Jordy van der Hoorn

The drift paradox in population

dynamics:

Conditions for population persistence and travelling waves

Master thesis

Thesis advisor: Dr. S.C. Hille

Date master exam: 28 September 2016

Mathematisch Instituut, Universiteit Leiden

Acknowledgements

First of all I wish to express my deepest appreciation and gratitude to my super-visor Dr. S.C. Hille from the Institute of Mathematics in Leiden. I want to thankhim for his guidance, critical comments and warm encouragement throughoutthe process of writing this thesis. Despite some hard times this thesis forced meto face, specially at the end of the thesis, I really appreciated working underhis supervision. Specially the humoristic moments when we were busy trying toget a good visual view of our 3-dimensional system.

The second person I really want to thank is Dr. O.W. van Gaans, also fromthe Institute of Mathematics in Leiden, who is always willing to take time foranswering my questions. The many discussions and suggestions that he raisedup were very helpful during writing this thesis.

Finally, the support of my family and friends has been invaluable to me. Itis safe to say that this thesis would not have been possible without the supportand help of all those people.

Preface

This thesis is founded on the paper [Pachepsky], which discusses the ”drift para-dox”: how can a population, living in a stream, survive without being washedaway? Initially it was the plan to elaborate on the paper in the direction ofinteracting populations living together in a stream. However, when startingwith a detailed study of [Pachepsky], several questions arose - same relating tomistakes or misprints, others to more mathematically oriented, eg. whether onecould rigorously prove the existence of travelling waves/ population fronts.

In this master thesis the reader will therefore find elaborations in detail onmathematical results that were simply stated in [Pachepsky] without proof, orthat provide corrections to results in this paper. The study of the travellingwave problem resulted in a novel approach to study the existence of a hetero-clinic orbit in the three dimensional system, by considering regions in a statespace and arguments for exclusion of existence of trajectories from one region toanother. A graphical representation of these relations allow to draw conclusionson possible behaviours in the 3D system.

Contents

Contents 6

1 Introduction 81.1 Outline. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 The model 11

3 Population persistence 133.1 Population persistence for µ ≥ 1 . . . . . . . . . . . . . . . . . . 13

3.1.1 Equivalence of two initial value problems . . . . . . . . . 153.1.2 A series expansion for the unforced system . . . . . . . . 163.1.3 A series expansion for the forced system . . . . . . . . . . 253.1.4 Condition for population persistence . . . . . . . . . . . . 27

3.2 Population persistence for µ < 1 . . . . . . . . . . . . . . . . . . 30

4 Persistence or extinction through a moving population front 334.1 Idealized model equations and homogeneous steady states . . . . 334.2 Travelling wave equations . . . . . . . . . . . . . . . . . . . . . . 344.3 Necessary conditions for the existence of a heteroclinic orbit by

analysing the phase portraits. . . . . . . . . . . . . . . . . . . . . 364.4 Dimensionality of the stable and unstable manifolds. . . . . . . . 52

4.4.1 Linear stability analysis. . . . . . . . . . . . . . . . . . . . 524.4.2 Dimensionality of the stable and unstable manifolds. . . . 53

4.5 The propagation speeds and necessary conditions for the exis-tence of a heteroclinic orbit by analysing the dimensionality ofthe manifolds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5 Conclusions 63

References 66

1 Introduction

On earth there are billion of ecosystems that are the habitats of many differ-ent organisms. Lots of these ecosystems contain unidirectional water currents,very strong or weak, because of the biased downstream flow. Examples of suchsystems are oceans, rivers, streams, creeks but also your intestines. Compar-ing these systems, it is understandable that the biotic environment in streamsand creeks is more conducive for growth and reproduction of aquatic organismsthan oceans and rivers. But how is it possible that aquatic organisms with a lowmotility can persist in ecological systems like streams and creeks without beingswept downstream into large rivers or oceans? In other words: How do creekand river populations avoid extinction? Without a mechanism that ensures thatindividuals of a population can move upstream, any small advection will ensurethat the average location of the population will move downstream. So this willlead to extinction of the population in his own habitat. This phenomenon ofpersistence for populations subjected to continuous advection is called ”the driftparadox”. In this thesis we look which factors play an important role in thedrift paradox and how they interact. Examples of such factors are difussion,advection speed, domain size and growth rate.

After this problem was recognized by Hans Robert Muller in 1954 [9], manymathematicians have done research about it. They all had their own hypothesesabout the mechanism of the upstream movement. Muller thought for examplethat adult insects balance out the downward drift of the insect larvae by flyingupstream to deposit eggs, while some others thought that insects crawl backupwards using the benthos (bottom of the stream) [4].

Speirs and Gurney were the first who came up with a system which was agood but simple representation of a population residing in a small river subjectto advection (stream flow) and diffusion (representing random movement). Thismodel is as follows:

∂n

∂t= f(n)n+D

∂2n

∂x2− v ∂n

∂x. (1)

Here, n(x, t) is the density of the population per unit area, f(n) the local percapita growth rate of the population, D the diffusion coefficient and v the ad-vection speed.

In this thesis, we examine the model proposed by [Pachepsky]. We start withextending the model of Speirs and Gurney [4] by dividing the population intotwo interacting compartments, individuals living on the benthos and individualsdrifting in the flow. We assume that the rate of entering the drift compartmentis constant. Secondly, we derive necessary and sufficient conditions for persis-tence of the population. Next, we transform the extended model using logisticgrowth and travelling wave coordinates and we assume that there exists a trav-elling wave between the two found steady states. We do this because we willstudy the existence of a heteroclinic orbit between the steady states. We will

8

find necessary conditions with the assumption there exists a heteroclinic orbitbetween the steady states. Finally, we calculate the propagation speeds of thetravelling waves.

1.1 Outline.

This thesis is organized as follows. In Section 2, we explain the extension ofthe model of Speirs and Gurney [5]. We also give a few assumptions that areapplicable during this entire thesis.

In Section 3 we study persistence of the population. In Section 2 we havedivided the population into two interacting compartment, individuals living onthe benthos and individuals drifting in the flow. The per capita rate at whichindividuals in the benthic population enter the drift is given by µ. In Section3.1 we consider the case µ ≥ 1. We start with deriving the explicit solution ofnb(x, t). Then we show in 3.1.1-3.1.4 that we will find the same condition forpopulation persistence if we assume that nb(x, 0) = 0. With this condition wewill also calculate the critical domain size with respect to the advection speednecessary for population persistence. In Section 3.2 we show that the popu-lation, irrespective of the domain length and the advection speed, will alwayspersist for µ < 1.

In Section 4 we consider spatial spread of the population in time. Becausethere is advection in our system, we need to distinguish between the propaga-tion speed downstream (in the direction of advection) and upstream (againstthe advection). In Section 4.1 we consider the system, defined in Section 3, withlogistic growth and we determine the steady states of the system that we ob-tained. In Section 4.2 we assume there exists a travelling wave between the twofound steady states, a zero and a non-zero steady state. We recast the currentsystem into travelling wave coordinates and then transform it into a system offirst-order equations.

In Section 4.3 we derive necessary conditions for the existence of a heteroclinicorbit between the steady states by studying the phase portraits. From Sections4.1 and 4.2 we found two steady states and a 3-dimensional system in which thenullclines all intersect each other. We consider the spaces that are separated bythe nullclines as regions and study, with the assumption there exists a hetero-clinic orbit between the steady states, through which regions an orbit must goto reach the other steady state. We split Section 4.4 up in two subsections. InSection 4.4.1 we linearise the system, found in Section 4.2, around the steadystates and in Section 4.4.2 we determine the possible dimension of the stableand unstable manifolds of the steady states.

In Section 4.5 we also derive necessary conditions for the existence of a het-eroclinic orbit between the steady states. Again with the assumption thereexists one. But in contrast to Section 4.3, we focus in this section on the dimen-

9

sion of the stable and unstable manifolds. We will also determine the criticalpropagation speeds of the travelling waves. With the critical propagation speedwe mean the transition propagation speed of the travelling waves in which wecertainly know there does not exist a heteroclinic orbit and in which there couldexist a heteroclinic orbit.

10

2 The model

We extend the system of Speirs and Gurney [5] by dividing the population intotwo interacting compartments, individuals living on the benthos and individu-als drifting in the flow, following [Pachepsky]. We consider a population whereindividuals can only reproduce on the benthos and enter the water column todrift until they settle on the benthos again. Reproduction only occurs on localscale. Transfer between individuals on the benthos and in the drift are modelledPoisson processes. This means that the number of individuals on the benthosthat enter the water column to drift is Poisson distributed. The same holdsfor individuals in the drift that settle on the benthos. The movement of anindividual is expressed as a combination of advection and diffusion. We assumethat the stream advection is uniform in the horizontal and vertical directions.The movement of an individual by advection represents the uniform stream flow.The movement of an individual by diffusion represents individual swimming andthe heterogeneous stream flow.

We assume the domain to be the one-dimensional interval (0, L), representinga stream reach. This yields the following system

∂nd∂t

=µnb − σnd +D∂2nd∂x2

− v ∂nd∂x

,

∂nb∂t

=f(nb)nb − µnb + σnd,

(2)

where x = 0 is the top of the stream reach and x = L the end. Note thatµ, σ, v,D, f(nb) > 0, are described as follows

• nb is the population density on the benthos.

• nd is the population density in the drift.

• f(nb) is the per capita rate of increase of the benthic population.

• µ is the per capita rate at which individuals in the benthic populationenter the drift.

• σ is the per capita rate at which the organisms return to benthic popula-tion from drifting.

• D is the diffusion coefficient.

• v is the advection speed.

Note that model (2) does not incorporate death for individuals drifting in theflow. If we want to add death for individuals drifting in the flow into the model,then we need to add −δnd into the first equation of (2), where δ is the deathfactor.

11

We assume no Allee effect in the population. The Allee effect is a phenomenonseen in population biology when a small population grows faster when the or-ganisms are at high population density than it would if the population was atlow density. This could come about through multiple mechanisms. In one exam-ple, you can imagine a population spread out over a large area compared to oneconcentrated in a small space. If the limit on reproduction is how often malesand females meet, then the concentrated population will have more encountersand therefore more reproduction.

This implies in our model that the maximum per capita population growthrate is found as the population density approaches zero, f(0) =max{f(nb)}

The boundary conditions for our model are given by

vnd(0, t)−D(∂nd∂x

)x=0

=0,

nd(L, t) =0,(3)

for all t ≥ 0. This means that no individuals enter at the top of the streamreach, and individuals in the stream cannot move beyond the top the stream.At the bottom of the stream reach, individuals that cross the boundary nevercome back.

12

3 Population persistence

In this section we derive a necessary and sufficient condition for populationpersistence in the model presented in the previous section, i.e. system (2). Bythis we mean that we will find a condition such that individuals of the populationwill always survive either on the benthos as in the drift. With the use of thiscondition we will also calculate the critical domain size with respect to theadvection speed necessary for population persistence. Since the maximum percapita growth rate is at low densities, population persistence is equivalent topopulation growth at small densities (Lewis and Kareiva, 1993). We thereforelinearize system (2) around the zero steady state and obtain conditions underwhich a small population grows. This linearized system is given by

∂nd∂t

=µnb − σnd +D∂2nd∂x2

− v ∂nd∂x

,

∂nb∂t

=rnb − µnb + σnd,

with r = f(0). We now rescale this system by setting

t = rt, µ =µ

r, σ =

σ

r, x =

x√Dr

, v =v√Dr

.

For simplicity we drop the tildes, such that system (2) becomes

∂nd∂t

=µnb − σnd +∂2nd∂x2

− v ∂nd∂x

, (4a)

∂nb∂t

=(1− µ)nb + σnd. (4b)

We will now consider the two cases µ ≥ 1 and µ < 1 separately.

3.1 Population persistence for µ ≥ 1

For µ ≥ 1 we have that the per capita rate at which individuals in the benthicpopulation enter the drift is higher than the per capita rate of increase of thebenthic population. So this means that the total growth rate of the benthicpopulation at each location is negative. We will show that persistence is possibleprovided that the domain L is large enough with respect to the advection speedv. We start with deriving expressions for nb(x, t) and nd(x, t) as explicit seriesexpansions.

Proposition 1 nb(x, t) is given by

nb(x, t) = nb(x, 0)e(1−µ)t + σe(1−µ)t∫ t

0

e(µ−1)τnd(x, τ)dτ. (5)

13

Proof. We apply the variation of constants formula to (4b) and write nb(x, t)in terms of nd(x, t). So first we need to find the homogeneous solution, whichwe write as (nb(x, t))h. We get

dnb(x, t)

dt=(1− µ)nb(x, t),∫

1

(nb(x, t))hdnb(x, t) =

∫(1− µ)dt,

ln |(nb(x, t))h| =(1− µ)t+ c0,

(nb(x, t))h =c1e(1−µ)t,

with c0, c1 ∈ R. To find the particular solution, which we denote by (nb(x, t))p,we set (nb(x, t))p = c1(t)e(1−µ)t. Substituting this into (4b) gives

c′1(t)e(1−µ)t + c1(t)(1− µ)e(1−µ)t =c1(t)(1− µ)e(1−µ)t + σnd(x, t),

c′1(t)e(1−µ)t =σnd(x, t),

c′1(t) =σe(µ−1)tnd(x, t),

c1(t) =σ

∫ t

0

e(µ−1)τnd(x, τ)dτ + c2,

with c2 ∈ R. So the particular solution becomes

(nb(x, t))p = e(1−µ)t(σ

∫ t

0

e(µ−1)τnd(x, τ)dτ + c2

)For the solution nb(x, t) we now find

nb(x, t) =(nb(x, t))h + (nb(x, t))p,

=c1e(1−µ)t + e(1−µ)t

(σ

∫ t

0

e(µ−1)τnd(x, τ)dτ + c2

),

=c3e(1−µ)t + σe(1−µ)t

∫ t

0

e(µ−1)τnd(x, τ)dτ,

with c3 = c1 + c2. Because nb(x, 0) = c3 for t = 0, we obtain (5). �.

At this point, [Pachepsky] simply say that without loss of generality we canassume that nb(x, 0) = 0. So this means that all individuals are initially inthe mobile class nd. While this does not reflect biologically realistic conditions,we will show however that this assumption is valid for deriving conditions forpopulation persistence.

To show that without loss of generality we can assume that nb(x, 0) = 0, we needto do three steps. First we show that system (4) for general initial conditions

14

generates the same solutions as the following system

∂nd∂t

=µnb − σnd +∂2nd∂x2

− v ∂nd∂x

+ µe(1−µ)ta, (6a)

∂nb∂t

=(1− µ)nb + σnd, (6b)

with a = nb(x, 0) = n0b and for initial conditions nd(x, 0) = n0d, nb(x, 0) = 0.Second, we determine the solutions of system (6) with a = 0 and a 6= 0 for initialconditions nd(x, 0) = n0d, nb(x, 0) = 0. Finally, we show that both solutionsgenerate the same condition for population persistence which implies that wecan assume a = nb(x, 0) = n0b = 0.

3.1.1 Equivalence of two initial value problems

The following proposition allows to simplify somewhat the initial conditionsunder which one studies persistence or extinction of the population.

Proposition 2 Let t 7→ n(x, t;n0d, n0b) ∈ R2 be the unique solution to system (4)

for initial conditions nd(x, 0) = n0d, nb(x, 0) = n0b , and t 7→ n(x, t;n0d, n0b) ∈ R2

be the unique solution to system (6) with a = n0b and for initial conditionsnd(x, 0) = n0d, nb(x, 0) = 0. Then

n(x, t;n0d, n0b) = n(x, t;n0d, n

0b) + e(1−µ)tn0b(x),

for all t ≥ 0.

Proof. Let us write n(x, t) = (nd(x, t), nb(x, t)) and n(x, t) = (nd(x, t), nb(x, t)).Now we put

nb(x, t) := nb(x, t)− e(1−µ)tnb(x, 0). (7)

from which it follows that

nb(x, t) = nb(x, t) + e(1−µ)tnb(x, 0). (8)

Note that from (7) it follows that nb(x, 0) = nb(x, 0) − nb(x, 0) = 0. From (7)and (8) we find

∂nb∂t

=∂nb∂t

+ (µ− 1)e(1−µ)tnb(x, 0),

=(1− µ)nb + σnd + (µ− 1)e(1−µ)tnb(x, 0),

=(1− µ)(nb + e(1−µ)tnb(x, 0)) + σnd + (µ− 1)e(1−µ)tnb(x, 0),

=(1− µ)nb + σnd.

Note that we substituted (4b) in the second line. From (4a) and (8) we find

∂nd∂t

=µ(nb(x, t) + e(1−µ)tnb(x, 0)

)− σnd +

∂2nd∂x2

− v ∂nd∂x

,

=µnb − σnd +∂2nd∂x2

− v ∂nd∂x

+ µe(1−µ)tnb(x, 0).

15

So we obtain the following system

∂nd∂t

=µnb − σnd +∂2nd∂x2

− v ∂nd∂x

+ µe(1−µ)tnb(x, 0),

∂nb∂t

=(1− µ)nb + σnd,

(9)

with nb(x, 0) = 0. So we conclude (nd(x, t), nb(x, t)) satisfies system (6) witha = nb(x, 0) = n0b and for initial conditions nd(x, 0) = n0d, nb(x, 0) = 0. So

n(x, t;n0d, n0b) = n(x, t;n0d, n

0b) + e(1−µ)tn0b(x),

for all t ≥ 0. �

Corollary 3.0.1 If 0 < µ ≤ 1, then the population persists according to model(2) for any initial condition n0b 6= 0 (positive).

Corollary 3.0.2 If µ > 1, then the population persists according to model (2)if and only if n(x, t;n0d, n

0b) does not converge to 0 as t→∞.

3.1.2 A series expansion for the unforced system

In the previous section we found that system (4) for general initial conditionsgenerates essentially the same information as ”forced system” (6) with a = n0band for initial conditions nd(x, 0) = n0d, nb(x, 0) = 0 when concerned with popu-lation persistence or extinction. As a next step we will show that assuming eventhat nb(x, 0) = n0b = 0 in system (4) yields sufficient information. This condi-tion only holds if a = 0 in system (6). In this section we derive the solution ofsystem (6) with a = 0, i.e. system (4), and for initial conditions nd(x, 0) = n0d,nb(x, 0) = 0.

Because system (6) with a = 0 is identical to system (4), we can just lookat system (4) with initial conditions nd(x, 0) = n0d, nb(x, 0) = 0. For simplicity,we drop the tildes. Now from Lemma 5 it follows that nb(x, t) is given by

nb(x, t) = σe(1−µ)t∫ t

0

e(µ−1)τnd(x, τ)dτ. (10)

Now we have expressed the solution of nb(x, t) in terms of nd(x, t), we can finda rather explicit expression for the solution of nd(x, t).

Lemma 3.1 Put u(x, t) := nd(x, t)e(µ−1)t. Then u satisfies:

∂u(x, t)

∂t=∂2u(x, t)

∂x2− v ∂u(x, t)

∂x− αu(x, t) + µσ

∫ t

0

u(x, τ)dτ, (11)

with α = σ − µ+ 1.

16

Proof. Substituting expression (10) into (4a), one obtains

∂nd(x, t)

∂t=µσe(1−µ)t

∫ t

0

e(µ−1)τnd(x, τ)dτ − σnd(x, t)

+∂2nd(x, t)

∂x2− v ∂nd(x, t)

∂x.

(12)

We now set u(x, t) = nd(x, t)e(µ−1)t, from which it follows that nd(x, t) =

u(x, t)e(1−µ)t. Substituting these expressions into (12), we find

∂u(x, t)

∂te(1−µ)t + u(x, t)(1− µ)e(1−µ)t =µσe(1−µ)t

∫ t

0

e(µ−1)τnd(x, τ)dτ

− σu(x, t)e(1−µ)t − ve(1−µ)t ∂u(x, t)

∂x

+ e(1−µ)t∂2u(x, t)

∂x2.

Dividing by e(1−µ)t gives

∂u(x, t)

∂t+ u(x, t)(1− µ) = µσ

∫ t

0

u(x, τ)dτ − σu(x, t) +∂2u(x, t)

∂x2− v ∂u(x, t)

∂x.

Setting α = σ − µ+ 1 yields

∂u(x, t)

∂t=∂2u(x, t)

∂x2− v ∂u(x, t)

∂x− αu(x, t) + µσ

∫ t

0

u(x, τ)dτ. �

A series expansion for u(x, t) can now be found by separation of variables.

Lemma 3.2 A non-zero solution of the form U(x, t) = X(x)T (t) to (11) sat-isfies

X ′′(x)− vX ′(x) + λX(x) = 0, (13a)

T ′(t) + (α+ λ)T (t)− µσ∫ t

0

T (τ)dτ = 0, (13b)

with λ ∈ C.

Proof. Substituting u(x, t) = X(x)T (t) into (11) yields

∂X(x)T (t)

∂t=∂2X(x)T (t)

∂x2− v ∂X(x)T (t)

∂x− αX(x)T (t) + µσ

∫ t

0

X(x)T (τ)dτ.

For X ′(x) :=dX(x)

dxand T ′(t) :=

dT (t)

dtit follows that

X(x)T ′(t) = T (t)X ′′(x)− vT (t)X ′(x)− αX(x)T (t) + µσX(x)

∫ t

0

T (τ)dτ.

17

Writing all the terms of T (t) on the left-hand side and all the terms of X(x) onthe right-hand side gives

T ′(t) + αT (t)− µσ∫ t0T (τ)dτ

T (t)=X ′′(x)− vX ′(x)

X(x).

Because the left-hand side of the equation is a function with respect to t and theright-hand side a function with respect to x, we must have that both functionsare equal to the same constant −λ ∈ C (separation constant). So, we must have

X ′′(x)− vX ′(x)

X(x)= −λ,

andT ′(t) + αT (t)− µσ

∫ t0T (τ)dτ

T (t)= −λ.

From these equations we obtain

X ′′(x)− vX ′(x) + λX(x) = 0,

and

T ′(t) + (α+ λ)T (t)− µσ∫ t

0

T (τ)dτ = 0. �

Now we will determine the solutions of (13a) and (13b).

Lemma 3.3 The general solution T (t) = Tλ(t) to (13b) for λ ∈ C is of theform

T (t) = c(m+e

m+t −m−em−t), (14)

with c ∈ C and

m± = m±(λ) =−(α+ λ)±

√(α+ λ)2 + 4µσ

2. (15)

Proof. Differentiating (13b) with respect to t gives

T ′′(t) + (α+ λ)T ′(t)− µσT (t) = 0. (16)

Suppose that T (t) = cemt is a solution of (16), with c,m ∈ C. SubstitutingT (t) = cemt into (16) gives

cm2emt + c(α+ λ)memt − µσcemt = 0.

cemt(m2 + (α+ λ)m− µσ

)= 0. (17)

From (17) we find m2 + (α+λ)m−µσ = 0, because m is finite. So T (t) = cemt

is a solution of (16) when m = m±(λ) given by

m±(λ) =−(α+ λ)±

√(α+ λ)2 + 4µσ

2.

18

The general solution of (16) is now given by

T (t) = c1em+t + c2e

m−t, (18)

with c1, c2 ∈ C arbitrary and m± as in (15).

We now substitute this solution into (13b). We find

c1m+em+t + c2m−e

m−t + (α+ λ)(c1em+t + c2e

m−t)

− µσ∫ t

0

c1em+τ + c2e

m−τdτ = 0.(19)

For the integral we have∫ t

0

c1em+τ + c2e

m−τdτ =c1m+

em+t +c2m−

em−t − c1m+− c2m−

.

So from (19) we now must have for all t ≥ 0:

c1m+em+t + c2m−e

m−t + (α+ λ)(c1em+t + c2e

m−t)

− µσ( c1m+

em+t +c2m−

em−t − c1m+− c2m−

)= 0

(20)

Since m± 6= 0 we can divide (17) by m. We find

c1m+em+t + (α+ λ)c1e

m+t − µσ c1m+

em+t = 0, (21a)

c2m−em−t + (α+ λ)c2e

m−t − µσ c2m−

em−t = 0. (21b)

Substituting (21a) and (21b) into (20), we obtain

c1m+

+c2m−

= 0. (22)

Now we substitute (22) into the general solution (18) and we find

T (t) =c1em+t − c1m−

m+em−t,

=c3(m+e

m+t −m−em−t), (23)

with c3 = c1m+∈ C. �

Remark. The solution that we found for T (t) (see (14)) differs from the onefound in [Pachepsky]. They found that T (t) = c1m+e

m+t + c2m−em−t.

We now turn to (13a) and determine the solution for X(x).

Lemma 3.4 Let λ ∈ C and λ 6= 14v

2. The general solution of X(x) is given by:

X(x) = a1ez+x + a2e

z−x, (24)

with a1, a2 ∈ C and z± ∈ C are the two distinct solutions to z2 − vz + λ = 0

19

Proof. Suppose that X(x) = aezx is a solution to (13a), with a, z ∈ C. Substi-tuting X(x) = aezx into (13a) gives

aezx(z2 − vz + λ) = 0. (25)

From (25) it follows that X(x) = aezx is a solution to (13a) with a1, a2 ∈ C andz± ∈ C the two distinct solutions to z2 − vz + λ = 0. �

If λ = 14v

2, then z+ = z− = 12v, which implies that (13a) might have another

solution.

Lemma 3.5 Let λ = 14v

2. The general solution of X(x) is now given by:

X(x) = a1e12 vx + a2xe

12 vx, (26)

with a1, a2 ∈ C.

Proof. Because λ = 14v

2, we get z+ = z− = 12v. Using the proof of Lemma (3.4)

we conclude that X1(x) = a1e12 vx is a solution to (13a). Now we suppose that

X2(x) = a2xe12vx is also a solution to (13a). If we substitute X2(x) = a2xe

12vx

into (13a) for λ = 14v

2, we find

a2ve12vx(1 +

1

4vx− 1− 1

2vx+

1

4vx)

= 0

So we conclude that X2(x) = a2xe12 vx is indeed a solution to (13a). �

To say something further about the solutions of X(x), we will first look tothe boundary conditions. We have that nd(x, t) = X(x)T (t)e(1−µ)t. So from(3) we obtain (

vX(0)−X ′(0))T (t)e(1−µ)t = 0,

X(L)T (t)e(1−µ)t = 0.

Because T (t) = 0 is the trivial solution, we must have that

vX(0)−X ′(0) = 0, (27a)

X(L) = 0. (27b)

Lemma 3.6 If λ = 14v

2, then the general solution (26) satisfies the boundaryconditions (27a)-(27b) if and only if a1 = a2 = 0.

Proof. (⇒) If we apply the boundary conditions (27a)-(27b) to (26), we get

1

2va1 − a2 = 0, (28a)

(a1 + a2L)e12vL = 0. (28b)

20

From (28a) we find a2 = 12va1 and from (28b) we find a1 + a2L = 0. If we

substitute a2 = 12va1 into a1 + a2L = 0, we obtain a1(1 + 1

2vL) = 0. Note thatthis must hold for all different values of v and L. So it implies that a1 = 0. Andfrom a1 = 0, we find a2 = 0.

(⇐) From a1 = a2 = 0, we get X(x) = 0 for all x ∈ (0, L). This meansthat the boundary conditions (27a)-(27b) are always satisfied. �

Lemma 3.7 Let λ ∈ C, λ 6= 14v

2 and z± ∈ C. The general solution (24)satisfies the boundary conditions (27a)-(27b) if and only if the solutions z+ andz− to the equation z2 − vz + λ = 0 satisfy the equation

z =v

1 + e(2z−v)L. (29)

Proof.(⇒) If we apply the boundary conditions (27a)-(27b) to (24) , we findMa = 0 with

M =

(v − z+ v − z−ez+L ez−L

)and a =

(a1 a2

)>.

There exists a non-trivial solution a = (a1, a2)> if and only if det(M)=0. Thisholds if

ez−L(v − z+)− ez+L(v − z−) = 0. (30)

Multiplying with ez+L gives

e(z++z−)L(v − z+)− e2z+L(v − z−) = 0. (31)

Because z± are the solutions to z2 − vz + λ = 0, it follows that z+ + z− = v.Substitute this into (31), obtaining

evL(v − z+)− e2z+Lz+ = 0.

Note that this equation also holds if we replace z+ by z−. To see that, firstmultiply (30) by ez−L and then again use that z+ + z− = v. So we have

evL(v − z)− e2zLz = 0, (32)

with z = z+ or z = z−. Now we find

−z(evL + e2zL) =− vevL,

z =vevL

evL + e2zL,

z =v

1 + e(2z−v)L.

(⇐) Multiplying (29) by evL and moving all the terms to the left-hand sidegives (32). Because (32) holds for both z±, we can just pick one, say z = z+.For z = z+ and knowing that z+ + z− = v, we end up with (31). Dividing byez+L gives (30). This means that X(x) satisfies the boundary conditions. �

21

Lemma 3.8 If Xλ(t) satisfies the boundary conditions (27a)-(27b), then λ =14 (v2 + y2) with y ∈ R. In particular, λ > 0.

Proof. From (29) we find

e(2z−v)L =v − zz

. (33)

For ω = 2z − v, we find z = 12v + 1

2ω. If we substitute this into (33), we obtain

eωL =v − 1

2v −12ω

12v + 1

2ω=v − ωv + ω

. (34)

Now we substitute ω = x + iy, x, y ∈ R, into the right-hand side of (34). Weobtain

θ =v − wv + w

=v − x− iyv + x+ iy

=v − x− iyv + x+ iy

· v + x− iyv + x− iy

=v2 − x2 − y2 − 2viy

(v + x)2 + y2.

For the modulus squared of θ we get

|θ|2 =(v2 − x2 − y2)2 + 4v2y2

((v + x)2 + y2)2,

=v4 − 2v2(x2 + y2) + (x2 + y2)2 + 4v2y2

(v + x)4 + 2y2(v + x)2 + y4,

=v4 + x4 + y4 − 2v2x2 + 2v2y2 + 2x2y2

(v + x)4 + 2y2(v + x)2 + y4,

=(v + x)4 − 8v2x2 − 4v3x− 4vx3 + 2v2y2 + 2x2y2 + y4

(v + x)4 + 2y2(v + x)2 + y4,

=(v + x)4 + 2y2(v + x)2 + y4 − 4vxy2 − 8v2x2 − 4v3x− 4vx3

(v + x)4 + 2y2(v + x)2 + y4,

=(v + x)4 + 2y2(v + x)2 + y4 − 4vx((v + x)2 + y2)

(v + x)4 + 2y2(v + x)2 + y4.

(35)

And for the modulus squared of eωL we find

|eωL|2 =|e(x+iy)L|2,=|eLx(cos(Ly) + i sin(Ly))|2,=e2Lx.

(36)

So if x > 0, then |θ|2 < 1, while |eωL|2 > 1. And if x < 0, then |θ|2 > 1, while|eωL|2 < 1. This means that there cannot be a solution with x 6= 0. For x = 0,we have |θ|2 = |eωL|2 = 1. So a solution may exists for ω = iy, y ∈ R. From thiswe find z = 1

2 (v + iy), y ∈ R. Let y± be the y corresponding to z±. Becausev = z+ + z− = v+ 1

2 i(y+ + y−), we must have y+ = −y−. This implies z+ = z−.Now from z+z− = λ, we obtain λ = 1

4 (v2 + y2) ∈ R>0. �

22

If λ = 14v

2, only the trivial solution satisfies the boundary conditions (Lemma3.6). So we know that λ > 1

4v2 > 0. We can now look at the solution of X(x)

in more detail.

Lemma 3.9 Let λ > 14v

2. Applying the boundary conditions (27a)-(27b) to(24) gives √

4λ− v2v

+ tan(L

2

√4λ− v2) = 0. (37)

Proof. Because z± are complex for λ > 14v

2, we obtain from (24),

X(x) = a1ev+i√

4λ−v22 x + a2e

v−i√

4λ−v22 x,

= a1evx2

(cos(

x

2

√4λ− v2) + i sin(

x

2

√4λ− v2)

)+ a2e

vx2

(cos(

x

2

√4λ− v2)− i sin(

x

2

√4λ− v2)

),

= a3evx2 cos(

x

2

√4λ− v2) + a4e

vx2 sin(

x

2

√4λ− v2), (38)

with a3 = a1 + a2 and a4 = i(a1 − a2), both in C.

Applying the boundary conditions for X(x) to (38) yields

evL2

(a3 cos(

L

2

√4λ− v2) + a4 sin(

L

2

√4λ− v2)

)= 0. (39)

and1

2a4√

4λ− v2 − a3v

2= 0. (40)

From (40) we find a3 =a4√

4λ− v2v

. Substituting this into (39) yields

a4evL2

(√4λ− v2v

cos(L

2

√4λ− v2) + sin(

L

2

√4λ− v2)

)= 0,

√4λ− v2v

cos(L

2

√4λ− v2) + sin(

L

2

√4λ− v2) = 0,

√4λ− v2v

+ tan(L

2

√4λ− v2) = 0. �

Lemma 3.10 There exists a strictly increasing sequence λn = λn(v, L) > 0,n ∈ N, such that λ > 0 is a solution to (37) if and only if λ = λn for some n.

Proof. (⇒) First we put x =√

4λ− v2 > 0, such that (37) becomes

tan(L

2x) = −1

vx. (41)

23

Figure 1 shows a plot of (41). We note that the vertical asymptotes of tan(L2 x)

are at L2 x = 1

2π + kπ, k ∈ Z. That is, x = 2L ( 1

2π + kπ) = 2k+1L π, k ∈ Z. We

also have that − 1vx is a decreasing line through 0 because v > 0. So on each

interval ( 2k+1L π, 2(k+1)+1

L π), we have exactly one intersection point with the liney = − 1

vx. This gives x1, x2, ... with 0 < x1 < x2 < ... and we find

λn = λn(v, L) =1

4(x2n + v2) > 0, n = 1, 2, 3, ... (42)

(⇐) We have argued that each solution λ to (37) is such that the associated xsolves (41). Hence, λ is of the form (42) for some n. On the other hand, λ ofthe form (42) has xn satisfying (41). So (37) holds. �

Figure 1: tan(L2 x) = − 1vx for L = 2 and v = 3 1

3 .

Corresponding to each λn we have m±(λn) = m±,n. So now we can write thesolution for u(x, t) and therefore for nd(x, t).

Proposition 3 The solution nd(x, t) is given by

nd(x, t) =∞∑n=1

[a4,n

(√4λn − v2v

evx2 cos(

x

2

√4λn − v2) + e

vx2 sin(

x

2

√4λn − v2)

)]×[c3,n

(m+,ne

(m+,n+1−µ)t −m−,ne(m−,n+1−µ)t)].

(43)

24

Proof.

nd(x, t) =

∞∑n=1

Xλn(x)Tλn(t)e(1−µ)t,

=

∞∑n=1

[a3,ne

vx2 cos(

x

2

√4λn − v2) + a4,ne

vx2 sin(

x

2

√4λn − v2)

]×[c3,n

(m+,ne

(m+,n+1−µ)t −m−,ne(m−,n+1−µ)t)],

=

∞∑n=1

[a4,n

(√4λn − v2v

evx2 cos(

x

2

√4λn − v2) + e

vx2 sin(

x

2

√4λn − v2)

)]×[c3,n

(m+,ne

(m+,n+1−µ)t −m−,ne(m−,n+1−µ)t)]. �

3.1.3 A series expansion for the forced system

So now we have convenient expressions for the solutions nd(x, t) and nb(x, t) forsystem (6) with a = 0, we can also determine the expressions for the solutionsto system (6) with a = nb(x, 0) = n0b .

Let St be the solution operator for system (4). In fact, if U0 = (n0d, 0)> then

St[U0] are the found solutions in the previous section. The solution of system(6) for a 6= 0 is now given by

U(x, t) = St[U0] + µ

∫ t

0

St−s[e(1−µ)snb(x, 0)~e1

]ds, (44)

with U(x, t) = (nd(x, t), nb(x, t))>. Writing U(x, t) as two components, we

obtain

nd(x, t) =(St[U0])1 + µ

∫ t

0

(St−s

[e(1−µ)snb(x, 0)~e1

])1ds,

nb(x, t) =(St[U0])2 + µ

∫ t

0

(St−s

[e(1−µ)snb(x, 0)~e1

])2ds.

Note that from the previous section we already found (St[U0])1 and (St[U0])2.For (St[U0])1 we found:

(St[U0])1 =

∞∑n=1

An(x)

(m+,ne

(m+,n+1−µ)t −m−,ne(m−,n+1−µ)t), (45)

with

An(x) = a4,nc3,n

(√4λn − v2v

evx2 cos(

x

2

√4λn − v2) + e

vx2 sin(

x

2

√4λn − v2)

).

25

Moreover, for (St[U0])2 we found (see (10)):

(St[U0])2 = σe(1−µ)t∫ t

0

e(µ−1)τnd(x, τ)dτ,

= σe(1−µ)t∫ t

0

e(µ−1)τ∞∑n=1

An(x)

(m+,ne

(m+,n+1−µ)τ −m−,ne(m−,n+1−µ)τ)dτ,

= σe(1−µ)t∫ t

0

∞∑n=1

An(x)

(m+,ne

m+,nτ −m−,nem−,nτ

)dτ,

= σe(1−µ)t∞∑n=1

An(x)

∫ t

0

m+,nem+,nτ −m−,nem−,nτdτ,

= σ

∞∑n=1

An(x)

(e(m+,n+1−µ)t − e(m−,n+1−µ)t

). (46)

To determine µ∫ t0

(St−s

[e(1−µ)snb(x, 0)~e1

])1ds, we note that St[nb(x, 0)~e1] has

the same form as St[nd(x, 0)~e1]. So (St[nb(x, 0)~e1])1 has the same form as (43)but with different coefficients. This is because the coefficients now depend onnb(x, 0) instead of nd(x, 0). We find

µ

∫ t

0

St−s[e(1−µ)snb(x, 0)~e1

]ds =µ

∫ t

0

e(1−µ)sSt−s[nb(x, 0)~e1

]ds,

=

∫ t

0

e(1−µ)s∞∑n=1

Bn(x)

(m+,ne

(m+,n+1−µ)(t−s)

−m−,ne(m−,n+1−µ)(t−s))ds,

=

∞∑n=1

Bn(x)

(m+,ne

(m+,n+1−µ)t∫ t

0

e−m+,nsds

−m−,ne(m−,n+1−µ)t∫ t

0

e−m−,nsds

),

=

∞∑n=1

Bn(x)

(e(m+,n+1−µ)t − e(m−,n+1−µ)t

),

(47)

with

Bn(x) = c3,na4,n

(√4λn − v2v

evx2 cos(

x

2

√4λn − v2) + e

vx2 sin(

x

2

√4λn − v2)

).

To determine µ∫ t0

(St−s

[e(1−µ)snb(x, 0)~e1

])2ds, we note that St[nb(x, 0)~e1] has

the same form as St[nd(x, 0)~e1]. So (St[nb(x, 0)~e1])2 has the same form as (10)but with different coefficients. With the use of (46) we find

26

µ

∫ t

0

St−s[e(1−µ)snb(x, 0)~e1

]ds =µ

∫ t

0

e(1−µ)sSt−s[nb(x, 0)~e1

]ds,

=µ

∫ t

0

e(1−µ)sσe(1−µ)(t−s)∫ t−s

0

e(µ−1)τnd(x, τ)dτds,

=µσ

∫ t

0

e(1−µ)se(1−µ)(t−s)∫ t−s

0

∞∑n=1

Bn(x)(m+,ne

m+,nτ −m−,nem−,nτ

)dτds,

=µσ

∫ t

0

e(1−µ)t∞∑n=1

Bn(x)(em+,n(t−s) − em−,n(t−s)

)ds,

=µσe(1−µ)t∞∑n=1

Bn(x)∫ t

0

em+,n(t−s) − em−,n(t−s)ds,

=µσe(1−µ)t∞∑n=1

Bn(x)(1

m+,n(1− em+,nt)− 1

m−,n(1− em−,nt)

)=µσ

∞∑n=1

Bn(x)

(1

m+,n(e(1−µ)t − e(m+,n+1−µ)t)

− 1

m−,n(e(1−µ)t − e(m−,n+1−µ)t)

)(48)

3.1.4 Condition for population persistence

Because we have found the solutions of the forced and unforced systems, we cannow look for which values of the parameters the population persists and whenit goes extinct for t→∞. This leads to the condition for persistence. Note thatif µ ≤ 1 there is population persistence (Corollary 3.0.1). The interesting caseis µ > 1.

For the following theorem, note that the sequence of positive real numbersλn = λn(v, L) is introduced in Lemma 3.10

27

Theorem 3.11 If µ > 1, then the population persists if and only if λ1(v, L) <σ

µ− 1.

Proof.(⇒) For system (6) with a = 0 we see in (43) that nd → 0 for t→∞ ifm+,n+1−µ < 0 and m−,n+1−µ < 0 for all n. Because m−,n < m+,n, the onlycondition for extinction of the population will be m+,n + 1−µ < 0. For system(6) with a 6= 0 we see in (47) that we also find the condition m+,n + 1− µ < 0for population extinction (see Corollary (3.0.2)). Note that we find the samecondition if we look to the population on the benthos.

Lemma 3.12 m+(λ) is a decreasing function of λ on (0,∞).

Proof. Differentiating m+ with respect to λ gives

m′+(λ) =− 1

2+

α+ λ

2√

(α+ λ)2 + 4µσ,

=α+ λ−

√(α+ λ)2 + 4µσ

2√

(α+ λ)2 + 4µσ.

Because µ, σ > 0 we have√

(α+ λ)2 + 4µσ > α+λ. And because 2√

(α+ λ)2 + 4µσ >0, we get m′+ < 0. So m+ is a decreasing function for λ > 0. �

This means that nd → 0 for t → ∞ if and only if m+,1 + 1 − µ < 0. Weget

−(α+ λ1) +√

(α+ λ1)2 + 4µσ

2+ 1− µ <0,

−(α+ λ1) +√

(α+ λ1)2 + 4µσ + 2− 2µ <0.

That is2(µ− 1) + α+ λ1 >

√(α+ λ1)2 + 4µσ. (49)

Now we first show that 2(µ− 1) + α+ λ1 > 0. We have

2(µ− 1) + α+ λ1 >0,

λ1 >2(1− µ)− α,>1− (µ+ σ).

And because µ + σ > 1 we have that λ1 > 1 − (µ + σ) is always satisfied. Sonow squaring both sides in (49) yields

4(µ− 1)2 + (α+ λ1)2 + 4(µ− 1)(α+ λ1) > (α+ λ1)2 + 4µσ,

(µ− 1)2 + (µ− 1)(α+ λ1) > µσ,

α+ λ1 >µσ − (µ− 1)2

µ− 1,

λ1 >µσ − (µ− 1)2 − α(µ− 1)

µ− 1.

28

And because α = σ − µ+ 1, we find that nd → 0 for t→∞ if

λ1(v, L) >µσ − (µ− 1)2 − (σ − µ+ 1)(µ− 1)

µ− 1=

σ

µ− 1.

So we conclude that a population only persists when

λ1(v, L) <σ

µ− 1. (50)

(⇐) We can immediately conclude that nd → 0 for t→∞ if λ1(v, L) >σ

µ− 1.

So this implies that the population persists if λ1(v, L) >σ

µ− 1. �

Now we know the necessary condition for population persistence, we can lookto the relationship between the domain size and the advection speed such thatthe population persists or goes extinct.

Corollary 3.12.1 The population persists if

L > L∗ =2√

4σµ−1 − v2

[π + arctan

(− 1

v

√4σ

µ− 1− v2

)]. (51)

Proof. From (37) we find

tan(L

2

√4λ1 − v2) =−

√4λ1 − v2v

,

L

2

√4λ1 − v2 =kπ + arctan

(− 1

v

√4λ1 − v2

), (52a)

L =2√

4λ1 − v2[kπ + arctan

(− 1

v

√4λ1 − v2

)]. (52b)

Note that in (52a) we had to add kπ, k ∈ Z≥1, because the period of tan(x) isπ and L

2

√4λ1 − v2 > 0 and − 1

v

√4λ1 − v2 < 0. Now from (52b), we see that L

can be as large as possible. But because we want to find the critical length L∗,

we need the smallest L that satisfies (52a). Because limx→−∞

arctan(x) = −1

2π, it

implies that k = 1. Now from (50) we find for the critical length

L∗ =2√

4σµ−1 − v2

[π + arctan

(− 1

v

√4σ

µ− 1− v2

)]. (53)

Note that L∗ becomes larger when λ1 becomes smaller. So this means that thepopulation persists for L > L∗. �

Remark. In [Pachepsky] the term π within the brackets in (51) is missing.

Corollary 3.12.2 If v ≥ v∗ := 2√

σµ−1 , the population cannot persist on a

domain of any size.

29

Proof. We have that L∗ is an increasing function with respect to v. This canbe found by differentiating L∗ with respect to v. Intiuitively, this can also beconcluded by imagining that a population with faster advection, will require alarger domain size to persist. But since arctan(x) is a bounded function, wehave that L∗ tends to infinity if 4σ

µ−1 − v2 goes to 0. In other words

L∗ →∞ for v ↑ v∗ = 2

√σ

µ− 1. �

Corollary 3.12.3 The population persist if 1 ≤ µ < µ∗ := 1 +4σ

9π2

L2 + v2.

Proof. From (42), we know that λ1 ∈(14 ( π

2

L2 +v2), 14 ( 9π2

L2 +v2)). So this implies

that if 14 ( 9π2

L2 + v2) < σµ−1 , the population persists, according to Theorem 3.11.

We find that the population persists for

µ < µ∗ =4σ

9π2

L2 + v2+ 1. �

Remark. We see that σµ−1 →∞ for µ ↓ 1, which implies that for given domain

size L, condition (50) is always satisfied for µ sufficient close to 1. Furthermore,we note that v∗ →∞ for µ→ 1. So this means that the population will alwayspersist for µ ↓ 1.

3.2 Population persistence for µ < 1

For µ < 1 we have that the per capita rate of increase of the benthic populationis higher than the per capita rate at which individuals in the benthic populationenter the drift. So this means that the population will always persist becausethe total growth rate of the benthic population at each location is positive. Wewill show that this is mathematically also correct.

Lemma 3.13 nd(x, t) is non-negative for non-negative initial values.

Proof. We will show that the maximum principle holds for (4a). Supposethat nd(x, t) first crosses from positive to negative at time t0 and at point x0.And suppose that nd(x, t) has a non-degenerate minimum at x0. Then we get∂nd∂x (x0, t0) = 0 and ∂2nd

∂x2 (x0, t0) > 0. Hence, from (4a) we get ∂nd∂t (x0, t0) > 0.

So nd(x, t) cannot evolve forward in time into the region nd(x, t) < 0. �

Now we will look at the dynamics of the benthic population, nb(x, t).

Lemma 3.14 nb(x, t) ≥ 0 for t ≥ 0.

30

Proof. From (4b) and Lemma 3.13 we obtain

∂nb∂t

= (1− µ)nb + σnd ≥ (1− µ)nb, (54)

From (54) we obtain

∂nb∂t− (1− µ)nb = q, q ∈ R≥0 (55)

For the homogeneous solution we find (nb(x, t))h = ce(1−µ)t, with c ∈ C. And

for the particular solution we find (nb(x, t))p =q

µ− 1. So we find

nb(x, t) = ce(1−µ)t +q

µ− 1, q ∈ R≥0.

And because we assume that nb(x, 0) = 0, we get c =q

1− µ. So the solution

becomesnb(x, t) =

q

1− µ(e(1−µ)t − 1

), q ∈ R≥0.

We note that e(1−µ)t > 1 for 0 < µ < 1. Thus, we conclude that nb(x, t) ≥ 0for t ≥ 0. �

We can also show this by using Gronwalls Inequality. A version of GronwallsInequality is as follows [13]:

Theorem 3.15 (Gronwall’s inequality [13]) Let I denote an interval on thereal line of the form [a,∞) or [a, b] or [a, b) with a < b. Let β and u be real-valued continious functions defined on I. If u is differentiable in the interiorI0 of I (the interval I without the end points a and possibly b) and satisfies thedifferential inequality

u′(t) ≤ β(t)u(t), t ∈ I0,

then u is bounded by the solution of the corresponding differential equationy′(t) = β(t)y(t):

u(t) ≤ u(a)e∫ taβ(s)ds,

for all t ∈ I.

Lemma 3.16 nb(x, t) ≥ 0 for t ≥ 0 using Gronwalls Inequality.

Proof. We haved

dtnb ≥ (1− µ)nb,

consequently, ddtnb − (1 − µ)nb ≥ 0. Multiplying by the positive ”integrating

factor” e−∫ t0(1−µ)ds = e(µ−1)t gives

e(µ−1)t( ddtnb − (1− µ)nb

)=

d

dt

(nbe

(µ−1)t) ≥ 0.

31

Integrating this inequality finally implies

nbe(µ−1)t ≥ c. (56)

So we find nb(x, t) ≥ ce(1−µ)t. From nb(x, 0) = 0, we get c = 0. So we concludethat nb(x, t) ≥ 0 for t ≥ 0.

We conclude that the population on the benthos, irrespective of the domainlength and the advection speed, will always persist at small densities. We alsoshowed that the population grows at least exponentially in this case.

32

4 Persistence or extinction through a movingpopulation front

In the previous section, we derived necessary and sufficient conditions for sys-tem (2) such that the population will persists or goes extinct. In this sectionwe consider spatial spread of the population in time, in particular by means ofa fixed front in population density that is moving up or downwards. When weintroduce advection into the system, we need to distinguish between the propa-gation speed downstream (in the direction of advection) and upstream (againstthe advection). With increasing advection, the propagation speed downstreamincreases, whereas the propagation speed upstream decreases.

Before we can determine the up- and downstream propagation speed, we re-cast the system in travelling wave coordinates and then transform it into asystem of first-order equations. This system is three dimensional. The exis-tence of a travelling wave is equivalent to the existence of a heteroclinic orbitbetween the steady states of this system. [Pachepsky] did not provide necessaryand sufficient conditions for the existence of the travelling waves, only necessaryconditions. In this section we make an analysis that goes beyond the resultsobtained by [Pachepky]. However, the three dimensionality of the system in theend prevented us from obtaining a full characterisation of existence of travellingwaves as well.

Next we assume there exist a travelling wave between the found steady statesand determine by phase portrait necessary conditions there actually could existone. Then we linearize the system around the steady states, which informs usabout the dimensions of the stable and unstable manifolds of the steady states.

4.1 Idealized model equations and homogeneous steadystates

We consider system (2) with logistic growth with carrying capacity K and in-trinsic growth rate r, that is

f(nb)nb = rnb(1−nbK

).

In order to arrive at a system similar to (4), we set nd = ndK and nb = nb

K . Wefind

∂nd∂t

=1

K

∂nd∂t

,

=1

K

(µnb − σnd +

∂2nd∂x2

− v ∂nd∂x

),

= µnb − σnd +∂2nd∂x2

− v ∂nd∂x

.

33

For ∂nb∂t we obtain

∂nb∂t

=1

K

∂nb∂t

,

=1

K

[rnb(1−

nbK

)− µnb + σnd],

= r(nb − n2b)− µnb + σnd,

= rnb(1− nb)− µnb + σnd.

Now we drop the tildes for convenience, and obtain

∂nd∂t

=µnb − σnd +∂2nd∂x2

− v ∂nd∂x

, (57a)

∂nb∂t

=rnb(1− nb)− µnb + σnd. (57b)

We are going to study travelling waves in (57), defined on the infinite domainR (instead of (0, L), as before). One should interpret this change of domainas an idealisation of travelling population fronts in the finite domain. In thelatter, one cannot expect fronts moving at fixed speed, with fixed shape. Nearthe boundary, propagation and shape must change substantially.

For the travelling wave problem, we are interested in the spatially homoge-neous steady states of (57), i.e. nd(x, t) = n∗d ∈ R>0 and nb(x, t) = n∗b ∈ R>0.From (57a) we find nd = µ

σnb and then from (57b) we get rnb(1−nb) = 0. Thisimplies that nb = 0 or nb = 1. For nb = 0 we find nd = 0 and for nb = 1 we findnd = µ

σ . So the spatially homogeneous steady states of system (57) are given by

(nb0 , nd0) = (0, 0) and (nb1 , nd1) = (1,µ

σ)

4.2 Travelling wave equations

Now assume that there exists a travelling wave connecting the two spatiallyhomogeneous steady states. This means that there is a solution of (57) of theform (nb, nd)(x, t) = (Nb, Nd)(x − ct) = (Nb, Nd)(z), for some fixed c ∈ R, thewave propagation speed. Substitution of this specific solution into (57) yields

−cN ′d =µNb − σNd − vN ′d +N ′′d ,

−cN ′b =Nb(1−Nb)− µNb + σNd.(58)

Note that the prime stands for the derivative with respect to z. System (58) isequivalent to

N ′b =N2b

c+

(µ− 1)Nbc

− σ

cNd,

N ′d =M,

M ′ =− µNb + σNd + (v − c)M.

(59)

34

Note that the steady states are now given by

x∗0 := (Nb0 , Nd0 ,M0) = (0, 0, 0) and x∗1 := (Nb1 , Nd1 ,M1) = (1,µ

σ, 0).

To find the behaviour at ±∞ of for Nb, Nd and M = N ′d, we need to takeinto account the fact that we have two types of waves, the upstream and thedownstream facing wave. For the upstream facing wave (see Figure 2), we havethe following boundary conditions at z = ±∞

Nb(∞) = 1, Nd(∞) =µ

σ, M(∞) = 0,

Nb(−∞) = 0, Nd(−∞) = 0, M(−∞) = 0.

Figure 2: The upstream facing wave sketch for either nb or nd.

In Figure 2 we see that the upstream facing wave with wave propagation speedc < 0 will imply population spread. This means that the population will persist.For a wave propagation speed c > 0, we see that the population goes extinct.

For a downstream facing wave (see Figure 3), we have the following bound-ary conditions

Nb(∞) = 0, Nd(∞) = 0, M(∞) = 0,

Nb(−∞) = 1, Nd(−∞) =µ

σ, M(−∞) = 0.

35

Figure 3: The downstream facing wave sketch for either nb or nd.

In Figure 3 we see that the downstream facing wave with a positive wave propa-gation speed will imply population spread and persistence, while for a negativewave propagation speed the population goes extinct.

To summarise:

Proposition 4 A travelling wave solution to (57) exists if and only if the threedimensional system (59) for some c 6= 0 admits a heteroclinic orbit connectingthe steady states x∗0 and x∗1.

c < 0 c > 0Upstream facing wave from x0 to x1 (pers.) from x0 to x1 (ext.)Downstream facing wave from x1 to x0 (ext.) from x1 to x0 (pers.)

Table 1: Population persistence and extinction for different values of the prop-agation speed c for the up- and downstream facing waves.

4.3 Necessary conditions for the existence of a hetero-clinic orbit by analysing the phase portraits.

To get conditions for the up- and downstream propagation speed c∗ of the trav-elling waves, we first need to understand that both steady states can only beconnected if there exists a heteroclinic orbit between them. In this section wewill study the phase portraits of system (59) to see for which values of the pa-rameters there could exist a heteroclinic orbit between the steady states.

The analysis and presentation that we provide in this section does not appearin [Pachepsky]. It is new as for as we could see in the literature.

The first step is to find the nullclines of system (59). From N ′b = 0, N ′d = 0 andM ′ = 0 we find

36

• Nb-nullcline: parabolic cylinder Nd = 1σNb(Nb + µ− 1).

• Nd-nullcline: plane M = 0.

• M -nullcline: plane −µNb + σNd + (v − c)M = 0.

Biologically relevant solutions are only there for which Nb, Nd ∈ R≥0. To makea sketch (see Figure 4 ) of how the nullclines are located in the space

S = {(Nb, Nd,M)|Nb ≥ 0, Nd ≥ 0,M ∈ R},

, we need to look more specific to the equations of these nullclines.

From N ′b = 0 we see that the location where the parabolic cylinder intersectsthe plane Nd = 0 in S depends on µ. From Nd = 0 we get Nb = 1 − µ. So forµ < 1 we have Nb = 1−µ > 0 and for µ ≥ 1 we get Nb = 0 because Nb ≥ 0. Tosee how the plane M ′ = 0 is located in the space S, we will study the normalvector of this plane. The normal vector of the plane −µNb+σNd+(v−c)M = 0is given by ~n = (−µ, σ, v− c). So the Nb-component of ~n is always negative andthe Nd-component always positive. For the M -component of ~n we have that itcan only be positive if v > c. So we always have v− c > 0 for c < 0, but if c > 0we must have v > c for v − c > 0. For v = c we get the plane Nd = µ

σNb whichdoes not depend on M any more. Note that this also means that for M = 0 weget the line Nd = µ

σNb.

37

(a) v − c > 0 and µ ≥ 1. (b) v − c > 0 and µ < 1.

(c) v − c < 0 and µ ≥ 1. (d) v − c < 0 and µ < 1.

Figure 4: Location of the nullclines for the travelling wave equations.

The Nb-nullcline is the parabolic cylinder (orange) and the M -nullcline is theplane (green). The Nd-nullcline is the plane M = 0, i.e. the (Nb, Nd)-coordinateplane. The bold lines lie in the respective surfaces and have M ≥ 0. For theNb-nullcline (orange), the parabolic intersection curve with the plane M = 0 isshown. Note that x∗0(0, 0, 0) and x∗1 = (1, µσ , 0) are the steady states.

Because the nullclines separate the relevant part of space S in different domainsof R3, we consider these domains as regions. Note that we find 23 = 8 differentregions because all the nullclines intersect each other. The regions are denotedby the Roman letters A to H. In the following definitions, ’above (or below) the

38

parabolic cylinder’ refers to a larger (or smaller) Nd-coordinate, while ’above(or below) the plane M ’=0’ refers to a larger (or smaller) M -coordinate.

• A:= The region above (v − c ≥ 0) or below (v − c < 0) the plane M ′ = 0and above the parabolic cylinder N ′b = 0 for M > 0.

• B:= The region below (v − c ≥ 0) or above (v − c < 0) the plane M ′ = 0and above the parabolic cylinder N ′b = 0 for M > 0.

• C:= The region below (v − c ≥ 0) or above (v − c < 0) the plane M ′ = 0and below the parabolic cylinder N ′b = 0 for M > 0.

• D:= The region above (v − c ≥ 0) or below (v − c < 0) the plane M ′ = 0and above the parabolic cylinder N ′b = 0 for M < 0.

• E:= The region below (v − c ≥ 0) or above (v − c < 0) the plane M ′ = 0and above the parabolic cylinder N ′b = 0 for M < 0.

• F := The region below (v − c ≥ 0) or above (v − c < 0) the plane M ′ = 0and below the parabolic cylinder N ′b = 0 for M < 0.

• G:= The region above (v − c ≥ 0) or below (v − c < 0) the plane M ′ = 0and below the parabolic cylinder N ′b = 0 for M > 0.

• H:= The region above (v − c ≥ 0) or below (v − c < 0) the plane M ′ = 0and below the parabolic cylinder N ′b = 0 for M < 0.

In Figure 4 we see that the form of the regions will change as µ changes fromµ ≥ 1 to µ < 1. Their geometric location will stay the same for µ ≥ 1 as forµ < 1 however.

Remark. The definitions of the regions are such that the normal vector ~nof the plane M ′ = 0 points into regions A,D,G and H, and out of regionsB,C,E and F . When c changes from v − c > 0 to v − c < 0, the direction ofchanges from pointing upwards to pointing downwards and the location of e.g.A and B changes accordingly. Be aware of this phenomenon when reading thesubsequent parts.

To get a better visual view of the location of the regions, in Figure 4 we sketchedthe regions in a 2D-sketch for M = −ε < 0, M = 0 and M = ε > 0 for the casev − c > 0 and µ ≥ 1. Note that ε is small.

39

(a) M = −ε < 0 (b) M = 0

(c) M = ε > 0

Figure 4: The location of the regions when intersected with the planes M =−ε < 0, M = 0 and M = ε > 0. The figure is drawn for v − c ≥ 0, µ ≥ 1.

Now we exactly know what the location of every different region is, we can con-tinue with finding the possible heteroclinic orbits. From now on we will considerthe cases c < 0, c > 0 separately. We start with c < 0.

In Tabel 2 we show the signs of N ′b, N′d and M ′ for c < 0 in every different

region.

40

XXXXXXXXXXRegionsSign

N ′b N ′d M ′

A + + +B + + −C − + −D + − +E + − −F − − −G − + +H − − +

Table 2: The signs of N ′b, N′d and M ′ for c < 0 in every region.

Proposition 5 Let c < 0 and µ ≥ 1. There does not exist a heteroclinic orbitfrom x∗1 to x∗0, through S.

Proof. Suppose, for the sake of contradiction, there exists a heteroclinic orbitfrom x∗1 to x∗0. Then there exist an orbit φ(t) : R → R3(= (Nb, Nd,M)) withφ(t) → x∗0 as t → ∞ and φ(t) → x∗1 as t → −∞. It is only possible for φ(t) toarrive in x∗0 through the regions A, B, C, D, E and F because these are theonly regions that are connected with x∗0, in the sense that x∗0 lies in their closure.

In A, B and C we have M > 0. This means that N ′d > 0. So φ(t) can neverreach x∗0 because the Nd-component of φ(t) will always become more positive.In D and E, the orbit will move in the direction of the parabolic cylinder N ′b = 0because N ′b > 0. So φ(t) can also not arrive in x∗0 through D and E becauseNb-component of φ(t) will always become more positive. So the only possibleoption for the orbit to arrive in x∗0 that is left is through F . But in F we haveM ′ < 0. And because M < 0, the M -component of the orbit will always becomemore negative which means that φ(t) can also never reach x∗0 through F . So wehave found a contradiction that there exists a heteroclinic orbit from x∗1 to x∗0.�

Proposition 6 Let c < 0 and µ < 1. There does not exist a heteroclinic orbitfrom x∗1 to x∗0, through S.

Proof. The intersection points of the parabolic cylinder N ′b = 0 with the planeNd = 0 are given by (Nb, Nd,M) = (0, 0,M) and (Nb, Nd,M) = (1 − µ, 0,M),M ∈ R. Note that 1 − µ > 0. We have that N ′b > 0 above N ′b = 0 and N ′b < 0below N ′b = 0. This means that the Nb-component of an orbit that starts in x∗1can never become less than 1−µ without crossing the plane Nd = 0, see Figure5. So an orbit that starts in x∗1 can never reach x∗0 through S.

41

Figure 5: Partial phase portrait for c < 0 and µ < 1, at some level M = M0 ∈R. Quantitatively, the situation is the same for each M0.

Proposition 7 Let c < 0. If there exists a heteroclinic orbit from x∗0 to x∗1,then this orbit moves from x∗0 to A to B to x∗1 possibly with multiple switchesbetween A and B.

Proof. Suppose there exists a heteroclinic orbit from x∗0 to x∗1. Then thereexists an orbit γ(t) : R → R3(= (Nb, Nd,M)) with γ(t) → x∗1 as t → ∞ andγ(t)→ x∗0 as t→ −∞. Because all the regions are connected to x∗1, it is possiblefor γ(t) to arrive in x∗1 through all these regions, in principle.

Lemma 4.1 Let c < 0. An orbit cannot reach x∗1 through A, E, F or G.

Proof. In A and G we have M ′ > 0. And because M > 0, the M -componentof an orbit will always become more positive. So an orbit can never reach x∗1through A or G. In E and F we have M ′ < 0. And because M < 0, theM -component of an orbit will always become more negative. So an orbit canalso never reach x∗1 through E and F . �

Lemma 4.2 Let c < 0. An orbit can only leave x∗0 through A.

Proof. The only possible regions for an orbit to leave x∗0 are A, B, C, D, Eand F , since these regions are connected to x∗0. In A we have N ′b > 0, N ′d > 0and M ′ > 0 which implies that an orbit could leave x∗0 through A. In B and Cwe have M ′ < 0, while M > 0. So an orbit starting from x∗0 and entering intoB or C cannot have increasing M -component, as it should. So an orbit can notleave x∗0 through B or C. In D, E and F we have M < 0, which means thatN ′d < 0. So the Nd-component can never become positive for an orbit startingfrom x∗0 and entering into D,E or F . This means that an orbit can also notleave x∗0 through D, E and F . So we conclude that the orbit can only leave x∗0through A. �

42

From Lemma 4.1 and Lemma 4.2 we find that γ(t) leaves x∗0 through A andthat it reaches x∗1 through B, C, D or H. Because we certainly know that γ(t)moves into A, we will look what happens with the orbit in A.

Lemma 4.3 Let c < 0. An orbit in A can only move out of A if it crosses theplane M ′ = 0.

Proof. In A we have M > 0, N ′b > 0, N ′d > 0 and M ′ > 0, see Table 2. So theonly possible way for an orbit to leave A is to cross the Nb-nullcline, that is theparabolic cylinder, or the plane M ′ = 0. But on the parabolic cylinder, on theboundary of A, the vectorfield points into A since for M > 0 we have N ′d > 0and M ′ > 0 on this Nb-nullcline. �

So after γ(t) moves into A, it has to cross the plane M ′ = 0 at least once. Butwhat happens on this plane? Because the orbit is still above the Nb-nullclineand Nd-nullcline, we get N ′b > 0 and N ′d > 0. This means that after γ(t) reachesthe plane M ′ = 0, it can move into B or A again.

Lemma 4.4 Let S0 := (N0b , N

0d ,M

0) be a point where an orbit reaches the

plane M ′ = 0, ~n the normal vector of M ′ = 0 and ρ = −µ((N0

b )2

c +(µ−1)N0

b

c −σcN

0d ) + σM0.

1. If ρ > 0, then an orbit in S0 is pointing in the direction of the same sideof the plane as ~n does.

2. If ρ < 0, then an orbit in S0 is pointing in the direction of the other sideof the plane as ~n does.

3. If ρ = 0, then there is no conclusion.

Proof. Suppose that an orbit reaches the plane M ′ = 0. The tangent vector~r of this orbit in S0 is parallel to the (Nb, Nd)-plane because M ′ = 0. Thistangent vector is given by ~r = (r1, r2, 0) with r1 = N ′b and r2 = N ′d = M0. Forthe normal vector of the plane M ′ = 0 we have ~n = (−µ, σ, v − c). Figure 5aand 5b show that if the angle between both vectors is less than 1

2π, then ~r ispointing in the direction of the same side of the plane as ~n does. If the anglebetween both vectors is greater than 1

2π, then ~r is pointing in the direction ofthe other side of the plane as ~n does. The angle between ~r and ~n can be foundby the following formula,

cos(α) =〈~n,~r〉‖~n‖ · ‖~r‖

.

43

Note that cos(α) > 0 for α < 12π and cos(α) < 0 for α > 1

2π. Furthermore, wehave that ‖~n‖ · ‖~r‖ > 0. So we only need to look to the sign of 〈~n,~r〉. We find

〈~n,~r〉 =− µr1 + σr2,

=− µr1 + σM0,

=− µ((N0

b )2

c+

(µ− 1)N0b

c− σ

cN0d ) + σM0.

So we conclude that if −µ((N0

b )2

c +(µ−1)N0

b

c − σcN

0d ) + σM0 > 0, then the orbit

in S0 is pointing in in the direction of the same side of the plane as ~n does. And

if −µ((N0

b )2

c +(µ−1)N0

b

c − σcN

0d ) + σM0 < 0, then the orbit in S0 is pointing in

the direction of the same side of the plane as ~n does. Note that for −µ((N0

b )2

c +(µ−1)N0

b

c − σcN

0d ) +σM0 = 0 we get Figure 5c and 5d. From these figures we see

that we cannot conclude which side of the plane the orbit will go. �

(a) ρ > 0. (b) ρ < 0.

(c) ρ = 0. (d) ρ = 0.

Figure 5: Direction of the tangent vector ~r for ρ > 0, ρ < 0 and ρ = 0.

44

Because v − c > 0 for c < 0 we conclude that after γ(t) reaches the plane

M ′ = 0, it will move into A if −µ((N0

b )2

c +(µ−1)N0

b

c − σcN

0d ) +σM0 > 0 or into B

if −µ((N0

b )2

c +(µ−1)N0

b

c − σcN

0d ) + σM0 < 0. Note that γ(t) could have multiple

touching events with the plane M ′ = 0 before it reaches a point where it maycross the plane.

Lemma 4.5 An orbit in B can only reach x∗1 if it reaches x∗1 through B withpossible multiple crossing events with A.

Proof. In B we have M > 0, N ′b > 0, N ′d > 0 and M ′ < 0. This means thatwe have the following possible options for an orbit in B (see Figure 6 ).

1. The orbit reaches the Nb-nullcline (the parabolic cylinder; orange).

2. The orbit reaches the plane M ′ = 0 (the upper plane; green).

3. The orbit reaches the plane M = 0 (the lower horizontal plane; white).

4. The orbit reaches the Nb-nullcline and the plane M = 0 at the same time.

5. The orbit reaches the planes M ′ = 0 and M = 0 at the same time.

6. The orbit reaches x∗1.

Figure 6: Case c < 0 and µ < 1.

The region B is the wedge-like shape enclosed by the green plane (M ′ = 0), theorange parabolic cylinder (N ′b = 0) and the (Nb, Nd)-coordinate plane (horizon-tal white plane.

Suppose an orbit reaches the Nb-nullcline. On this Nb-nullcline we have N ′d > 0and M ′ < 0 because M > 0. This means that the orbit will move into B again.If an orbit reaches the plane M ′ = 0, then we have found that the orbit pointsinto A or B. If it points into A, the orbit can come back in B or it will stay inA forever. The third option is that the orbit reaches the plane M = 0. Becausethe orbit is below the plane M ′ = 0, we have M ′ < 0. This means that theorbit will move into E. This cannot give a heteroclinic orbit because the orbitwill eventually move out of S, according to Lemma 4.6.

Lemma 4.6 Let c < 0. An orbit starting in F will leave S. An orbit startingin E will move into F .

45

Proof. In E we have M < 0, N ′b > 0, N ′d < 0 and M ′ < 0. This means thatthe orbit eventually will reach the Nb-nullcline after which it will move into F .But because M becomes more negative in F , the orbit eventually will cross thethe plane Nd = 0. So this leads to no biologically relevant solutions. �

If an orbit reaches the Nb-nullcline and the plane M = 0 at the same time,we get that the M -component of the orbit becomes more negative which im-plies that it will move into F . And in F , we saw that we get no biologicallyrelevant solutions eventually. The last option is that an orbit reaches the planesM ′ = 0 and M = 0 at the same time. If this happens, then the orbit stays onthe plane M = 0 in the beginning because N ′b > 0. Then it will move into Ebecause M ′ becomes negative. This will lead to no biologically relevant solu-tions according to Lemma 4.6. So we conclude that an orbit in B can only reachx∗1 if it reaches x∗1 through B with possible multiple crossing events with A. �

A schematic representation of the possible paths of an orbit that starts in x∗0,is given in Figure 7.

Figure 7: Representation of the possible paths of an orbit that starts in x∗0 forthe case c < 0.

46

Note that the circles stand for regions and the squares for steady states. Alsonotice that n.b.r.s stands for ’no biologically relevant solutions’ and that thisis the region where Nb < 0 or Nd < 0. To define the straight arrows, letX be a region with x0 ∈ X, x(t;x0) the solution which starts at x0 andτX(x0) = inf{t > 0|x(t, x0) ∈ Xc}, i.e. the time in which the solution touches anullcline for the first time. If inf{t > 0|x(t, x0) ∈ Xc} = ∅, then τX(x0) = −∞by definition. Now we can define the straight arrows. If there exists an arrowbetween X and itself, then there exists x0 ∈ X such that x(t, x0) ∈ X for allt ≥ 0. Let Y be another region. If there exists an straight arrow from X to Y ,then there exists a x0 ∈ X such that x(t, x0) ∈ Y for t > τX(x0) and t− τX(x0)sufficiently small. A dotted arrow means that an orbit will move from the steadystate into the indicated region or that there is an orbit starting in that regionconverging to the steady state.

So looking at the graph in Figure 7, we conclude that if there exists a hete-roclinic orbit from x∗0 to x∗1 for c < 0, then this orbit moves from x∗0 to A to Bto x∗1 possibly with multiple switches between A and B. �

Now we will look if there also could exist a heteroclinic orbit for c > 0. Notethat for c > 0 we have two cases: 0 < c < v and c > v. But because of thedefinitions we used for the regions A to H, the conclusions stay the same forboth of the cases.

Because c is positive, we have that N ′b will be negative in the regions abovethe parabolic cylinder N ′b = 0 and positive in the regions below, see Table 3.

XXXXXXXXXXRegionsSign

N ′b N ′d M ′

A − + +B − + −C + + −D − − +E − − −F + − −G + + +H + − +

Table 3: The signs of N ′b, N′d and M ′ in every region for c > 0.

Proposition 8 Let c > 0 and µ ≥ 1. There does not exist a heteroclinic orbitfrom x∗0 to x∗1 in S.

Proof. Suppose, for the sake of contradiction, there exists a heteroclinic orbitfrom x∗0 to x∗1. Then there exist an orbit φ(t) : R → R3(= (Nb, Nd,M)) withφ(t) → x∗1 as t → ∞ and φ(t) → x∗0 as t → −∞. It is only possible for φ(t) to

47

leave x∗0 through the regions A, B, C, D, E or F.

In A, B, D and E we have N ′b < 0. Because φ(t) starts in x∗0, the Nb-componentof the orbit can never become positive in these regions. So φ(t) cannot leavex∗0 through A, B, D and E. In C we have M ′ < 0. So the M -component ofthe orbit can never become positive which means that φ(t) can also never leavex∗0 through C. The last option to leave x∗0 is through F . But in F we havethat N ′d < 0. So the Nb-component of φ(t) can never become positive, whichimplies that the orbit can also not leave x∗0 through F . So we have found acontradiction that there exists a heteroclinic orbit from x∗0 to x∗1 contained inS. �

Proposition 9 Let c > 0 and µ < 1. There does not exist a heteroclinic orbitfrom x∗0 to x∗1 in S.

Proof. The intersection points of the parabolic cylinder N ′b = 0 with the planeNd = 0 are given by (Nb, Nd,M) = (0, 0,M) and (Nb, Nd,M) = (1 − µ, 0,M).Note that 1 − µ > 0. We have that N ′b < 0 above N ′b = 0 and N ′b > 0 belowN ′b = 0. This means that the Nb-component of an orbit that starts in x∗0 cannever become greater than 0 without getting a negative Nd-component, see Fig-ure 8. So an orbit that starts in x∗0 can never reach x∗1 which means there doesnot exist a heteroclinic orbit from x∗0 to x∗1. �

Figure 8: Partial phase portrait for c > 0 and µ < 1, on a plane M = M0,M0 ∈ R. For each M0 ∈ R the situation is quantitatively similar.

Lemma 4.7 Let c > 0. If there exists a heteroclinic orbit from x∗1 to x∗0, thenthis orbit moves from x∗1 to E to D to x∗0 possibly with multiple switches betweenE and D.

48

Proof. Suppose there exists a heteroclinic orbit from x∗1 to x∗0. Then thereexists an orbit γ(t) : R → R3(= (Nb, Nd,M)) with γ(t) → x∗0 as t → ∞ andγ(t)→ x∗1 as t→ −∞. It is only possible for γ(t) to reach x∗0 through A, B, C,D, E and F .

Lemma 4.8 Let c > 0. An orbit can only reach x∗0 through D.

Proof. In A, B and C we have M > 0. This means that N ′d > 0. So an orbitin A, B or C can never reach x∗0 because the Nd-component of the orbit willalways become more positive. In D we have N ′b < 0, N ′d < 0 and M ′ > 0 whichmeans that all the components of the orbit are pointing in the direction of x∗0.So an orbit could reach x∗0 through D. In E we have M < 0 and M ′ < 0. So anorbit can never reach x∗0 through E because the M -component becomes morenegative. It is also not possible to reach x∗0 through F because in F we haveN ′b > 0 which means that the Nb-component becomes more positive. So weconclude that an orbit can only reach x∗0 through D. �

Lemma 4.9 Let c > 0. An orbit that reaches x∗0, cannot leave x∗1 through A,B, C, D, F , G or H.

Proof. In A we have M ′ > 0, N ′b < 0 and N ′d > 0. This implies that forv − c ≥ 0, an orbit eventually will cross the plane Nb = 0. So this leads to nobiologically relevant solutions. For v − c < 0, an orbit can also reach the planeM ′ = 0. Because M > 0 and the orbit is still above N ′b = 0, we get N ′b < 0 andN ′d > 0. So after the orbit reaches the plane M ′ = 0, it will move into A again.And because N ′b stays negative, it will also lead to no biologically relevant solu-tions. In B and C we have M ′ < 0, which implies that the M -component of anorbit that starts in x∗1 never becomes positive. So it is impossible for an orbitto leave x∗1 through B or C. An orbit can also not leave x∗1 through D or Hbecause there we have M ′ > 0. So the M -component never becomes negative.In F we have M < 0, N ′b > 0, N ′d < 0 and M ′ < 0. This means that the orbiteventually will cross the the plane Nd = 0 for v − c ≥ 0. So this leads to nobiologically relevant solutions. For v − c < 0, an orbit can also reach the planeM ′ = 0. Because M < 0 and the orbit is still below N ′b = 0, we get N ′b > 0 andN ′d < 0. So after the orbit reaches the plane M ′ = 0, it will move into F again.And because N ′d stays negative, it will also lead to no biologically relevant solu-tions. In G we have M > 0, N ′b > 0, N ′d > 0 and M ′ > 0. So the only possibleway for an orbit to leave G is through the Nb-nullcline or M -nullcline.

Suppose that an orbit in G reaches the Nb-nullcline. On this nullcline we haveN ′d > 0 and M ′ > 0. This means that the orbit will move into A what leads tono biologically relevant solutions. Now suppose that an orbit in G reaches theM -nullcline. On this nullcline we have N ′b > 0 and N ′d > 0. So from here, theorbit can move into C or G again. Suppose the orbit will move into C. In Cwe have M > 0, N ′b > 0, N ′d > 0 and M ′ < 0. This means that the orbit canonly leave C through the M -nullcline or Nd-nullcline. We found that the orbitwill move into C or G after it reaches the M -nullcline. So suppose it reaches

49

the Nd-nullcline. On this nullcline we have N ′b > 0 and M ′ < 0 which meansthat the orbit will move into F . And we found that this leads to no biologicallyrelevant solutions. So we conclude that an orbit that could reach x∗0, can notleave x∗1 through A, B, C, D, F , G or H. � .

From Propostions 4.8 and 4.9 it follows that γ(t) must leave x∗1 through Eand that it reaches x∗0 through D. Now we will look what happens with anorbit in E.

Proposition 10 Let c > 0. An orbit in E can only reach D if it crosses theplane M ′ = 0 from E.

Proof. In E we have M < 0, N ′b < 0, N ′d < 0 and M ′ < 0. This means thatan orbit can only leave E if it reaches the Nb-nullcline or the plane M ′ = 0.On the Nb-nullcline, we have N ′d < 0 and M ′ < 0. So the orbit will move intoF . And we found that this leads to no biologically relevant solutions. On theM -nullcline, we have N ′b < 0 and N ′d < 0. So on this nullcline, an orbit willmove into D or E again. Note that for v − c ≥ 0, an orbit needs to reach theplane M ′ = 0 before Nb becomes negative. We conclude that an orbit in E canonly reach D if it crosses the plane M ′ = 0 from E. �

So after γ(t) moves into E, it has to cross the plane M ′ = 0 at least onceto get into D. The conditions that γ(t) crosses the plane M ′ = 0 can be foundin Lemma 4.4. Note that γ(t) could have multiple touching events with theplane M ′ = 0 before it reaches a point where it will cross the plane.

Proposition 11 Let c > 0. An orbit in D can only reach x∗0 if it reaches x∗0through D with possible multiple crossing events with E.

Proof. In D we have M < 0, N ′b < 0, N ′d < 0 and M ′ > 0. This means thatwe have the following possible options for an orbit in D.

1. The orbit reaches the plane M ′ = 0.

2. The orbit reaches the plane M = 0.

3. The orbit reaches the plane M ′ = 0 and the plane M = 0 at the sametime.

4. The orbit reaches the plane Nb = 0 without reaching x∗0.

5. The orbit reaches x∗0

Suppose an orbit reaches the plane M ′ = 0. On this plane, we found that anorbit will move into E or D again. If it points into E, then we have seen thatit will return to D or it will lead to no biologically relevant solutions. On theplane M = 0, we have N ′d < 0 and M ′ > 0. So an orbit that reaches the planeM = 0 from D will move into A. And here we found that it can never reachx∗0 again. The third option is that an orbit reaches the planes M ′ = 0 and

50

M = 0 at the same time. Then we find that the orbit will also move into Abecause first we only get N ′b < 0 and then also M ′ > 0. If an orbit reachesthe plane Nb = 0, then we get no biologically relevant solutions because onthis plane we have N ′b < 0. So we conclude that an orbit in D can only reachx∗0 if it reaches x∗0 through D with possible multiple crossing events with E. �

A schematic representation of the possible paths of an orbit that starts in x∗1,is given in Figure 9.

Figure 9: Representation of the possible paths of an orbit that starts in x∗1 forthe case c > 0.

51

So looking to Figure 9 we conclude that if there exists a heteroclinic orbit fromx∗1 to x∗0 for c > 0, then this orbit moves from x∗1 to E to D to x∗0 possibly withmultiple switches between E and D. �

Conclusion: From Propositions 5, 6, 8 and 9 it seems that extinction is notrealized by means of ”washout”, nor a downwards moving extinction front noran upwards moving extinction front (see Table 1). The decay of the populationseems to be more complicated.

4.4 Dimensionality of the stable and unstable manifolds.

In Section 4.3 we have seen that there could only exist a heteroclinic orbit fromx∗0 to x∗1 for c < 0 and from x∗1 to x∗0 for c > 0. The conclusion of the previoussection does not imply that there exists a heteroclinic orbit for any c indicated.A necessary condition for the existence of a heteroclinic orbit is that one steadystate must have a non-trivial unstable manifold and the other one a non-trivialstable manifold. These should ’point’ into the right region, as analysed in theprevious section. In this section we will analyse the stability of the steady statesafter which we can determine the possible dimension of the stable and unstablemanifolds. In this we follow [Pachepsky], but elaborate substantially in detailon the exposition provided there.

4.4.1 Linear stability analysis.

To analyse the stability of the both steady states, we first linearize system (59)around both steady states and then we determine the characteristic polynomialsto find the eigenvalues.

Lemma 4.10 Linearization around x∗0 = (0, 0, 0) yields the characteristic poly-nomial

P0(λ) = A1λ3 +A2λ

2 +A3λ+A4, (60)

where

A1 =− 1,

A2 =µ− 1

c+ v − c,

A3 =(µ− 1)(1− v

c) + σ,

A4 =σ

c.

Proof. The Jacobian Df(Nb, Nd,M) of the right-hand side of (59) is given by

Df(Nb, Nd,M) =

2Nb+µ−1c −σc 00 0 1−µ σ v − c

.

Substituting x∗0 = (0, 0, 0) gives

52

Df(0, 0, 0) =

µ−1c −σc 00 0 1−µ σ v − c

.

For the characteristic polynomial P0(λ), we obtain

P0(λ) =det(Df(0, 0, 0)− λI

),

=− λ(v − c− λ)(µ− 1

c− λ) +

σµ

c− σ(

µ− 1

c− λ),

=v(1− µ)

cλ+ vλ2 + (µ− 1)λ− cλ2 +

µ− 1

cλ2 − λ3 +

σµ

c+σ(1− µ)

c+ σλ,

=− λ3 +(µ− 1

c+ v − c

)λ2 +

((µ− 1)(1− v

c) + σ

)λ+

σ

c. �

Lemma 4.11 Linearization around x∗1 = (1, µσ , 0) yields the characteristic poly-nomial

P1(λ) = B1λ3 +B2λ

2 +B3λ+B4, (61)

where

B1 =− 1,

B2 =µ+ 1

c+ v − c,

B3 =(µ+ 1)(1− v

c) + σ,

B4 =− σ

c.

Proof. The Jacobian Df(Nb, Nd,M) for x∗1 = (1, µσ , 0) is given by

Df(1, µσ , 0) =

µ+1c −σc 00 0 1−µ σ v − c

.

For the characteristic polynomial, we obtain

P1(λ) =det(Df(1,

µ

σ, 0)− λI

),

=− λ(v − c− λ)(µ+ 1

c− λ) +

σµ

c− σ(

µ+ 1

c− λ),

=− v(µ+ 1)

cλ+ vλ2 + (µ+ 1)λ− cλ2 +

µ+ 1

cλ2 − λ3 +

σµ

c− σ(µ+ 1)

c+ σλ,

=− λ3 +(µ+ 1

c+ v − c

)λ2 +

((µ+ 1)(1− v

c) + σ

)λ− σ

c. �

4.4.2 Dimensionality of the stable and unstable manifolds.

To determine the possible number of positive and negative real roots of P0(λ) =0 and P1(λ) = 0, we will use Descarte’s Rule of Signs (see Theorem 4.12, [2]).We consider the cases c < 0 and c > 0 separately.

53

Theorem 4.12 (Descarte’s rule of signs, [2]) If the terms of a single-variablepolynomial with real coefficients are ordered by descending variable exponent,then the number of positive roots of the polynomial is either equal to the numberof sign differences between nonzero coefficients, or is less than it by an evennumber. The number of negative roots is the number of sign changes after mul-tiplying the coefficients of odd-power terms by -1, or fewer than it by an evennumber.

The case c < 0.

First we will look to the possible dimensions of the stable and unstable man-ifolds of the zero steady state. Using Descartes rule of signs, we need to lookto the sign changes of the coefficients of P0(λ) and P0(−λ). In P0(λ) we havethat A1 < 0, A2 > 0 and A4 < 0 for µ < 1. For µ > 1 we have that A1 < 0,A3 > 0 and A4 < 0. And for µ = 1 we have that A1 < 0, A2 > 0, A3 > 0 andA4 < 0. This means that the coefficients of P0(λ) always have two sign changes.So P0(λ) = 0 has zero or two positive real roots. For P0(−λ) we have

P0(−λ) = −A1λ3 +A2λ

2 −A3λ+A4. (62)

We see that −A1 > 0, A2 > 0 and A4 < 0 for µ < 1. For µ > 1 we have−A1 > 0, −A3 < 0 and A4 < 0. And for µ = 1 we have −A1 > 0, A2 > 0,−A3 < 0 and A4 < 0. This implies that P0(−λ) = 0 has exactly one positivereal root, which means that P0(λ) = 0 has exactly one negative real root. Nowthere are two possible cases,

1. P0(λ) = 0 has two positive real roots and one negative real root.

2. P0(λ) = 0 has one negative real root and two conjugated non-real roots,both with either Re > 0 or Re < 0.

We know that if P0(λ) = 0 has two complex roots, the real part of these rootshas the same sign because non-real roots of a polynomial with real coefficientsmust occur in conjugate pairs. Now we know all the possible roots of P0(λ) = 0,we conclude that the possible dimension of the stable manifold of the zero steadystate, W s

0 , is 1 or 3. For the unstable manifold Wu0 we have that the possible

dimension is 2 or 0, respectively.

Now we will determine the possible dimensions of the stable and unstable man-ifolds of the non-zero steady state. In P1(λ), we have that B1 < 0, B3 > 0 andB4 > 0. This means that P1(λ) = 0 has exactly one positive real root. ForP1(−λ) we have

P1(−λ) = −B1λ3 +B2λ

2 −B3λ+B4. (63)

We see that −B1 > 0, −B3 < 0 and B4 > 0. This implies that P1(−λ) = 0has zero or two positive real roots, which means that P1(λ) = 0 has zero or twonegative real roots. Now there are again two possible cases,

54

1. P1(λ) = 0 has one positive real root and two negative real roots.

2. P1(λ) = 0 has one positive real root and two conjugated non-real roots,both with either Re > 0 or Re < 0.

So we conclude that dim(W s1 ) = 0 or 2 and dim(Wu

1 ) = 3 or 1, respectively.

The case c > 0.

In P0(λ), we have that A1 < 0 and A4 > 0 (For µ = 1 we also have A3 > 0).This means that P0(λ) = 0 has one or three positive real roots. In P0(−λ),we have that −A1 > 0 and A4 > 0 (For µ = 1 we also have −A3 < 0) Thisimplies that P0(−λ) = 0 has zero or two positive real roots, which means thatP0(λ) = 0 has zero or two negative real roots. Now there are three possiblecases,

1. P0(λ) = 0 has three positive real roots.

2. P0(λ) = 0 has one positive real root and two negative real roots.

3. P0(λ) = 0 has one positive real root and two conjugated non-real roots,both with either Re > 0 or Re < 0.

So we conclude that dim(W s0 ) = 0 or 2 and dim(Wu

0 ) = 3 or 1, respectively.

In P1(λ), we have that B1 < 0 and B4 < 0. This means that P1(λ) = 0has zero or two positive real roots. In P1(−λ), we have that −B1 > 0 andB4 < 0. This implies that P1(−λ) = 0 has one or three positive real roots,which means that P1(λ) = 0 has one or three negative real roots. Now thereare again three possible cases,

1. P1(λ) = 0 has three negative real roots.

2. P1(λ) = 0 has two positive real roots and one negative real root.

3. P1(λ) = 0 has one negative real root and two conjugated non-real roots,both with either Re > 0 or Re < 0.

So we conclude that dim(W s1 ) = 1 or 3 and dim(Wu

1 ) = 2 or 0, respectively.

A summary of the results on dimensions of the stable and unstable manifold ofx∗0 and x∗1 are given in Tables 4 and 5.

c < 0 c > 0dim(W s

0 ) 1 or 3 0 or 2dim(Wu

0 ) 2 or 0 3 or 1

Table 4: The possible dimensions of the stable and unstable manifold of x∗0.

55

c < 0 c > 0dim(W s

1 ) 0 or 2 1 or 3dim(Wu

1 ) 3 or 1 2 or 0

Table 5: The possible dimensions of the stable and unstable manifold of x∗1.

Remark (1). In [Pachepsky], the possibility of non-real roots had not beentaken into consideration.

Remark (2). Tables 4 and 5 do not provide very insightful necessary con-ditions for the existence of a heteroclinic orbit. One needs to deal with with thesolutions to the characteristic polynomials properly, to exclude existence

4.5 The propagation speeds and necessary conditions forthe existence of a heteroclinic orbit by analysing thedimensionality of the manifolds.

For the existence of a heteroclinic orbit connecting the steady states x∗0 and x∗1,it is necessary that one steady state has an unstable manifold and the other onea stable manifold. However, this does not prove the existence of a heteroclinicorbit. It only says that it could be possible there exists one. In this section weonly provide necessary conditions for the existence of a heteroclinic orbit, basedon consideration of the dimensionality of stable and unstable manifolds. In thiswe follow [Pachepsky] essentially, but elaborate on their exposition.

In the previous section we saw that P0(λ) always has one negative real rootfor c < 0. We also saw that P1(λ) always has one positive real root for c < 0.So there is no obstruction to the existence of a heteroclinic orbit starting fromthe non-zero steady state to the zero steady state because both manifolds havedimension of at least one.

Note that P0(λ) and P1(λ) both could have non-real roots. For the unstablemanifold of the non-zero steady state this will not give any problems but for thestable manifold of the zero steady state it will. The approach of the orbit to thesteady states would be oscillatory what in the case of the zero steady state willlead to negative solutions. So this would not give a biologically relevant solution.

Now we know that it could be possible there is a heteroclinic orbit startingfrom the non-zero steady state to the zero steady state for c < 0, we also wouldlike to find conditions on the propagation speed such that there exists a hetero-clinic orbit starting from the zero steady state to the non-zero steady state forc < 0.

Proposition 12 Let c < 0. If there exists an upstream travelling wave from x∗0

56

to x∗1 with speed c, then R0+(c) > 0 and R1

−(c) < 0, where

R0+(c) = −(λ0+)3 +A2(λ0+)2 +A3λ

0+ +A4, (64)

λ0+ =A2 +

√A2

2 + 3A3

3,

and

R1−(c) =− (λ1−)3 +B2(λ1−)2 +B3λ

1− +B4, (65)

λ1− =B2 −

√B2

2 + 3B3

3.

Proof. To make it possible to have a heteroclinic orbit from the zero steadystate to the non-zero steady state, we need that P0(λ) has at least one realpositive root and P1 at least one root with Re < 0 (it is allowed that the orbitoscillates around the non-zero steady state). We will first look to the roots ofP0(λ).

From the previous section we see that the only case in which we for certainknow that P0(λ) has at least one positive real root is the case where P0(λ) hastwo positive real roots and one negative real root. The transition point betweenzero or two positive real roots is exactly when the local maximum of P0(λ) is apositive real root, see Figure 10. This means that P ′0(λ) = 0 and P0(λ) = 0 forλ > 0.

Figure 10: The shape of P0(λ) when its two roots switch from complex to real.

From P ′0(λ) = 0 we find

−3λ2 + 2A2λ+A3 = 0. (66)

With the abc-formula we find that the solutions to (66) are

λ0± =A2 ±

√A2

2 + 3A3

3. (67)

57

Note that λ0± is a function of c.

We have that P ′0(λ) has one positive and one negative real root. Here λ0+ isthe positive real root which corresponds to the local maximum of P0(λ) and λ0−the negative real root which corresponds to the local minimum of P0. Note thatP0(λ) has exactly one real positive root if P0(λ0+) = 0. So we conclude thatif P0(λ0+) > 0, then P0(λ) has two positive real roots. So for P0(λ0+) > 0, thedimension of the unstable manifold of the zero steady state is two.

For the propagation speed, we know that P0(λ0+) is a function of c. We de-fine R0

+(c) := P0(λ0+). The critical propagation speed c∗0 can then be found bysolving the equation,

R0+(c∗0) = −(λ0+)3 +A2(λ0+)2 +A3λ

0+ +A4 = 0. (68)

This means that for all values of c with c < 0 and R0+(c) > 0, it could be possible

there is an upstream travelling wave starting starting from the zero steady state.

For P1(λ), we need at least one root with Re < 0. The only case in whichwe certainly know we have at least one root with Re < 0 is the case whereP1(λ) = 0 has one positive real root and two negative real roots. The transitionpoint between zero negative real roots and two negative real roots is exactlywhen the local minimum of P1(λ) is a negative real root, see Figure 11. Thismeans that P ′1(λ) = 0 and P1(λ) = 0 for λ < 0.

Figure 11: The shape of P1(λ) when it’s two roots switch from complex toreal.

From P ′1(λ) = 0 we find

−3λ2 + 2B2λ+B3 = 0. (69)

With the abc-formula we find that the solutions to (69) are

λ1± =B2 ±

√B2

2 + 3B3

3. (70)

58

Note that λ1± is a function of c.

We have that P ′1(λ) has one positive and one negative real root. Here λ1− isthe negative real root which corresponds to the local minimum of P1(λ) andλ1+ the positive real root which corresponds to the local maximum of P1. Notethat P1(λ) has exactly one negative real root if P1(λ1−) = 0. So we concludethat if P1(λ11) < 0, then P1 has two negative real roots. So for P1(λ1−) < 0, thedimension of the stable manifold of the non-zero steady state is two.

For the propagation speed, we know that P1(λ1−) is a function of c. We de-fine R1

−(c) := P1(λ1−(c)). The critical propagation speed c∗1 can then be foundby solving the equation,

R1−(c∗1) = −(λ1−)3 +B2(λ1−)2 +B3λ

1− +B4 = 0. (71)

This means that for all values of c with c < 0 and R1−(c) < 0, it could be possible

there is a upstream travelling wave to the non-zero steady state. �

Remark.The critical speeds c∗0 and c∗1 defined in (68) and (71) provide boundson possible wave propagation speeds c that these need to satisfy. One cannotconclude of course that for c that satisfy the bounds, there exists a heteroclinicorbit.

In the previous section we also saw that P0(λ) always has al least one posi-tive real root and P1(λ) always has at least one negative real root for c > 0. Sothere is no obstruction to the existence of a heteroclinic orbit starting from thezero steady state to the non-zero steady state from dimensionality considerationbecause both manifolds have dimension of at least one.

Now we know that it could be possible there is a heteroclinic orbit startingfrom the zero steady state to the non-zero steady state for c > 0, we also wantto find conditions on the propagation speed in this case, similar to the onesestablished before.

Proposition 13 Let c > 0. If there exists a downstream travelling wave fromx∗1 to x∗0 with speed c, then R0

−(c) < 0 and R1+(c) > 0, with

R0−(c) =− (λ0−)3 +A2(λ0−)2 +A3λ

0− +A4,

λ0− =A2 −

√A2

2 + 3A3

3,

and

R1+(c) =− (λ1+)3 +B2(λ1+)2 +B3λ

1+ +B4,

λ1+ =B2 +

√B2

2 + 3B3

3.

59

Proof. To make it possible to have a heteroclinic orbit from the non-zero steadystate to the zero steady state, we need that P0(λ) has at least one negative realroot and P1(λ) at least one root with Re > 0.

The only case that we know for sure we have at least one negative real rootis the case where P0(λ) = 0 has one positive real root and two negative realroots. The transition point between zero or two negative real roots is exactlywhen the local minimum of P0(λ) is a negative real root, see Figure 12. Thismeans that P ′0(λ) = 0 and P0(λ) = 0 for λ < 0.

Figure 12: The shape of P0(λ) when its two roots switch from complex to real.

From P ′0(λ) = 0 we find the same solutions as in (69). Here λ0− is the negativereal root which corresponds to the local minimum of P0(λ) and λ0+ the positivereal root which corresponds to the local maximum of P0(λ). Note that P0(λ) hasexactly one negative real root if P0(λ0−) = 0. So we conclude that if P0(λ0−) < 0,then P0(λ) has two negative real roots. So for P0(λ−) < 0, the dimension of thestable manifold of the zero steady state is two.

For the propagation speed, we know that P0(λ0−) is a function of c. We de-fine R0

−(c) := P0(λ0−(c)) The critical propagation speed c∗0 can then be foundby solving the equation,

R0−(c∗0) = −(λ0−)3 +A2(λ0−)2 +A3λ

0− +A4 = 0.

This means that for all values of c with c > 0 and R0−(c) < 0, it could be possible

there is a downstream travelling wave to the zero steady state.

For P1(λ), we need at least one root with Re > 0. The only case in whichwe certainly know we have at least one root with Re > 0 is the case whereP1(λ) = 0 has two positive real roots and one negative real root. The transitionpoint between zero positive real roots and two positive real roots is exactly whenthe local maximum of P1(λ) is a positive real root, see Figure 13. This meansthat P ′1(λ) = 0 and P1(λ) = 0 for λ > 0.From P ′1(λ) = 0 we find the same solutions as in (70). Here λ1+ is the positive

60

Figure 13: The shape of P1(λ) when its two roots switch from complex to real.

real root which corresponds to the local maximum of P1(λ) and λ1− the negativereal root which corresponds to the local minimum of P1(λ). Note that P1(λ) hasexactly one positive real root if P1(λ1+) = 0. So we conclude that if P1(λ1+) > 0,then P1(λ) has two positive real roots. So for P1(λ1+) > 0, the dimension of theunstable manifold of the non-zero steady state is two.

For the propagation speed, we know that P1(λ1+) is a function of c. We de-fine R1

+(c) := P1(λ1+(c)) The critical propagation speed c∗1 can then be foundby solving the equation,

R1+(c∗1) = −(λ1+)3 +B2(λ1+)2 +B3λ

1+ +B4 = 0.

This means that for all values of c with c > 0 and R1+(c) > 0, it could be possible

there is a downstream travelling wave starting from non-zero steady state. �

We would like to see how c∗0 and c∗1 depend on v (and L) such that one may ob-tain conditions that exclude a propagation speed c that satisfies the conditionsof Proposition 12 (for c < 0) and Proposition 13 (for c > 0), this excluding atravelling wave solution. To that end we have:

Proposition 14 R0+(c) is given by

R0+(c) = λ0+(

2

9A2

2 +2

3A3) +

1

9A2A3 +A4.

Proof. Because λ0+ satisfies (66), we find

(λ0+)2 =2

3A2λ

0+ +

1

3A3. (72)

61

Substituting (72) into (64) yields

R0+(c) =− (λ0+)3 +A2(λ0+)2 +A3λ

0+ +A4,

=− λ0+(2

3A2λ

0+ +

1

3A3) +A2(

2

3A2λ

0+ +

1

3A3) +A3λ

0+ +A4,

=− 2

3A2(λ0+)2 − 1

3A3λ

0+ +

2

3A2

2λ0+ +

1

3A2A3 +A3λ

0+ +A4,

=− 2

3A2(

2

3A2λ

0+ +

1

3A3) +

2

3A3λ

0+ +

2

3A2

2λ0+ +

1

3A2A3 +A4,

=− 4

9A2

2λ0+ −

2

9A2A3 +

2

3A3λ

0+ +

2

3A2

2λ0+ +

1

3A2A3 +A4,

=2

9A2

2λ0+ +

2

3A3λ

0+ +

1

9A2A3 +A4,

=λ0+(2

9A2

2 +2

3A3) +

1

9A2A3 +A4. �

Note that the calculations for determining R0−(c), R1

+(c) and R1−(c) are almost

exactly the same. We find

R0−(c) =λ0−(

2

9A2

2 +2

3A3) +

1

9A2A3 +A4,

R1+(c) =λ1+(

2

9B2

2 +2

3B3) +

1

9B2B3 +B4,

R1−(c) =λ1−(

2

9B2

2 +2

3B3) +

1

9B2B3 +B4.

Unfortunately, the expressions for the Ri±(c) are such that they do not providemuch insight into exclusion of travelling wave solutions. It should be noted thatin [Pachepsky] for a very particular choice of parameters (µ = 0.8, σ = 0.8),c∗0 and c∗1 are numerically solved, providing an indication of the range in whichpropagation speeds should be in that particular case. Generally results seemhard to obtain (analytically) though.

62

5 Conclusions

In this thesis we studied a model for a population residing in a stream or smallriver, subject to advection (stream flow) and diffusion (representing randommovement). The individuals of the population can live on the benthos or driftin the flow, while reproduction can only occur on the benthos.

For the case µ < 1 we showed that persistence of the population is alwaysguaranteed, irrespective of the domain length and the advection speed, becausethe total growth rate of the benthic population at each location is always posi-tive.

For the case µ ≥ 1, we first derived a necessary condition for persistence ofthe population. We found that this condition is given by

λ1(v, L) <σ

µ− 1.

For µ ≥ 1 we also calculated that persistence is possible provided that thedomain L is large enough with respect to the advection speed v. This conditionis given by

L > L∗ =2√

4σµ−1 − v2

[π + arctan

(− 1

v

√4σ

µ− 1− v2

)].

This critical length contains the additional term π that is missing in [Pachep-sky]. We also found, that for any L, the population will persist provided µ ≥ 1sufficiently close to 1.

In the next section we considered spatial spread of the population in time,by means of a travelling population front. In mathematically terms this meansconsidering the existence of a travelling wave solution to an idealized modeldefined on R instead of the original model on (0, L). The considerations werelimited to a logistic growth model for the population on the benthos.

We distinguished between the propagation speed downstream and upstreambecause with increasing advection the propagation speed downstream increases,whereas the propagation speed upstream decreases. Before we determined theup- and downstream propagations speeds, we recast the three dimensional sys-tem into travelling wave coordinates and then transformed it into a system offirst-order equations, see (59).

Next we assumed there exists a travelling wave between the found steady statesand we determined by phase portrait analyse necessary conditions there ac-tually could exist one. Because all the nullclines of the 3-dimensional foundsystem intersect each other, we ended up with eight different regions which wedenoted by A to H. We found that if there exists a heteroclinic orbit from the

63

zero-steady state to the non-zero steady state, then c < 0 and this orbit movesfrom the zero steady state to A to B to the non-zero steady state possibly withmultiple switches between A and B, see Figure 7. We also found that if thereexists a heteroclinic orbit from the non-zero steady state to zero steady state,then c > 0 and this orbit moves from the non-zero steady state to E to D to thezero steady state possibly with multiple switches between E and D, see Figure9. The dimensionality of the problem prevented as from providing sufficientconditions for the existence of travelling waves, though. In [Pachepsky], theaspect of existence is not treated either. We obtained a more detailed view ofhow the heteroclinic orbits need to run through the biologically relevant part ofstate space.

We also found that extinction is not realized by means of ”washout”, nor adownwards moving extinction front nor an upwards moving extinction front(see Table 1). The decay of the population seems to be more complicated.

Then we linearized the system around the steady states, which informed usabout the stability of the manifolds around the steady states. The characteris-tic polynomials belonging to the zero and non-zero steady state are given by

P0(λ) = −λ3 +(µ− 1

c+ v − c

)λ2 +

((µ− 1)(1− v

c) + σ

)λ+

σ

c,

and

P1(λ) = −λ3 +(µ+ 1

c+ v − c

)λ2 +

((µ+ 1)(1− v

c) + σ

)λ− σ

c.

With the use of these polynomials and Descarte’s rule of Signs we determinedthe possible dimensionality of the stable and unstable manifolds of the steadystates, see Table 4 and Table 5.

In the last part of this thesis we determined conditions on the propagationspeed of the travelling wave that necessarily need to hold if the wave exists.This provides further necessary conditions for the existence of a heteroclinicorbit between the steady states. We found that if there exists an upstreamtravelling wave with speed c < 0 starting from the zero steady state to thenon-zero steady state, then R0

+(c) > 0 and R1−(c) < 0. And if there exists a

downstream travelling wave with speed c > 0 starting from the non-zero steadystate to the zero steady state, then R0

−(c) < 0 and R1+(c) > 0. Where R0

±(c)and R1

±(c) are given by

R0±(c) =λ0±(

2

9A2

2 +2

3A3) +

1

9A2A3 +A4,

R1+(c) =λ1±(

2

9B2

2 +2

3B3) +

1

9B2B3 +B4,

64

where

λ0± =A2 ±

√A2

2 + 3A3

3,

λ1± =B2 ±

√B2

2 + 3B3

3,

and with A2, A3, B2 and B3 as in (60) and (61). The critical propagationspeed for the upstream travelling wave can be found by solving the equationsR0

+(c∗0) = 0 and R1−(c∗1) = 0. And for the downstream travelling wave R0

−(c∗0) =0 and R1

+(c∗1) = 0

65

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[7] Humphries, S., Ruxton, G.D., (2002). Is there really a drift paradox? J.Anim. Ecol. 71, 151-154.

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[13] Pachpatte, B.G. (1998). Inequalities for differential and integral equations.San Diego: Academic Press.

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66

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µ en σ en µ∗(σ)

67