblem,to find Nm,we needto know
Nm-I and Nm-z. But to calculatethesevalues,we needto knowNm-3'etc.
Therefore,if m is very large,then the calculationsinvolvedfor the beetle
problemareeasierthanthosefor therabbitproblem.However,in question6 of
Worksheet1 we will seethatthereis an easierwayto calculateNmfor the
rabbitproblem,whenm is large.
1. ConsidertheFibonaccinumbers1,1,2,3,5,8,13,21,34,55,89,144,233,...
(a) Calculate(to3 decimalplaces)thefractionsobtainedfromdividingeach
b b h
. b h . 1 Z 3 5 8 13 21 34 55 89 144 233num er y t e prevIousnum er, t at IS, T, T, 2' 3' 5' g, 13'21'34"'55'89' 144'
(b) What do you notice about the valuesyou calculatedin (a)?
(c) The calculations in (a) and (b) suggestthat for two consecutivenumbers
N
Nm+I' Nm, in the Fibonacci sequence,~::= r for large m. What is thatNm
value of r?
I
W~
I I
W~
11 &i~W~ .
I
w~
I
W~
I
w~
W~
w~
Fig. 1. ProofofTheorem1(seetextfordetails)
92
DiscreteMathematicsanditsApplicationtoEcology DiscreteMathematicsanditsApplicationtoEcology 93
, -
,~'-
L ' .- I
I r-
--------
5. CalculateNp Nz, N3,N4andNs (eachto 2 decimalplaces)for equation(3)
in the casewherer =3.1,No=0.70.What do you think are thevaluesof
N6,N7,Ns,Ng,...?
6. In question1,we madean approximatecalculationfor r. We can makea
Nm+I Nm
moreexactcalculationasfollows.Supposethat - = - =r.
Nm Nm- I
(a) From equation (1) showt;at Nm+I = (1 + ~)Nm.
(b) From (a), show that 1 + - = r.
r
(c) From (b), calculate r.
-
I
2. The logistic map
In Worksheet1,question3,wesawthatthemodelspresentedin Section1have
theirlimitations,andwe introduceda newmodel
Nm+1= rNm(1.0 - Nm).
This modelis calledthe logisticmodel,or the logisticmap.Beforewe analyse
thismodelfurtherweneedthefollowingdefinition.
(4)
Fig.2. The Beetleproblemfor thecaser =2.NotethedifferencesbetweenthisfigureandFig. 1
(d) Is theRabbitproblemreallythatdifferentfromtheBeetleproblem?
2. Fromequation(2)calculateNI, Nz, N3andN4 in thefollowingcases.
(a) No=2.0,r= 2.0.
(b) No=2.0,r =0.5.
3. Fromequation(2)whatsizewill Nmbe for m largein thefollowingcases.(a) r > 1.0.
(b) r = 1.0.
(c) r < 1.0.
(d) Do youthinkthatequations(1)and(2)aregoodmathematicalmodels?
For thefollowingexercisesassumethattherecurrencerelationhastheform
Nm+1=rNm(1.0- Nm). (3)
(Notethatif Nmis small,equation(3) looksa bit like equation(2».
4. CalculateNI, Nz, N3andN4 in thefollowingcases.
(a) No=0.6,r =0.5.
(b) No=0.4,r =2.0.
(c) No=0.6,r =2.0.
(d) No=0.5,r =2.0.
(e) In (b),(c)and(d)whatdoyouthinkthevalueof Nmwouldbefor m verylarge?
(f) Is thisabettermathematicalmodelthanequations(1)and(2)?
Definition
If thesuccessivevaluescalculatedfromarecurrencerelationtendtowardssome
fixedvalue,we call thisvaluethe steadystatevalue.It hasthe propertythat
Nm+1=Nm =N*, thesteadystatevalue,for largem.
The steadystatesfor equation(4)aregivenby
r-1
N* =0 or N* = -.
r
(5)
For 0 <r < 1thesecondsteadystatedoesnotmakesensebecauseit isnegative
andif westartoff with0~No~1 in equation(4)then0~Nm~ 1for everym.
So theonlysensiblesolutionis thefirst solutionof equation(5)(comparethis
withWorksheet1,question4(a».For 1<r <3 thesecondsolutionmakessense
andit appearsfromWorksheet1,question4(e),thatthisis thesteadystate,at
leastfor r =2.0.Hence,althoughtherearenowtwopossibilitiesfor the steady
state,we havemovedoff the first solutiononto the second.This is calleda
bifurcation.For r =3.1,the secondsolutionof equation(5) is still a sensible
steadystatebutWorksheet1,question5 suggeststhatwehavea2-cyclesolution.
We maysummarizetheseresultsonthebifurcationdiagramin Fig. 3.Note that
wehavebeencarefulnottogotoofar beyondr =3 in casethereareyet more
surprisesin store!
We cangetan ideaof whatthesolutiontoequation(4)will looklikeby using
a graphicaltechniquecalledcobwebbing.The techniqueis illustratedin Fig. 4
andtheresultsrepresentedgraphicallyin Fig.5.
DiscreteMathematicsanditsApplicationtoEcology
1.0
DiscreteMathematicsanditsApplicationtoEcology
1.0
95
N* Nrn
0.8
0.6
0.4
0.2
0.0
0.5 1.0 1.5 2.0 2.5
r
3.0 3.5 4.0
Fig. 3. Bifurcationdiagramfor equation(4) showinghowthesteadystateN* varies
withr. The dashedlinesareunstablesolutions
0.8
0.6
0.4
8 10 12 14
0.2
0.0
2 64
m
Fig.5. Graphicalrepresentationof thesolutionobtainedbythecobwebbingtechnique
illustratedin Fig.4 (Comparethissolutionwith'thatfromWorksheet1,question4)
We drawthe curveof equation(4) (with r =0.5)andalsothe straightline
Nm+I =Nm.Wefirstfindthevalue0.6alongthehorizontalxis.Thisis No.Now
NI =0.5X NoG - No)'But thisis exactlythelengthof theverticalstraightline
fromthepointNo to thecurve.If wethendrawfromtherea lineparallelto the
horizontalaxis,thenthedistancethisline is abovethehorizontalaxisis always
NI, If we drop a perpendicularline onto the horizontalaxisfrom the point
wherethis line intersectsthe Nm+I=Nm line,we havethatthedistancefrom
thispoint to the origin is also NI, That is, givenNo, we havebeenable to
calculate,graphically,thevalueof NI, In a similarwaywecan nowgo on to
calculateN2,N3,..., etc.1.0
Nm+1
0.8
0.6
0.4
0.2
0.0
0.2 0.4 0.6
No
0.8 1.0
N2 NI Nm
Fig.4. Illustrationof thecobwebbingtechniquefor equation(4)with No=0.6,r =0.5
N =Nm+J m
0.0 0.6 0.80.2 0.4
Nm
Fig.6. CobwebforWorksheet2,question1
1.0
Nm+1
0.8
0.6
0.4
0.2
96 DiscreteMathematicsanditsApplicationtoEcology
Worksheet2
1. Use Fig. 6 to cobwebthe solutionto equation(4) for the caseNo=0.7,
r= 2.0.
2. Illustrateyour resultsto question1 graphically.Doesthissolutioncycleor
doesit go to a steadystate?
3. Useyourcalculatorto find Nj, Nz, N3, N4, Ns, N6,N7, and N8 (eachto 3
decimalplaces)for equation(4) in the caseNo=0.501,r =3.5.Draw the
cobwebdiagramon thefollowingfigure(Fig.7).
1.0
Nm+l
0.8
0.6
0.4
0.2
Nm+l=3.5Nm(1.0-Nm)
0.0 0.2 0.4 0.6 0.8 1.0
N
III
Fig.7. CobwebforWorksheet2,question3
4. Illustrateyourresultsto question3 graphically.
5. Fromequation(4),whatis theequationfor Nm+zin termsof Nm?Whatare
thesteadystatesof thisequation?
3. Chaos
FromWorksheet2,question3,wesawthatif r =3.5,thenequation(4)hasa
4-cyclesolution.As r increasesfurther,wegetan 8-cyclesolutionandthena
16-cyclesolution,etc.We canrepresentheseresults(Fig.8)on thebifurcation
diagramwebeganto drawin Fig.3.
If weincreaser beyondacriticalvaluere:::::3.83welosetheperiodicstructure
of the solution.Figure 9 illustratesthe casewhen r =3.9.In this casethe
solutiondoesnothaveaperiodicstructure.It issaidtobechaotic.If wewereto
changethevalueof No slightlywewouldget a completelydifferentlooking
solution.Anotherchapterin thisbook,'A Little Bit of Chaos',introducesyouto
otheraspectsof chaos.
DiscreteMathematicsanditsApplicationtoEcology
1.0
N*
0.8
0.6
0.4
0.2
0.0 0.5
97
4.01.0 1.5 2.5 3.0 3.52.0
r
Fig. 8. The completedbifurcationdiagramillustrating,schematically,the cascadeof
perioddoublingbifurcations
Equation(4) is a goodmodelfor insectgrowth- manyinsectpopulations
eitherreacha steadystateor oscillatein a 2-cycle(seeFig. 10).
Nm
1.0
0.8
0.6
5 10 15 20 25
0.4
0.2
m
Fig. 9. The solutionto equation(4) for thecaser =3.9,No=0.2.Notethatfor some
generationsthepopulationgetsverylargethencrashesbackin thefollowinggeneration
98 DiscreteMathematicsanditsApplicationtoEcology
1.5
1.0 1.0
0.5
I I I I " I I
V\.A-~
0.5
0.0
(a)
0.0
(b)
105 5 10
Fig. 10. Examples of population growth of two types of beetles. The vertical axis is the
number of beetles (thousands), the horizontal axis is the generation number. (Redrawn
from Hassel (1976))
Note thatfor r> 4 themodelpredictsnegativepopulationsandthusis not
realisticin thiscase.You maywantto doa fewiterationsor drawsomecobweb
mapsto seewhy(hint:startoff with r =4.1,No=0.5).
The analysisof themodelequation(4)suggeststhatif welet r gettoo large
thenwewill getcyclingpopulationswithverylargenumbersor chaoticpopula-
tionswiththepossibilityof verylargenumbers.Therefore,in orderto control
thepopulationnumbers,we reallyneedto keepr <3.83andpreferablymuch
smaller.We candecreaser by,for example,releasingsterileinsectsinto the
population,so thattheaveragebirthrateperinsectdecreases.
We canwritedownmorecomplicatedrelations,or iteratedmaps.For example,
wecouldhave
rNm
b
Nm+ 1= 0.0+aNm)
(6)
where r, a and b are positiveconstants.Figure 11 showsthe resultsfrom
equation(6) for thecaser =20,a =0.01,No=0.01anddifferentvaluesof b.
4. Summary
In this chapterwe haveillustratedhowmathematicalmodelscanbe usedto
studypopulationgrowth.Theanalysisof thesemodelsdeterminesif theassump-
tionswe madein formingthemodelaresensible.If not,wehaveto reformulate
themodel.Thishelpsustounderstandmoredeeplytheprocessesof population
growththatwe are tryingto modelandalsopredictshowthepopulationwill
reactif wecarryoutcertainexperiments.The roleof a predictivemathematical
model is very important.For example,it can aid our understandingof how
certaindiseasespreadwithin a populationand help us to choosethe best
controlstrategy.Mathematicalmodelsare importantnot only in ecology,but
alsoin thefieldsof biology,physics,chemistryandmedicine.
DiscreteMathematicsanditsApplicationtoEcology
99
40 ,.-.- 4
3
30
';?,E<"'>
0- 20
';i?"E
"b- 2
10
15
m
20 25
0
100._-e-.A I
2 4 6 8 10 12 14
m
5
(b)
(a)
20
15
E
';i?"
;:; 10
5
0 I I I I
5 10 15 20 25
m
(c)
Fig. 11. The solutionto equation(6) for the case r =20, a =0.01,No=0.01and
differentvaluesof b: (a)b =0.5,(b) b =2.5,(c) b =5.0.Thesolutionshownin (a) isvery
similarto thebehaviourof growingbacteriaandofyeast
SolutionstoWorksheet1
1. (a) 1,2, 1.500,1.667,1.600,1.625,1.615,1.619,1.618,1.618,1.618,1.618.
(b) The valuescalculatedin (a) tendto 1.618to 3 decimalplaces.(In fact,
1+/5
thisis thegoldenmean ).2
Nm+l
(c) -::::: r,wherer:::::1.618to3decimalplaces.
Nm
(d) The Rabbitproblem,for largern,is a specialcaseof theBeetleproblem
with r = 1.618to 3 decimalplaces.
2. (a) 4.0,8.0,16.0,32.0.
(b) 1.0,0.5,0.25,0.125.
3. (a) Nmwill beverylarge.
(b) N",=No.
100 DiscreteMathematicsanditsApplicationtoEcology
(c) Nmwill beverysmall.
(d) Equation(1) predictsthatthe populationof rabbitswill growforever;
equation(2)predictsthatthepopulationof beetleswill eithergrowandgrow
(if r> 1.0),or die downto zero(if r < 1.0)or stayat its initial value(if
r =1.0).Neitherof thesemodelsis verygoodbecausepopulationsdo not
behavein thisway.For example,deathhasbeenignored.
4. (a) 0.120,0.053,0.025,0.012to 3 decimalplaces.
(b) 0.480,0.499,0.500,0.500to 3 decimalplaces.
(c) 0.480,0.499,0.500,0.500to 3 decimalplaces.
(d) 0.500,0.500,0.500,0.500.
(e) Nmtendsto 0.5for largem.
(f) This seemsa bettermodelbecausefor certainr-valuesthe population
reachesa steadyvalue.
5. 0.65,0.71,0.65,0.71,0.65,0.71,0.65,0.71,..
6. (a) From the assumption,Nm-I =Nmlr, thereforeequation(1) becomes
Nm+I =Nm+Nmlr.
(b) From (a), we have Nm+1=(1+1/r)Nm. The assumptionsays that
Nm+I=rNm.Therefore1+ 1/r =r.
(c) From (b), rZ- r - 1=O.Therefore2r =1:i:{5. The negativesolution
doesnotmakesensebecauser isassumedpositive,thereforer =(1+ {5)/2.
Solutionsto Worksheet2
1.
1.0
0.8
Nm+l
0.6
0.4 t /1:___m,Y--~r ,'" ,'" ,'" ,'" ,'" ,
t' ", ", ", ", ", ", ", ", ", "
0.2
0.0
0.2 0.4 \ 0.6 0.8 1.0
NI NZN3 NO Nm
Fig.12 Cobwebdiagramfor Worksheet2,question1
DiscreteMathematicsanditsApplicationtoEcology 101
2.
1.0
~
Nm
0.8
.
0.6
./-.-.---.-.-.--.---0.4
0.2
2 4 6 8 10 12 14m
Fig. 13 Graphicalrepresentationof thesolutionobtainedby the cobwebbingtech-
niquefor Worksheet2,question1.The solutiongoesto a steadystate
3. 0.875,0.383,0.827,0.501,0.875,0.383,0.827,0.501
1.0
0.8
Nm+!
0.6
0.4 V, l :\t'T ": um_~m--m_--~u i" ," ," ,
" ," ," ,
" ," ," ,
" ," ,
" ,, , .,
0.2
,,
:\N +J=3.5N (I.O-N ),m m m
,
,
,,
0.0
0.2 0.4 0.6
NZ NO Nm
0.8
N3 NI
Fig 14 Cobwebdiagramfor Worksheet2,question3
lLJ:L DiscreteMathematicsanditsApplicationtoEcology
4. 1.0
Nm
0.8
0.6
0.4
0.2
2 4 6 8 10 12 14m
Fig 15 Graphicalrepresentationof the solutionobtainedby the cobwebbingtech-
niquefor Worksheet2,question3
5. Fromequation(4)wecanwriteNm+2=rNm+10.0- Nm+1)'Usingequation(3)again,
we haveNm+1=rNmO.O- Nm),sowecanwrite
Nm+2=r2NmO.0- N",)(1.0- rNm(1.0- Nm)).
The steadystatesof equation(7)aregivenby Nm+2=Nm=N*. Thuswe have
N* =r2N*(1.0- N*)(1.0 - rN*(I.O - N*))
whichhassolutionN* =0 or
(7)
(8)
(
1
)
1 1
(N*)3 - 2(N*)2+ 1+ - N* + - - - =O.
r r3 r
1
We knowthat1- - is alsoa steadystatesowecanfactorthisoutfromequation(9)r
to get
(9)
(N*-1+~)((N*)2-(1+~)N*+~+ :2)=0. (10)
Thus the two further solutions are given by the roots of ((N*)2 - 0 +
l/r)N* +1/r - 1/r3)=0, thatis,2N* =1+1/r:t [0 +1/r)2- (4Ir)(1+1/r)]I/2.
This maybewrittenas2N* = 1+(l/r):t O/r)[(r - 1)2- 4]1/2.Note thatthissolu-
tion will makesenseonlyif r> 3,sowehavea 2-cyclesolutionpossibleif r> 3.
References
R. AndersonandR. M. May,'The logicof vaccination,'NewScientist(18thNovember,
1982),410-15
O. M. Hassel,Thedynamicsof competitionandpredation(EdwardArnold Publishers,
1976)
R. M. May(editor),TheoreticalEcology,PrinciplesandApplications(2ndedition)(Black-
well ScientificPublications,1981)


Discrete mathematics and its application to ecology

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