A Two-Host Shared Macro Parasite System With Spatial Heterogeneity

UNIVERSIT
`
A DEGLI STUDI DI TRENTO
Facolt`a di Scienze Matematiche, Fisiche e Naturali
Corso di Laurea Specialistica in Matematica
Final Thesis
A Two-Host Shared Macroparasite System
with Spatial Heterogeneity
Relatori: Laureando:
prof. Pugliese Andrea Manica Mattia
dr. Ros`a Roberto
Anno Accademico 2009-2010
ii
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1.1 Condition for Equilibrium Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 General Condition for Diffusion-Driven Instability . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 Fertility Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Single-Species Equation 19
3.1 The Best Location for a Favorable Habitat Patch . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.2 Parasite and Larvae Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4 Two Host Shared Macroparasite System 33
4.1 Space Independent Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.1.1 Direct Competition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.2 Spatially Dependent Habitat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5 Results 55
Appendix 57
Bibliography 67
iii
iv
Introduction
Observations in wildlife supported by knowledge deriving from theoretical studies highlighted
the role of parasites in host population dynamics. Although single host-parasite systems have
been extensively studied [4] [14] the case when more hosts share a common parasite is not
fully understood. It has been demonstrated that transmission of shared parasites among
species can deeply affect host population especially when a host can act as a reservoir for
parasites 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 complexity
of 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 correlations
between parasite infection and components of host fitness. Until now one of the most studied
case 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 decline
caused by competition mediated via a caecal nematode with the ring-necked pheasant. This
hypothesis has been supported by a two-host macroparasite model and an accurate work
of controlled comparison of the macroparasite’s impact on the two hosts [30][31][32]. All of
these 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 parasite
but 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 only
one species persists may avoid the extinction of either species.
The biological system considered as case-study is the interaction between two galliform
species in the province of Trento (northern Italy); specifically, rock partridge (Alectoris
graeca saxatilis) and black grouse (Tetrao tetrix). Here two species share a common spatial
domain, the mountain, but have a different habitat on it. It is believed that for these species
a 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 was
featured, 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 grouse’s habitat. As a
consequence 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.
v
Competition is a common interaction between organisms or species and generally leads to a
lowered fitness of one of them due to other’s presence. In nature it arise when both require
a 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 as
competition 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 hand
indirect 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 that
affect differently the two competing species (apparent competition).
An important principle related to competition is the Gause’s Principle of competitive exclu-
sion (CEP), which states that two species competing for the same resources cannot coexist if
other ecological factors are constant. The key concept is that more similar the shared needs
are, 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 it
does 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 Lazaro’s species,
those who disappear from fossil registers and suddenly appear many million years, urges a
deep consideration of CEP’s meaning and definition. Some specialists simply states that
CEP ”doesn’t work” [13]. Other [24] defined the CEP as the instance where two species
make their livings in identical ways being unable to coexist in a stable fashion, therefore
when possible must segregate to avoid extinction.
As a matter of fact, spatial heterogeneity is a key feature in competition. Lopez [22] shows
that 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 simulations
are 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 a
powerful 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 would
survive.
This work will focus on the behaviour of a dynamical system in which an invading species
introduces a parasite that’s very aggressive against resident species. In the first part a com-
parison between results from previous work for a single-host parasite interaction and results
of the model analysis with spatial diffusion is given, then we focus on what effects may have
spatial dependent parameters on the system behaviour. At last a model featuring spatial
diffusion and apparent competition between two species is introduced and studied in order
to understand the role of spatial heterogeneity in host exclusion.
vi
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 doesn’t
feature any other structures than the basic ones even if they may be nevertheless biologically
interesting.
Since Anderson and May works [1][2][3][4][5][6][7] onward the most used method to define
these kind of models is to start from one ordinary differential equations for each host class
and 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 how
parasites are distributed among hosts. All these steps lead to system of equations involving
only three populations: hosts, parasites and larvae.
A host class is defined by Anderson as hosts that carry the same number of parasite under
the 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 compare
it with one that is not.
Let h
(j)
= h
(j)
(x, t), j ∈ N
0
be the spatial density of hosts having j parasites at a point
x and time t.
Define the total number of hosts in the region Ω with a burden of j parasites and the spatial
density of the total number of hosts as
H
(j)
= H
(j)
(t) =
_
Ω
h
(j)
(x, t)dx and H = H(x, t) =
+∞

j=0
h
(j)
(x, t) (1.1)
Therefore the total number of parasite is
P =
+∞

j=0
jh
(j)
1
Define now the following parameters modeling the dynamical behaviour of hosts, parasites
and larvae:
Symbol Description
b(H, x, t) is the natural density dependent fertility rate in (x, t) ;
γ(j) is a factor modulating the fertility of hosts in the class h
(j)
; 0 < γ(j) < 1;
Host µ
h
(H, x, t) is the intrinsic density dependent mortality in (x, t) ;
α(j) is the induced mortality of hosts due to the burden of having j parasites;
D
h
∇h
(j)
is the spatial flux of host population;
σ(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 δ(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 host’s death. Host death leads to parasite death. Note that any influence of
parasites on hosts is modeled by α and γ.
A key concept, whose importance would be explained later on, is the total birth rate of
hosts, defined as
∞

j=0
γ(j)b(H, x, t)H
(j)
(1.2)
More definitions of both b(H, x, t) and γ 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)
;
• There’s not spatial flux of the larvae population;
• D
h
≥ 0 ;
• The natural mortality rate of host is just a density dependent rate: µ
h
(H, x, t) = d+vH;
• The recruitment of larvae on hosts is spatially homogeneous and doesn’t depend on the
distribution of parasites among the host class: R
(j)
(H, L, x, t) = βL ;
Consider now hosts free from parasites (h
(0)
). The rate of change of this class involves the
fertility of hosts, the natural mortality rate, the recruitment rate of larvae and the natural
mortality rate of parasites from adults in the class h
(1)
.
So the equation for hosts having no parasites is
∂h
(0)
∂t
−D∆h
(0)
=
+∞

j=0
γ(j)b(H, x, t)h
(j)
−µ
h
(h
(0)
, x, t)h
(0)
−R
0
h
(0)
+ σ(1)h
(1)
(1.3)
2
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 host’s birth term
and the presence of an added mortality term due to the parasite burden.
∂h
(j)
∂t
−D∆h
(j)
= σ(j+1)h
(j+1)
−[µ
h
(h
(j)
, x, t)+R
j
+α(j)+σ(j)]h
(j)
+R
j−1
h
(j−1)
= F
j
(1.4)
The global dynamics of hosts can be described as follows. Assume that X is a random
variable 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
−D∆H = b(H, x, t)
+∞

j=0
γ(j)h
(j)
−
+∞

j=0
µ
h
(h
(j)
, x, t)h
(j)
−
+∞

j=1
α(j)h
(j)
= H
_
b(H, x, t)
+∞

j=0
γ(j)
h
(j)
H
−µ
h
(H, x, t) −
+∞

j=0
α(j)
h
(j)
H
_
= H[b(H, x, t)E(γ(X)) −µ
h
(H, x, t) −E(α(X))] (1.5)
If another host species H
2
is present and there is a direct competition between them a term
θ(x, t)H
2
H has to be added. This will be useful in the next chapter when a two host model
will be considered.
The global dynamics for parasites is determined multipling (1.4) by j the summing over
all j. This gives
∂P
∂t
−D
+∞

j=1
j∆h
(j)
=
+∞

j=1
jF
j
Recalling the definition of the spatial density of parasites, the hypotesis one and HP(X =
j) = h
(j)
it’s possible to express the dynamics as follows
∂P
∂t
−D∆P =
+∞

j=1
j[σ(j + 1)h
(j+1)
−[µ
h
(h
(j)
, x, t) + R
j
+ α(j) + σ(j)]h
(j)
+ R
j−1
h
(j−1)
]
= H[−µ
h
(H, x, t)E(X) −
+∞

j=1
α(j)jP(X = j)] +
+∞

j=1
j[σ(j + 1)h
(j+1)
+
+R
j−1
h
(j−1)
−σ(j)h
(j)
−R
j
h
(j)
] =
= H[−µ
h
(H, x, t)E(X) −E(Xα(X)) +
+∞

j=0
(R
j
P(X = j) −σ(j)P(X = j))]
= H[R(H, L) −µ
h
(H, x, t)E(X) −E(Xα(X)) −E(σ(X))] (1.6)
3
Larvae dynamics is slightly different, in fact it involves four major phenomena.
Hypotesis 2 :
• There’s 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: δ(L, x, t) =
δL.
One models the dynamic of larvae as
∂L
∂t
= HE(h(X)) −R(H, L) −δL (1.7)
Putting equations (1.5), (1.6) and (1.7) together, the dynamics for the host-parasite system
is described by:
_
¸
¸
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
¸
¸
_
∂H
∂t
= D∆H + [b(H, x, t)E(γ(X)) −µ
h
(H, x, t) −E(α(X))]H
∂P
∂t
= D∆P + [R(H, L) −µ
h
(H, x, t)E(X) −E(Xα(X)) −E(σ(X))]H
∂L
∂t
= HE(h(X)) −R(H, L) −δL
(1.8)
For biological reasons all parameters are nonnegative.
It’s 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 and
natural death rate of parasites are density independent. This means α(j) = jα and
σ(j) = jσ.
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 α(j) = j
2
α and
σ(j) = jσ.
3. Parasite induced mortality of hosts is density independent and natural death rate of
parasites is density dependent. This means α(j) = jα and σ(j) = j
2
σ
4. Parasite induced mortality of hosts and natural death rate of parasites are density
dependent. This means α(j) = j
2
α and σ(j) = j
2
σ.
4
The natural mortality rate of host has been defined as a density dependent rate: µ
h
(H, x, t).
To model the effects of a logistic growth the mortality rate term becomes
µ
h
(H, x, t) = d + vH, where v =
b −d
H
K
with H
K
representing the carrying capacity
Finally to have a specific model is necessary to fix the distribution of parasite in hosts. Note
that 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 mean
and a clumping parameter k. When fixing an aggregation parameter k one has to take in
account that this parameter is a population statistic and does not correspond to a biological
concept, 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 parasite
burden while others have few parasites.
Fixed the distribution one can evaluate E(γ(X)), E(α(X)), E(Xα(X)) and E(σ(X)).
Under the hypotesis that γ(j) = γ
j
, 0 < γ ≤ 1 (1.9)
E(γ(X)) =
∞

x=0
γ
x
_
−k
x
_
p
k
(−q)
x
= p
k
∞

x=0
_
−k
x
_
(−qγ)
x
= p
k
(1 −qγ)
−k
=
_
p
1 −qγ
_
k
It’s known that
z = P/H , p =
k
z + k
and q = 1 −p =
z
z + k
so when parasites have negative effects on host fertility their average fertility reduction may
be expressed as
E(γ(X)) =
_
kH
kH + (1 −γ)P
_
k
(1.10)
so the new fertility rare would be
b(x)
_
kH
kH + (1 −γ)P
_
k
(1.11)
Similarly for host mortality induced by parasites, under the hypothesis that α(X) = αX
E(α(X)) = E(αX) = α
P
H
and E(σ(X)) = σ
P
H
and finally
E(Xα(X)) = αE(X
2
) = α(V ar(X)+E(X)
2
) = α
_
kq
p
2
+
k
2
q
2
p
2
_
=
P
H
_
α + α
_
1 +
1
k
_
P
H
_
5
Hence the resulting space structured model is
_
¸
¸
¸
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
¸
¸
¸
_
∂H
∂t
= D∆H + H(b(x)
_
kH
kH + (1 −γ)P
_
k
−d −vH) −αP
∂P
∂t
= D∆P + βψHL −P(σ + d + vH + α + α(1 +
1
k
)
P
H
)
∂L
∂t
= hP −βHL −δL
(1.12)
This will be the single host parasite model that is going to be analyzed in the next chapter.
6
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 relevant
issues have to be explained in such case and not often explanations can be achieved in details
by algebraic methods. Undoubtedly one of these issues is spatial heterogeneity, another one
may be density dependence, however to understand how spatial heterogeneity affects single
host-parasite interaction again a reductionist approach is followed.
At first, to carry out the analysis for model (1.8) only the biological processes necessary
to a simple description will be considered. This means to study the model without having
any spatial dependent parameters, neither fertility reduction, in order to understand clearly
host-macroparasite interactions. In the process, models with and without diffusion will be
compared, then the fertility reduction will be added, too.
Spatial dependency, specifically a spatial dependent fertility rate, will be discussed in the
next chapter due its the complexity.
2.1 Constant hosts fertility with no reduction
As previously stated the straightforward hypothesis to begin with are
• Parasites don’t 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
= D∆H + H(b −d −vH) −αP
∂P
∂t
= D∆P + βψHL −P
_
σ + d + vH + α + α
_
1 +
1
k
_
P
H
_
∂L
∂t
= hP −βHL −δL
(2.1)
7
The same model without diffusion has been already studied in various articles. A comparison
between these models allows to investigate the effect of spatial dispersion on host-parasite
interaction.
Focus at first on the model without diffusion. A comprehensive mathematical analysis is
possible but not simple. Start looking for the system’s equilibrium points . They are solu-
tions of the system of equations
_
¸
_
¸
_
H(b −d −vH) −αP = 0
βψHL −P(σ + d + vH + α + α(1 +
1
k
)
P
H
) = 0
hP −βHL −δL = 0
(2.2)
It’s quite easy to identify two equilibrium point, respectively E
1
= (0, 0, 0) and E
1
=
(H
K
, 0, 0) where H
K
=
b−d
v
is the carrying capacity.
However exist other equilibrium points in which parasite is present. Their value is given by
_
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
_
¯
H =
k
2βv
_
B ±
_
B
2
+
4δAβv
k
_
¯
P =
1
α
¯
H(b −d −v
¯
H)
¯
L =
h
¯
P
β
¯
H + δ
(2.3)
where
B = β(A −ψ) −
δv
k
, A = σ + b + α +
b −d
k
Since A > 0 only one value of
¯
H is positive, only one equilibrium is feasible under the condi-
tion that
¯
H < H
K
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 of
the equilibrium points E
0
and E
1
.
Jac((H, P, L)) =
_
_
_
b −d −2vH −α 0
ψβL −vP + α(1 +
1
k
)
P
H
2
−(σ + d + vH + α + 2α(1 +
1
k
)
P
H
) ψβH
−βL h −(βH + δ)
_
_
_
8
Uninfected Equilibrium
First of all study the stability of uninfected equilibrium. It’s easy to understand that if it is
stable there can’t be parasite persistence. On the other hand, if it’s not, parasite invasion
and persistence may be possible. Remind that H
k
=
b−d
v
so in E
1
= (H
k
, 0, 0)
Jac((H
k
, 0, 0)) =
_
_
b −d −2vH
k
−α 0
0 −(σ + d + vH
k
+ α) ψβH
k
0 h −(βH
k
+ δ)
_
_
That is
Jac((H
k
, 0, 0)) −λI =
_
_
b −d −2vH
k
−λ −α 0
0 −(σ + d + vH
k
+ α + λ) ψβH
k
0 h −(βH
k
+ δ + λ)
_
_
The determinant of this specific matrix can be written as (b−d−2vH
k
−λ)det|A−λI| = 0
where
A =
_
−(σ + d + vH
k
+ α) ψβH
k
h −(βH
k
+ δ)
_
An eigenvalue is λ = b −d −2vH
k
and others are the solutions of the equation λ
2
−trAλ +
+det|A| = 0
The steady state will be stable under the following conditions:
trA = −(σ + b + α + βH
k
+ δ) < 0 (2.4)
det|A| = (σ + b + α)(βH
k
+ δ) −hψβH
k
> 0
note that condition (2.4) is always true because parameters are non negative by definition.
Therefore the stability conditions are:
b −d −2vH
k
= d −b < 0 (2.5)
(σ + b + α)(βH
k
+ δ) −hψβH
k
> 0 (2.6)
Condition (2.5) is necessary to avoid host population’s extinction due to its natural growth
rate. When further analysis featuring diffusion and spatial heterogeneity will be conducted
on the model, it will turn out that this condition is slightly different. Under certain condition
hosts 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 and
can be expressed as
R
0
> 1, or as one should better say R
0
> 1 ⇔
¯
H < H
k
where
R
0
=
_
hψ
σ + b + α
−1
__
βH
k
δ
_
(2.7)
9
It’s possible to assign a biological meaning to (2.7):
h
σ + b + α
expresses the average number of larvae that originate from an adult parasite,
βH
k
δ + βH
k
is the probability that a larvae is recruited by an host before dying and ψ 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. Rearranging
the factor in a different order it can express the number of free-living stages produced from
a single free-living stage.
Therefore if condition R
0
> 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 more
useful 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 boundaries
were shown in parameter regions.
In fact solving the determinant of |Jac((
¯
H,
¯
P,
¯
L)) −λI| =
_
_
_
b −d −2vH −λ −α 0
ψβL −vP + α(1 +
1
k
)
P
H
2
−(λ + σ + d + vH + α + 2α(1 +
1
k
)
P
H
) ψβH
−βL h −(βH + δ + λ)
_
_
_
is indeed quite complex so in this case one define the determinant as
|Jac((
¯
H,
¯
P,
¯
L)) −λI| =
_
_
a −λ b 0
d e −λ f
g h i −λ
_
_
so by definition it will be
λ
3
−λ
2
(a + e + i) + λ(ei + a(e + i) −fh −bd) + bdi + afh −bfg −aei = 0 (2.8)
Using Routh-Hurwitz conditions [27] one obtains the necessary and sufficient conditions to
have Reλ < 0. These are:
1. −(a + e + i) > 0
2. bdi + afh −bfg −aei > 0
3. −(a + e + i)(ei + a(e + i) −fh −bd) −(bdi + afh −bfg −aei) > 0
Given the parameter’s 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 can
appreciate how a change in natural mortality of larvae (δ) affects the system behaviour.
10
(a) Infected equilibrium is stable when δ is above stability
threshold
(b) Instability of infected equilibrium when δ is under sta-
bility threshold
Figure 2.1: Host-Parasite behaviour when infected equilibrium is stable or unstable for model (2.1) without
diffusion using the same set of parameter except for δ
The simulations show two different examples: in the first one the endemic equilibrium is
stable and the solutions converge to it with damped oscillations, in the second the infected
equilibrium is unstable and the solutions tend to a limit cycle.
Table 2.1 shows the minimum values of δ for which the infected equilibrium is stable, given
this set of parameters: b = 0.6, d = 0.5, α = 0.05, σ = 4, b = 0.6, β = 0.12, φ = 0.5,
h = 100, v = 0.00083.
It has been estimate [29] that varying the parasite induced host mortality (α) between 0
Table 2.1: Stability Region
if k = 3 5 8 10
then δ > 5.25 8.74 11.38 12.46
and 0.1, while keeping other parameter fixed, doesn’t substantially modify this threshold
values.
It is possible to see from simulations that the solutions may oscillate quite far from the
equilibrium just when δ is not close to the boundary values.
11
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 is
present. In other words the purpose is to check what effect may produce on system dynamic
the introduction of spatial diffusion in host population.
The most interesting case will be to linearize about the endemic equilibrium E
1
:= (
ˆ
H,
ˆ
P,
ˆ
L),
under the condition that is stable, and then look for any spatial instability. Unfortunately
this case is algebraically way too complicated. So as an example a less complicated case is
presented.
Consider the reaction diffusion system and linearize about the steady state E
1
:= (H
K
, 0, 0).
Set
w = (H −H
K
, P, L) (2.9)
and (2.1) becomes, for |w| small,
w
t
= J(E
1
)w + D∆w (2.10)
where
J(E
1
) =
_
_
b −d −2vH
K
−α 0
0 −(σ + d + vH
K
+ α) ψβH
K
0 h −(βH
K
+ δ)
_
_
and D is the matrix
_
_
D
1
0 0
0 D
2
0
0 0 0
_
_
As one can see two different values of diffusion coefficients are considered in matrix D. It is
well know [27] Turing instability may occur only if diffusion coefficients are different, thus
to test whether spatial pattern may arise we allow for different diffusion, although parasites
are 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 define
W(r) such that is the time independent solution of the spatial eigenvalue problem defined
by
∆W + φ
2
W = 0, (n·∇)W = 0 for r on ∂Ω (2.11)
where φ is the eigenvalue.
In this model the domain is one-dimensional so W ∝ cos(nπx/L) where n is an integer.
So in this case φ = nπ/L is the eigenvalue and 1/φ is a measure of the wave-like pattern.
Usually the eigenvalue φ is called wave-number and with finite domains there is a discrete
set of possible wave-numbers.
It can be easily shown that boundary condition are satisfied.
12
Let W
φ
(r) be the corresponding eigenfunction to the wave-number φ; each of them satisfy
zero flux boundary condition. Look for solution w(r, t) of (2.10) in the form
w(r, t) =

φ
c
φ
e
λt
W
φ
(r) (2.12)
where c
φ
are constants determined by a Fourier expansion of initial conditions in terms of
W
φ
(r).
λ is the eigenvalue which determines temporal growth. So substituting the form (2.12)
into (2.10) one obtains, for each k,
λW
φ
= J(E
1
)W
φ
+ D∆W
φ
= J(E
1
)W
φ
−Dφ
2
W
φ
(2.13)
and one can determine λ by calculating the roots of the characteristic polynomial
|J(E
1
) −Dφ
2
−λI| = 0 (2.14)
where J(E
1
) −Dφ
2
−λI =
_
_
b −d −2vH
K
−D
1
φ
2
−λ −α 0
0 −(σ + d + vH
K
+ α + D
2
φ
2
+ λ) ψβN
k
0 h −(βH
K
+ δ + λ)
_
_
Evaluate the determinant to get the eigenvalues: one is λ = b −d −2vH
k
−D
1
φ
2
,
the other two are given by the solution of:
λ
2
+λ(σ +d+vN
k
+α+D
2
φ
2
+βN
k
+δ) +(σ +d+vN
k
+α+D
2
φ
2
)(βN
k
+δ) −hψβN
k
= 0
It is easy to see that Reλ < 0 if these conditions are true:
1. b −d −2vH
k
< D
1
φ
2
2. −(σ + d + vH
k
+ α + D
2
φ
2
+ βH
k
+ δ) < 0
3. (σ + b + α + D
2
φ
2
)(βH
k
+ δ) −hψβH
k
> 0
If it’s know by hypothesis that the equilibrium is stable when there is not any spatial effect,
Reλ(φ
2
= 0) < 0, the preceding conditions (2.4) (2.6) hold. Therefore, being D
i
φ
2
> 0, the
conditions with spatial diffusion are satisfied too and that implies there isn’t any diffusion
driven instability.
13
(a) Host behaviour when δ is above the stability threshold
(b) Host behaviour when δ is under the stability threshold
Figure 2.2: Host behaviour when infected equilibrium is stable or unstable for model (2.1), given other
parameters values as Tab 2.1
14
Host-Parasite Coexistence Equilibrium
As already stated, conditions for the stability of endemic equilibrium are quite complex to
write down explicitly, so the model has to be simulated numerically.
The results obtained suggest that hots’s diffusion doesn’t play any relevant role if none of the
parameters depend on space. Again the stability of the host-parasite coexistence equilibrium
depends on the exact values of parameter when both stabilizing and destabilizing processes
are relevant. Similarly as in the previous section is possible to compute the threshold for
the natural mortality of larvae which determines a stability switch. The simulations seems
to confirm that this stability boundary are similar to the boundary computed for the model
without diffusion. One can expect that they are exactly the same, but the exact value of
the stability threshold is not easily determined because in its proximity solutions tend to the
equilibrium very slowly, so an huge amount of iterations is needed to check if solutions are
not oscillating around a limit cycle.
As shown in Figure 2.2 time oscillations are coherent in space, then Figure 2.3 shown the
behaviour 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) using
the same set of parameter except for δ
Figure 2.4: Host behaviour when δ is above the
stability threshold
Figure 2.5: Host behaviour when δ is under the
stability threshold
As one can see comparing Figure 2.3 and Figure 2.1 the behaviour is quite identical for
models with or without diffusion under the condition that none of parameters depend on
space and that the diffusion coefficient is equal for host and parasite.
15
2.3 Fertility Reduction
It is reasonable to assume that parasite reduce host fitness not only by increasing host
mortality but decreasing host fertility, too. How to model the fertility reduction has been
discussed previously (1.11). We made the assumption that γ(j) = γ
j
, but different choices
are possible. Again not considering any spatial dependency on parameters the model studied
is
_
¸
¸
¸
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
¸
¸
¸
_
∂H
∂t
= D∆H + (b
_
kH
kH + (1 −γ)P
_
k
−d −vH)H −αP
∂P
∂t
= D∆P + βψHL −P
_
σ + d + vH + α + α
_
1 +
1
k
_
P
H
_
∂L
∂t
= hP −βHL −δL
(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 solutions
tend to a limit cycle.
Figure 2.6: Host behaviour when a small fertility reduction is present. Parameter’s values are such that the infected equilibrium
would be in the stability region near the threshold if there isn’t any fertility reduction.
16
To compensate the destabilizing effect of fertility reduction and again achieve stability for the
coexistence equilibrium one has to enforce stabilizing processes or decreasing other destabi-
lizing forces, for instance increasing the natural mortality of free-living stage. To determine
the stability conditions is quite complex, none the less the simulations confirm the struggle
between 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. Parameter’s value are as Figure 2.6 except for α. While
at first the infected equilibrium was in the stability region, then with fertility reduction has become unstable, now increasing
the parasite induced mortality is stable again
γ and α, it’s possible to find a stability boundary. Its value depending on the value of k.
Numerical simulations show that decreasing γ (increasing the fertility reduction strength)
keeping α fixed lead to instability, viceversa increasing α may lead to stability of infected
equilibrium.
Host diffusion doesn’t seem to affect the model prediction if all the parameters do not depend
on space.
17
18
Chapter 3
Single-Species Equation
In the previous chapter a comparison between the model with and without diffusion was
given. It has been shown that analysis is quite complex and it requires extensive numerical
simulations. Next step to further study how host-parasite interactions affects population
dynamics is to introduce a space dependent fertility rate. Therefore the habitat of the host
species 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 model
we 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 Ω ×(0, ∞)
d(x)
∂u
∂
− →
n
+ β(x)u = 0 on ∂Ω ×(0, ∞)
(3.1)
In general, solutions to equations of the form (3.1) need not be bounded nor exist for all
time. However in this study f(x, u) is in the common logistic form f(x, u) = a(x)u −c(x)u
2
and we assume that there is positive number H
k
such that if u > H
k
then f(x, u) < 0 for all
x. H
k
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 H
k
will be a supersolution.
Define
d(x) = D
f(x, H) = H(b(x) −d −vH)
Moreover, in this case study, Ω is a one dimensional space domain, so our single species
equation
∂H
∂t
= D∆H + H(b(x) −d −vH) in Ω ×(0, ∞)
D
∂H
∂
− →
n
= 0 on ∂Ω ×(0, ∞)
(3.2)
may be written as (3.1)
Now focus on the single species equation (3.1), to study its behaviour.
19
By our definition this model shares these two key feature of the logistic model: f(x, 0) = 0
and f(x, H) < 0 for some sufficiently large H. f(x, H) is continuos and differentiable with
respect 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 for
extinction and persistence.
Proposition 1 Suppose that f(x, H) ≤ g
0
(x)H for x ∈ Ω where g
0
(x) is a bounded measur-
able function g
0
(x) ∈ C
α
(
¯
Ω). If the principal eigenvalue λ
1
of
_
∇· D∇ϕ + g
0
(x)ϕ = λϕ in Ω
D
∂ϕ
∂
− →
n
= 0 on ∂Ω
(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 = ce
λ
1
t
ϕ
1
, where ϕ
1
> 0 is the eigenfunction corresponding to λ
1
in (3.3) and
c > 0 is a constant. We have
∂¯ u
∂t
−D∇¯ u −f(x, ¯ u) =
= λ
1
¯ u −[D∇¯ u + g
0
(x)¯ u] + g
0
(x)¯ u −f(x, ¯ u)
= g
0
(x)¯ u −f(x, ¯ u) ≥ 0
so that ¯ u is a supersolution to (3.2). If u(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 λ
1
< 0
we have ¯ u(x, t) →0 exponentially as t →∞, so u(x, t) must decay toward zero exponentially
as 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 large
enough at some point x and some density H so that the corresponding spatially averaged
growth rate given by the eigenvalue λ
1
for g
0
(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 C
2
in u and C
α
in
x, and there exists a M > 0 such that g(x, H) < 0 for H > M. If the principal eigenvalues
λ
1
is positive in the problem
_
∇· D∇ϕ + g(x, 0)ϕ = λϕ in Ω
D
∂H
∂
− →
n
= 0 on ∂Ω
(3.4)
then (3.1) has a minimal positive equilibrium H
∗
and all solutions to (3.4) which are initially
positive on an open subset of Ω are eventually bounded below by orbits which increase toward
H
∗
as t →∞
20
Proof The assumptions on f(x, H) imply that we can write f(x, H) = [g(x, 0)+g
1
(x.H)H]H
where g
1
(x, H) is C
1
in H. Let ϕ
1
be an eigenfunction for (3.2) with ϕ
1
> 0 on Ω. For > 0
sufficiently small
∇· D∇(ϕ
1
) + f(x, ϕ
1
) =
= [D∆ϕ
1
+ g(x, 0)ϕ
1
] + g
1
(x.ϕ
1
)
2
ϕ
2
1
= λ
1
ϕ
1
+ g
1
(x.ϕ
1
)
2
ϕ
2
1
= ϕ
1
[λ
1
+ g
1
(x, ϕ
1
)ϕ
1
] > 0
Since g
1
and ϕ
1
are bounded, it follows that for > 0 small, ϕ
1
is a sub solution for the
elliptic problem (3.4) corresponding to (3.2). If
¯
H(x, t) is a solution to (2) with
¯
H(x, 0) = ϕ
1
,
then
∂
¯
H
∂t
|
t=0
> 0 on Ω 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 a
solution to (3.2) which is initially nonnegative and is positive on an open subset of Ω, then
the strong maximum principles implies H(x, t) > 0 on
¯
Ω for t > 0. Choosing any t
0
> 0 we
can take > 0 so small that ϕ
1
< H(x, t
0
) on
¯
Ω; then
¯
H(x, t −t
0
) < H(x, t) for t = t
0
and
thus by the maximum principles for t > t
0
. Hence, H(x, t) is bounded below by
¯
H(x, t −t
0
)
and
¯
H(, x, t
0
) ↑ 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).
It’s important to note that the eigenvalue problem used to predict extinction may be different
from the problem used to predict persistence.
Define g
0
and g as in Proposition 1 and in Proposition 2, then the two eigenvalue problems
coincide when the following relation is true.
g
0
(x) = max
H≥0
[f(x, H)/H] = lim
H→0
+
[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 eigenvalue’s 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 eigenvalue
problem as
D∆u + m(x)u = λu in Ω ×(0, ∞)
D
∂u
∂
− →
n
= 0 on ∂Ω ×(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 λ
1
.
A fundamental results is the following (see [12], [23] for a proof and further details, also
non pure Neumann boundary conditions are considered):
Theorem 1 Suppose that ∂Ω is piecewise of class C
2+α
and that Ω satisfies the interior
cone condition. Suppose that D ∈ C
1+α
(
¯
Ω) with D > 0, that m(x) ∈ L
∞
(Ω). The principal
21
eigenvalue of (3.5) is given by
λ
1
= max
u∈W
1,2
(Ω),u=0
_
¸
¸
_
¸
¸
_
−
_
Ω
D|∇u|
2
dx +
_
Ω
m(x)u
2
dx
_
Ω
u
2
dx
_
¸
¸
_
¸
¸
_
(3.6)
or, alternatively
λ
1
= max
_
−
_
Ω
D|∇u
2
| dx +
_
Ω
m(x)u
2
dx : where u ∈ W
1,2
(Ω) ,
_
Ω
u
2
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 λ
1
(D, m) then λ
1
(D, m) is
increasing with respect to m in the sense that if m
1
≥ m
2
then λ
1
(D, m
1
) ≥ λ
1
(D, m
2
), and
if m
1
> m
2
on a subset of positive measure then λ
1
(D, m
1
) > λ
1
(D, m
2
). Similarly, λ
1
(D, m)
is decreasing with respect to D in the same sense unless m(x) is constant, in this case the
eigenfunction u
1
turns out to be a constant so the terms involving D in (3.7) will have no
effect.
Proof Let u
1
the eigenfunction associated with λ
1
(D, m
2
). Then u
1
is the maximizer of the
quotient (3.6) for λ
1
(D, m
2
) and we have
λ
1
(D, m
2
) =
−
_
Ω
D|∇u
2
1
| dx +
_
Ω
m
2
(x)u
2
1
dx
_
Ω
u
2
1
dx
≤
−
_
Ω
D|∇u
2
1
| dx +
_
Ω
m
1
(x)u
2
1
dx
_
Ω
u
2
1
dx
≤ max
u∈W
1,2
(Ω),u=0
−
_
Ω
D|∇u
2
1
| dx +
_
Ω
m
1
(x)u
2
1
dx
_
Ω
u
2
1
dx
= λ
1
(D, m
1
)
Since u
1
> 0 in Ω the first inequality is strict if m
1
(x) > m
2
(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 increase
the average growth rate as measured by λ
1
, and it does, confirming the reasonable conjecture
that a better growth rate improves the chance of survival.
22
Consider now a related eigenvalue problem to (3.5)
D∆φ + σm(x)φ = 0 in Ω
D
∂φ
∂
− →
n
= 0 on ∂Ω
(3.8)
If m(x) is strictly positive then the principal eigenvalue σ
1
for (3.8) could be characterized
as in (3.6), but corresponding to −λ
1
σ
1
= min
φ∈W
1,2
(Ω),φ=0
_
¸
¸
_
¸
¸
_
_
Ω
D|∇φ
2
| dx
_
Ω
m(x)φ
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/σ
1
as the maximum of the
reciprocal form in (3.8).
Theorem 2 Suppose that Ω and the coefficients D and m satisfy the hypothesis of theorem
1. Assume further that ∂Ω is of class C
1
, and that m(x) is positive on an open subset of Ω.
The problem (3.8) admits a positive principal eigenvalue σ
+
1
determined by
1
σ
+
1
= max
φ∈W
1,2
(Ω),φ=0
_
¸
¸
_
¸
¸
_
_
Ω
m(x)φ
2
dx
_
Ω
D|∇φ|
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) with
respect to L
p
(Ω) for any p ∈ (1, ∞] in the case of one space dimension. It shall be noted
that problem (3.8) also admits a negative principal eigenvalue if m(x) is negative on an open
subset of Ω.
The denominator on the right side of (3.10) would vanish if φ where constant, and hence the
maximum may not exist. For this kind of problem it has been proved that the existence of
a positive or negative principal eigenvalue depends on the integral of m(x) over Ω.
Theorem 3 In the case of Neumann boundary condition the problem (3.8) admits a positive
principal eigenvalue if and only
_
Ω
m(x) dx < 0 (3.11)
In that case the positive principal eigenvalue is the only positive eigenvalue which admits a
positive 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
23
possible when there are places where the local growth rate is positive but negative on the
average. In such a case the proximity of bad habitat to good is the mechanism leading to a
critical patch size.
Theorem 4 Suppose that r is a positive parameter and that (3.11) holds, The principal
eigenvalue λ
1
of
D∆H + rm(x)H = λH in Ω
D
∂H
∂
− →
n
= 0 on ∂Ω
(3.12)
is positive if and only if 0 < σ
+
1
< r where σ
+
1
is the positive principal eigenvalue of (3.8).
If the inequality (3.11) is reversed then λ
1
> 0 for all r > 0
Note that for (3.5) r = 1.
Proof Let H
1
be the eigenfunction for λ
1
. Multiplying by H
1
in (3.12), integrating over Ω,
and applying Green’s formula and the boundary conditions yields:
−
_
Ω
D|∇H
1
|
2
dx + r
_
Ω
m(x)H
2
1
dx = λ
1
_
Ω
H
2
1
dx
By (3.10)
_
Ω
m(x)H
2
1
dx ≤
1
σ
+
1
_
Ω
D|∇H
1
|
2
dx
so
_
r
σ
+
1
−1
__
Ω
D|∇H
1
|
2
dx ≥ λ
1
_
Ω
H
2
1
dx
so λ
1
< 0 if r < σ
+
1
. On the other hand, if φ
1
is the eigenfunction for σ
+
1
we can multiply
(3.8) by φ
1
and integrate to obtain
−
_
Ω
D|∇φ
1
|
2
dx = −σ
+
1
_
Ω
m(x)φ
2
1
dx
so that by theorem 1 as it applies to (3.12) one has
λ
1
≥
−
_
Ω
D|∇φ
1
|
2
dx + σ
_
Ω
m(x)φ
2
1
dx
_
Ω
φ
2
1
dx
= (σ −σ
+
1
)
_
Ω
m(x)φ
2
1
dx
_
Ω
φ
2
1
dx
By one has
_
Ω
m(x)φ
2
1
dx > 0 since σ
+
1
> 0, so λ
1
> 0 for σ > σ
+
1
If condition (3.11) is
reversed then one uses the test function u ≡ 1 in (3.6,3.7) and obtain λ
1
> 0 immediately if
σ > 0.
24
Because there is not any dispersal of individuals across the border, the only mechanism that
might cause a loss of population is dispersal into regions where the average local growth rate
is negative. When condition (3.11) is satisfied it describes this scenario. But individuals
to effectively average the local growth rate have to disperse sufficiently fast; as a result
population will decline (λ
1
< 0) if (3.11) holds and D is large.
To demonstrate this intuition rewrite (3.12) to obtain ∆u +
rm(x)
D
u =
λ
D
u. If (3.11) holds
σ
+
1
is positive and it s characterized by (3.10). By Theorem 4, since (3.11) holds the principal
eigenvalue λ
1
is positive if and only if r/D > σ
+
1
, where σ
+
1
is the principal eigenvalue of
∆φ + σm(x)φ = 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 eigenfunction
corresponding to the principal eigenvalue could be too complicated.
The habitat could be represented by different values of growth rate. This assumption means
that there are more favorable areas for species reproduction and life, but nevertheless a single
individual could move through the whole space domain. This distinction may play a role
after, when another species is introduced. In fact this imply that there are not any refugee
patches, thus the individuals of the two species will be potentially always in contact due to
their random movement.
However if we define the local growth rate m(x) as a piecewise constant we can obtain a
transcendental 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 one
dimensional.
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 the
principals eigenvalue λ
1
of the problem
D∆H + m(x)H = λH in (0, L)
D
∂H
∂
− →
n
= 0
(3.13)
is positive. It follows from Theorem 4 that λ
1
/D > 0, and hence λ
1
> 0 if and only if
1/D > σ
+
1
(m(x)) where σ
+
1
(m(x)) is the positive principal eigenvalue of
∆φ + σm(x)φ = 0 in Ω
∂H
∂
− →
n
= 0 on ∂Ω
(3.14)
A smaller value of σ
+
1
(m(x)) reflects a more favorable environment than a larger one, because
25
the condition 1/D > σ
+
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 σ
+
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
_
Ω
m(x) dx < 0 and hence σ
+
1
(m(x))
will exist.
The parameter a determines the location of the favorable patch, so we need to determine
how σ
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 φ in (3.14). If we let α =
_
σ
+
1
(m(x)) we must construct the eigenfunction as
φ(x) =
_
¸
_
¸
_
cosh(αx) for x ∈ [0, a)
Acos(α
√
k(x −c)) for x ∈ [a, a + l)
Bcosh(α(1 −x)) for x ∈ [a + l, 1)
Matching φ(x) and φ

(x) across the interface and solving as before yields
cot(α
√
kl) =
k cosh(αa) cosh(α(1 −a −l)) −sinh(αa) sinh(α(1 −a −l))
√
k sinh(α(1 −l))
=
(k −1) cosh(α(1 −l)) + (k + 1) cosh(α(1 −2a −l))
√
k sinh(α(1 −l))
= G(a, α)
It turns out that G(a, α) is decreasing in α, with G(a, α) → (k − 1)
√
k as α → ∞, and
G(a, α) → ∞ as α → 0
+
with order
√
k/(α(1 − l)). Since cot(α
√
kL) → ∞ with order
1/(
√
kαl) as α →0
+
, it follows from the assumption kl < 1−l that cot(α
√
kL) is decreasing
in α, increasing G(a, α) moves the graph of G(a, α) (with respect to α) upward and hence
causes the curves to intersect at a smaller value of α. Computing ∂G(a, α)/∂a yields a
fraction with a positive denominator and numerator equal to −2α(k1) sinh(α(1 − 2a − l)).
Thus, ∂G(a, α)/∂a is negative for all α if 0 < a < (1 − l)/2 and positive for (1 − l)/2 <
a < 1 − l. Thus, the graph of G(a, α) is lowest, leading to the largest value of α at the
intersection point, if a = (1 − l)/2. It follows that this is the worst patch under Neumann
condition. The cases a = 0 and a = 1 −l give the same value for α at the intersection point.
That value corresponds to the optimal location of the favorable patch, which in this case is
at either end of interval [0, 1].
26
3.2 Parasite and Larvae Equation
Analysis of the stability of the uninfected equilibria may be done as follow. Consider the
parasite and larvae equations linearized in the equilibrium (
ˆ
H, 0, 0),
_
¸
¸
¸
_
¸
¸
¸
_
∂P
∂t
= D∆P + (βψ
ˆ
H(x))L −P(η + v
ˆ
H(x))
∂L
∂t
= hP −(β
ˆ
H(x) −δ)L
(3.15)
Where, to simplify, η := σ +α +d To solve this system of equations subject to the zero flux
boundary conditions first look for the time independent solution of the spatial eigenvalue
problem defined by
_
_
_
D∆P + L(βψ
ˆ
H(x)) −P(η + v
ˆ
H(x)) = λP
hP −(βH(x) + δ)L = λL
(3.16)
substituting the value of L in the second equation one obtains
_
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
_
L =
P
β
ˆ
H(x) + δ + λ
D∆P + P
_
βψ
ˆ
H(x)
β
ˆ
H(x) + δ + λ
−η −vH(x) −λ
_
= 0
(3.17)
It’s possible to use the powerful tool of Sturm-Liouville theory to study the system.
In fact, if one recalls the Sturm-Liouville equation
d
dx
[p(x)u

(x)] −q(x)u(x) + µw(x)u(x) = 0
∂u
∂
− →
n
= 0
(3.18)
it easy to see that equation (3.17) is in the Sturm-Liouville form given these definition:
µ = −λ
p(x) = D
w(x) = 1
q(x) = η + v
ˆ
H(x) −
βψ
ˆ
H(x)
β
ˆ
H(x) + δ + λ
(3.19)
It has already been proved that
−
_
l
0
p(x)(u

(x))
2
dx −
_
l
0
q(x)u

(x) +
_
l
0
µw(x)u

(x) = 0 (3.20)
27
µ =
_
Ω
p(x)(u

(x))
2
dx +
_
Ω
q(x)u
2
(x)dx
_
Ω
w(x)u

(x)dx
(3.21)
and the eigenvalue can be ordered such that
µ
0
< µ
1
< µ
2
< ... < µ
k
< ... (3.22)
if we normalize the eigenfunction
_
Ω
w(x)u

(x)dx = 1 (3.23)
then it’s known that
µ
0
= min
__
Ω
p(x)(u

(x))
2
dx +
_
Ω
q(x)u
2
(x) dx
_
(3.24)
With u satisfying (3.23) and the boundary condition
Since q(x) depends on λ, all eigenvalues µ
0
< µ
1
< µ
2
< ... < µ
k
depends on λ
There is a solution of (3.17) when λ = −µ
k
(λ).
Now look for solution λ = −µ
0
(λ), if λ
1
> λ
2
then
µ
0
(λ
1
) = min
_
_
Ω
D(u

(x))
2
dx +
_
Ω
η + v
ˆ
H(x) −
βψ
ˆ
H(x)
β
ˆ
H(x) + δ + λ
1
u
2
(x) dx
_
>
µ
0
(λ
2
) = min
_
_
Ω
D(u

(x))
2
dx +
_
Ω
η + v
ˆ
H(x) −
βψ
ˆ
H(x)
β
ˆ
H(x) + δ + λ
2
u
2
(x) dx
_
so when λ is increasing −µ
0
(λ) is decreasing. Hence if −µ
0
(λ) < 0 one has λ > −µ
0
(λ) >
−µ
k
(λ) so there is no solution with λ > 0 of λ = −µ
k
(λ), k = 0, 1, ...
If −µ
0
(λ) > 0 there is a unique λ
∗
such that λ
∗
= −µ
k
(λ
∗
).
Therefore for parasite establishment −µ
0
(0) < 0 is a necessary and sufficient condition
Proposition 3 To have parasite establishment −µ
0
(0) < 0 is a necessary and sufficient
condition, i. e. the eigenvalues of
D∆P + P
_
βψ
ˆ
H(x)
β
ˆ
H(x) + δ
−η −vH(x)
_
+ µP = 0 , with
∂P
∂
− →
n
= 0
have to be negative.
28
3.3 Discussion
These last two chapters focus on the behaviour of a single species, at first when parameters
are space independent. The absence of diffusion driven instability has been pointed out for
the 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, none
of which has the whole domain (the side of a mountain) as habitat, but instead they have
different favorable patches corresponding on which altitude they are. In order to model this
property 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 computing
the 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, α = 0.05,
σ = 4, β = 0.12, ψ = 1, h = 100, v = 0.00083, δ = 60, b(x) = (3.5 −x)(x −6.5), if x ∈ [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 which
may be the system behaviour. The first result is that the distribution of the host population
tends to follow the shape of its fertility function; otherwise, if the average fertility is too low
and diffusion too large, it may gets extinct even when parasite is absent . When parasite is
present, an infected equilibrium is feasible. It may be stable or unstable, in the latter case
29
(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, α = 0.1,, σ
i
= 4, β = 0.12, φ = 1.,
h = 100, v = 0.00083, δ = 4, γ = 0.1, b(x) = (3.5 −x)(x −6.5)
“
kH
2
kH
2
+(1−γ
2
)P
2
”
k
, if x ∈ [3.5, 6.5], 0 otherwise.
solutions tend to cycle. To have parasite establishment a sufficiently high host growth rate
at 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 is
that now the average growth rate may be even negative on the spatial domain but still allow
for 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 infected
equilibrium may be feasible and unstable. However decreasing the value of fertility on the
whole domain tend to stabilize the host-parasite coexistence equilibrium.
The instability of host-parasite coexistence equilibrium can be achieved enforcing fertility
reduction or decreasing free living state natural mortality (or increasing k, the aggregation
parameter, which corresponds to decreasing aggregation). The dynamics depends on the
exact values of parameters. In all the numerical simulations performed in which coexistence
equilibrium was unstable, solutions oscillated in time tending to a limit cycle. In contrast to
the 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 large
part of the space domain and an overwhelming effect of destabilizing forces. The condition
on average growth is quite subtle, indeed. In fact it seems (Figure 3.2) that a sufficiently
high average value is not sufficient to have spatial instability if in wide regions of the domain
there is a strong detrimental effect on host. In that case the loss of individuals prevails on
population dispersal and damps down oscillations.
If conditions are favorable for spatial instability to arise, the effect on host is that after
30
(a) Host’s density (b) Parasites density
(c) Long Interval Host density
Figure 3.3: Effects of spatial instability. Parameter values are: d = 0.5, α = 0.1,, σ
i
= 4, β = 0.12, φ = 1., h = 100,
v = 0.00083, δ = 60, γ = 0.1, b(x) = 2.5e
−
(x−4)
2
4
“
kH
kH+(1−γ
2
)P
”
k
.
reaching a peak, when they are decreasing due to parasite effect on mortality and fertility
rates, a portion of them move to the border (in this context the region with a lower average
growth rate, therefore where there were fewer presence of larvae and parasite). However if
one looks at a long time interval this behaviour is not so evident because of the oscillations
in time that may hide the side dispersal. This pattern is pictured on Figure 3.3
31
32
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 been
studied using numerical simulations and is now known in its main properties. Moreover
it’s clear that the stability of those system depends on the balance between stabilizing and
destabilizing forces. Parasite induced mortality, reduction of host fertility and persistence of
free living stages natural mortality are some of them.
The model is going to describe now two host species sharing a macroparasite, with infection
spreading 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 a
reservoir for the parasite, maintaining a high number of infective stages resulting in high
level of infection in the less tolerant species.
Greenman and Hudson [17] investigate such models adopting a geometric approach based
on bifurcation theory in order to sidestep algebraic intractability. In their articles they focus
on three different form of competition: direct competition, apparent competition and intra
specific 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 address
if the destabilizing effect of parasite induced reduction in host fecundity extends to the
two species system, if the microparasite phenomena of host exclusion inversion and domi-
nance reversal occurs also for macroparasite system, if threshold conditions can be expressed
in 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 by
Greenman and Hudson. One is to make enough accurate assumption to substantially sim-
plify the model equation, another is to maintain complexity but carrying out extensive
numerical simulations to assess the validity of conjectured rules about the feasibility and
33
stability structure of system’s equilibria. The downside of the first method is the possible
loss of information or the oversimplification of the model, while for the other is the extensive
amount 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 spatial
heterogeneity is present.
So, recall the generic model (2.1) and add the equations for host two.
_
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
_
∂H
1
∂t
= D
1
∆H
1
+ H
1
(b
1
(x) −d
1
−v
1
H
1
−θ
1
H
2
) −α
1
P
1
∂H
2
∂t
= D
2
∆H
2
+ H
2
(b
2
(x) −d
2
−v
2
H
2
−θ
2
H
1
) −α
2
P
2
∂P
1
∂t
= D
1
∆P
1
+ β
1
ψ
1
H
1
L −P
1
(σ
1
+ d
1
+ v
1
H
1
+ θ
1
H
2
+ α
1
+ α
1
(1 +
1
k
1
)
P
1
H
1
)
∂P
2
∂t
= D
2
∆P
2
+ β
2
ψ
2
H
2
L −P
2
(σ
2
+ d
2
+ v
2
H
2
+ θ
2
H
1
+ α
2
+ α
2
(1 +
1
k
2
)
P
2
H
2
)
∂L
∂t
= h
1
P
1
+ h
2
P
2
−(β
1
H
1
+ β
2
H
2
+ δ)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
_
¸
¸
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
¸
¸
_
dH
i
dt
= H
i
(b
i
−d
i
−v
i
H
i
−θ
i
H
j
) −α
i
P
i
dP
i
dt
= β
i
ψ
i
H
i
L −P
i
(σ
i
+ d
i
+ v
i
H
i
+ θ
i
H
j
+ α
i
+ α
i
(1 +
1
k
i
)
P
i
H
i
)
dL
dt
= h
i
P
i
+ h
j
P
j
−(β
i
H
i
+ β
j
H
j
+ δ)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.
34
Suppose that host j, with(j = i) is absent, the threshold condition for parasite invasion
can be expressed similarly to (2.7) so
R
0i
=
_
h
i
ψ
i
σ
i
+ α
i
+ b
i
−1
__
β
i
ψ
i
H
Ki
δ
_
> 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 direct
competition is absent, is given by the upper branch of the hyperbola
(R
−1
01
−1)(R
−1
02
−1) = 1 (4.4)
This is also the stability boundary for the uninfected coexistence equilibrium. Above this
curve the uninfected equilibrium is stable against invasion by the parasite and below unsta-
ble. These properties can be established by standard techniques [27].
It’s possible to define a Tolerance Index S
0i
as follows: if host j is absent and density
dependence can be ignored, so v
i
= 0. then from (4.2)
_
_
_
H
i
(b
i
−d
i
) −α
i
P
i
= 0
β
i
ψ
i
H
i
L −P
i
_
σ
i
+ d
i
+ α
i
+ α
i
_
1 +
1
k
i
_
P
i
H
i
_
= 0
so
_
¸
¸
_
¸
¸
_
H
i
(b
i
−d
i
) = α
i
P
i
β
i
ψ
i
H
i
L =
H
i
(b
i
−d
i
)
α
_
σ
i
+ d
i
+ α
i
+ α
i
_
1 +
1
k
i
_
H
i
(b
i
−d
i
)
αH
i
_
and then L
i0
denotes the equilibrium value of L where
L
i0
=
(σ
i
+ α
i
+ b
i
+
b
i
−d
i
k
i
)(b
i
−d
i
)
β
i
ψ
i
α
i
(4.5)
Define now Tolerance Index as
S
0i
=
L
i0
L
j0
If S
0i
> 1 then host i is more tolerant to the the infection so it will coexist a higher population
of 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 threshold
condition for host exclusion. This boundary value, defined as R
∗
0i
, is computed as follows.
35
Table 4.1: Parameters and Coefficient List
Parameters
b(x) is the natural density dependent fertility rate in (x, t) ;
γ is the factor reducing the fertility of hosts; 0 < γ(j) < 1;
d is the intrinsic density dependent mortality;
v is the intraspecific competition index
θ is the direct competition index
α is the induced mortality of hosts due to the parasite burden;
D
i
is the diffusion coefficient of host population;
σ is the natural mortality rate for parasites in the class h
(j)
;
h is the laying rate per host
δ is the natural death rate of larvae;
β is the recruitment rate of parasite on hosts;
Derived Parameters
r
i
= b
i
−d
i
s
i
= σ
i
+ α
i
+ b
i
n
i
= s
i
+ r
i
/k
i
L
i0
= (n
i
r
i
)/(β
i
α
i
)
S
0i
=
L
i0
L
j0
R
0i
=
_
h
i
ψ
i
σ
i
+ α
i
+ b
i
−1
__
β
i
H
Ki
δ
_
Dimensionless Coefficients
c
00
= r
2
/r
1
c
3i
= 1/k
i
c
1i
= s
i
/r
i
c
4i
= δ/r
1
c
2i
=
h
i
ψ
i
r
i
c
5i
= δ/(β
i
H
Ki
)
36
Analysis of Equilibria
The model defined in terms of parasite intensity Z
i
= P
i
/H
i
becomes
_
¸
¸
¸
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
¸
¸
¸
_
dH
i
dt
= H
i
(b
i
−d
i
−v
i
H
i
−θ
i
H
j
−α
i
Z
i
)
dZ
i
dt
= β
i
ψ
i
L −Z
i
(σ
i
+ b
i
+ α
i
) −
α
i
k
i
Z
2
i
dL
dt
= h
i
H
i
Z
i
+ h
j
H
j
Z
j
−(β
i
H
i
+ β
j
H
j
+ δ)L
(4.6)
The equations for host one-parasite coexistence equilibrium, when host two is absent, are
given by
1 = z
1
+ u
1
(4.7)
w = z
1
(c
11
+ c
31
z
1
)/(c
11
+ c
31
) (4.8)
wS
01
= z
2
(c
12
+ c
32
z
2
)/(c
12
+ c
32
) (4.9)
w(c
51
+ u
1
) = c
21
u
1
z
1
/(c
11
+ c
31
) (4.10)
where u
i
= H
i
/H
Ki
, z
i
= (α
i
Z
i
)/r
i
, w = L/L
10
, v
i
= (b
i
− d
i
)/H
Ki
and the dimensionless
coefficients listed in Table 4.1.
Proof
To obtain (4.7) consider
H
1
(b
1
−d
1
−v
1
H
1
−θ
1
H
1
−α
1
Z
1
) = 0
that becomes, under the hypotesis that host two is absent,
(b
1
−d
1
−v
1
H
1
−α
1
Z
1
) = 0
1 −
v
1
b
1
−d
1
H
1
−
α
1
b
1
−d
1
Z
1
= 0
then with u
i
= H
i
/H
Ki
, z
i
= (α
i
Z
i
)/r
i
one has (4.7)
To obtain (4.8) consider
β
1
ψ
1
L −Z
1
(σ
1
+ b
1
+ α
1
) −
α
1
k
1
Z
2
1
= 0
that may be written as
L =
Z
1
(σ
1
+ b
1
+ α
1
+
α
1
k
1
Z
1
)
β
1
ψ
1
37
define r
i
= b
i
−d
i
then
L =
r
2
1
α
1
Z
1
α
1
r
1
_
σ
1
+ b
1
+ α
1
r
1
+
α
1
k
1
r
1
Z
1
_
β
1
ψ
1
so by previously definition of dimensionless coefficients this becomes
L =
r
2
1
z
1
(c
11
+ c
31
z
1
)
β
1
ψ
1
α
1
recall (4.5) so
w =
L
L
10
=
r
2
1
z
1
(c
11
+ c
31
z
1
)
α
1
β
1
ψ
1
α
1
β
1
ψ
1
(σ
1
+ α
1
+ b
1
+
b
1
−d
1
k
1
)r
1
=
z
1
(c
11
+ c
31
z
1
)
c
11
+ c
31
To obtain (4.9) consider the equation for parasite two. In fact, even if it seems unobivous in
the equilibrium for H
i
, in the absence of other host (H
j
= 0), one has that Z
j
> 0. Stated
this, retrace the previous work to obtain
L =
r
2
2
z
2
(c
12
+ c
32
z
2
)
β
2
ψ
2
α
2
now recall the tolerance index definition and write the equation using dimensionless coeffi-
cients
wS
01
=
L
L
10
L
10
L
20
=
z
2
(c
12
+ c
32
z
2
)
c
12
+ c
32
To obtain (4.10) consider
h
1
H
1
Z
1
−(β
1
H
1
+ δ)L = 0
so divide for L
10
then
(β
1
H
1
+ δ)w =
h
1
H
1
Z
1
L
10
=
h
1
H
1
Z
1
α
1
β
1
ψ
1
(σ
1
+ α
1
+ b
1
+
b
1
−d
1
k
1
)r
1
therefore
β
1
H
K
_
β
1
H
1
β
1
H
K
+
δ
β
1
H
K
_
w =
α
1
Z
1
r
1
β
1
ψ
1
r
1
r
1
h
1
H
1
(σ
1
+ α
1
+ b
1
+
b
1
−d
1
k
1
)
β
1
H
K
(u
1
+ c
51
)w = z
1
β
1
ψ
1
h
1
H
1
r
1
(c
11
+ c
31
)
38
Finally one obtains the last equilibrium equation
(u
1
+ c
51
)w =
c
21
u
1
z
1
(c
11
+ c
31
)
Then, having the equations for the equilibrium populations when host one coexists with
the parasite (with host 2 absent), it is possible to have a quadratic equation for u
1
. In fact
recalling equation (4.7)
z
1
= 1 −u
1
and substituting in (4.8), it becomes w =
(1 −u
1
)(c
11
+ c
31
(1 −u
1
))
c
11
+ c
31
then one may write (4.10) as w =
c
21
u
1
(1 −u
1
)
(c
11
+ c
31
)(c
51
+ u
1
)
and again substituting w one has
c
21
u
1
(c
51
+ u
1
)
= (c
11
+ c
31
(1 −u
1
))
now it is easy to rearrange and obtain the quadratic equation
c
31
u
2
1
−[c
11
+ c
31
−c
21
−c
31
c
51
]u
1
−c
51
(c
11
+ c
31
) = 0 (4.11)
To be feasible solutions of (4.11) must be positive and equal or less than one since u
1
is the
fraction between the number of hosts and their carrying capacity.
Provided that condition for parasite invasion holds (R
01
> 1) there is a unique positive root
which is feasible for (4.11). The proof is straightforward. Note that
R
01
=
_
h
1
ψ
1
σ
1
+ α
1
+ b
1
−1
__
β
1
H
K1
δ
_
=
_
c
21
c
11
−1
__
1
c
51
_
If u
1
= 0 then (4.11) is negative, so there is a unique positive solution. Moreover if u
1
= 1
then (4.11) is positive if and only if c
21
> c
11
(1 + c
51
) that is R
01
> 1.
If c
31
and c
32
are not positive there can be up to four positive real and feasible solutions of
the equilibrium equations (4.7), (4.8), (4.9), (4.10), but for this model inequality c
3i
> 0 is
always true, in fact c
3i
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 boundary
value between infected host exclusion and invasion.
The infected host one equilibrium is stable against host 2 invasion provided that
1. c
12
+ 2c
32
z
2
> 0
2. 1 −θ
2
u
1
−z
2
< 0
These are, respectively, the first-order conditions for stability of equations for Z
2
and for
H
2
in (4.6). The first one is always satisfied under the assumptions made. In the second
define the inequality slack as q = θ
2
u
1
+z
2
−1, then similarly as how we found the quadratic
equation for u
1
from this and from (4.7) (4.8) (4.10) substitute for z
2
and w to obtain
Q(u
1
, θ
2
) = qE(u
1
, θ
2
, q) (4.12)
39
Proof
Recall (4.7): z
1
= 1 −u
1
then (4.8) becomes w =
(1 −u
1
)(c
11
+ c
31
(1 −u
1
))
c
11
+ c
31
,
z
2
= q + 1 −θ
2
u
1
then (4.9) becomes w =
(q + 1 −θ
2
u
1
)(c
12
+ c
32
(q + 1 −θ
2
u
1
))
S
01
(c
12
+ c
32
)
then substituting w one has
S
01
(c
12
+c
32
)(1−u
1
)(c
11
+c
31
(1−u
1
)) = (q+1−θ
2
u
1
)(c
12
+c
32
(q+1−θ
2
u
1
))(c
11
+c
31
) (4.13)
rearrange the right side of the equation to obtain
q(c
12
+ 2c
32
−2c
32
θ
2
u
1
+ qc
32
)(c
11
+ c
31
) + (c
12
−θ
2
c
12
u
1
+ c
32
−2c
32
θ
2
u
1
+ c
32
θ
2
2
u
2
1
)(c
11
+ c
31
) =
q(c
12
+ 2c
32
−2c
32
θ
2
u
1
+ qc
32
)(c
11
+ c
31
) + (c
12
+ c
32
+ θ
2
u
1
(c
12
+ c
32
+ c
32
(1 −θ
2
u
1
)))(c
11
+ c
31
)
Define E(u
1
, θ
2
, q) = (qc
32
+c
12
+2c
32
(1 −θ
2
u
1
))(c
11
+c
32
) so the right side may be written
qE(u
1
, θ
2
, q) + (c
12
+ c
32
+ θ
2
u
1
(c
12
+ c
32
+ c
32
(1 −θ
2
u
1
)))(c
11
+ c
31
)
Similarly for the left side
S
01
(c
12
+ c
32
)(1 −u
1
)(c
11
+ c
31
(1 −u
1
)) =
S
01
(c
31
u
2
1
−(c
11
+ 2c
31
)u
1
+ c
11
+ c
31
)(c
12
+ c
32
)
Now define
A
1
= S
01
(c
12
+ c
32
)c
31
B
1
= −S
01
(c
12
+ c
32
)(c
11
+ 2c
31
)
therefore (4.13) becomes
A
1
u
2
1
+ B
1
u
1
+ S
01
(c
11
+ c
31
)(c
12
+ c
32
) −(c
12
+ c
32
+ θ
2
u
1
(c
12
+ c
32
+ c
32
(1 −θ
2
u
1
)))(c
11
+ c
31
)
= qE(u
1
, θ
2
, q)
So at last define
C
1
= (S
01
−1)(c
12
+ c
32
)(c
11
+ c
31
)
Q(u
1
, θ
2
) = A
1
u
2
1
+ B
1
u
1
+ C
1
+ θ
2
u
1
(c
12
+ c
32
+ c
32
(1 −θ
2
u
1
))(c
11
+ c
31
)
and finally one obtains (4.12)
Q(u
1
, θ
2
) = qE(u
1
, θ
2
, q)
40
The boundary between infected host exclusion or invasion is given by condition 2 (q =
1 −θ
2
u
1
−z
2
< 0), exactly when q = 1 −θ
2
u
1
−z
2
= 0. In that case (4.12) becomes
Q(u
1
, θ
2
) = 0.
Define now ¯ u
1
as the value of u
1
such that Q(u
1
, θ
2
) = 0 then substitute ¯ u in (4.11) to obtain
c
51
=
c
31
¯ u
2
1
−(c
11
+ c
31
−c
21
)¯ u
1
¯ u
1
(1 −c
31
) + c
11
(4.14)
Define ¯ c
51
as that value of c
51
.
Now the threshold value ( R
∗
01
) is given by the value of R
01
given by this value of ¯ c
51
.
Stated that u
1
∈ [0, 1] it’s possible to determine the existence and position of the threshold
R
∗
01
by comparing the values of Q at the end points of h
1
.
Since
Q(0, θ
2
) = (S
01
−1)(c
11
+ c
31
)(c
12
+ c
32
)
Q(1, θ
2
) = (θ
2
−1)(c
12
+ c
32
(1 −θ
2
))
the study of the properties of the solutions can be split in the cases depending on the strength
of direct competition.
So, suppose that species one is more tolerant, or equivalently S
01
> 1, and that its direct
competitive force against hot two is low, θ
2
< 1.
Then it’s easy to check that Q(0, θ
2
), Q(1, θ
2
) have different signs, since all dimensionless
coefficients are positive. Therefore must exists a unique root (¯ u
1
) for Q(u
1
, θ
2
) = 0 in
u
1
∈ [0, 1]. Substituting in equation (4.11) this root value for u
1
gives
c
51
=
¯ u
1
(c
31
¯ u
1
−(c
11
+ c
31
−c
21
))
c
11
+ 2c
31
then there is an unique value of c
51
and hence of R
01
, provided that (c
11
+ c
31
− c
21
) < 0,
that is exploiting the dimensionless coefficients s
1
< s
1
+ r
1
/k
1
< h
1
.
If, however, s
1
< h
1
< s
1
+r
1
/k
1
there is a positive lower bound on the possible values of u
1
and if the root value calculated before is lower there is not any threshold R
∗
01
.
If S
01
< 1 then both value of Q in the end points are negative, so there may be no so-
lution for Q(u
1
, θ
2
) = 0 unless the curvature of Q is negative. In that case a double root
is generated and there will be two value of R
∗
01
, equivalently two threshold. The curvature
Cur of Q is measured by
Cur = A
1
−θ
2
2
(c
11
+ c
31
)c
32
so to be negative, exploiting the dimensionless coefficients, the ratio S
01
/θ
2
2
has to be suffi-
ciently small.
The result is that S
01
> 1 seems to be a necessary condition to have host exclusion even if
there are some special cases when exclusion does not occur.
41
As one would expect increasing θ
2
increases the value of Q over the feasible interval of
u
1
, so, if it exists, the region of host 2 exclusion would expand. This could be noted by
looking at Q(u
1
, θ
2
) = A
1
u
2
1
+ B
1
u
1
+ C
1
+ θ
2
u
1
(c
12
+ c
32
+ c
32
(1 −θ
2
u
1
))(c
11
+ c
31
).
Recall
Q(0, θ
2
) = (S
01
−1)(c
11
+ c
31
)(c
12
+ c
32
)
Q(1, θ
2
) = (θ
2
−1)(c
12
+ c
32
(1 −θ
2
))
If 1 < θ
2
< 1 + c
12
/c
32
and the less tolerant species is host one (S
01
< 1) then Q(0, θ
2
),
Q(1, θ
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 but
enduring a stronger competition by host two.
On the other hand if apparent competition against host two is strong (S
01
> 1) then both
value for Q are positive and the equilibrium is stable unless the curvature is negative, again.
In that case there are two roots of Q(u
1
, θ
2
) = 0 generating two value of R
∗
01
between which
the equilibrium is unstable.
If competition is particularly strong θ
2
> 1 + c
12
/c
32
then Q(1, θ
2
) becomes negative, hence
there is a root in the interval 1/v
2
< u
1
< 1 since substituting u
1
= 1/v
2
in Q(u
1
, θ
2
)
still gives a positive value. However any root in this interval is not relevant since z
2
be-
comes negative due to the definition of the inequality slack q. By the way, over the interval
0 < u
1
< 1/v
2
, when v
2
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
invasion
z
2
< 1 −θ
2
u
1
(4.15)
z
2
is implicitly defined by (4.9) that is possible to write as
F(z
2
) :=
z
2
(c
12
+ c
32
z
2
)
c
12
+ c
32
=
S
01
(1 −u
1
)(c
11
+ c
31
(1 −u
1
))
c
11
+ c
31
(4.16)
similarly as was done before for (4.13).
F(z
2
) is clearly an increasing function, so one can apply it at (4.15) to obtain
F(z
2
) < F(1 −θ
2
u
1
) (4.17)
Then by (4.16) and the definition of F one has that
S
01
(1 −u
1
)(c
11
+ c
31
(1 −u
1
))
c
11
+ c
31
<
(1 −θ
2
u
1
)(c
12
+ c
32
(1 −θ
2
u
1
))
c
12
+ c
32
(4.18)
Hence
S
01
(1 −u
1
)(c
11
+ c
31
(1 −u
1
))(c
12
+ c
32
) −(1 −θ
2
u
1
)(c
12
+ c
32
(1 −θ
2
u
1
))(c
11
+ c
31
) < 0
42
and this is exactly Q(u
1
, θ
2
) < 0. As stated previously there is a unique positive root (¯ u
1
)
for Q(u
1
, θ
2
) = 0 if S
01
> 1 and θ
2
< 1. Substituting in equation (4.11) this root value for
u
1
gives ¯ c
51
=
¯ u
1
(c
31
¯ u
1
−(c
11
+c
31
−c
21
))
c
11
+2c
31
then there is an unique value of c
51
and hence of R
01
,
provided that (c
11
+ c
31
−c
21
) < 0.
Condition R
01
> R
∗
01
imply c
51
< ¯ c
51
therefore u
1
< ¯ u
1
by the implicit function theorem
applied to (4.11). Moreover Q(0, θ
2
) > 0, hence Q(u
1
, θ
2
) > 0 and as a consequence host two
can not invade. However, as stated above if θ
2
= 0 or sufficiently small, if S
01
< 1 then Q is
negative, then invasion is feasible.
Figured out all these threshold it’s possible to summarize them on maps of Figure 4.1 (in
which 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 the
cross section of parameter space defined by the inverse s of the basic reproductive numbers R
0
i
−1
. The shaded are indicates
feasibility, the numbers refers to how many equilibria of that type are present in state space and how many of these are stable.
Coefficient values: c
11
= 1.75, c
12
= 1.8, c
21
= 2.19, c
22
= 2.25, c
31
= 0.25, c
23
= 0.2, c
40
= 2, c
00
= 1, S
01
= 2, θ
i
= 0
R
∗
= R
−1
01
To a better understanding of the bifurcation pattern shown refer to Greenman and Hudson
43
[17]. For our purpose is sufficient to have a brief description of the threshold structure. The
maps are constructed on the cross section of parameter space defined by the inverses of R
0i
to
ensure that the region of parasite persistence is next to the origin, because condition R
0i
> 1
is necessary for parasite persistence. Figure 4.1(a,b) picture the infected equilibrium which
describe 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 in
Figure 4.1 (b) because the parameter’s choice gives S
01
> 1 and therefore S
02
< 1. If S
0i
→1
then R
∗−1
0i
moves from coincidence with the line R
−1
0i
= 1 to coincidence with the vertical
axis. If S
0i
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 species
coexist with each other and the parasite. The shaded area is now the feasibility region for
the infected coexistence equilibrium and its feasibility region is bounded by the line R
∗
and
the upper branch of the hyperbola (4.4). The infected equilibrium is stable when feasible.
Figure 4.2: Host two exclusion when δ is under
stability threshold for single species host one en-
demic equilibrium
Figure 4.3: Host two exclusion when δ is above
stability threshold for single species host one en-
demic equilibrium
These conjectures about the feasibility and stability structure of the system’s 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 if
diffusion 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 (α) that it’s 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. S
01
> 1 : verified under the hypothesis that α
2
= 2α
1
while all other parameters are
equals between host 1 and host 2
44
2. R
01
> R
∗
01
> 1 : it’s possible to vary R
01
keeping R
∗
01
fixed just increasing or decreasing
the value of δ, H
K1
or β.
An interesting feature of the single host system was that the infected equilibrium may be
stable or unstable depending on larvae natural mortality (δ), keeping fixed all others param-
eters. Since increasing the value of δ has a stabilizing effect on the single species equilibrium
but simultaneously decrease the value of the basic reproductive number it would be less
probable to obtain coexistence when both single host infected equilibrium are unstable.
As expected exclusion of host 2 doesn’t 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 < θ
i
< 1 with i = 1, 2) the system will tend to species coexistence,
when one or both are strong (θ
i
> 1 with i = 1, 2) one species is excluded.
In our models both competing forces are present and not surprisingly the analysis becomes
extremely difficult. Moreover the system may tend to very different solutions. Fortunately
enough the work of Greenman and Hudson [17] again provides some conjectures on system
behaviour. 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 numerical
simulations 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 is
algebraically intractable, but its properties seem related to the equilibria representing single
host coexistence with the parasite and uninfected hosts coexistence. Greenman and Hudson
achieved a brilliant result describing how apparent and direct competition interact.
Their result are summarized in Figure 4.4 and are indexed by direct competition indices
θ
1
(horizontal), θ
2
(vertical) from weak to strong competition. Each cell in the array cor-
responds to a particular choice of the indices θ
1
, θ
2
and contains the maps for the three
infected equilibria for this choice of index values.
The bifurcation maps displayed are obviously incomplete, since all the regions where there is
more than one equilibrium of a given type are left out, but is still a first step to understand
system 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 θ
1
increase the vertical asymptote of the hyperbola moves to
the vertical axis, the horizontal asymptote moves to the horizontal line R
o2
= 1 and the
hyperbola collapses onto its asymptotes. Recall that the hyperbola is the boundary for the
45
stability of uninfected coexistence equilibrium and for the feasibility of the infected coexis-
tence equilibrium. The vertical line L
1
= R
∗
01
, namely the threshold condition for host two
exclusion, stays fixed while L
2
= R
∗
02
moves to R
02
= 1, if they exist.
Figure 4.4: An example of bifurcation maps for the model with direct and apparent competition. Maps are indexed by direct
competition indices θ
1
(horizontal), θ
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 and
unstable components.
Suppose now a weak competition against host one (Fig 4.4(g)(h)) and a strong competition
46
Figure 4.5: Example of behaviour for Map a of
Figure 4.4
Figure 4.6: Example of behaviour for Map c of
Figure 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 L
1
and L
2
intersect and then there is fragmentation of the feasible region of the infected coexistence
equilibrium. 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 a
stable infected coexistence equilibrium.
47
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 the
threshold but not the structure.
In the absence of direct competition the more tolerant species will exclude the other if its
parasite productivity is sufficiently high (condition R
0i
> R
∗
0i
> 1). That means the host
has to be a reservoir for parasite without being too much affected by their presence and
simultaneously allow the parasite to spread enough to destabilize the less tolerant host. If
the two host directly compete host exclusion inversion may happen when the competitive
force 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 be
important 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 host
exclusion and to determine some criterion for this to happen. If one focus only on direct
competition it has been already shown that the presence of refugee patches may avoid hosts
exclusion [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 the
presence of a more favorable habitat for the species. The following discussion will focus only
on the two host interaction since system dynamic for a single host has been already studied
and 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. S
01
> 1
2.
_
h
i
ψ
i
σ
i
+ α
i
+ b
i
−1
__
β
i
H
Ki
δ
_
= R
01
> 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 more larvae because of the
presence of larvae laid by the reservoir species. Then he supposes that the recruitment rate
depend somehow on diffusion rate and habitat overlap to obtain an estimate of how great
spatial separation between the two species has to be to allow persistence of both of them.
Define
¯
β as the value of β such that one of the conditions does not hold anymore, so exclusion
of the less tolerant host is avoided,
Under such hypothesis, the level of spatial separation at which the model predicts host
coexistence is the proportion between the given value of β
2
and
¯
β. Therefore first determine
the value of
¯
β as the value of β
2
(the recruitment rate) such that S
01
= 1, (remember to check
that condition R
1
> R
∗
> 1 holds, too) then compute the proportion between this value and
48
Figure 4.8: Spatial segregation defined as 1 −
p
q
the value of β
2
used in the model. If the model predicted host exclusion, obviously the β
2
exclusion value is proportionally bigger than
¯
β by definition of S
01
. That proportion could
be interpreted as a rough estimation of the spatial separation needed to allow coexistence.
Note that there is a boundary value for both β
i
, but to find a threshold value for one β
i
one
has to fix all other parameters, so there is an implicit clumsy hypothesis that one ingestion
rate is affected by habitat overlapping while the other is not.
It’s possible to use numerical simulations to test if this prediction is correct or if it is a too
wide approximation. To do so one define the value of overlapping habitat as the proportion
between the shared habitat (p) and the total habitat (q) covered by hosts. Therefore spatial
segregation is 1 −
p
q
. The results do not fit the conjecture. The threshold values of spatial
Table 4.2: Spatial separation needed to have hosts coexistence. Parameters are: d
i
= 0.5, α
1
= 0.05, σ
i
= 4,
β
1
= 0.12, φ
i
= 1., h
i
= 100, v
i
= 0.00083, δ = 6, b
i
= 0.6.
α
2
Simulations β
2
Conjecture
0.1 74 % 49 %
0.15 78 % 65 %
0.2 80 % 74 %
separation given by this conjecture differs from the values computed by simulations, as one
can see in the example given in Tab. 4.2. Maybe a deeper knowledge on how the recruitment
rate is related to spatial diffusion may help to improve this conjecture.
Having no clue on system behaviour when spatial dependency is introduced one runs exten-
sive numerical simulations varying host habitat in order to establish if apparent competition
is the cause of host exclusion and if spatial separation plays a role in it. It has to be said
that the exact parameter’s value is a key factor for determining system behaviour; however
49
Figure 4.9: The fertility rate b
1
(x) = 2.5e
−
(x−3)
2
4 as host 1 habitat
for this case study there are not enough biological data to give reasonable values. Therefore
a more general approach is given than a prediction on what may happen between Black
Grouse and Rock Partridge.
First define the space dependent natural density dependent fertility rate b
i
(H
i
, x, t) in order
to model the habitat of host. Figure 4.9.
Suppose then that host two has the same fertility rate and the same parameters value as
host one, the only difference being a higher induced mortality due to parasites (α). Consider
the case where both hosts share the same habitat (identical fertility rate) and both species
when alone can coexist with the parasite. For the sake of clarity suppose that direct com-
petition is absent. Now define parameter values to have host two extinction. Obviously a
Figure 4.10: Host two is driven to extinction when there is a total overlapping of habitats. Parameters are
defined as follow: d
i
= 0.5, α
1
= 0.05, α
2
= 0.2, σ
i
= 4, β
i
= 0.12, φ
i
= 1., h
i
= 100, v
i
= 0.00083, δ = 60,
b
i
(x) = 2.5e
−
(x−3)
2
4
with i = 1, 2
50
different choice may let the two hosts to coexist, then spatial separation will only improve
host persistence. In this case, to have host exclusion, one has to improve the detrimental
Figure 4.11: The shaded area is the fertility rate b
1
(x) = 2.5e
−
(x−3)
2
4 as host 1 habitat, while the other line is host two
fertility rate b
2
(x) = 2.5e
−
(x−4)
2
4
effects of parasite on less tolerant host by increasing or parasite induced mortality on host
two or the total amount of larvae laid by host one.
Note that the parameter values used are such that for host one the single-host infected equi-
librium is stable. The next step is to change the distance between species habitat (Figure
4.11) in order to establish whether host exclusion may take place and if it’s possible to give
or conjecture a criterion to predict when it will happen.
One would expect that moving away host one habitat may reduce the larvae burden on
host two allowing the two species to coexist and, in fact it does. It is important to note
Figure 4.12: Host two coexists with host one when there is a partial overlapping of habitats. The graph displays the value
of population peak. Parameters are defined as follow: d
i
= 0.5, α
1
= 0.05, α
2
= 0.2, σ
i
= 4, β
i
= 0.12, φ
i
= 1., h
i
= 100,
v
i
= 0.00083, δ = 60, with i = 1, 2 and b
i
(x) as (4.11)
51
(a) Host two, when host one is present
(b) Host two, when other host is absent
Figure 4.13: Different final distribution for host two, the less tolerant.
52
that, even if spatial separation between the two hosts permits coexistence, a detrimental ef-
fect on host two is still present. Apparent competition pushes the less tolerant species away
from other host and its parasite burden, modifying the final distribution. It’s then possible
to state that a change in the most favorable habitat occurs due to the increased presence of
free living stages laid by the reservoir host.
Figure 4.14 displays the value of spatial segregation necessary to avoid host exclusion in
four cases. On the y-axis there are the average growth rate for host two, that not surpris-
ingly displays even negative values. As explained in previous chapter this is because, even
if host two may have a positive growth rate only in a small portion of the domain this may
be sufficient for its survival.
Figure 4.14: The growth rates are m
2
(x) = Ae
−
(x−3)
2
4 −0.5 and m
1
(x) = 2.5e
−
(x−B)
2
4 −0.5. Changing coefficient value A
changes the growth rate for less tolerant host, changing coefficient value B changes habitat for the other host.When fertility re-
duction is present the growth rate is m
2
(x) = Ae
−
(x−3)
2
4
“
kH
2
kH
2
+(1−γ
2
)P
2
”
k
−0.5 or m
1
(x) = 2.5e
−
(x−B)
2
4
“
kH
1
kH
1
+(1−γ
1
)P
1
”
k
−
0.5.
Parameters are defined as follow: d
i
= 0.5, α
1
= 0.05, α
2
= 0.2, σ
i
= 4, β
i
= 0.12, φ
i
= 1.,
h
i
= 100, v
i
= 0.00083, δ = 60 with i = 1, 2.
53
In all cases except for ”Equal Growth Rate” the fertility rate of host one is fixed:
b(x) = Ae
−
(x−B)
2
4
_
kH
2
kH
2
+(1−γ
2
)P
2
_
k
, but for parameter B that allows to shift the habitat. A is
fixed on 2.5. On the other hand for host two, the less tolerant, B = 3 while A varies.
The values of fertility reduction are:
γ
1
= 0.9, γ
1
= 1: in the ”Fertility reduction on less tolerant host” case,
γ
1
= 0.8, γ
1
= 0.95: in the ”Fertility reduction on both hosts” case.
γ
1,2
= 1: in the other two cases,
As one shall immediately notice decreasing the average growth rate of host two imply a
minor habitat overlap to be sufficient to cause extinction.
Fertility reduction, too, if affects only the less tolerant host is a powerful mechanism that
enforces host exclusion.
The infected coexistence equilibrium may be stable or unstable depending on the stabil-
ity of host one infected equilibrium. The stability of this last one depends on the maximum
value of fertility, the habitat shape and the strength of destabilizing forces.
54
Chapter 5
Results
One Host
The study of single host-parasite system highlighted the role of spatial heterogeneity and
helped to identify which biological features are relevant to stabilize or destabilize host par-
asite coexistence. This has been done mostly using numerical simulations.
It has been observed that fertility reduction is a powerful destabilizing force and when fer-
tility is space dependent, under certain condition, can cause spatial instability, too.
In the space independent case, the larvae natural mortality is a key parameter. Generally
an increase of larvae mortality matches an increase of stability for the infected equilibrium;
the exact value of the threshold depending on the value of parasite aggregation (k).
When the infected equilibrium is unstable solutions tend to oscillate in time.
With spatial dependence, habitat is interpreted as the region where local growth rate is
positive and the importance of where is located has been explicitly shown for a simple case.
Under the condition of a strictly reflecting boundary the result is that the ends of the spatial
domain are the most favorable patches. If the local growth rate is positive in some places
but negative on the average the mechanism to a critical patch size is related to the proximity
of bad habitat to good. This is due to the chance of dispersal into unfavorable region. Since
the diffusion coefficient plays a role in letting individuals effectively average the local growth
rate, under the condition that the average growth rate is negative, increasing the diffusion
coefficient may lead to host extinction.
Then, studying the stability of the uninfected equilibrium condition, we succeeded in defin-
ing a condition for parasite establishment (Proposition 3). Using Sturm-Liouville theory, it
has been pointed out that parasite establishment will occur if the eigenvalues of the parasite
and larvae equations linearized in the uninfected equilibrium are negative.
The stability of infected equilibrium depends on parameter value. However is still possible to
identify some characteristic patterns. For instance an higher value of aggregation correspond
to an higher stability threshold for the free-living stages mortality rate. A stronger fertility
reduction may lead to instability, even spatially if fertility rate is sufficiently high and only
small regions have a negative local growth rate, so spatial oscillation are not dampen out.
55
Two Hosts
Although algebraic intractability constantly arise in these kind of models the behaviour and
the underlying mechanisms that affects systems dynamic of these two-host shared macropar-
asite systems are understood. The conclusions highlighted for a single host-parasite interac-
tion hold even when another host is introduced. The only difference being that the result of
the interaction between the more tolerant host and the parasite influences the behaviour of
the less tolerant host.
It has been shown that apparent competition may play a major role in host exclusion and
that spatial separation between species may prevent this phenomena. Moreover it is now
clear that various forces contribute to determine the spatial spread necessary. Further study
on finding an analytical threshold for this has to be made, however all the simulation showed
the same pattern. If the common pool of infection exclude the less tolerant species when
habitats are overlapping hosts coexistence may nevertheless arise if one species moves suffi-
ciently away. Therefore the driving force behind host exclusion is the presence of a reservoir
host and not the shared parasite by itself.
The result of this study suggests that, in the specific interaction between rock partridge and
black grouse held as case study, the actual space separation between the two species may be
sufficient to avoid one species extinction.
At the moment there are not accurate parameters estimations for the magnitude of parasite
deleterious effects on hosts, neither for other key parameters. A precise prediction on what
would happen if the habitats will start to overlap is not possible, yet. However the risk
that a shared macroparasite could act as a deadly competing weapon, if spatial separation is
lost, could not be dismissed and preservative efforts should be made to maintain the actual
species habitat.
56
Appendix
Program
C++ code. Using CRANK-NICHOLSON Algorithm as suggested on the book by Burden
and Faires [11]
#include <fstream>
#include <iomanip>
#include <iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
#define true 1
#define false 0
/*Functions declaration*/
double F1(double X);
double NL1(double H1,double P1,double H2,int I,double N);
double F2(double X);
double NL2(double H2,double P2,double H1,int I,double N);
double PF1(double X);
double PNL1(double H1,double P, double H2, double L);
double PF2(double X);
double PNL2(double H2,double P,double H1, double L);
double LF(double X);
double LNL(double H,double P, double L);
void INPUT(int *, double *, double *, double *, double *, int *, int *);
void OUTPUT(double, double, int, double *, double, double *, double *);
57
main()
{
/*Variables declaration*/
double V1[10000], L1[10000], U1[10000], Z1[10000];
double PV1[10000], PL1[10000], PU1[10000], PZ1[10000];
double V2[10000], L2[10000], U2[10000], Z2[10000];
double PV2[10000], PL2[10000], PU2[10000], PZ2[10000];
double LV[10000], LZ[10000];
/*Coefficients declaration*/
double FT,FX,D1,D2,H,K,lambda1,lambda2,T,X,prob,TL,KL,a,check,check2,ctrl,ctrl2;
int N,M,N1,N2,M1,FLAG,I1,I,J;
int OK;
double beta1,phi1, h1, beta2,phi2, h2, delta;
/* Define values for Larvae Equation */
h1= ; beta1= ; phi1= ;
h2= ; beta2= ; phi2= ;
delta= ;
/*Call to input function*/
INPUT(&OK, &FX, &FT, &D1, &D2, &N, &M);
if (OK) {
/* Index */
N1 = N - 1;
N2 = N - 2;
/* STEP 1: Grid Building */
H = FX / N;
K = FT / M;
lambda1 = (D1 * K) / ( H * H );
lambda2 = (D2 * K) / ( H * H );
/* set dV(N)/dx = dV(0)/dx=0 boundary condition*/
V1[N] = F1(N1*H);
V1[0] = F1(H);
PV1[N] = PF1(N1*H);
PV1[0] = PF1(H);
/*then for host and parasite 2, too.*/
/* STEP 2 : Initial condition*/
for (I=1; I<=N1; I++) {V1[I] = F1(I*H);}
for (I=1; I<=N1; I++) {PV1[I] = PF1(I*H);}
/*then for host 2, parasite 2 and larvae, too*/
58
/*Saving on file initial condition, just one point on the grid */
char file1[30];
cout << "Enter the name of the file where you want to save data HOST ONE: ";
cin >> file1;
ofstream HOST_ONE(file1, ios::app);
HOST_ONE <<" H_1 "<< " P_1 " <<" L "<<endl;
HOST_ONE <<V1[25]<< " "<<PV1[25]<<" " <<LV[25]<<endl;
/*Saving on file initial condition, or all the spatial domain */
char file1[30];
cout << "Enter the name of the file where you want to save data HOST ONE: ";
cin >> file1;
ofstream host1_3d(file1, ios::app);
for (I=0; I<=N; I++) {
host1_3d << fixed << setprecision( 8 )<<V1[I]<<" ";}
host1_3d <<endl;
/* Same for all other species*/
/* STEP 3: time iteration, code only for host 1 parasite 1 and larvae */
for (J=1; J<=M; J++) {
/* current t(j) */
T = J * K;
/* STEPS 4: Solve a tridiagonal linear system using Algorithm 6.7 */
/* Host 1 */
L1[0] = 1.0 + lambda1/2.0 - K*NL1(V1[1],PV1[1],V2[1],1,H)/2.0;
U1[0] = -lambda1 / ( 2.0 * L1[0] );
/* STEP 5 */
for (I=2; I<=N2; I++) {
L1[I-1] = 1.0 + lambda1 - K*NL1(V1[I],PV1[I],V2[I],I,H)/2.0 +
lambda1 * U1[I-2]/ 2.0;
U1[I-1] = -lambda1 / ( 2.0 * L1[I-1] ); }
L1[N2] = 1.0 + lambda1/2.0 - K*NL1(V1[N1],PV1[N1],V2[N1],N1,H)/2.0 +
59
lambda1 * U1[N2-1]/2.0;
Z1[0] = ((1.0 -lambda1/2.0 + K*NL1(V1[1],PV1[1],V2[1],1,H)/2.0)*V1[1]+
lambda1*V1[2]/2.0)/L1[0];
for (I=2; I<=N2; I++){
Z1[I-1] = ((1.0 - lambda1 + K*NL1(V1[I],PV1[I],V2[I],I,H)/2.0)*V1[I] +
0.5*lambda1*(V1[I+1]+V1[I-1]+Z1[I-2]))/L1[I-1];}
Z1[N2] = ((1.0 - lambda1/2.0 + K*NL1(V1[N1],PV1[N1],V2[I],N1,H)/2.0)*V1[N1] +
0.5*lambda1*(V1[N2]+Z1[N2-1]))/L1[N2];
V1[N1] = Z1[N2];
for (I1=1; I1<=N2; I1++) {
I = N2 - I1 + 1;
V1[I] = Z1[I-1] - U1[I-1] * V1[I+1];}
V1[0]=V1[1];
V1[N]=V1[N1];
/*Parasite 1*/
PL1[0] = 1.0 + lambda1/2.0 - K*PNL1(V1[1],PV1[1],V2[1],LV[1])/2.0;
PU1[0] = -lambda1 / ( 2.0 * PL1[0] );
for (I=2; I<=N2; I++) {
PL1[I-1] = 1.0 + lambda1 - K*PNL1(V1[I],PV1[I],V2[I],LV[I])/2.0+
lambda1 * PU1[I-2]/ 2.0;
PU1[I-1] = -lambda1 / ( 2.0 * PL1[I-1] ); }
PL1[N2] = 1.0 +( lambda1 - K*PNL1(V1[N1],PV1[N1],V2[N1],LV[N1]))/2.0 +
lambda1 * PU1[N2-1]/2.0;
PZ1[0] = ((1.0-lambda1/2.0 + K*PNL1(V1[1],PV1[1],V2[1],LV[1])/2.0)*PV1[1]+
lambda1*PV1[2]/2.0 + K*beta1*phi1*LV[1]*V1[1])/PL1[0];
for (I=2; I<=N2; I++){
PZ1[I-1] = ((1.0 - lambda1 + K*PNL1(V1[I],PV1[I],V2[I],LV[I])/2.0)*PV1[I]+
0.5*lambda1*(PV1[I+1]+PV1[I-1]+PZ1[I-2])+K*beta1*phi1*LV[I]*V1[I])/PL1[I-1];}
PZ1[N2] = ((1.0 - lambda1/2.0 + K*PNL1(V1[N1],PV1[N1],V2[N1],LV[N1])/2.0)*
PV1[N1]+ 0.5*lambda1*(PV1[N2]+PZ1[N2-1])+K*beta1*phi1*LV[N1]*V1[N1])/PL1[N2];
PV1[N1] = PZ1[N2];
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for (I1=1; I1<=N2; I1++) {
I = N2 - I1 + 1;
PV1[I] = PZ1[I-1] - PU1[I-1] * PV1[I+1];
}
PV1[0]=PV1[1];
PV1[N]=PV1[N1];
/*Larve*/
for (I=0; I<=N; I++) {
LZ[I] = LV[I];
LV[I] = (LZ[I]*(2.0 - K*beta1*V1[I] - K*beta2*V2[I] - K*delta) +
2.0*K*(h1*PV1[I]+h2*PV2[I]))/(2.0+K*beta1*V1[I]+K*beta2*V2[I]+K*delta);
}
for(I=1;I<=N1;I++){
double test;
test=(K*NL1(V1[I],PV1[I],V2[I], I,H)-lambda1*U1[I])/2.0 -1.0;
if(lambda1<test) {
check=1.;
printf("Host Tridiagonal system condition violated.\n");}}
if(check==1.) break;
for(I=1;I<=N1;I++){
double testp;
testp= (K*PNL1(V1[I],PV1[I],V2[I],LV[I])-lambda1*PU1[I])/2.0 -1.0;
if(lambda1 < testp) {
check=2.;
printf("Parasite Tridiagonal system condition violated.\n");}}
if(check2==2.) break;
for(I=1;I<=N1;I++){
double testl;
testl=LV[I]*(2.0 - K*beta1*V1[I] - K*beta2*V2[I] - K*delta) +
2.0*K*(h1*PV1[I]+h2*PV2[I]);
if(testl< 0.) {
check=3.;
printf("Larve condition violated.\n");}}
if(check==3.) break;
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/*STEP 5: Save results of iteration*/
if (J % 100==0 ){
/*one point of the space domain*/
HOST_ONE << fixed << setprecision( 8 )<<V1[25]<<" "<< PV1[25] <<
" "<<LV[25]<<endl;
/*the whole space domain*/
for (I=0; I<=N; I++) {
host_1 << fixed << setprecision( 8 )<<V1[I]<<" ";}
host_1 << endl;
}
/* STEP 6: Call output function */
OUTPUT(FT, X, N, V1, H, PV1,LV);
}
/* STEP 12: End */
return 0;
}
/* Change HOST initial condition for a new problem */
double F1(double X)
{
double f;
f= ;
return f;
}
/* Similar for others*/
/* Define host function nl=b-d-v*H1-mu*H2-alfa*P1/H1 */
double NL1(double H1, double P1,double H2, int I, double N )
{
double d,v, alfa,b,nl,gamma,k,mu;
k= ; gamma=; b= ; d= ; v= ; alfa= ; mu= ;
if(H1<=0.) nl=0.;
else {
if(I*N /*set condition if wanted*/) nl= ;;
else nl= ;}
return nl;}
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/* Define host function nl2 */
/* Define parasite function PNL for a new problem */
double PNL1(double H1,double P1,double H2, double L)
{
double b, phi,sigma,alfa,k,pnl,beta,d,v,mu;
sigma= ; alfa= ; k= ; d= ; v= ; mu= ;
if(H1<=0.) pnl=0.;
else pnl= -sigma-d-v*H1-mu*H2-alfa-alfa*(1.+1./k)*P1/H1;
return pnl;}
/* Define parasite function pnl2 */
/*Function input permits to define space and time domain,
the steps and the diffusion coefficients*/
void INPUT(int *OK, double *FX, double *FT, double *D1, double *D2, int *N, int *M)
{
int FLAG;
char AA;
\
printf("Using the Crank-Nicolson Method.\n");
printf("Ready to start? Answer Y or N.\n");
scanf("\n%c", &AA);
if ((AA == ’Y’) || (AA == ’y’)) {
printf("The lefthand endpoint on the X-axis is 0.\n");
*OK =false;
while (!(*OK)) {
printf("Input the righthand endpoint on the X-axis.\n");
scanf("%lf", FX);
if (*FX <= 0.0)
printf("Must be positive number.\n");
else *OK = true;
}
*OK = false;
while (!(*OK)) {
printf("Input the maximum value of the time variable T.\n");
scanf("%lf", FT);
if (*FT <= 0.0)
printf("Must be positive number.\n");
else *OK = true;
63
}
printf("Input the diffusion coefficient for HOST ONE:\n");
scanf("%lf", D1);
*OK = false;
printf("Input the diffusion coefficient for HOST TWO:\n");
scanf("%lf", D2);
*OK = false;
while (!(*OK)) {
printf("Input integer N = number of intervals on X-axis, at least 2\n");
scanf("%d",N);
printf("and M = number of time intervals.\n");
scanf("%d", M);
if ((*N <= 1) || (*M <= 0))
printf("Numbers are not within correct range.\n");
else *OK = true;
}
}
else {
printf("The program will end so that the function F can be created.\n");
*OK = false;
}
}
/*funtion output save the last iteration*/
void OUTPUT(double FT, double X, int N, double *V1, double H,
double *PV1, double *LV)
{
int I, J, FLAG;
char NAME[30];
FILE *OUP;
printf("Choice of output method:\n");
printf("1. Output to screen\n");
printf("2. Output to text file\n");
printf("Please enter 1 or 2.\n");
scanf("%d", &FLAG);
if (FLAG == 2) {
printf("Input the file name in the form - drive:name.ext\n");
printf("for example: A:OUTPUT.DTA\n");
64
scanf("%s", NAME);
OUP = fopen(NAME, "w");
}
else OUP = stdout;
fprintf(OUP, " I X H P L\n", FT);
for (I=0; I<=N; I++) {
X = I * H;
fprintf(OUP, "%3d %11.8f %13.8f %13.8f %13.8f\n", I, X, V1[I], PV1[I], LV[I]);
}
fclose(OUP);
}
65
66
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