a two-host shared macro parasite system with spatial heterogeneity

8/3/2019 A Two-Host Shared Macro Parasite System With Spatial Heterogeneity

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UNIVERSITA DEGLI STUDI DI TRENTOFacolta di Scienze Matematiche, Fisiche e Naturali

Corso di Laurea Specialistica in Matematica

Final Thesis

A Two-Host Shared Macroparasite Systemwith Spatial Heterogeneity

Relatori: Laureando:prof. Pugliese Andrea Manica Mattiadr. Rosa Roberto

Anno Accademico 2009-2010


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Contents

Introduction v

1 Description of the Spatially Structured Model 1

2 Single Host Macroparasite System 7

2.1 Constant hosts fertility with no reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.1 Condition for Equilibrium Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 General Condition for Diffusion-Driven Instability . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Fertility Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Single-Species Equation 19

3.1 The Best Location for a Favorable Habitat Patch . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Parasite and Larvae Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Two Host Shared Macroparasite System 33

4.1 Space Independent Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.1.1 Direct Competition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Spatially Dependent Habitat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5 Results 55

Appendix 57

Bibliography 67

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Introduction

Observations in wildlife supported by knowledge deriving from theoretical studies highlightedthe role of parasites in host population dynamics. Although single host-parasite systems havebeen extensively studied [4] [14] the case when more hosts share a common parasite is notfully understood. It has been demonstrated that transmission of shared parasites amongspecies can deeply affect host population especially when a host can act as a reservoir forparasites that cause severe mortality and morbidity in the other host species (less tolerant).

However, clear evidence in nature of such phenomena is limited because of the complexityof interactions and the number of factors involved.In the vast majority of cases the role of a non specific parasite in the decline of host pop-ulation is suggested by descriptive or correlational work, for instance negative correlationsbetween parasite infection and components of host fitness. Until now one of the most studiedcase where evidences suggested that apparent competition has played a key role is the de-cline of wild grey partridge population in the UK from 1950. It is suspected that the declinecaused by competition mediated via a caecal nematode with the ring-necked pheasant. Thishypothesis has been supported by a two-host macroparasite model and an accurate workof controlled comparison of the macroparasites impact on the two hosts [30][31][32]. All ofthese studies neglect to consider if spatial heterogeneity in hosts habitat plays any role when

species compete through a common parasite.The importance of spatial heterogeneity arise when two species that share a common parasitebut are not in contact begin, for any reasons (e.g. climate change or a competitive pressure),to overlap their habitat. It would be useful to study if the presence of refugees where onlyone species persists may avoid the extinction of either species.

The biological system considered as case-study is the interaction between two galliformspecies in the province of Trento (northern Italy); specifically, rock partridge (Alectorisgraeca saxatilis) and black grouse (Tetrao tetrix). Here two species share a common spatialdomain, the mountain, but have a different habitat on it. It is believed that for these speciesa key factor in habitat selection is vegetation, which strongly depends on altitude. So when

considering spatial diffusion only a one dimensional domain characterizing the altitude wasfeatured, in order to keep the math and the numerical simulation accessible.It is reasonable to think that changes in the ecology, mainly because of climate change [15][21]or more direct human interferences, will lead to an upward shift of grouses habitat. As aconsequence there would be increasing contacts between species due to overlapping habitats.If that happens the result would be a stronger competition between the two species.

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Competition is a common interaction between organisms or species and generally leads to alowered fitness of one of them due to others presence. In nature it arise when both requirea limited resources, as food, water or habitat, to live. Ecology defines intraspecific competi-tion as competition among members of the same species, while interspecific competition ascompetition between individuals of different species. Furthermore the mechanisms by which

it occurs could be divided into direct and indirect.

Direct competition implies aggression or interference between individuals, on the other handindirect competition may be due to a limited resource that will depletes for one species (ex-ploitation competition) or due to another species, for example predators or parasites thataffect differently the two competing species (apparent competition).

An important principle related to competition is the Gauses Principle of competitive exclu-sion (CEP), which states that two species competing for the same resources cannot coexist ifother ecological factors are constant. The key concept is that more similar the shared needsare, the more intense the competition is. CEP is predicted by a number of mathematical

and theoretical models, such as the Lotka-Volterra models of competition, but nevertheless itdoes not appear to occur in nature, where high biodiversity in seemingly homogeneous habi-tats is commonly observed. As a matter of fact the existence of so called Lazaros species,those who disappear from fossil registers and suddenly appear many million years, urges adeep consideration of CEPs meaning and definition. Some specialists simply states thatCEP doesnt work [13]. Other [24] defined the CEP as the instance where two speciesmake their livings in identical ways being unable to coexist in a stable fashion, thereforewhen possible must segregate to avoid extinction.As a matter of fact, spatial heterogeneity is a key feature in competition. Lopez [22] showsthat for a class of spatially heterogeneous direct-competing models species may indeed sur-vive if they can segregate in some refuge patches .In this controversy the importance of

mathematical models arise because they provide an idealized behaviour against which obser-vation and experimental data can be compared and judge, similarly numerical simulationsare a tool that can be used to explore model behaviour.

As mentioned before, it has already been shown that a non-specific parasite may act as apowerful competitive weapon [18] and various studies confirmed that the presence of a reser-voir species may lead to the exclusion of a more vulnerable species that otherwise wouldsurvive.

This work will focus on the behaviour of a dynamical system in which an invading speciesintroduces a parasite thats very aggressive against resident species. In the first part a com-

parison between results from previous work for a single-host parasite interaction and resultsof the model analysis with spatial diffusion is given, then we focus on what effects may havespatial dependent parameters on the system behaviour. At last a model featuring spatialdiffusion and apparent competition between two species is introduced and studied in orderto understand the role of spatial heterogeneity in host exclusion.

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Chapter 1

Description of the Spatially

Structured Model

The standard host-parasite model in literature describes the evolution in time of a number

of hosts and parasites by a system of ordinary differential equations. Generally it doesntfeature any other structures than the basic ones even if they may be nevertheless biologicallyinteresting.Since Anderson and May works [1][2][3][4][5][6][7] onward the most used method to definethese kind of models is to start from one ordinary differential equations for each host classand then sum up all the classes to have a single equation for the total number of hosts, simi-larly for the total number of parasites. Then to have a closed system one has to assume howparasites are distributed among hosts. All these steps lead to system of equations involvingonly three populations: hosts, parasites and larvae.A host class is defined by Anderson as hosts that carry the same number of parasite underthe hypotesis that a class could increase by the death of one parasite within the next higher

class or by establishment of one parasite into a host belonging to the prior lower class.Initially the goal of our efforts is to study a system that is spatially structured and compareit with one that is not.

Let h(j) = h(j)(x, t), j N0 be the spatial density of hosts having j parasites at a pointx and time t.Define the total number of hosts in the region with a burden of j parasites and the spatialdensity of the total number of hosts as

H(j) = H(j)(t) =

h(j)(x, t)dx and H = H(x, t) =+

j=0h(j)(x, t) (1.1)

Therefore the total number of parasite is

P =+j=0

jh(j)

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Define now the following parameters modeling the dynamical behaviour of hosts, parasitesand larvae:

Symbol Description

b(H,x,t) is the natural density dependent fertility rate in (x, t) ;(j) is a factor modulating the fertility of hosts in the class h(j); 0 < (j) < 1;

Host h(H,x,t) is the intrinsic density dependent mortality in (x, t) ;(j) is the induced mortality of hosts due to the burden of having j parasites;

Dhh(j) is the spatial flux of host population;(j) is the natural mortality rate for parasites in the class h(j) ;

Parasite h(j) is the laying rate per host in the class h(j);and Larvae (L,x,t) is the natural death rate of larvae in (x, t);

R(j)(H,L,x,t) is the recruitment rate of larvae on hosts of class h(j).

Table 1.1: Summary of parameters.

It should be taken in account that in this case, as many others in nature, parasite can not

survive at hosts death. Host death leads to parasite death. Note that any influence ofparasites on hosts is modeled by and .

A key concept, whose importance would be explained later on, is the total birth rate ofhosts, defined as

j=0

(j)b(H,x,t)H(j) (1.2)

More definitions of both b(H,x,t) and will be held during this work to test different host-parasite and host-habitat interactions.Finally to describe the dynamics of host population make these assumptions on the biological

features of our case study:Hypotesis 1 :

All hosts are born free from parasites, so all newborn hosts belong to the class h(0); Theres not spatial flux of the larvae population; Dh 0 ; The natural mortality rate of host is just a density dependent rate: h(H,x,t) = d+vH; The recruitment of larvae on hosts is spatially homogeneous and doesnt depend on the

distribution of parasites among the host class: R(j)(H,L,x,t) = L ;

Consider now hosts free from parasites (h(0)

). The rate of change of this class involves thefertility of hosts, the natural mortality rate, the recruitment rate of larvae and the naturalmortality rate of parasites from adults in the class h(1).So the equation for hosts having no parasites is

h(0)

t Dh(0) =

+j=0

(j)b(H,x,t)h(j) h(h(0), x , t)h(0) R0h(0) + (1)h(1) (1.3)

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Consider now hosts having parasites. The equation for the class h(j) is similar to (1.3).As it may be evident, the main differences consist in the absence of the hosts birth termand the presence of an added mortality term due to the parasite burden.

h(j)t

Dh(j) = (j +1)h(j+1)[h(h(j), x , t)+Rj+(j)+(j)]h(j) +Rj1h(j1) = Fj (1.4)The global dynamics of hosts can be described as follows. Assume that X is a randomvariable representing the number of parasites. The probability for an host to have exactly j

parasites is P(X = j) =h(j)

H.

Summing (1.3) and (1.4) over all classes one obtains

H

t DH = b(H,x,t)

+j=0

(j)h(j) +j=0

h(h(j), x , t)h(j)

+j=1

(j)h(j)

= H

b(H,x,t)

+j=0

(j) h(j)

H h(H,x,t)

+j=0

(j)h(j)

H

= H[b(H,x,t)E((X)) h(H,x,t) E((X))] (1.5)

If another host species H2 is present and there is a direct competition between them a term(x, t)H2H has to be added. This will be useful in the next chapter when a two host modelwill be considered.

The global dynamics for parasites is determined multipling (1.4) by j the summing overall j. This gives

P

t D

+j=1

jh(j) =+j=1

jFj

Recalling the definition of the spatial density of parasites, the hypotesis one and HP(X =j) = h(j) its possible to express the dynamics as follows

P

t DP =

+j=1

j[(j + 1)h(j+1) [h(h(j), x , t) + Rj + (j) + (j)]h(j) + Rj1h(j1)]

= H[

h(H,x,t)E(X)

+

j=1

(j)jP(X = j)] ++

j=1

j[(j + 1)h(j+1) +

+Rj1h(j1) (j)h(j) Rjh(j)] =

= H[h(H,x,t)E(X) E(X(X)) ++j=0

(RjP(X = j) (j)P(X = j))]

= H[R(H, L) h(H,x,t)E(X) E(X(X)) E((X))] (1.6)3


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Larvae dynamics is slightly different, in fact it involves four major phenomena.

Hypotesis 2 :

Theres not spatial flux of the larvae population, Hosts lay larvae through excrements at a certain rate h depending on their parasite

burden so the total number of larvae laid is+

j=0 h(j)h(j),

There is a loss term due to hosts recruitment, The natural mortality rate of larvae is spatially and temporally homogeneous: (L,x,t) =

L.

One models the dynamic of larvae as

L

t= HE(h(X)) R(H, L) L (1.7)

Putting equations (1.5), (1.6) and (1.7) together, the dynamics for the host-parasite systemis described by:

H

t= DH+ [b(H,x,t)E((X)) h(H,x,t) E((X))]H

P

t= DP + [R(H, L) h(H,x,t)E(X) E(X(X)) E((X))]H

L

t= HE(h(X))

R(H, L)

L

(1.8)

For biological reasons all parameters are nonnegative.

Its possible to have many different assumptions on the definition of parameter [10],[26].For example:

1. Parasite induced mortality of hosts depends linearly on the number of parasites andnatural death rate of parasites are density independent. This means (j) = j and(j) = j.

2. Parasite induced mortality of hosts is dependent linearly on the number of parasites

and natural death rate of parasites is density independent. This means (j) = j

2

and(j) = j.

3. Parasite induced mortality of hosts is density independent and natural death rate ofparasites is density dependent. This means (j) = j and (j) = j2

4. Parasite induced mortality of hosts and natural death rate of parasites are densitydependent. This means (j) = j2 and (j) = j2.

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The natural mortality rate of host has been defined as a density dependent rate: h(H,x,t).To model the effects of a logistic growth the mortality rate term becomes

h(H,x,t) = d + vH, where v =b dHK

with HK representing the carrying capacity

Finally to have a specific model is necessary to fix the distribution of parasite in hosts. Notethat in the dynamics of host-parasite interaction an important role is played by host hetero-genicity especially in causing most parasite to be aggregate in few hosts.Some classical assumptions [4] are that parasites are distributed in the host population ac-cording to a Poisson law with mean z = P/H or a negative binomial with the same meanand a clumping parameter k. When fixing an aggregation parameter k one has to take inaccount that this parameter is a population statistic and does not correspond to a biologicalconcept, so different values of k may correspond to different biological mechanisms.

To the best of our knowledge on this interaction, it seems plausible to chose for the fol-lowing models the negative binomial distribution, as only few hosts carry a high parasiteburden while others have few parasites.

Fixed the distribution one can evaluate E((X)), E((X)), E(X(X)) and E((X)).

Under the hypotesis that (j) = j, 0 < 1 (1.9)

E((X)) =x=0

xk

x

pk(q)x = pk

x=0

kx

(q)x = pk(1 q)k =

p

1 qk

Its known that

z = P/H , p =k

z + k

and q = 1

p =

z

z + kso when parasites have negative effects on host fertility their average fertility reduction maybe expressed as

E((X)) =

kH

kH + (1 )Pk

(1.10)

so the new fertility rare would be

b(x)

kH

kH + (1 )Pk

(1.11)

Similarly for host mortality induced by parasites, under the hypothesis that (X) = X

E((X)) = E(X) = P

Hand E((X)) =

P

H

and finally

E(X(X)) = E(X2) = (V ar(X)+E(X)2) =

kq

p2+

k2q2

p2

=

P

H

+

1 +

1

k

P

H

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Hence the resulting space structured model is

H

t= DH+ H(b(x)

kH

kH + (1 )Pk

d vH) P

Pt = DP + HL P( + d + vH + + (1 + 1k ) PH)

L

t= hP H L L

(1.12)

This will be the single host parasite model that is going to be analyzed in the next chapter.

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Chapter 2

Single Host Macroparasite System

Before studying the interaction between both hosts and their shared macroparasite is funda-mental to understand the interaction between a single host and its parasite. Many relevantissues have to be explained in such case and not often explanations can be achieved in detailsby algebraic methods. Undoubtedly one of these issues is spatial heterogeneity, another onemay be density dependence, however to understand how spatial heterogeneity affects singlehost-parasite interaction again a reductionist approach is followed.At first, to carry out the analysis for model (1.8) only the biological processes necessaryto a simple description will be considered. This means to study the model without havingany spatial dependent parameters, neither fertility reduction, in order to understand clearlyhost-macroparasite interactions. In the process, models with and without diffusion will becompared, then the fertility reduction will be added, too.Spatial dependency, specifically a spatial dependent fertility rate, will be discussed in thenext chapter due its the complexity.

2.1 Constant hosts fertility with no reduction

As previously stated the straightforward hypothesis to begin with are

Parasites dont have any role in reducing hosts fertility; The natural fertility rate of host is constant in time, space and density of hosts. Parasites are negative binomially distributed in host.These assumptions will lead to the model

H

t = DH+ H(b d vH) PP

t= DP + HL P

+ d + vH + +

1 +

1

k

P

H

L

t= hP H L L

(2.1)

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The same model without diffusion has been already studied in various articles. A comparisonbetween these models allows to investigate the effect of spatial dispersion on host-parasiteinteraction.

Focus at first on the model without diffusion. A comprehensive mathematical analysis ispossible but not simple. Start looking for the systems equilibrium points . They are solu-tions of the system of equations

H(b d vH) P = 0HL P( + d + vH + + (1 + 1

k)

P

H) = 0

hP H L L = 0(2.2)

Its quite easy to identify two equilibrium point, respectively E1 = (0, 0, 0) and E1 =(HK, 0, 0) where HK =

bd

v

is the carrying capacity.However exist other equilibrium points in which parasite is present. Their value is given by

H =k

2v

B

B2 +

4Av

k

P =1

H(b d vH)

L =hP

H+

(2.3)

where

B = (A ) v

k , A = + b + +

b

d

k

Since A > 0 only one value ofH is positive, only one equilibrium is feasible under the condi-tion that H < HK otherwise P will be negative. This is called the coexistence equilibrium.

2.1.1 Condition for Equilibrium Stability

The study of the sign of the real part of eigenvalues will give conditions for the stability ofthe equilibrium points E0 and E1.

Jac((H,P,L)) =

b d 2vH 0L vP + (1 + 1

k)

P

H2( + d + vH + + 2(1 + 1

k)

P

H) H

L h (H + )

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Uninfected Equilibrium

First of all study the stability of uninfected equilibrium. Its easy to understand that if it isstable there cant be parasite persistence. On the other hand, if its not, parasite invasionand persistence may be possible. Remind that Hk =

bdv

so in E1 = (Hk, 0, 0)

Jac((Hk, 0, 0)) =

b d 2vHk 00 ( + d + vHk + ) Hk0 h (Hk + )

That is

Jac((Hk, 0, 0)) I = b d 2vHk 00 ( + d + vHk + + ) Hk

0 h (Hk + + )

The determinant of this specific matrix can be written as (b

d

2vHk

)det

|A

I

|= 0

where

A =

( + d + vHk + ) Hkh (Hk + )

An eigenvalue is = b d 2vHk and others are the solutions of the equation 2 trA ++det|A| = 0

The steady state will be stable under the following conditions:

trA = ( + b + + Hk + ) < 0 (2.4)

det|A| = ( + b + )(Hk + ) hHk > 0note that condition (2.4) is always true because parameters are non negative by definition.Therefore the stability conditions are:

b d 2vHk = d b < 0 (2.5)( + b + )(Hk + ) hHk > 0 (2.6)

Condition (2.5) is necessary to avoid host populations extinction due to its natural growthrate. When further analysis featuring diffusion and spatial heterogeneity will be conductedon the model, it will turn out that this condition is slightly different. Under certain conditionhosts survival may be feasible even if the average growth rate is negative.

The second condition (2.6), when reversed, gives the threshold to have parasite invasion andcan be expressed as

R0 > 1, or as one should better say R0 > 1 H < Hkwhere

R0 =

h

+ b + 1

Hk

(2.7)

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Its possible to assign a biological meaning to (2.7):h

+ b + expresses the average number of larvae that originate from an adult parasite,

Hk + Hk

is the probability that a larvae is recruited by an host before dying and is the

probability that it may growth and evolve in an adult parasite.So it is the number of adult parasites produced from a single adult parasite. Rearrangingthe factor in a different order it can express the number of free-living stages produced froma single free-living stage.Therefore if condition R0 > 1 is satisfied one can expect an increase of parasites.

Endemic Equilibrium

Conditions for the stability of the endemic equilibrium are quite complex so it seems moreuseful to show the method used to achieve the stability boundaries than writing down ex-plicitly the conditions. A similar approach can be found in [29] where stability boundarieswere shown in parameter regions.

In fact solving the determinant of |Jac((H, P , L)) I| =

b d 2vH 0L vP + (1 + 1

k)

P

H2( + + d + vH + + 2(1 + 1

k)

P

H) H

L h (H + + )

is indeed quite complex so in this case one define the determinant as

|Jac((H, P , L)) I| =

a b 0

d e fg h i

so by definition it will be

3 2(a + e + i) + (ei + a(e + i) f h bd) + bdi + afh bfg aei = 0 (2.8)Using Routh-Hurwitz conditions [27] one obtains the necessary and sufficient conditions tohave Re < 0. These are:

1. (a + e + i) > 02. bdi + afh bf g aei > 03. (a + e + i)(ei + a(e + i) f h bd) (bdi + afh bfg aei) > 0

Given the parameters values is now possible to compute the stability of the endemic equi-librium.

Moreover using a computer it is possible to evaluate the stability threshold for one pa-rameter keeping the others fixed. In this case, chosen an aggregation parameter k, one canappreciate how a change in natural mortality of larvae () affects the system behaviour.

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(a) Infected equilibrium is stable when is above stabilitythreshold

(b) Instability of infected equilibrium when is under sta-bility threshold

Figure 2.1: Host-Parasite behaviour when infected equilibrium is stable or unstable for model (2.1) withoutdiffusion using the same set of parameter except for

The simulations show two different examples: in the first one the endemic equilibrium isstable and the solutions converge to it with damped oscillations, in the second the infectedequilibrium is unstable and the solutions tend to a limit cycle.

Table 2.1 shows the minimum values of for which the infected equilibrium is stable, giventhis set of parameters: b = 0.6, d = 0.5, = 0.05, = 4, b = 0.6, = 0.12, = 0.5,

h = 100, v = 0.00083.It has been estimate [29] that varying the parasite induced host mortality () between 0

Table 2.1: Stability Region

if k = 3 5 8 10then > 5.25 8.74 11.38 12.46

and 0.1, while keeping other parameter fixed, doesnt substantially modify this thresholdvalues.

It is possible to see from simulations that the solutions may oscillate quite far from theequilibrium just when is not close to the boundary values.

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2.2 General Condition for Diffusion-Driven Instability

Consider now the model (2.1), with zero flux boundary conditions and given initial condi-tions. It would be interesting to look for diffusion driven instability or Turing instability.This mechanism happens when an homogeneous steady state is stable to small perturbations

in the absence of diffusion, but is unstable to small spatial perturbations when diffusion ispresent. In other words the purpose is to check what effect may produce on system dynamicthe introduction of spatial diffusion in host population.

The most interesting case will be to linearize about the endemic equilibrium E1 := (H, P , L),under the condition that is stable, and then look for any spatial instability. Unfortunatelythis case is algebraically way too complicated. So as an example a less complicated case ispresented.Consider the reaction diffusion system and linearize about the steady state E1 := (HK, 0, 0).Set

w = (H HK, P , L) (2.9)and (2.1) becomes, for |w| small,

wt = J(E1)w + Dw (2.10)

where

J(E1) =

b d 2vHK 00 ( + d + vHK + ) HK

0 h (HK + )

and D is the matrix

D1 0 00 D2 00 0 0

As one can see two different values of diffusion coefficients are considered in matrix D. It iswell know [27] Turing instability may occur only if diffusion coefficients are different, thusto test whether spatial pattern may arise we allow for different diffusion, although parasitesare carried inside hosts. So this assumption imply the same diffusion coefficient.To solve this system of equations subject to the zero flux boundary conditions first defineW(r) such that is the time independent solution of the spatial eigenvalue problem definedby

W + 2W = 0, (n)W = 0 for r on (2.11)

where is the eigenvalue.

In this model the domain is one-dimensional so W cos(nx/L) where n is an integer.So in this case = n/L is the eigenvalue and 1/ is a measure of the wave-like pattern.Usually the eigenvalue is called wave-number and with finite domains there is a discreteset of possible wave-numbers.It can be easily shown that boundary condition are satisfied.

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Let W(r) be the corresponding eigenfunction to the wave-number ; each of them satisfyzero flux boundary condition. Look for solution w(r, t) of (2.10) in the form

w(r, t) =

cetW(r) (2.12)

where c are constants determined by a Fourier expansion of initial conditions in terms ofW(r).

is the eigenvalue which determines temporal growth. So substituting the form (2.12)into (2.10) one obtains, for each k,

W = J(E1)W + DW = J(E1)W D2W (2.13)and one can determine by calculating the roots of the characteristic polynomial

|J(E1) D2 I| = 0 (2.14)where J(E1) D2 I = b d 2vHK D12 00 ( + d + vHK + + D22 + ) Nk

0 h (HK + + )

Evaluate the determinant to get the eigenvalues: one is = b d 2vHk D12,the other two are given by the solution of:

2 + ( + d + vNk + + D22 + Nk + ) + ( + d + vNk + + D2

2)(Nk + ) hNk = 0

It is easy to see that Re < 0 if these conditions are true:

1. b d 2vHk < D12

2. ( + d + vHk + + D22 + Hk + ) < 03. ( + b + + D2

2)(Hk + ) hHk > 0If its know by hypothesis that the equilibrium is stable when there is not any spatial effect,Re(2 = 0) < 0, the preceding conditions (2.4) (2.6) hold. Therefore, being Di

2 > 0, theconditions with spatial diffusion are satisfied too and that implies there isnt any diffusiondriven instability.

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(a) Host behaviour when is above the stability threshold

(b) Host behaviour when is under the stability threshold

Figure 2.2: Host behaviour when infected equilibrium is stable or unstable for model (2.1), given otherparameters values as Tab 2.1

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Host-Parasite Coexistence Equilibrium

As already stated, conditions for the stability of endemic equilibrium are quite complex towrite down explicitly, so the model has to be simulated numerically.The results obtained suggest that hotss diffusion doesnt play any relevant role if none of theparameters depend on space. Again the stability of the host-parasite coexistence equilibrium

depends on the exact values of parameter when both stabilizing and destabilizing processesare relevant. Similarly as in the previous section is possible to compute the threshold forthe natural mortality of larvae which determines a stability switch. The simulations seemsto confirm that this stability boundary are similar to the boundary computed for the modelwithout diffusion. One can expect that they are exactly the same, but the exact value ofthe stability threshold is not easily determined because in its proximity solutions tend to theequilibrium very slowly, so an huge amount of iterations is needed to check if solutions arenot oscillating around a limit cycle.

As shown in Figure 2.2 time oscillations are coherent in space, then Figure 2.3 shown thebehaviour of solutions in one point of the spatial domain. Again as Figure 2.1, Figure 2.3

gives two example, one for the stability, one for the instability of the infected equilibrium.

Figure 2.3: Host-Parasite behaviour when infected equilibrium is stable or unstable for model (2.1) usingthe same set of parameter except for

Figure 2.4: Host behaviour when is above thestability threshold

Figure 2.5: Host behaviour when is under thestability threshold

As one can see comparing Figure 2.3 and Figure 2.1 the behaviour is quite identical formodels with or without diffusion under the condition that none of parameters depend onspace and that the diffusion coefficient is equal for host and parasite.

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2.3 Fertility Reduction

It is reasonable to assume that parasite reduce host fitness not only by increasing hostmortality but decreasing host fertility, too. How to model the fertility reduction has beendiscussed previously (1.11). We made the assumption that (j) = j, but different choices

are possible. Again not considering any spatial dependency on parameters the model studiedis

H

t= DH+ (b

kH

kH + (1 )Pk

d vH)H P

P

t= DP + HL P

+ d + vH + +

1 +

1

k

P

H

L

t= hP H L L

(2.15)

Parasite-induced reduction of host fertility has a destabilizing effect on model behaviour.As a consequence the stability threshold given previously change. Again unstable solutionstend to a limit cycle.

Figure 2.6: Host behaviour when a small fertility reduction is present. Parameters values are such that the infected equilibriumwould be in the stability region near the threshold if there isnt any fertility reduction.

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To compensate the destabilizing effect of fertility reduction and again achieve stability for thecoexistence equilibrium one has to enforce stabilizing processes or decreasing other destabi-lizing forces, for instance increasing the natural mortality of free-living stage. To determinethe stability conditions is quite complex, none the less the simulations confirm the strugglebetween stabilizing and destabilizing processes.

An interesting feature is that now increasing the mortality caused by parasite has a stabi-lizing effect on infected equilibrium. In fact keeping fixed all parameters value apart from

Figure 2.7: Host behaviour when a small fertility reduction is present. Parameters value are as Figure 2.6 except for . Whileat first the infected equilibrium was in the stability region, then with fertility reduction has become unstable, now increasingthe parasite induced mortality is stable again

and , its possible to find a stability boundary. Its value depending on the value of k.Numerical simulations show that decreasing (increasing the fertility reduction strength)keeping fixed lead to instability, viceversa increasing may lead to stability of infectedequilibrium.Host diffusion doesnt seem to affect the model prediction if all the parameters do not dependon space.

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Chapter 3

Single-Species Equation

In the previous chapter a comparison between the model with and without diffusion wasgiven. It has been shown that analysis is quite complex and it requires extensive numericalsimulations. Next step to further study how host-parasite interactions affects populationdynamics is to introduce a space dependent fertility rate. Therefore the habitat of the hostspecies is not the whole domain, but is modeled through the fertility function.First of all is better to analyze the dynamics of the host species alone. Consider the modelwe are studying in this form that is analogous to the one analyzed by Cantrell and Costern[12]. We refer to that book for several concepts.

u

t= d(x) u + f(x, u) in (0, )

d(x)u

n + (x)u = 0 on (0, )(3.1)

In general, solutions to equations of the form (3.1) need not be bounded nor exist for alltime. However in this study f(x, u) is in the common logistic form f(x, u) = a(x)u c(x)u2and we assume that there is positive number Hk such that ifu > Hk then f(x, u) < 0 for allx. Hk is analogous to the classical carrying capacity. In this case all positive solution of (3.1)will be bounded as t , because any constant larger than Hk will be a supersolution.Define

d(x) = Df(x, H) = H(b(x) d vH)

Moreover, in this case study, is a one dimensional space domain, so our single speciesequation

H

t

= DH+ H(b(x)

d

vH) in

(0,

)

DH

n = 0 on (0, )(3.2)

may be written as (3.1)

Now focus on the single species equation (3.1), to study its behaviour.

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By our definition this model shares these two key feature of the logistic model: f(x, 0) = 0and f(x, H) < 0 for some sufficiently large H. f(x, H) is continuos and differentiable withrespect to H at H = 0 so it implies that f(x, H) = g(x, H)H for some function g(x, H)bounded above for all H for a function of x. Cantrell [12] gives the following criterions forextinction and persistence.

Proposition 1 Suppose that f(x, H) g0(x)H for x where g0(x) is a bounded measur-able function g0(x) C(). If the principal eigenvalue 1 of D + g0(x) = in

D

n = 0 on (3.3)

is negative, then (3.1) has no positive equilibria and all nonnegative solutions decay expo-nentially to zero as t Proof Let u = ce1t1, where 1 > 0 is the eigenfunction corresponding to 1 in (3.3) andc > 0 is a constant. We have

ut

Du f(x, u) =

= 1u [Du + g0(x)u] + g0(x)u f(x, u)

= g0(x)u f(x, u) 0so that u is a supersolution to (3.2). Ifu(x, t) is any nonnegative solution to (3.2) we may

choose c so large that u(x, 0) > u(x, 0). Then u(x, t) > u(x, t) for all t > 0, and since 1 < 0we have u(x, t) 0 exponentially as t , so u(x, t) must decay toward zero exponentiallyas well. This also rules out any positive equilibria for (3.2).

The Proposition one means that the pro capita growth rate f(x, H)/H must be largeenough at some point x and some density H so that the corresponding spatially averagedgrowth rate given by the eigenvalue 1 for g0(x) = sup[f(x, H)/H : H > 0] is positive.Moreover, logistic model attains its highest growth rate at low densities; if it allows for pop-ulation growth at any density, at low density hence, it predicts invasibility.

Next proposition shows that invasibility implies persistence in model (3.2)

Proposition 2 Suppose that f(x, H) = g(x, H)H with g(x, H) of class C2 in u and C inx, and there exists a M > 0 such that g(x, H) < 0 for H > M. If the principal eigenvalues1 is positive in the problem D + g(x, 0) = in

DH

n = 0 on (3.4)

then (3.1) has a minimal positive equilibriumH and all solutions to (3.4) which are initiallypositive on an open subset of are eventually bounded below by orbits which increase towardH as t

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Proof The assumptions on f(x, H) imply that we can write f(x, H) = [g(x, 0)+g1(x.H)H]Hwhere g1(x, H) is C

1 in H. Let 1 be an eigenfunction for (3.2) with 1 > 0 on . For > 0sufficiently small

D(1) + f(x,1) == [D1 + g(x, 0)1] + g1(x.1)

221

= 11 + g1(x.1)

2

2

1= 1[1 + g1(x,1)1] > 0

Since g1 and 1 are bounded, it follows that for > 0 small, 1 is a sub solution for theelliptic problem (3.4) corresponding to (3.2). IfH(x, t) is a solution to (2) with H(x, 0) = 1,

then Ht

|t=0 > 0 on and general properties of sub- and supersolutions imply that H(x, t)is increasing in t. Since M > H is a supersolution to (2) we must have H(x, t) H(x)as t , where H is the minimal positive equilibrium solution of (3.2). If H(x, t) is asolution to (3.2) which is initially nonnegative and is positive on an open subset of , thenthe strong maximum principles implies H(x, t) > 0 on for t > 0. Choosing any t0 > 0 wecan take > 0 so small that 1 < H(x, t0) on ; then H(x, t t0) < H(x, t) for t = t0 andthus by the maximum principles for t > t0. Hence, H(x, t) is bounded below by H(x, t

t0)

and H(, x , t0) H as t , as desired.The principal eigenvalues in (3.3) and (3.4) allow to understand how spatial heterogeneity,

drift, patch size and boundary conditions affects model (3.1).Its important to note that the eigenvalue problem used to predict extinction may be differentfrom the problem used to predict persistence.Define g0 and g as in Proposition 1 and in Proposition 2, then the two eigenvalue problemscoincide when the following relation is true.

g0(x) = maxH0

[f(x, H)/H] = limH0+

[f(x, H)/H] =f

H(x, 0) = g(x, 0)

In the logistic case, as in this thesis, these equivalence hold.

This raises a new issue, to determine the principal eigenvalues sign of (3.3), or (3.4). Re-calling f(x, H) one can define m(x) := g(x, 0) = b(x) d and then rewrite the eigenvalueproblem as

Du + m(x)u = u in (0, )

Du

n = 0 on (0, )(3.5)

So the population has a dispersal rate D and a local growth rate m(x), thus increasing m(x)

should increase the average growth rate, as measured by the principal eigenvalue 1.

A fundamental results is the following (see [12], [23] for a proof and further details, alsonon pure Neumann boundary conditions are considered):

Theorem 1 Suppose that is piecewise of class C2+ and that satisfies the interiorcone condition. Suppose that D C1+() with D > 0, that m(x) L(). The principal

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eigenvalue of (3.5) is given by

1 = maxuW1,2(),u=0

D|u|2 dx +

m(x)u2 dx

u2 dx

(3.6)

or, alternatively

1 = max

D|u2| dx +

m(x)u2 dx : where u W1,2() ,

u2 dx = 1

(3.7)

Theorem 1 permits comparisons between the principal eigenvalues of different problems.

Corollary 1 If we denote the principal eigenvalue of (3.5) as 1(D, m) then 1(D, m) isincreasing with respect to m in the sense that if m1 m2 then 1(D, m1) 1(D, m2), andifm1 > m2 on a subset of positive measure then 1(D, m1) > 1(D, m2). Similarly, 1(D, m)is decreasing with respect to D in the same sense unless m(x) is constant, in this case theeigenfunction u1 turns out to be a constant so the terms involving D in (3.7) will have noeffect.

Proof Let u1 the eigenfunction associated with 1(D, m2). Then u1 is the maximizer of thequotient (3.6) for 1(D, m2) and we have

1(D, m2) =

D|u21| dx +

m2(x)u21 dx

u

2

1 dx

D|u21| dx +

m1(x)u21 dx

u21 dx

maxuW1,2(),u=0

D|u21| dx +

m1(x)u21 dx

u21 dx

= 1(D, m1)

Since u1 > 0 in the first inequality is strict ifm1(x) > m2(x) on a set of positive measure.The argument for D is similar.

The parameter m(x) describes the local growth rate, so increasing its value should increasethe average growth rate as measured by 1, and it does, confirming the reasonable conjecturethat a better growth rate improves the chance of survival.

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Consider now a related eigenvalue problem to (3.5)

D + m(x) = 0 in

D

n = 0 on (3.8)

Ifm(x) is strictly positive then the principal eigenvalue 1 for (3.8) could be characterizedas in (3.6), but corresponding to 1

1 = minW1,2(),=0

D|2| dx

m(x)2 dx

(3.9)

With this formulation when m(x) change sign a problem arises, in fact this formulation may

not make sense. This problem can be addressed by looking at 1/1 as the maximum of thereciprocal form in (3.8).

Theorem 2 Suppose that and the coefficients D and m satisfy the hypothesis of theorem1. Assume further that is of class C1, and that m(x) is positive on an open subset of .The problem (3.8) admits a positive principal eigenvalue +1 determined by

1

+1= max

W1,2(),=0

m(x)2 dx

D||2 dx

(3.10)

The principal eigenvalue is the only positive eigenvalue admitting a positive eigenfunction,and it is a simple eigenvalue. The principal eigenvalue depends continuously on m(x) withrespect to Lp() for any p (1, ] in the case of one space dimension. It shall be notedthat problem (3.8) also admits a negative principal eigenvalue if m(x) is negative on an opensubset of .

The denominator on the right side of (3.10) would vanish if where constant, and hence themaximum may not exist. For this kind of problem it has been proved that the existence ofa positive or negative principal eigenvalue depends on the integral of m(x) over .

Theorem 3 In the case of Neumann boundary condition the problem (3.8) admits a positive

principal eigenvalue if and only

m(x) dx < 0 (3.11)

In that case the positive principal eigenvalue is the only positive eigenvalue which admits apositive eigenfunction, and it is a simple eigenvalue. Moreover it is characterized by (3.10).What happens to the principal eigenvalue of (3.5) when inequality (3.11) holds? This is

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possible when there are places where the local growth rate is positive but negative on theaverage. In such a case the proximity of bad habitat to good is the mechanism leading to acritical patch size.

Theorem 4 Suppose that r is a positive parameter and that (3.11) holds, The principaleigenvalue 1 of

DH+ rm(x)H = H in

DH

n = 0 on (3.12)

is positive if and only if 0 < +1 < r where +1 is the positive principal eigenvalue of (3.8).

If the inequality (3.11) is reversed then 1 > 0 for all r > 0

Note that for (3.5) r = 1.

Proof Let H1 be the eigenfunction for 1. Multiplying by H1 in (3.12), integrating over ,and applying Greens formula and the boundary conditions yields:

D|H1|2 dx + r

m(x)H21 dx = 1

H21 dx

By (3.10)

m(x)H21 dx 1

+1

D|H1|2 dxso

r+1 1

D|H1|

2

dx 1 H21 dxso 1 < 0 if r <

+1 . On the other hand, if1 is the eigenfunction for

+1 we can multiply

(3.8) by 1 and integrate to obtain

D|1|2 dx = +1

m(x)21 dx

so that by theorem 1 as it applies to (3.12) one has

1

D|1|2 dx +

m(x)21 dx

21 dx= (

+

1

)

m(x)21 dx

21 dx

By one has

m(x)21 dx > 0 since +1 > 0, so 1 > 0 for >

+1 If condition (3.11) is

reversed then one uses the test function u 1 in (3.6,3.7) and obtain 1 > 0 immediately if > 0.

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Because there is not any dispersal of individuals across the border, the only mechanism thatmight cause a loss of population is dispersal into regions where the average local growth rateis negative. When condition (3.11) is satisfied it describes this scenario. But individualsto effectively average the local growth rate have to disperse sufficiently fast; as a resultpopulation will decline (1 < 0) if (3.11) holds and D is large.

To demonstrate this intuition rewrite (3.12) to obtain u + rm(x)D u = D u. If (3.11) holds

+1 is positive and it s characterized by (3.10). By Theorem 4, since (3.11) holds the principaleigenvalue 1 is positive if and only if r/D >

+1 , where

+1 is the principal eigenvalue of

+ m(x) = 0. Increasing D this is impossible for any r, and in particular for r = 1.

3.1 The Best Location for a Favorable Habitat Patch

It would be extremely useful to locate the most favorable habitat patch for a species. Unfortu-nately the most of the times is not possible because the characterization of the eigenfunctioncorresponding to the principal eigenvalue could be too complicated.

The habitat could be represented by different values of growth rate. This assumption meansthat there are more favorable areas for species reproduction and life, but nevertheless a singleindividual could move through the whole space domain. This distinction may play a roleafter, when another species is introduced. In fact this imply that there are not any refugeepatches, thus the individuals of the two species will be potentially always in contact due totheir random movement.However if we define the local growth rate m(x) as a piecewise constant we can obtain atranscendental equation which determines the principal eigenvalue, by matching the eigen-function and its derivative across the jumps in m(x). Note that the domain is still onedimensional.By studying that equation we can determine how the principal eigenvalue depends on various

features of the growth rate.

Consider the model similar to (3.2), this model predicts growth for the population if theprincipals eigenvalue 1 of the problem

DH+ m(x)H = H in (0, L)

DH

n = 0(3.13)

is positive. It follows from Theorem 4 that 1/D > 0, and hence 1 > 0 if and only if1/D > +1 (m(x)) where

+1 (m(x)) is the positive principal eigenvalue of

+ m(x) = 0 in

H

n = 0 on (3.14)

A smaller value of+1 (m(x)) reflects a more favorable environment than a larger one, because

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the condition 1/D > +1 (m(x)) is more easily satisfied. Thus the way that changes in m(x)affect the quality of the environment can be measured by how they affect +1 (m(x)).

Select only a single interval of length l < 1 which is favorable for population growth(m(x) > 0) while the rest of the interval (0, 1) is unfavorable, what is the optimal loca-

tion for the favorable interval?

Assume

m(x) =

1 for x [0, a)k for x [a, a + l)1 for x [a + l, 1)

for some k > 0 such that kl (1 l) < 0; in this way

m(x) dx < 0 and hence +1 (m(x))

will exist.The parameter a determines the location of the favorable patch, so we need to determinehow 1

+

(m(x)) depends on a.

On each of the subintervals [0, a), [a, a + T) and [a + T, 1] we can solve explicitly for the

eigenfunction in (3.14). If we let =

+1 (m(x)) we must construct the eigenfunction as

(x) =

cosh(x) for x [0, a)A cos(

k(x c)) for x [a, a + l)

B cosh((1 x)) for x [a + l, 1)Matching (x) and (x) across the interface and solving as before yields

cot(

kl) =k cosh(a) cosh((1 a l)) sinh(a) sinh((1 a l))

k sinh((1 l))=

(k 1) cosh((1 l)) + (k + 1) cosh((1 2a l))k sinh((1 l))

= G(a, )

It turns out that G(a, ) is decreasing in , with G(a, ) (k 1)

k as , andG(a, ) as 0+ with order

k/((1 l)). Since cot(

kL) with order

1/(

kl) as 0+, it follows from the assumption kl < 1 l that cot(

kL) is decreasingin , increasing G(a, ) moves the graph of G(a, ) (with respect to ) upward and hencecauses the curves to intersect at a smaller value of . Computing G(a, )/a yields afraction with a positive denominator and numerator equal to

2(k1) sinh((1

2a

l)).

Thus, G(a, )/a is negative for all if 0 < a < (1 l)/2 and positive for (1 l)/2 2 then

0(1) = min D(u(x))2 dx + + vH(x)

H(x)

H(x) + + 1u2(x) dx >

0(2) = min

D(u(x))2 dx +

+ vH(x) H(x)H(x) + + 2

u2(x) dx

so when is increasing 0() is decreasing. Hence if0() < 0 one has > 0() >k() so there is no solution with > 0 of = k(), k = 0, 1, ...If0() > 0 there is a unique such that = k().

Therefore for parasite establishment 0(0) < 0 is a necessary and sufficient conditionProposition 3 To have parasite establishment

0(0) < 0 is a necessary and sufficient

condition, i. e. the eigenvalues of

DP + P

H(x)

H(x) + vH(x)

+ P = 0 , with

P

n = 0

have to be negative.

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3.3 Discussion

These last two chapters focus on the behaviour of a single species, at first when parametersare space independent. The absence of diffusion driven instability has been pointed out forthe first case and theoretical and numerical results agree on the fact that system behaviour

with or without diffusion are qualitatively similar.However the biological interactions that give rise to this study features two species, noneof which has the whole domain (the side of a mountain) as habitat, but instead they havedifferent favorable patches corresponding on which altitude they are. In order to model thisproperty properly, the natural fertility rate has been defined as space dependent.If a space dependent fertility rate is introduced then new issues arise. First of all computingthe exact value of stability threshold is not possible, so one mostly rely on numerical simu-lations.It has been shown that if the average growth rate is positive then there is host persistence,

Figure 3.1: Host behaviour when infected equilibrium is stable for model (1.12). Parameter values are: d = 0.5, = 0.05, = 4, = 0.12, = 1, h = 100, v = 0.00083, = 60, b(x) = (3.5 x)(x 6.5), ifx [3.5, 6.5], 0 otherwise.

while if it is negative there could be extinction if conditions of Theorem 4 are not satisfied.

Obviously host persistence is a necessary condition for parasite establishment.The stability of infected equilibrium depends on the same stabilizing and destabilizing pro-cesses that were analyzed in the spatial independent problem, so one has a idea of whichmay be the system behaviour. The first result is that the distribution of the host populationtends to follow the shape of its fertility function; otherwise, if the average fertility is too lowand diffusion too large, it may gets extinct even when parasite is absent . When parasite ispresent, an infected equilibrium is feasible. It may be stable or unstable, in the latter case

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(a) Host population when parasites have a detrimental ef-

fect on host fertility

(b) Same simulations from another perspective, it is pos-

sible to note that there is no spatial instability

Figure 3.2: Destabilizing effects of host fertility reduction. Parameter values are: d = 0.6, = 0.1,, i = 4, = 0.12, = 1.,

h = 100, v = 0.00083, = 4, = 0.1, b(x) = (3.5 x)(x 6.5)

kH2kH2+(12)P2

k, ifx [3.5, 6.5], 0 otherwise.

solutions tend to cycle. To have parasite establishment a sufficiently high host growth rateat least on a region is needed, similarly as condition (2.7) which in this case is given by con-dition expressed in Proposition 3. Indeed, an interesting feature that arise with diffusion isthat now the average growth rate may be even negative on the spatial domain but still allowfor host survival and host parasite coexistence. When a species has a very high fertility rate

only on a small region of the spatial domain while outside is strongly negative, the infectedequilibrium may be feasible and unstable. However decreasing the value of fertility on thewhole domain tend to stabilize the host-parasite coexistence equilibrium.The instability of host-parasite coexistence equilibrium can be achieved enforcing fertilityreduction or decreasing free living state natural mortality (or increasing k, the aggregationparameter, which corresponds to decreasing aggregation). The dynamics depends on theexact values of parameters. In all the numerical simulations performed in which coexistenceequilibrium was unstable, solutions oscillated in time tending to a limit cycle. In contrast tothe case of space independent parameter, spatial instability may arise in this model (1.12).An analytical explanation of which conditions are required to have a stable infected equi-librium was not achieved: however looking at simulations one may infer that the conditions

needed to have spatial instability are two: a positive local growth rate on sufficient largepart of the space domain and an overwhelming effect of destabilizing forces. The conditionon average growth is quite subtle, indeed. In fact it seems (Figure 3.2) that a sufficientlyhigh average value is not sufficient to have spatial instability if in wide regions of the domainthere is a strong detrimental effect on host. In that case the loss of individuals prevails onpopulation dispersal and damps down oscillations.If conditions are favorable for spatial instability to arise, the effect on host is that after

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(a) Hosts density (b) Parasites density

(c) Long Interval Host density

Figure 3.3: Effects of spatial instability. Parameter values are: d = 0.5, = 0.1,, i = 4, = 0.12, = 1., h = 100,

v = 0.00083, = 60, = 0.1, b(x) = 2.5e(x4)2

4

kH

kH+(12)P

k.

reaching a peak, when they are decreasing due to parasite effect on mortality and fertilityrates, a portion of them move to the border (in this context the region with a lower averagegrowth rate, therefore where there were fewer presence of larvae and parasite). However ifone looks at a long time interval this behaviour is not so evident because of the oscillationsin time that may hide the side dispersal. This pattern is pictured on Figure 3.3

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Chapter 4

Two Host Shared Macroparasite

System

In the previous chapters a basic understanding has been provided of single species system.

Even if a complete algebraic understanding was not possible, the system behaviour has beenstudied using numerical simulations and is now known in its main properties. Moreoverits clear that the stability of those system depends on the balance between stabilizing anddestabilizing forces. Parasite induced mortality, reduction of host fertility and persistence offree living stages natural mortality are some of them.The model is going to describe now two host species sharing a macroparasite, with infectionspreading through a common pool of free living stages. It has been observed [30][31][32]that when two or more species may be infected by a macroparasite the more tolerant one(i.e., the one whose demographic parameters are less affected by parasites) could act as areservoir for the parasite, maintaining a high number of infective stages resulting in highlevel of infection in the less tolerant species.

Greenman and Hudson [17] investigate such models adopting a geometric approach basedon bifurcation theory in order to sidestep algebraic intractability. In their articles they focuson three different form of competition: direct competition, apparent competition and intraspecific competition, moreover they investigate the case in which parasite reduce host fertil-ity adding therefore more complexity at the system.

Obviously this approach does not exhaust the analysis but is a powerful tool to addressif the destabilizing effect of parasite induced reduction in host fecundity extends to thetwo species system, if the microparasite phenomena of host exclusion inversion and domi-nance reversal occurs also for macroparasite system, if threshold conditions can be expressedin model-independent form and how parasite-mediated and direct competition interact to

bring about host exclusion.

As already stated, a complete algebraic analysis is not possible because of the high di-mensionality of the system. Two other method besides the clever geometric approach byGreenman and Hudson. One is to make enough accurate assumption to substantially sim-plify the model equation, another is to maintain complexity but carrying out extensivenumerical simulations to assess the validity of conjectured rules about the feasibility and

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stability structure of systems equilibria. The downside of the first method is the possibleloss of information or the oversimplification of the model, while for the other is the extensiveamount of simulations needed in order to explore system dynamics.

The approach chosen in this work is to retrace Greenman and Hudson work using numerical

simulations, making the most of their result to conjecture the system behaviour when spatialheterogeneity is present.So, recall the generic model (2.1) and add the equations for host two.

H1t

= D1H1 + H1(b1(x) d1 v1H1 1H2) 1P1

H2t

= D2H2 + H2(b2(x) d2 v2H2 2H1) 2P2

P1t

= D1P1 + 11H1L P1(1 + d1 + v1H1 + 1H2 + 1 + 1(1 + 1k1

)P1H1

)

P2t

= D2P2 + 22H2L P2(2 + d2 + v2H2 + 2H1 + 2 + 2(1 + 1k2

)P2H2

)

L

t= h1P1 + h2P2 (1H1 + 2H2 + )L

(4.1)

where i = 1, 2, j = i and with Neumann boundary condition

4.1 Space Independent Parameters

Initially consider no reduction on host space independent fertility and no host diffusion.Therefore the model is

dHidt

= Hi(bi di viHi iHj) iPi

dPidt

= iiHiL Pi(i + di + viHi + iHj + i + i(1 + 1ki

)PiHi

)

dL

dt= h

iPi

+ hj

Pj

(iH

i+

jHj

+ )L

(4.2)

where i = 1, 2, and j = i

As for the single species model the threshold condition for parasite invasion and host exclu-sion can be established by the Routh-Hurwitz method.

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Suppose that host j, with(j = i) is absent, the threshold condition for parasite invasioncan be expressed similarly to (2.7) so

R0i =

hii

i + i + bi 1

iiHKi

> 1 (4.3)

that is again the basic reproductive number for parasite i.The threshold that bounds the feasibility region for any coexistence equilibrium, when directcompetition is absent, is given by the upper branch of the hyperbola

(R101 1)(R102 1) = 1 (4.4)

This is also the stability boundary for the uninfected coexistence equilibrium. Above thiscurve the uninfected equilibrium is stable against invasion by the parasite and below unsta-ble. These properties can be established by standard techniques [27].

Its possible to define a Tolerance Index S0i as follows: if host j is absent and densitydependence can be ignored, so vi = 0. then from (4.2)

Hi(bi di) iPi = 0iiHiL Pi

i + di + i + i

1 +

1

ki

PiHi

= 0

so

Hi(bi di) = iPi

iiHiL =Hi(bi

di)

i + di + i + i

1 +1

ki Hi(bi di)Hi

and then Li0 denotes the equilibrium value of L where

Li0 =(i + i + bi +

bidiki

)(bi di)iii

(4.5)

Define now Tolerance Index as

S0i = Li0Lj0

IfS0i > 1 then host i is more tolerant to the the infection so it will coexist a higher populationof larvae. This condition is necessary for host j to be excluded, but not sufficient.Now study conditions for parasite invasibility. Is then possible to obtain another thresholdcondition for host exclusion. This boundary value, defined as R0i, is computed as follows.

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Table 4.1: Parameters and Coefficient List

Parameters

b(x) is the natural density dependent fertility rate in (x, t) ; is the factor reducing the fertility of hosts; 0 < (j) < 1;d is the intrinsic density dependent mortality;v is the intraspecific competition index is the direct competition index is the induced mortality of hosts due to the parasite burden;

Di is the diffusion coefficient of host population; is the natural mortality rate for parasites in the class h(j) ;h is the laying rate per host is the natural death rate of larvae; is the recruitment rate of parasite on hosts;

Derived Parameters

ri = bi di si = i + i + bini = si + ri/ki Li0 = (niri)/(ii)

S0i =Li0

Lj0R0i =

hii

i + i + bi 1iHKi

Dimensionless Coefficients

c00 = r2/r1 c3i = 1/kic1i = si/ri c4i = /r1

c2i =hii

ric5i = /(iHKi)

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Analysis of Equilibria

The model defined in terms of parasite intensity Zi = Pi/Hi becomes

dHidt

= Hi(bi di viHi iHj iZi)

dZidt

= iiL Zi(i + bi + i) iki

Z2i

dL

dt= hiHiZi + hjHjZj (iHi + jHj + )L

(4.6)

The equations for host one-parasite coexistence equilibrium, when host two is absent, aregiven by

1 = z1 + u1 (4.7)

w = z1(c11 + c31z1)/(c11 + c31) (4.8)wS01 = z2(c12 + c32z2)/(c12 + c32) (4.9)

w(c51 + u1) = c21u1z1/(c11 + c31) (4.10)

where ui = Hi/HKi, zi = (iZi)/ri, w = L/L10, vi = (bi di)/HKi and the dimensionlesscoefficients listed in Table 4.1.

Proof

To obtain (4.7) consider

H1(b1 d1 v1H1 1H1 1Z1) = 0that becomes, under the hypotesis that host two is absent,

(b1 d1 v1H1 1Z1) = 01 v1

b1 d1 H1 1

b1 d1 Z1 = 0

then with ui = Hi/HKi , zi = (iZi)/ri one has (4.7)


11L Z1(1 + b1 + 1) 1k1

Z21 = 0

that may be written as

L =Z1(1 + b1 + 1 +

1k1

Z1)

11

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define ri = bi di then

L =

r211Z11r1

1 + b1 + 1

r1+

1k1r1

Z1

11

so by previously definition of dimensionless coefficients this becomes

L =r21z1(c11 + c31z1)

111

recall (4.5) so

w =L

L10=

r21z1(c11 + c31z1)

111

111

(1 + 1 + b1 +b1d1k1

)r1=

z1(c11 + c31z1)

c11 + c31

To obtain (4.9) consider the equation for parasite two. In fact, even if it seems unobivous inthe equilibrium for Hi, in the absence of other host (Hj = 0), one has that Zj > 0. Statedthis, retrace the previous work to obtain

L =r22z2(c12 + c32z2)

222

now recall the tolerance index definition and write the equation using dimensionless coeffi-cients

wS01 =L

L10

L10L20

=z2(c12 + c32z2)

c12 + c32


h1H1Z1 (1H1 + )L = 0so divide for L10 then

(1H1 + )w =h1H1Z1

L10=

h1H1Z1111

(1 + 1 + b1 +b1d1k1

)r1

therefore

1HK

1H11HK

+

1HK

w =

1Z1r1

11r1

r1h1H1

(1 + 1 + b1 +b1d1k1

)

1HK(u1 + c51)w = z111h1H1

r1(c11 + c31)

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Finally one obtains the last equilibrium equation

(u1 + c51)w =c21u1z1

(c11 + c31)

Then, having the equations for the equilibrium populations when host one coexists withthe parasite (with host 2 absent), it is possible to have a quadratic equation for u1. In factrecalling equation (4.7)

z1 = 1 u1 and substituting in (4.8), it becomes w = (1 u1)(c11 + c31(1 u1))c11 + c31

then one may write (4.10) as w =c21u1(1 u1)

(c11 + c31)(c51 + u1)and again substituting w one has

c21u1(c51 + u1)

= (c11 + c31(1 u1))

now it is easy to rearrange and obtain the quadratic equation

c31u21 [c11 + c31 c21 c31c51]u1 c51(c11 + c31) = 0 (4.11)

To be feasible solutions of (4.11) must be positive and equal or less than one since u1 is thefraction between the number of hosts and their carrying capacity.Provided that condition for parasite invasion holds (R01 > 1) there is a unique positive rootwhich is feasible for (4.11). The proof is straightforward. Note that

R01 =

h11

1 + 1 + b1 1

1HK1

=

c21c11

1

1

c51

If u1 = 0 then (4.11) is negative, so there is a unique positive solution. Moreover if u1 = 1

then (4.11) is positive if and only if c21 > c11(1 + c51) that is R01 > 1.If c31 and c32 are not positive there can be up to four positive real and feasible solutions ofthe equilibrium equations (4.7), (4.8), (4.9), (4.10), but for this model inequality c3i > 0 isalways true, in fact c3i equals the inverse measure of aggregation that is positive. So there is

just one feasible solution and hence just one infected host one equilibrium above threshold.Stated that solutions of (4.11) are equilibrium points one wants to investigate the boundaryvalue between infected host exclusion and invasion.

The infected host one equilibrium is stable against host 2 invasion provided that

1. c12 + 2c32z2 > 0

2. 1 2u1 z2 < 0These are, respectively, the first-order conditions for stability of equations for Z2 and forH2 in (4.6). The first one is always satisfied under the assumptions made. In the seconddefine the inequality slack as q = 2u1 + z2 1, then similarly as how we found the quadraticequation for u1 from this and from (4.7) (4.8) (4.10) substitute for z2 and w to obtain

Q(u1, 2) = qE(u1, 2, q) (4.12)

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Proof

Recall (4.7): z1 = 1 u1 then (4.8) becomes w = (1 u1)(c11 + c31(1 u1))c11 + c31

,

z2

= q + 1

2u

1then (4.9) becomes w =

(q + 1

2u1)(c12 + c32(q + 1

2u1))

S01(c12 + c32)

then substituting w one has

S01(c12 +c32)(1u1)(c11 +c31(1u1)) = (q +12u1)(c12 +c32(q +12u1))(c11 +c31) (4.13)rearrange the right side of the equation to obtain

q(c12 + 2c32 2c322u1 + qc32)(c11 + c31) + (c12 2c12u1 + c32 2c322u1 + c3222u21)(c11 + c31) =q(c12 + 2c32 2c322u1 + qc32)(c11 + c31) + (c12 + c32 + 2u1(c12 + c32 + c32(1 2u1)))(c11 + c31)Define E(u1, 2, q) = (qc32 + c12 + 2c32(1

2u1))(c11 + c32) so the right side may be written

qE(u1, 2, q) + (c12 + c32 + 2u1(c12 + c32 + c32(1 2u1)))(c11 + c31)Similarly for the left side

S01(c12 + c32)(1 u1)(c11 + c31(1 u1)) =S01(c31u

21 (c11 + 2c31)u1 + c11 + c31)(c12 + c32)

Now define

A1 = S01(c12 + c32)c31

B1 =

S01(c12 + c32)(c11 + 2c31)

therefore (4.13) becomes

A1u21 + B1u1 + S01(c11 + c31)(c12 + c32) (c12 + c32 + 2u1(c12 + c32 + c32(1 2u1)))(c11 + c31)

= qE(u1, 2, q)

So at last define

C1 = (S01 1)(c12 + c32)(c11 + c31)Q(u1, 2) = A1u

21 + B1u1 + C1 + 2u1(c12 + c32 + c32(1 2u1))(c11 + c31)

and finally one obtains (4.12)

Q(u1, 2) = qE(u1, 2, q)

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The boundary between infected host exclusion or invasion is given by condition 2 (q =1 2u1 z2 < 0), exactly when q = 1 2u1 z2 = 0. In that case (4.12) becomes

Q(u1, 2) = 0.

Define now u1 as the value ofu1 such that Q(u1, 2) = 0 then substitute u in (4.11) to obtain

c51 =c31u

21 (c11 + c31 c21)u1u1(1 c31) + c11 (4.14)

Define c51 as that value of c51.Now the threshold value ( R01 ) is given by the value of R01 given by this value of c51.

Stated that u1 [0, 1] its possible to determine the existence and position of the thresholdR01 by comparing the values of Q at the end points of h1.Since

Q(0, 2) = (S01

1)(c11 + c31)(c12 + c32)

Q(1, 2) = (2 1)(c12 + c32(1 2))the study of the properties of the solutions can be split in the cases depending on the strengthof direct competition.So, suppose that species one is more tolerant, or equivalently S01 > 1, and that its directcompetitive force against hot two is low, 2 < 1.Then its easy to check that Q(0, 2), Q(1, 2) have different signs, since all dimensionlesscoefficients are positive. Therefore must exists a unique root (u1) for Q(u1, 2) = 0 i nu1 [0, 1]. Substituting in equation (4.11) this root value for u1 gives

c51 =u1(c31u1 (c11 + c31 c21))

c11 + 2c31

then there is an unique value of c51 and hence of R01, provided that (c11 + c31 c21) < 0,that is exploiting the dimensionless coefficients s1 < s1 + r1/k1 < h1.If, however, s1 < h1 < s1 + r1/k1 there is a positive lower bound on the possible values of u1and if the root value calculated before is lower there is not any threshold R01.

If S01 < 1 then both value of Q in the end points are negative, so there may be no so-lution for Q(u1, 2) = 0 unless the curvature of Q is negative. In that case a double rootis generated and there will be two value of R01, equivalently two threshold. The curvatureCur of Q is measured by

Cur = A1

22(c11 + c31)c32

so to be negative, exploiting the dimensionless coefficients, the ratio S01/22 has to be suffi-

ciently small.

The result is that S01 > 1 seems to be a necessary condition to have host exclusion even ifthere are some special cases when exclusion does not occur.

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As one would expect increasing 2 increases the value of Q over the feasible interval ofu1, so, if it exists, the region of host 2 exclusion would expand. This could be noted bylooking at Q(u1, 2) = A1u

21 + B1u1 + C1 + 2u1(c12 + c32 + c32(1 2u1))(c11 + c31).

Recall

Q(0, 2) = (S01

1)(c11 + c31)(c12 + c32)

Q(1, 2) = (2 1)(c12 + c32(1 2))If 1 < 2 < 1 + c12/c32 and the less tolerant species is host one (S01 < 1) then Q(0, 2),Q(1, 2) have opposite signs. so there is a boundary value R

01 and the equilibrium will be

stable below that threshold. This is the phenomenon of host exclusion inversion.In the simulation this case is presented having host one more tolerant to the parasite butenduring a stronger competition by host two.

On the other hand if apparent competition against host two is strong ( S01 > 1) then bothvalue for Q are positive and the equilibrium is stable unless the curvature is negative, again.In that case there are two roots of Q(u1, 2) = 0 generating two value of R

01 between which

the equilibrium is unstable.

If competition is particularly strong 2 > 1 + c12/c32 then Q(1, 2) becomes negative, hencethere is a root in the interval 1/v2 < u1 < 1 since substituting u1 = 1/v2 in Q(u1, 2)still gives a positive value. However any root in this interval is not relevant since z2 be-comes negative due to the definition of the inequality slack q. By the way, over the interval0 < u1 < 1/v2, when v2 increases so does Q, hence increase the region of host exclusion.

A more straightforward analysis is possible.

Recall condition 2 for the stability of the infected host one equilibrium against host 2

invasionz2 < 1 2u1 (4.15)

z2 is implicitly defined by (4.9) that is possible to write as

F(z2) :=z2(c12 + c32z2)

c12 + c32=

S01(1 u1)(c11 + c31(1 u1))c11 + c31

(4.16)

similarly as was done before for (4.13).F(z2) is clearly an increasing function, so one can apply it at (4.15) to obtain

F(z2) < F(1 2u1) (4.17)

Then by (4.16) and the definition of F one has that

S01(1 u1)(c11 + c31(1 u1))c11 + c31

1 and 2 < 1. Substituting in equation (4.11) this root value for

u1 gives c51 =u1(c31u1(c11+c31c21))

c11+2c31then there is an unique value of c51 and hence of R01,

provided that (c11 + c31 c21) < 0.Condition R01 > R

01 imply c51 < c51 therefore u1 < u1 by the implicit function theorem

applied to (4.11). Moreover Q(0, 2) > 0, hence Q(u1, 2) > 0 and as a consequence host twocan not invade. However, as stated above if2 = 0 or sufficiently small, if S01 < 1 then Q isnegative, then invasion is feasible.

Figured out all these threshold its possible to summarize them on maps of Figure 4.1 (inwhich there is not direct competition) and 4.4.

Figure 4.1: The bifurcation maps for the infected equilibria when there is no direct competition. The map is shown in thecross section of parameter space defined by the inverse s of the basic reproductive numbers R0i

1. The shaded are indicatesfeasibility, the numbers refers to how many equilibria of that type are present in state space and how many of these are stable.Coefficient values: c11 = 1.75, c12 = 1.8, c21 = 2.19, c22 = 2.25, c31 = 0.25, c23 = 0.2, c40 = 2, c00 = 1, S01 = 2, i = 0R = R101

To a better understanding of the bifurcation pattern shown refer to Greenman and Hudson

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[17]. For our purpose is sufficient to have a brief description of the threshold structure. Themaps are constructed on the cross section of parameter space defined by the inverses ofR0i toensure that the region of parasite persistence is next to the origin, because condition R0i > 1is necessary for parasite persistence. Figure 4.1(a,b) picture the infected equilibrium whichdescribe the state of the system where a single-host species coexist with parasite in the ab-

sence of other host. The shaded area is its feasibility region. There is not the R

threshold inFigure 4.1 (b) because the parameters choice gives S01 > 1 and therefore S02 < 1. IfS0i 1then R10i moves from coincidence with the line R

10i = 1 to coincidence with the vertical

axis. IfS0i is decreased further the line R will no longer intersect the positive orthant of

the plane.The third map of Figure 4.1 describes the state of the system where the two host speciescoexist with each other and the parasite. The shaded area is now the feasibility region forthe infected coexistence equilibrium and its feasibility region is bounded by the line R andthe upper branch of the hyperbola (4.4). The infected equilibrium is stable when feasible.

Figure 4.2: Host two exclusion when is understability threshold for single species host one en-demic equilibrium

Figure 4.3: Host two exclusion when is abovestability threshold for single species host one en-demic equilibrium

These conjectures about the feasibility and stability structure of the systems behaviour pre-dict that the more tolerant host is able to exclude other host through apparent competition.The numerical simulations conducted confirm the system behaviour predicted, no matter ifdiffusion is present, until none of the parameters depend on space.

Suppose to chose the same parameters values for both host except for parasite induced

mortality () that its supposed to be doubled in the less tolerant host. Chose the parame-ter values as in previous chapter to compare hosts behaviours.

The thresholds necessary for host 2 exclusion are

1. S01 > 1 : verified under the hypothesis that 2 = 21 while all other parameters areequals between host 1 and host 2

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2. R01 > R01 > 1 : its possible to vary R01 keeping R

01 fixed just increasing or decreasing

the value of , HK1 or .

An interesting feature of the single host system was that the infected equilibrium may bestable or unstable depending on larvae natural mortality (), keeping fixed all others param-eters. Since increasing the value of has a stabilizing effect on the single species equilibriumbut simultaneously decrease the value of the basic reproductive number it would be lessprobable to obtain coexistence when both single host infected equilibrium are unstable.

As expected exclusion of host 2 doesnt affect the behaviour of the more tolerant species.

4.1.1 Direct Competition

Models featuring direct competition but not parasite infection, so neither apparent compe-tition, were widely studied and their behaviour is well know.Summarizing briefly the stability of the system depends on the strength of competitive forces.

When both are weak (0 < i < 1 with i = 1, 2) the system will tend to species coexistence,when one or both are strong (i > 1 with i = 1, 2) one species is excluded.

In our models both competing forces are present and not surprisingly the analysis becomesextremely difficult. Moreover the system may tend to very different solutions. Fortunatelyenough the work of Greenman and Hudson [17] again provides some conjectures on systembehaviour. As before the best way to proceed is to refer to the case without spatial hetero-geneity, presented in the work of Greenman and Hudson, then studying through numericalsimulations what changes may be observed introducing at first diffusion and at last space de-pendent fertility rate. The equilibrium representing the state of infected host coexistence isalgebraically intractable, but its properties seem related to the equilibria representing single

host coexistence with the parasite and uninfected hosts coexistence. Greenman and Hudsonachieved a brilliant result describing how apparent and direct competition interact.

Their result are summarized in Figure 4.4 and are indexed by direct competition indices1 (horizontal), 2 (vertical) from weak to strong competition. Each cell in the array cor-responds to a particular choice of the indices 1 , 2 and contains the maps for the threeinfected equilibria for this choice of index values.

The bifurcation maps displayed are obviously incomplete, since all the regions where there ismore than one equilibrium of a given type are left out, but is still a first step to understandsystem behaviour. In fact the number of parameter involved and the structure of the model

with two sources of competition can lead to any result.

Suppose that host one, that is more tolerant to infection, endure an increasing competi-tion from host two without being directly competitive itself. As show in the cell (a), (b),(c), (d) of Figure 4.4 when 1 increase the vertical asymptote of the hyperbola moves tothe vertical axis, the horizontal asymptote moves to the horizontal line Ro2 = 1 and thehyperbola collapses onto its asymptotes. Recall that the hyperbola is the boundary for the

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stability of uninfected coexistence equilibrium and for the feasibility of the infected coexis-tence equilibrium. The vertical line L1 = R

01, namely the threshold condition for host two

exclusion, stays fixed while L2 = R02 moves to R02 = 1, if they exist.

Figure 4.4: An example of bifurcation maps for the model with direct and apparent competition. Maps are indexed by directcompetition indices 1 (horizontal), 2 (vertical) from value 0 to 1.2 in steps of 0.6. Coefficients are as in Figure 4.1

Therefore it may happen that the hyperbola intersects one of these threshold so the fea-sible region so then cause the infected coexistence region to be fragmented into stable andunstable components.

Suppose now a weak competition against host one (Fig 4.4(g)(h)) and a strong competition

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Figure 4.5: Example of behaviour for Map a ofFigure 4.4

Figure 4.6: Example of behaviour for Map c ofFigure 4.4

against host two, the competitive forces combined against host two leads to its exclusion.When the competitive forces are reversed (Fig 4.4(c)(f)) a balance between direct and ap-parent competition is possible thus the hosts may coexist. In such case lines L1 and L2intersect and then there is fragmentation of the feasible region of the infected coexistenceequilibrium. Moreover it may happen an host exclusion inversion

Figure 4.7: Example of behaviour for Map e of Figure 4.4

Last but not least the case when both hosts strongly compete against each other (Fig 4.4(i)).The parasite-free coexistence state is feasible but never stable, instead is possible to have astable infected coexistence equilibrium.

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The tolerance index and the strength of direct competition are the key indexes explain-ing such interaction and consider hosts diffusion in the model changes the values of thethreshold but not the structure.In the absence of direct competition the more tolerant species will exclude the other if itsparasite productivity is sufficiently high (condition R0i > R

0i > 1). That means the host

has to be a reservoir for parasite without being too much affected by their presence andsimultaneously allow the parasite to spread enough to destabilize the less tolerant host. Ifthe two host directly compete host exclusion inversion may happen when the competitiveforce on the reservoir host is strong and the parasite productivity not too high.

4.2 Spatially Dependent Habitat

What happens when spatial heterogeneity are considered in these models?Apparent competition has been proven to be an important mechanism in species interaction,moreover is known that many causes can influence and change a species habitat. it would beimportant to investigate what could happen if a species more tolerant to a parasite begins

to overlap the habitat of a less tolerant species, if it is possible that this results into hostexclusion and to determine some criterion for this to happen. If one focus only on directcompetition it has been already shown that the presence of refugee patches may avoid hostsexclusion [22], is interesting to know if the same happens for apparent competition.

In this section the fertility of both hosts will depend on space. This feature model thepresence of a more favorable habitat for the species. The following discussion will focus onlyon the two host interaction since system dynamic for a single host has been already studiedand presented the previous chapter.

For a model without spatial heterogeneity and direct competition it has been shown that

exclusion will happen if the following condition are satisfied:

1. S01 > 1

2.

hii

i + i + bi 1

iHKi

= R01 > R

01 > 1

Tompkins [30] conjectured that a competing species would affect the ingestion rate of lar-vae by host, for example the less tolerant species would ingest mor

a two-host shared macro parasite system with spatial heterogeneity

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