Population growth � analysis of an age

structure population model

Nina Håkansson

LITH-MAT-EX�05/16�SE

Examensarbete: 20p

Level: D

Examiner: Vladimir KozlovDepartment of mathematicsApplied mathematicsLinköpings universitet

Supervisors: Vladimir Kozlov andBengt Ove TuressonDepartment of mathematicsApplied mathematicsLinköpings universitet

Uno WennergrenDepartment of biologyTheory and ModelingLinköpings universitet

Linköping May 2005

Matematiska Institutionen581 83 LINKOPINGSWEDEN

may 2005

x x

http://www.ep.liu.se/exjobb/mai/2005/tm/016/

LiTH - MAT - EX - - 05/16 - - SE

Population growth — analysis of an age structure population model

Nina Hakansson

This report presents an analysis of a partial differential equation, resulting from pop-ulation model with age structure. The existence and uniqueness of a solution to theequation are proved. We look at stability of the solution. The asymptotic behaviourof the solution is treated. The report also contains a section about the connection be-tween the solution to the age structure population model and a simple model withoutage structure.

Age structure, population model, partial differential equation, asymptotics, stability.Nyckelord

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ISSN0348-2960

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Svenska/Swedish

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Abstract

This report presents an analysis of a partial di�erential equation, resulting frompopulation model with age structure. The existence and uniqueness of a solutionto the equation are proved. We look at stability of the solution. The asymptoticbehaviour of the solution is treated. The report also contains a section aboutthe connection between the solution to the age structure population model anda simple model without age structure.

Contents

1 Introduction 91.1 Population models . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.1.1 Population models without age structure . . . . . . . . . 91.1.2 Population models with age structure � time-independent

case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.3 Population models with age structure � time-dependent

case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.4 Other population models with age structure . . . . . . . 10

1.2 Presentation of the model to study . . . . . . . . . . . . . . . . . 111.2.1 What to study . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Existence and uniqueness of solutions 132.1 The integral equation . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Existence and uniqueness of the solution . . . . . . . . . . . . . . 15

3 Stability of the model 173.1 Stability, considering m . . . . . . . . . . . . . . . . . . . . . . . 17

4 Asymptotics of N� the time-independent case 184.1 Representation of n(0, · ) . . . . . . . . . . . . . . . . . . . . . . 19

4.1.1 Domain of convergence . . . . . . . . . . . . . . . . . . . . 194.1.2 The Laplace transform of n(0, · ) . . . . . . . . . . . . . 204.1.3 Asymptotic behaviour of Ln(0, · ) . . . . . . . . . . . . 224.1.4 The representation of n(0, · ) . . . . . . . . . . . . . . . . 22

4.2 Asymptotics of N . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5 Asymptotics of N�the time-dependent case 265.1 Upper bounds for n(0, · ) . . . . . . . . . . . . . . . . . . . . . . 265.2 Lower bounds for n(0, · ) . . . . . . . . . . . . . . . . . . . . . . 295.3 Upper and lower bounds for N . . . . . . . . . . . . . . . . . . . 315.4 Comparison with the time-independent case . . . . . . . . . . . . 32

6 Comparison with the model without age structure 336.1 Biological analysis of the requirement σ0 > −1 . . . . . . . . . 34

7 Conclusions 35

A Banach theory 37A.1 De�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37A.2 The Banach �xed point theorem . . . . . . . . . . . . . . . . . . 37

1 Introduction

This section presents population models of di�erent types. In the followingsections we will analyse the age structure population models.

1.1 Population models

The number of individuals N in a population at time t, from a given start timet = 0, can be described with di�erential equations of varying complexity andaccuracy. The following models are presented in [1].

1.1.1 Population models without age structure

The change of the population can be described by the conservation equation:

dN(t)dt

= births− deaths + migration.

The simplest model occurs when there is no migration and the death andbirth rates are proportional to N . This model is frequently used in physics,chemistry and biology:

dN(t)dt

= bN(t)− dN(t) with solution N(t) = N0e(b−d)t, (1)

where N0 = N(0) is the initial population, b the constant birth rate and d theconstant death rate. This model predicts that N will increase as t →∞ if b > d.

It is reasonable to assume that a population will not increase forever; theenvironment is probably limiting N . At a certain population size there will, forexample, be insu�cient resources to support a larger population. The logisticgrowth (suggested by Verhulst 1836) includes such a limiting part:

dN(t)dt

= rN(t)(

1− N(t)K

).

Assume r > 0. If N0 < K the population will increase, but it will never exceedN = K. If N0 > K the population will instead decrease. The constant K iscalled the carrying capacity of the environment and is the largest N that theenvironment can support. This model predicts that N → K as t →∞.

A delay model compensates for the time it takes for new individuals to reachmaturity. It has the form:

dN(t)dt

= f(N(t), N(t− T )),

where T is a constant called the delay. This model describes a population witha maturation period and then a birth rate independent of the age distributionof the population.

1.1.2 Population models with age structure � time-independentcase

In a model with age structure, the death rate µ and birth rate m depends on theage structure of the population. First we present a model where the birth and

9

death rate are time-independent. Consider n(a, t), the number of individuals ofage a at time t. If we look at an age group at age a, from time t to t + dt thechange of individuals in that age group is

dn(a, t) = −µ(a)n(a, t)dt.

The left hand side can be expressed:

dn(a, t) = n(a + da, t + dt)− n(a, t)= n(a + da, t + dt)− n(a + da, t) + n(a + da, t)− n(a, t)

=∂n(a, t)

∂tdt +

∂n(a, t)∂a

da.

Our a represents chronological age, therefore da = dt. If the equation is dividedwith dt the result is a partial di�erential equation:

∂n(a, t)∂t

+∂n(a, t)

∂a= −µ(a)n(a, t), a, t ≥ 0.

The newborn part of the population contributes only to n(0, t), which gives theboundary condition at a = 0:

n(0, t) =∫ ∞

0

m(a)n(a, t) dt, t ≥ 0.

For t = 0, we have a start population with a certain age distribution f(a) andthe boundary condition at t = 0 is

n(a, 0) = f(a), a ≥ 0.

This model will be studied in the following sections.

1.1.3 Population models with age structure � time-dependent case

In a population where the environment changes over time, resulting from forexample pollution, the death and birth rates for the population will probablychange over time too. The model

∂n(a, t)∂t

+∂n(a, t)

∂a= −µ(a, t)n(a, t), a, t ≥ 0,

describes this. Its boundary conditions are

n(0, t) =∫ ∞

0

m(a, t)n(a, t) dt, t ≥ 0

n(a, 0) = f(a), a ≥ 0.

This model will be examined more closely in the following sections.

1.1.4 Other population models with age structure

In a population there can be competition and rivalry among individuals of thesame species and µ may also depend on N =

∫∞0

n(a, t) da, the total number ofindividuals. The model

∂n(a, t)∂t

+∂n(a, t)

∂a= −µ(a, t, N)n(a, t)

10

describes this. Its boundary conditions are:

n(0, t) =∫ ∞

0

m(a, t, N)n(a, t) dt, t ≥ 0

n(a, 0) = f(a), a ≥ 0.

A population can be in�uenced by other populations and the birth rate forn, mn, may also depend on P =

∫∞0

p(a, t) da, the total number of individualsin another population. The model

∂n(a, t)∂t

+∂n(a, t)

∂a= −µn(a, t, P )n(a, t)

∂p(a, t)∂t

+∂p(a, t)

∂a= −µp(a, t, N)p(a, t)

describes this. Its boundary conditions aren(0, t) =

∫ ∞

0

mn(a, t)n(a, t) dt, n(a, 0) = fn(a),

p(0, t) =∫ ∞

0

mp(a, t)p(a, t) dt, p(a, 0) = fp(a).

1.2 Presentation of the model to study

We will for the most part study the di�erential equation from section 1.1.3:

∂n(a, t)∂t

+∂n(a, t)

∂a= −µ(a, t)n(a, t), (2)

with the boundary conditions

n(0, t) =∫ ∞

0

m(a, t)n(a, t) dt, n(a, 0) = f(a). (3)

The function n(a, t), the number of individuals of age a at time t, is de�nedfor 0 ≤ t < ∞. It is positive and assumed to be continuous. The growth ofn(a, t) is assumed to be at most exponential.

The death rate µ(a, t), de�ned for a, t ≥ 0, is at the lowest no deaths andat most all individuals dead, so µ(a, t) ∈ [0, 1] for alla a, t. Since no individualcan became in�nitely old, the death rate becomes equal to 1 for large a. Thatis, µ(a, t) = 1, a > Aµ for some constant Aµ.

The birthrate m(a, t), de�ned for a, t ≥ 0, is a positive function. It hascompact support since the birthrate for old individuals are 0. That is m(a, t) =0, a > Am for some constant Am. We also asume that m is bounded by someconstant M .

Since f describes the start population, it is positive, bounded and has com-pact support. We will assume that the start population contains some individ-uals young enough to eventually have children, otherwise we get n(0, t) = 0 forall t.

From a biological perspective it is interesting to study the increase and de-crease of the total population. Therefore we will also study N(t) =

∫∞0

n(a, t) da,the number of individuals at time t.

11

1.2.1 What to study

In section 2 we will prove that the di�erential equation (2) has exactly onesolution with at most exponential growth. This is proved with help from theBanach �xed point theorem. Therefore we will have to transform the di�erentialequation to an integral equation. We will also consider the stability of the modelin section 3, using a corollary to Banach's theorem.

The asymptotics of the solution can be studied in two ways. In section 4we will use the Laplace transform. This method requires extra assumptions onthe functions and is too di�cult for the time-dependent case, so we will use iton the time-independent case. For the time-dependent case, in section 5, wewill use the second method. The asymptotics will be estimated using upper andlower bounds. Section 6 compares the solutionof the model with agestructurewith the solution to the model without age structure (1).

12

2 Existence and uniqueness of solutions

We will prove that the age structure model has exactly one solution, with at mostexponential growth. The di�erential equation (2),(3) described in section 1.2,will be transformed into an integral equation of the form

n(0, t) = (Kn)(t) + F (t) = (Tn)(t), t ≥ 0,

so that we can apply the Banach �xed point theorem (Appendix A.2) to showthat there exists a unique solution to the integral equation.

2.1 The integral equation

In this section we will prove the following theorem.

Theorem 1 A function n(a, t), that solves the problem (2),(3), satis�es

n(a, t) ={

f(a− t)e−R a

a−tµ(v,v+t−a) dv, a ≥ t

n(0, t− a)eR a0 µ(v,v+t−a) dv, a < t

(4)

where n(0, t) solves the integral equation:

n(0, t) =∫ t

0

m(a, t)n(0, t− a)e−R a0 µ(v,v+t−a) dv da

+∫ ∞

t

m(a, t)f(a− t)e−R a

a−tµ(v,v+t−a) dv da. (5)

If m and f are di�erentiable, then n(a, t) described by (4),(5) solves problem(2),(3).

Proof: To obtain this result, we start by transforming the equation into anordinary di�erential equation. Let a and t be functions of x and y:

a = x + y, t = x− y.

Then the partial di�erential equation can be rewritten as an ordinary di�erentialequation since

dn(x + y, x− y)dx

=∂n(a, t)

∂t

dt

dx+

∂n(a, t)∂a

da

dx=

∂n(a, t)∂t

+∂n(a, t)

∂a.

This di�erential equation has the solution

n(x + y, x− y) = C(2y)e−R x

yµ(s+y,s−y)ds,

where C is to be determined. With a and t as variables and v = s+ a−t2 , we get

n(a, t) = C(a− t)e−R a

a−tµ(v,v+t−a) dv. (6)

To determine the function C, let t = 0 in the expression above:

n(a, 0) = C(a) = f(a).

13

For a ≥ t, we get

n(a, t) = f(a− t)e−R a

a−tµ(v,v+t−a) dv.

This proves equation (4) for a ≥ t. For the case t < a, we need to �nd C(t) fornegative t. Let a = 0 in equation (6). This gives us

C(−t) = n(0, t)eR 0−t

µ(v,v+t) dv.

If we now replace t with t− a, we get

C(a− t) = n(0, t− a)eR 0

a−tµ(v,v+t−a) dv.

So for a < t, we get the following representation of n(a, t):

n(a, t) = n(0, t− a)eR a

a−tµ(v,v+t−a) dv

and equation (4) is proved also for a < t.All functions are known in these two expressions for n(a, t) except n(0, t−a).

To �nd n(0, · ), we use the boundary condition at a = 0,

n(0, t) =∫ ∞

0

m(a)n(a, t) da. (7)

If we insert the expression for n(a, t) from equation (4) into (7), we obtain theintegral equation (5).

Now let us consider the last part of Theorem 1. If m and f are di�erentiablefunctions, then substituting t− a = x we have

n(0, t) =∫ t

0

m(t− x, t)n(0, x)e−R t−x0 µ(v,v+x) dv dx

+∫ Am

0

m(x + t, t)f(x)e−R x+t

xµ(v,v−x) dv dx

is clearly di�erentiable since the right side is di�erentiable. From (4) it followsthat n is di�erentiable. �

To make the equations more readable, we will sometimes write only µ whenwhat we really mean is µ(v, v+t−a). The integral equation (5) can be rewrittenas n(0, t) = (Tn)(t), if T is de�ned as

(Tn)(t) = (Kn)(t) + F (t), t ≥ 0,

where

(Kn)(t) =∫ t

0

m(a, t)n(0, t− a)e−R a0 µ(v,v+t−a) dv da

and

F (t) =∫ ∞

t

m(a, t)f(a− t)e−R a

a−tµ(v,v+t−a) dv da.

14

2.2 Existence and uniqueness of the solution

In this section, we will prove that the integral equation (5) has exactly onesolution of at most exponential growth.

Let the space BΛ be the space of all continuous functions u de�ned on [0,∞)such that u(t) = O(eΛt) as t →∞. The norm in BΛ is de�ned by

‖u‖Λ = supt≥0

|u(t)|e−Λt.

Theorem 2 The equation n(0, t) = (Tn)(t), t ≥ 0 has exactly one solutionin BΛ for a su�ciently large Λ.

The theorem follows from Banach's theorem (Therorem 34 in the appendix)if we establich that (i) BΛ is a Banach space and (ii) that T is a contractionon BΛ.

Proposition 3 The space BΛ is a Banach space for all Λ.

Proof: To prove that BΛ is a Banach space we need to prove that ‖ · ‖Λ isin fact a norm and that the space is complete with respect to the metric de�nedby the norm.

We start by checking that ‖ · ‖Λ is a norm. The norm is obviously non-negative. Only zero has the norm equal to zero, since if ‖u‖Λ = 0, then u mustbe identically zero. Conversely ‖0‖Λ = 0. The norm is homogeneous, since

‖au‖Λ = supt≥0

|au(t)|e−Λt = supt≥0

|a||u(t)|e−Λt = |a| supt≥0

|u(t)|e−Λt = |a|‖u‖Λ.

The triangle inequality holds, since

‖u + v‖Λ = supt≥0

|u(t) + v(t)|e−Λt ≤ supt≥0

(|u(t)|+ |v(t)|)e−Λt

≤ supt≥0

|u(t)|e−Λt + supt≥0

|v(t)|e−Λt = ‖u‖Λ + ‖v‖Λ.

So ‖ · ‖Λ is a norm.To prove that BΛ is a complete space, set vi(t) = e−Λtui(t), where (u)i ⊂ BΛ

is a Cauchy sequence. Then (v)i is a Cauchy sequence in B0. Indeed,

‖um − un‖Λ = supt≥0

|um(t)− un(t)|e−Λt = supt≥0

|vm(t)− vn(t)| = ‖vm − vn‖0.

So BΛ with the norm ‖ · ‖ is complete since B0 = C[0,∞) with the ordinarysupremum norm is complete. �

Proposition 4 The operator T is a contraction on BΛ for Λ su�ciently large.

Proof: Since m and µ are bounded and non-negative, we have

|m(a, t)e−R a0 µ(v,v+t−a) dv| ≤ M.

The following inequality comes in handy:

|n(0, t)|e−Λ(t) ≤ supt≥0

|n(0, t)|e−Λ(t) = ‖n(0, · )‖.

15

For the mapping T we observe:

|(Tn)(t)|e−Λt =∣∣∣∣∫ t

0

m(a, t)e−R a0 µ dvn(0, t− a) da

+∫ ∞

t

m(a, t)f(a− t)e−R a

a−tµ dv da

∣∣∣∣e−Λt

≤∫ t

0

M |n(0, t− a)|e−Λ(t−a)e−Λa da +∫ ∞

t

M |f(a− t)|e−Λt da

≤∫ t

0

M‖(n(0, · )‖Λe−Λa da + MN(0).

This gives us

‖(Tn)‖Λ = supt≥0

|(Tn)(t)|e−Λt = supt≥0

M‖n(0, · )‖Λ(

1Λ− e−Λt

Λ

)+ MN(0)

≤ α‖n(0, · )‖Λ + MN(0),

where α = M/Λ. We have proved that T : BΛ → BΛ. If we choose Λ > M thenα < 1. We have

‖(Tn1 − Tn2)‖Λ = supt≥0

∣∣∣∣∫ t

0

m(a, t)e−R a0 µ dv(n1(0, t− a)− n2(0, t− a)) da

∣∣∣∣≤ α‖n1(0, · )− n2(0, · )‖Λ.

This shows that that T is a contraction. �

Finally, since BΛ is a Banach space and T is a contraction for Λ > MTheorem 2 is proved.

16

3 Stability of the model

To examine the stability of the model (2),(3), we use Corollary 35 to the �xed-point theorem in Appendix A.

3.1 Stability, considering m

If m is considered as a parameter we have the integral equation

nm(0, t) = (Tmnm)(t).

We will show that the model is stable with respect to small disturbances in m.Let P be the space of continuous and bounded functions with compact supportde�ned on [0,∞)× [0,∞), with the norm ‖m‖P = supa,t≥0 m(a, t). We want toprove the following theorem:

Theorem 5 The integral equation nm(0, t) = (Tmnm)(t) has exactly one solu-tion nm and nm tends to nm0 in BΛ as m tends to m0 in P .

Proof: We have to check conditions (i)-(iii) described in the corollary to theBanach �xed point theorem.

(i) P is a metric space.

(ii) The operator Tm is a contraction on BΛ with bound depedent on M andindependent of m.

(iii) Let ε > 0. Then

‖Tmn− Tm0n‖Λ ≤ supt≥0

∫ t

0

|m(a, t)−m0(a, t)||n(0, t− a)e−Λ(t−a)|e−Λa da

+supt≥0

∫ ∞

t

|m(a, t)−m0(a, t)||f(a− t)e−Λ(t−a)|e−Λa da

≤ supt≥0

∫ t

0

‖m−m0‖P ‖n(0, · )‖Λe−Λa da

+supt≥0

∫ ∞

t

‖m−m0‖P ‖f‖−Λe−Λa da

≤ ‖m−m0‖P

Λ(‖n(0, · )‖Λ + ‖f‖−Λ) < ε

if‖m−m0‖P < δ =

Λε

‖n(0, · )‖Λ + ‖f‖−Λ.

This proves the theorem. �

17

4 Asymptotics of N� the time-independent case

To determine the asymptotic behaviour of N , N(t) =∫∞0

n(a, t) da, the numberof individuals in a population at time t, in the time-independent case, we willuse the Laplace transform. This method will require the extra assumption ofbounded derivatives on µ, m and f .

If m and µ are independent of time we have the di�erential equation:

∂n(a, t)∂t

+∂n(a, t)

∂a= −µ(a)n(a, t)

with the solution

n(a, t) ={

f(a− t)e−R a

a−tµ(v) dv, a ≥ t

n(0, t− a)e−R a0 µ(v) dv, a < t

where n(0, t) satis�es the integral equation

n(0, t) =∫ t

0

m(a)n(0, t− a)e−R a0 µ(v) dv da +

∫ ∞

t

m(a)f(a− t)e−R a

a−tµ(v) dv da.

We write A(t) ∼ B(t) if limt→∞A(t)/B(t) = 1.

Theorem 6 The function N has the following asymptotic behaviour:

N(t) ∼ Ceσ0t if σ0 > −1N(t) ∼ Cte−t if σ0 = −1N(t) ∼ Ce−t if σ0 < −1

where σ0 satis�es ∫ ∞

0

m(a)e−σ0a−R a0 µ(v) dv da = 1.

We �rst prove a special case of Riemann-Lebesgues lemma that will be usefullater.

Lemma 7 Let R(a) be a continuous function on R with compact support andσ ∈ R. Then ∫ ∞

0

R(a)e−σae−iωa da → 0, when |ω| → ∞

and ∫ 0

−∞R(a)e−σae−iωa da → 0, when |ω| → ∞.

Proof: The function a 7→ R(a)e−σa is uniformly continuous. For ε > 0, letα = ε

A+2π . Choose ω0 so that for ω > ω0 |R(a1)e−σa1 −R(a2)e−σa2 | < α when|a1 − a2| ≤ 2π

ω . Choose M ∈ N so that Aω2π < M ≤ Aω0

2π + 1.We divide the integral into M integrals. Suppose that ω ≥ ω0. Then∫ ∞

0

R(a)e−σa−iωa da =M−1∑k=0

∫ 2π(k+1)ω

2πkω

R(a)e−σa(cos(aω)− sin(aω)) da.

18

For an arbitrary integral in the sum, we have∣∣∣∣∫ 2π(k+1)ω

2πkω

R(a)e−σa cos(aω) da

∣∣∣∣ ≤ ∣∣∣∣∫ 2π(k+1)ω

2πkω

R(2πk

ω)e−σ 2πk

ω cos(aω) da

∣∣∣∣+

∣∣∣∣∫ 2π(k+1)ω

2πkω

(R(a)e−σa −R(2πk

ω)e−σ 2πk

ω ) cos(aω) da

∣∣∣∣ ≤ 2πα

ω.

The integral is bounded by ε:∣∣∣∣∫ ∞

0

R(a)e−σa cos(aω) da

∣∣∣∣ ≤ M−1∑0

2πα

ω<

(Aω + 2π)αω

< (A + 2π)α = ε.

The same calculations can be used for the term with sin(aω). With thechange of variables a = −x the same calculations also hold for the secondintegral. This proves Lemma 7. �

4.1 Representation of n(0, · )In this section, we will �nd an asymptotic representation for n(0, · ). Set

g(t) = m(t)e−R t0 µ(v) dt, t ≥ 0, (8)

and

F (t) =∫ ∞

t

m(a)f(a− t)e−R a

a−tµ(v) dv da, t ≥ 0. (9)

Theorem 8 The function n(0, · ) has the asymptotic representation:

n(0, t) = Keσ0t + O(e(σ0−δ)t) as t →∞

where

K =LF (σ)

∂∂σ (1− Lg(σ))

∣∣∣∣σ=σ0

,

where L denotes the one-sided Laplace transform.

4.1.1 Domain of convergence

We start by examining where the Laplace transform of n(a, · ) converges. Weknow that n is a function with at most exponential growth:

‖n(a, · )‖Λ = supt≥0

|n(a, t)|e−Λt = B < ∞,

which gives us an upper bound for n:

n(a, t) ≤ BeΛt.

The Laplace transform of n, Ln(a, s) =∫∞0

n(a, t)e−st dt, converges if Re(s) ≥Λ.

19

4.1.2 The Laplace transform of n(0, · )

We now �nd the Laplace transform of n(0, · ) by transforming the integral equa-tion for n(0, · ):

n(0, t) =∫ t

0

m(a)n(0, t− a)e−R a0 µ(v) dv da +

∫ ∞

t

m(a)f(a− t)e−R a

a−tµ(v) dv da.

To transform n(0, t), use (8) and (9) to rewrite the equation as

n(0, t) = g(t) ∗ n(0, t) + F (t).

Observe that F is not identically zero, since some individuals are young enoughto eventually have children. Therefore LF is not identically zero. Using theLaplace transform, we get

Ln(0, s) =LF (s)

1− Lg(s)(10)

whereLg(s) =

∫ ∞

0

m(t)e−st−R t0 µ(v) dv dt

andLF (s) =

∫ ∞

0

∫ ∞

t

m(a)f(a− t)e−st−R a

a−tµ(v) dv da dt.

The term Ln(0, s) is de�ned where Re(s) > Λ but the right hand side of (10)has an extension to the whole complex plane. In [5], page 423, the followingsu�cient condition for inversion of the Laplace transformation is found. Ifthere exist positive constants h, R0 and k such that |Ln(0, s)| ≤ h/|s|k when|s| > R0, we can �nd n(0, · ) by using the inversion formula for the Laplacetransformation:

n(0, t) =1

2πi

∫ σ+i∞

σ−i∞Ln(0, s)estds,

where σ > Λ. In Lemma 12 we will see that the condition is satis�ed. Theintegral can be calculated by integration along a rectangle and calculations ofthe residues of the poles. Therefore we need to determine the poles of the righthand side of (10) inside the area of integration.

Lemma 9 The functions LF and Lg are entire.

Proof: Since m has compact support, we have

LF (s) =∫ Am

0

∫ Am

t

m(a)f(a− t)e−R a

t−aµ(v) dve−st da dt.

By di�erenting under the integral signs using the fact that e−st is entire, itfollows that LF is entire. The same argument applies to Lg:

Lg(s) =∫ Am

0

m(t)e−R t0 µ(v) dve−st dt. �

Now we examine where Lg(s)=1.

20

Lemma 10 The function 1− Lg has exactly one real zero, σ0, and the zero isof order one.

Proof: Let s = σ + iω. Recall that m is not allowed to be 0 everywhere. Fors = σ, Lg is a real-valued, continuous and decreasing function of σ:

Lg(σ1)− Lg(σ2) =∫ ∞

0

m(t)e−R t0 µ(v,) dv(e−σ1t − e−σ2t) dt > 0 if σ1 < σ2.

If σ > M = supa≥0 m(a), then Lg(s) < 1:

Lg(σ) ≤∫ ∞

0

Me−σt dt =M

σ< 1. (11)

There exists σ so that Lg(σ) =∫∞0

m(a)e−σt−R a0 µ(v) dv dt > 1, since m is not

identically zero and the last factor is greater than e−st.We want to show that the zero at s = σ0 is of order 1. We have

d

dσ(1− Lg(σ)) =

∫ Am

0

tm(t)e−σt−R t0 µ(v) dv dt

and the last integral is not 0 since the integrand is non-negative and not iden-tically zero. �

Notice that (11) gives us the information σ0 ≤ M . We need to prove thatthere are no complex zeroes in the area of integration.

Lemma 11 There exists a δ such that the function 1−Lg has no complex zeroin the half plane Re(s) > σ0 − δ.

Proof: We consider �ve cases.For s = σ + iω and σ > σ0, we have

Re(Lg(s)) =∫ ∞

0

m(a)e−R a0 µ(v) dve−σa cos(aω) da

<

∫ ∞

0

m(a)e−R a0 µ(v) dve−σ0a da = 1

For s = σ0 + iω and ω 6= 0, we have, since cos(aw) < 1 almost everywhere,

Re(Lg(s)) =∫ ∞

0

m(a)e−R a0 µ(v) dve−σ0a cos(aω) da

<

∫ ∞

0

m(a)e−R a0 µ(v) dve−σ0a da = 1.

For s = σ0 − 1 + iω there exist a ω0 such that Re(Lg(s)) < 1 for |ω| > |ω0|.This follows from Lemma 7 and the fact that m has compact support.

For s = σ + iω, where σ0 − 1 ≤ σ ≤ σ0 we can use Lemma 7 and the factthat Lg(s) is a integral over e−σt to obtain that Re(Lg(s)) < 1 for |ω| > |ω0|.

The function 1 − Lg has no zeros on the segment Re(s) = σ0, |ω| ≤ |ω0|.Therefore there exists open disks with radious < 1/2, where 1/(1 − Lg) isanalytic, around each point on the line segment. It follows from Heine-Boreltherorem that the segment can be covered by a �nite number of these disks.Therefore there exists a constant δ such that the function is analytic on σ0−δ <Re(s) ≤ σ0, |ω| ≤ |ω0|. �

Choose δ such that δ 6= σ0 + 1.

21

4.1.3 Asymptotic behaviour of Ln(0, · )

We will use integration by parts to �nd a representation of Ln(0, · ) that willmake it possible to calculate the integral in the inversion formula. To be ableto integrate by parts we must make the extra assumptions on f , m and µ.

Lemma 12 Let m and µ be di�erentiable with bounded derivatives and f twotimes di�erentiable with bounded derivatives. Then

Ln(0, s) =n(0, 0)

s+ O

(1s2

)as |Re(s)| → ∞.

Proof: We start with Lg(s):

Lg(s) =∫ ∞

0

m(t)e−st−R t0 µ(v,) dv dt =

[− m(t)e−

R t0 µ(v) dv

se−st

]Am

0

+1s

∫ Am

0

(m′(t)−m(t)µ(t))e−R t0 µ(v) dve−st dt = O

(1s

).

This gives us1

1− Lg(s)= 1 + O

(1s

).

Now look at LF (s):

LF (s) =∫ Am

0

∫ Am

t

m(a)f(a− t)e−st−R a

a−tµ(v) dv da dt =

∫ Am

0

F (t)e−st dt,

where F (t) has bounded derivatives. So we have

LF (s) =[−F (t)

se−st +

F ′(t)s2

e−st

]Am

0

+1s2

∫ Am

0

F ′′(t)e−st dt

=1s

∫ Am

0

m(a)f(a) da + O

(1s2

)=

n(0, 0)s

+ O

(1s2

).

The result for Ln(0, s) is

Ln(0, s) =(

n(0, 0)s

+ O

(1s2

) )(1 + O

(1s

) )=

n(0, 0)s

+ O

(1s2

). �

4.1.4 The representation of n(0, · )

Proof of Theorem 8: To �nd n(0, t) we will use Laplace inversion formulaand integrate along a line in the half plane σ > Λ:

n(0, t) =1

2πi

∫ σ+i∞

σ−i∞Ln(0, s)estds.

To calculate this we will integrate along a rectangle and use the fact that s = σ0

is the only pole inside the area of integration.

22

- σ

6

ω

-L4

6L1

�L2

?

L3

rσ0

We have

n(0, t) = − limω→∞

12πi

∫L2

Ln(0, s)estds− limω→∞

12πi

∫L3

Ln(0, s)estds

− limω→∞

12πi

∫L4

Ln(0, s)estds + Res(Ln(0, σ0)eσ0t

).

According to Lemma 12, the vertical integral along L3 can be written

n(0, 0)2πi

∫ σ0−δ+i∞

σ0−δ−i∞

e−st

sds +

12πi

∫ σ0−δ+i∞

σ0−δ−i∞

e−st

s2O(1)ds.

If we change the integration variable to ω and use the fact that the integral ofan odd function is zero,we see that these integrals equal

n(0, 0)π

e(σ0−δ)t

( ∫ ∞

0

(σ0 − δ) cos(ωt)ω2 + (σ0 − δ)2

dω +∫ ∞

0

ω sin(ωt)ω2 + (σ0 − δ)2

dω

)+ie(σ0−δ)t

∫ ∞

−∞

e−iωt

(σ0 − δ + iω)2O(1) dω.

Observe that the last integral is bounded since∫ ∞

−∞

1(σ0 − δ)2 + ω2

dω < ∞.

So with help from [4], page 181, we get

12πi

∫ σ0−δ+i∞

σ0−δ−i∞Ln(0, s)estds = 2n(0, 0)H(σ0 − δ) + O(e(σ0−δ)t) = O(e(σ0−δ)t),

where H denotes the heaviside-function. The residue of the pole at σ0 is

Res(Ln(0, σ0)eσ0t

)=

LF (σ0)eσ0t

∂∂σ (1− Lg(σ))σ=σ0

= Keσ0t.

For the two horizontal integrals along L2 and L4, Lemma 1 shows that Lg(s) → 0and LF (s) → 0 when ω →∞, since

m(t)e−σt−R t0 µ(v) dv

23

and ∫ ∞

t

m(a)f(a− t)e−σt−R a

a−tµ(v) dv da

are continuous and have compact support. Therefore Ln(0, s) → 0 when ω →∞and the two horizontal integrals tend to 0. The result is

n(0, t) = O(e(σ0−δ)t) + Keσ0t

and Theorem 8 is proved. �

4.2 Asymptotics of N

We are now ready to look at the asymptotics of N , which is de�ned by:

N(t) =∫ ∞

0

n(a, t) da =∫ t

0

n(0, t−a)e−R a0 µ(v) dv da+

∫ ∞

t

f(a−t)e−R a

a−tµ(v) dv da.

Proof Theorem 6: Let us start by inserting our new expression for n(0, t),found in the previous section, into the equation for N(t):

N(t) = K

∫ t

0

eσ0(t−a)−R a0 µ(v) dv da +

∫ t

0

e(σ0−δ)(t−a)−R a0 µ(v) dvO(1) da

+∫ ∞

t

f(a− t)e−R a

a−tµ(v) dv da = I1 + I2 + I3.

To estimate these integrals, we will use the fact that µ(a) = 1 for a ≥ Aµ; noindividual can be in�nitely old. Consider N(t) for t > Aµ. We will use ci todenote constants. We start with I1:

I1 = K

∫ t

0

eσ0(t−a)−R a0 µ(v) dv da

= eσ0tK

∫ Aµ

0

e−σ0a−R a0 µ(v) dv da + eσ0tK

∫ t

Aµ

e−σ0a−(a−Aµ)−R Aµ0 µ(v) dv da.

So for σ0 6= −1 we get

I1 = c1(σ0)eσ0t − c2(σ0)e−t

and for σ0 = −1I1 = c3e

−tt + c4(σ0)e−t.

For I2 we can use the same calculations as for I1. We have chosen δ so thatσ0 − δ 6= −1. We have

I2 =∫ t

0

e(σ0−δ)(t−a)−R a0 µ(v) dvO(1) da = O(1)e(σ0−δ)t

∫ t

0

e−(σ0−δ)a−R a0 µ(v) dv da

= O(1)e(σ0−δ)t

( ∫ t

Aµ

e−(σ0−δ+1)a−R Aµ0 µ(v) dv+Aµ da +

∫ Aµ

0

e−(σ0−δ)a−R a0 µ(v) dv

)= O(1)e(σ0−δ)t −O(1)e−t,

where O(1) denotes a bounded function.

24

Finally, we calculate the integral I3:

I3 =∫ ∞

t

f(a− t)e−R a

a−tµ(v) dv da =

∫ ∞

0

f(x)e−R Aµ

xµ(v) dv−

R x+tAµ

µ(v) dvdx

= e−t

∫ ∞

0

f(x)e−x−R Aµ

xµ(v) dv+Aµ dx = c5e

−t.

We have the following expressions for N(t). For σ0 6= −1:

N(t) = c1(σ0)eσ0t − c2(σ0)e−t + O(1)e(σ0−δ)t −O(1)e−t + c5e−t

and for σ0 = −1:

N(t) = c3e−tt− c4(σ0)e−t + O(1)e(−1−δ)t −O(1)e−t + c5e

−t. �

Notice that the population is increasing if σ0 > 0.

25

5 Asymptotics of N�the time-dependent case

In the previous section we found the asymptotic behaviour of N , in the time-independent case, by using the Laplace transform. This method does not workwhen m or µ depend on time, and it requires that m, f and µ are di�erentiablefunctions. In this section we will instead try to �nd upper and lower boundsfor N to get information about the asymptotic behaviour. We need �rst toconsider bounds n+(t) and n−(t) to n(0, t).

Recall that n(0, t) satis�es the integral equation (5):

n(0, t) =∫ t

0

m(a, t)n(0, t− a)e−R a0 µ dv da +

∫ ∞

t

m(a, t)f(a− t)e−R a

a−tµ dv da

= (Kn)(t) + F (t).

For a function n we use the notation E(n)(t) = n(0, t) − (Kn)(t) − F (t). Wewill �rst prove two lemma that describe su�cient conditions for such boundsn+ and n−.

Lemma 13 If n+ satis�es E(n+)(t) ≥ 0 for all t ≥ 0, and n, n+ is of atmost exponential growth, then n+ ≥ n(0, · ), where n(0, · ) is the solution to theintegral equation (5).

Proof: Let n1, n2 . . . denote the sequence of approximations to n obtainedfrom the Picard iteration scheme with n+ as the initial approximation. Thenn1 = (Kn+) + F ≤ n+. Suppose that nk−1 ≤ n+ for some k ≥ 2. Thennk = (Knk−1) + F ≤ (Kn+) + F ≤ n+. By induction, nk ≤ n+ for all k, andthus by letting k →∞, n ≤ n+. �

In an analog way, we can prove the following lemma.

Lemma 14 If n− satis�es E(n−)(t) ≤ 0 for all t ≥ 0, and n+ is of at mostexponential growth, then n− ≤ n(0, · ) where n(0, · ) is the solution to the integralequation (5).

Let σ0(t1) = −∞ if m(a, t1) ≡ 0 and otherwise, for a given t, let σ0(t) be thefunction that satis�es the following equation:∫ ∞

0

m(a, t)e−σ0(t)a−R a0 µ(v,v+t−a) dv da = 1.

Because of Lemma 10, there exists a unique σ0(t) for every t. We will seekbounds under certain conditions on σ0.

5.1 Upper bounds for n(0, · )We will seek upper bounds for n(0, · ). Let

Ff1(t) =∫ ∞

t

m(a, t)f1(a− t)e−R a

a−tµ dv da.

We will need a lemma.

26

Lemma 15 Suppose that nf1(0, · ) solve the integral equation (5), with f1 asstart population, where f1(a) > 0 for a ∈ [0, Am]. Then for the solution nf (0, · )to (5) with another start population f , there exists a constant C such that

nf (0, t) ≤ Cnf1(0, t) for all t ≥ 0.

Proof: We have

nf1(0, t) =∫ t

0

m(a, t)nf1(0, t− a)e−R a0 µ(v,v+t−a) dv da + Ff1(t).

Since f1(a) > 0 for x ∈ [0, Am] and f(a) is bounded, there exists a constant Cso that, for another f ,

CFf1(t)−Ff (t) =∫ ∞

t

m(a, t)(Cf1(a− t)− f(a− t))e−R a

a−tµ(v,v+t−a) dv da ≥ 0.

Therefore, for the other start population f , we get

E(Cnf1)(t) = Cnf1(0, · )−K(Cnf1)−Ff = C(nf1(0, · )−K(nf1)−Ff1)+CFf1−Ff .

Sincenf1(0, · )−K(nf1)− Ff1 = 0,

we have E(Cnf1) ≥ 0. Thus Cnf1(0, · ) is an upper bound for nf (0, · ). This istrue for an arbitrary f and therefore Cnf1(0, · ) ≥ n(0, · ). �

Now we can seek upper bounds for n under some conditions on σ0(t). Westart with σ0(t) that eventually has a constant upper boundary.

Lemma 16 If σ0(t) ≤ c for all t > T , then there exists constants, D1, D2, cand σ, such that

n(0, t) ≤ n+(t) = D1eσtH(T − t) + D2e

ctH(t− T )

where n is the solution to (5) and H denotes the Heaviside function.

Proof: Because of Lemma 15, it is enough to prove that nf1 ≤ n+, for

f1(a) = D1e−σa−

R a0 µ(v,v−a) dv

for a ∈ [0, Am]. Choose the constants so that D2ect ≥ D1e

σt for t ∈ [0, T ]. Then

E(n+) = D1eσtH(T − t)−D1e

σt

∫ t

0

m(a, t)e−σae−R a0 µ dvH(T − t + a) da

+D2ectH(t− T )−D2e

ct

∫ t

0

m(a, t)e−ca−R a0 µ dvH(t− T − a) da

+∫ ∞

t

m(a, t)f1(a− t)e−R a

a−tµ dv da.

For t > T , we get

E(n+) = D2ect −

∫ ∞

t−T

m(a, t)D1eσ(t−a)e−

R a0 µ dv da

−D2ect

∫ t−T

0

m(a, t)e−cae−R a0 µ dv da.

27

Using the fact that D1eσ(t−a) ≤ D2e

c(t−a), we get

E(n+) ≥ D2ect −D2e

ct

∫ t

0

m(a, t)e−cae−R a0 µ dv da ≥ 0.

Since, by assumtion, the last integral is not greater than 1.For t ≤ T , we get

E(n+) = D1eσt −D1e

σt

∫ t

0

m(a, t)e−σae−R a0 µ dv da− Ff1(t).

There exists a σ such that E(n+) ≥ 0, σ ≥ M will always do. Since

E(n+) ≥ D1eMt −D1e

σt

∫ ∞

0

Me−Mae−R a0 µ(v,v+t−a) dv da︸ ︷︷ ︸≤1

≥ 0. �

We can prove the following theorem.

Theorem 17 If σ0(t) ≤ c for all t > T , then there exists constants, D2 and c,such that

n(0, t) ≤ n+(t) = D2ect

where n is the solution to (5) and H denotes the Heaviside function.

Proof: Since we chose D2ect ≥ D1e

σt for t ∈ [0, T ], we have D1eσtH(T − t) +

D2ectH(t − T ) ≤ D2e

ct and n+(t) = D2ect is also a upper bound for n(0, · ).

�We will continue to look for a bound for n under some other conditions on

σ0. For every t let m(a, t) 6= 0 for some interval on a. This will mean thatσ0(t) 6= −∞. Under this condition we can �nd an upper bound for n(0, · ) for aσ0 that eventually becomes a decreasing function.

Lemma 18 If σ0 is a decreasing function on [T,∞], and if, for every �xed t,m( · t) is positive on some interval, then there exists constants D1, D2 and σsuch that

n(0, t) ≤ n+(t) = D1eσtH((T + Am)− t) + D2e

R t0 σ0(τ)d τH(t− (T + Am)).

Proof: Because of lemma 15, it is also here enough to prove that nf1 ≤ n+, for

f1(a) = D1e−σa−

R a0 µ(v,v−a) dv for a ∈ [0, Am].

Choose D1, D2 and σ such that D2eR t0 σ0(τ)d τ ≥ D1e

σt for t ∈ [0, T + Am]. Fort ≤ T + Am we get E(n+) ≥ 0 using the same calculations as in the previouscase for t ≤ T . Consider E(n+) for t ≥ T + Am:

E(n+) = D2eR t0 σ0(τ)d τ −

∫ ∞

t−(T+Am)

m(a, t)D1eσ(t−a)e−

R a0 µ dv da

−D2

∫ t−(T+Am)

0

m(a, t)eR t−a0 σ0(τ)d τe−

R a0 µ dv da.

28

If we now use the fact that D2eR t0 σ0(τ)d τ ≥ D1e

σt for t ∈ [0, T + Am], we get

E(n+) ≥ D2eR t0 σ0(τ)d τ −D2

∫ ∞

0

m(a, t)eR t−a0 σ0(τ)d τe−

R a0 µ dv da

= D2eR t0 σ0(τ)d τ

(1−

∫ ∞

0

m(a, t)e−R t

t−aσ0(τ)d τe−

R a0 µ dv da

).

Since σ0 is a decreasing function, we know that e−R t

t−aσ0(τ)d τ ≤ e−aσ0(t). Using

this, we obtain

E(n+) ≥ D2eR t0 σ0(τ)d τ

(1−

∫ ∞

0

m(a, t)e−aσ0(t)e−R a0 µ dv da

)= 0. �

Theorem 19 If σ0 is a decreasing function on [T,∞], and if, for every �xed t,m( · , t) is positive on some interval, then there exists a constant D2 such that

n(0, t) ≤ n+(t) = D2eR t0 σ0(τ)d τ .

Proof: We chose D1, D2 and σ such that D2eR t0 σ0(τ)d τ ≥ D1e

σt for t ∈ [0, T +Am]. �

5.2 Lower bounds for n(0, · )We will now seek lower bounds for n(0, · ). We seek bounds under the conditionthat m(a, t) ≥ 0 for an interval of length ε for all t and a small constant ε.

Lemma 20 Suppose that f(x) > 0 for x in an interval of length δ and g(u, x) >0 for every x for u in an interval of length ε, where ε and δ are positive. Alsosuppose that both g(u, x) and f(x) ≥ 0 for all x. Then∫ x

0

g(u, x)f(x− u)du > 0

for x in an interval of length δ + ε.

Proof: Suppose that f(x) > 0 for x ∈ [x1, x1 + δ] and and g(u, x) > 0 foru ∈ [u2, u2 + ε]. Let x ∈ (x1 + u2, x1 + u2 + δ + ε). For x ≥ u2 + ε weknow that g(u, x) > 0 when we integrate over the interval u ∈ [u2, u2 + ε].When we integrate over the interval u ∈ [u2, u2 + ε], x − u will vary in theinterval [x − u2 − ε, x − u2], which contains an interval where f(u − x) > 0.For x = x1 + u2 + ρ < u2 + ε we know that g(u, x) > 0 when we integrateover the interval u ∈ [u2, u2 + x1 + ρ]. When we integrate over the intervalu ∈ [u2, u2 + x1 + ρ], x − u will vary in the interval [x − u2 − x1 − ρ, x − u2],which contains an interval where f(u− x) > 0. �

Consider (KpF )(t) = K(Kp−1F )(t), where p is a positiv integer. Notice thatF (t) > 0 for an interval of length δ otherwise we get the solution n(0, t) = 0.We also know that g(a, t) = m(a, t)e−

R a0 µ dv > 0 for an interval of length ε.

Therefore, using the lemma above, we can choose p such that (KpF )(t) > 0 foran interval of length Am.

We will start to look for a lower bound for n, for σ0(t) that eventually isbounded from below.

29

Lemma 21 Suppose that σ0(t) ≥ c for all t > T . Then there exists a positiveinteger p and constants C and TF such that

n(0, t) ≥ n−(t) = H(t− TF )Cect + F (t) + (KF )(t) + . . . + (Kp−1F )(t)

Proof: Choose TF > T such that (KpF )(t) > 0 for t ∈ [TF , TF + Am] this ispossible according to Lemma 20. Consider �rst E(n−)(t) for t < TF :

E(n−) = H(t− TF )Cect︸ ︷︷ ︸=0

−(KH(t− TF )Cect)(t)− (KpF )(t) ≤ 0.

For t ≥ TF ,

E(n−) = Cect − Cect

∫ t

0

m(a)e−cae−R a0 µ(v) dvH(t− TF − a) da− (KP )F (t),

which can be rewritten:

E(n−) = Cect

(1−

∫ ∞

0

m(a, t)e−ca−R a0 µ dv da

)(12)

+Cect

∫ ∞

t−TF

m(a, t)e−ca−R a0 µ dv da− (Kp)F (t). (13)

Notice that the term in (12) is negative since σ0(t) ≥ c for t ≥ T . The �rstterm in (13) equals 0 for t ≥ TF + Am since m(a, t) = 0 for a ≥ Am. Noticethat (KpF )(t) > 0 for t ∈ [TF , TF + Am] and the �rst term in (13) is bounded.Therefore we can choose C such that E(n−)(t) ≤ 0.

Theorem 22 Suppose that σ0(t) ≥ c for all t > T . Then there exists constantsC and TF such that

n(0, t) ≥ n−(t) = H(t− TF )Cect.

Proof: We know that

H(t− TF )Ceσ0t ≤ H(t− TF )Cect + F (t) + (KF )(t) + . . . + (Kp−1F )(t).

Therefore H(t− TF )Cect is a lower bound as well. �There is a lower bound for an eventually increasing σ0.

Lemma 23 Suppose that σ0 is increasing on [T,∞]. Then there exists a positiveinteger p and constants TF and C such that

n(0, t) ≥ H(t− TF )Ce−R t0 σ0(τ)dτ + F (t) + (KF )(t) + . . . + (KpF )(t).

Proof: Choose TF > T +Am such that (KpF )(t) > 0 for t ∈ [TF , TF +Am] thisis possible according to Lemma 20. As in the previous lemma, E(n−)(t) ≤ 0 fort ≤ TF . For t > TF , we get

E(n−) = Ce−R t0 σ0(τ)dτ

(1−

∫ ∞

0

m(a, t)e−R t

t−aσ0(τ)dτ−

R a0 µ dv da

)+C

∫ ∞

t−TF

m(a, t)e−R t

t−aσ0(τ)dτ−

R a0 µ dv da− (Kp+1F )(t).

30

Since σ0(t) is increasing for t ≥ T we know that e−R t

t−aσ0(τ)dτ ≥ e−aσ0(t). For

t ≥ TF , this implies that

E(n−) ≤ Ce−R t0 σ0(τ)dτ

(1−

∫ ∞

0

m(a, t)e−σ0(t)a−R a0 µ dv da

)+C

∫ ∞

t−TF

m(a, t)e−R t

t−aσ0(τ)dτ−

R a0 µ dv da− (Kp+1)F (t)

= C

∫ ∞

t−TF

m(a, t)e−R t

t−aσ0(τ)dτ−

R a0 µ dv da︸ ︷︷ ︸

bounded

−(Kp+1)F (t).

As in the proof for Lemma 21 we can choose C and p so that E(n−) ≤ 0. �Clearly the following theorem is true.

Theorem 24 Suppose that σ0 is increasing on [T,∞]. Then there exists con-stants C an TF such that

n(0, t) ≥ H(t− TF )Ce−R t0 σ0(τ)d τ .

5.3 Upper and lower bounds for N

Now we are ready to look at N , which has the following equation:

N(t) =∫ ∞

0

n(a, t) da =∫ t

0

n(0, t− a)e−R a0 µ dv da +

∫ ∞

t

f(a− t)e−R a

a−tµ dv da.

We will �nd bounds for some combinations of conditions on σ0(t).

Theorem 25 Suppose that σ0(t) ≤ c for all t ≥ T . Then there exsists constantsC1 and C2 such that, for t > T + Aµ, we have

N(t) ≤ C1ect + C2e

−t, if c 6= −1,

andN(t) ≤ C1e

−t + C2te−t, if c = −1.

Proof: Insert our �rst upperbound n+(t), from Theorem 17, which was calcu-lated under the given condition on σ0, into the equation for N . Then

N(t) ≤∫ t

0

D2ec(t−a)e−

R a0 µ(v,v+t−a) dv da +

∫ ∞

t

f(a− t)e−R a

a−tµ(v,v+t−a) dv da.

For t > T + Aµ, using the fact that there exists a constant F such that f < F ,we get

N(t) ≤∫ Aµ

0

D2ec(t−a)e−

R a0 µ dv da +

∫ t

Aµ

D2ec(t−a)e−a+Aµ−

R Aµ0 µ(v,v+t−a) dv da

+∫ ∞

0

f(x)e−R Aµ

xµ(v,v+x) dv−

R x+tAµ

µ(v,v+x) dvdx

≤∫ Aµ

0

D2ec(t−a) da +

∫ t

Aµ

D2ec(t−a)e−a+Aµ da + F

∫ ∞

0

e−(x+t−Aµ) dx.

This proves the theorem. �

31

Theorem 26 Suppose that σ0(t) ≥ c for all t ≥ T . Then there exists constantsC1, C2 and TF such that, for t > TF + Aµ, we have

N(t) ≥ C1ect + C2e

−t, if c 6= −1

andN(t) ≥ C1e

−t + C3te−t, if c = −1.

Proof: Insert n−(0, t) from Theorem 22 into the equation for N .

N(t) ≥∫ t

0

H(t− a− TF )Cec(t−a)e−R a0 µ dv da +

∫ ∞

t

f(a− t)e−R a

a−tµ dv da.

Since e−R a0 µds ≥ e−a, we have

N(t) ≥∫ t−TF

0

Cec(t−a)e−a da +∫ ∞

t

f(a− t)e−t da. �

Theorem 27 Suppose that σ0(t) ≥ c and that σ0 is decreasing on [T −Am,∞]Then there exists constants C1, C2 and C3 such that, for t > T + Aµ, we have

N(t) ≤ C1eR t0 σ0(τ)d τ + C2e

R t0 σ0(τ)d τe−(c+1)t + C3e

−t, if c 6= −1,

and

N(t) ≤ C1eR t0 σ0(τ)d τ + C2e

R t0 σ0(τ)d τ (t−Aµ) + C3e

−t, if c = −1.

Proof: We use the fact that eR t−a

tσ0(τ)d τ ≤ e−σ0(t)a ≤ e−ca. We obtain

N(t) =∫ Aµ

0

D2eR t−a0 σ0(τ)d τe−

R a0 µ(v,v+t−a) dv da +

∫ ∞

t

f(a− t)e−R a

a−tµ dv da

+∫ t

Aµ

D2eR (t−a)0 σ0(τ)d τe+Aµ−a−

R Aµ0 µ(v,v+t−a) dv da

≤ D2eR t0 σ0(τ)d τ

( ∫ Aµ

0

e−R t

t−aσ0(τ)dτ da +

∫ t

Aµ

e−R t(t−a) σ0(τ)d τe−a da

)+C3e

−t.

For t ≥ T + Aµ this i bounded from above by

D2eR t0 σ0(τ)d τ

∫ Aµ

0

e−aσ0(t) da + D2eR t0 σ0(τ)d τ

∫ t

Aµ

e−(σ0(t)+1)a da + C4e−t

≤ D2eR t0 σ0(τ)d τ

∫ Aµ

0

e−ac da + D2eR t0 σ0(τ)d τ

∫ t

Aµ

e−(c+1)a da + C4e−t.

This proves the theorem. �

5.4 Comparison with the time-independent case

In the time-independent case we have σ0(t) = σ0 for all t ≥ 0. This means thatwe have the following estimate:

C1eσ0t ≤ n(0, t) ≤ C2e

σ0t.

This gives the same asymptotic behaviour for N as we found in Section 4.

32

6 Comparison with the model without age struc-

ture

Here we will compare the age structure population model, time-independentcase, with the model without age structure (1). First we examine what happensif we have m and µ constant. Let m(a) = b and µ(a) = d. We know that σ0

satis�es:

1 =∫ ∞

0

be−σ0a−R a0 d dv da =

∫ ∞

0

be−(σ0+d)a da =b

σ0 + d.

Nicely enough σ0 = b − d. So for m and µ constant and σ0 > −1 the solutionto the age structure population model, for the time-independent case, has thesame asymptotics, N(t) v eσ0t, as the most simple model.

For non constant m and µ σ0 is the asymptotic birthrate minus death ratefor the total population:

b− d = limt→∞

(∫∞0

m(a)n(a, t) da−∫∞0

µ(a)n(a, t) da

N(t)

)

= limt→∞

( ∫ t

0m(a)Keσ0(t−a)e−

R a0 µ(v) dv da−

∫ t

0µ(a)Keσ0(t−a)e−

R a0 µ(v) dv da∫ t

0Keσ0(t−a)e−

R a0 µ(v) dv da +

∫ t

0O(1)e(σ0−δ)(t−a) da +

∫∞t

n(a, t) da

+

∫ t

0O(1)e(σ0−δ)(t−a) da−

∫∞t

f(a− t))e−R a

a−tµ(v) dv da∫ t

0Keσ0(t−a)e−

R a0 µ(v) dv da +

∫ t

0O(1)e(σ0−δ)(t−a) da +

∫∞t

n(a, t) da

)

= limt→∞

(1−

∫ t

0e−σ0a(e

R a0 µ(v) dv)′ da∫ t

0e−σ0ae−

R a0 µ(v) dv da

)= lim

t→∞

e−σ0t−R t0 µ(v) dv∫ t

0e−σ0a−

R a0 µ(v) dv da

+ σ0 = σ0

It is interesting to �nd out when the simplest model has the same asymptoticsolution as the solution to the age structure population model. That is, whatconditions on m and µ correspond to σ0 > −1?

If σ0 > −1 we have:

1 =∫ ∞

0

m(a)e−σ0a−a+a−R a0 µ(v) dv da <

∫ ∞

0

m(a)ea−R a0 µ(v) dv da,

and if∫∞0

m(a)ea−R a0 µ(v) dv da > 1, we have for σ ≤ −1:

Lg(σ) =∫ ∞

0

m(a)e−(1+σ0)a+a−R a0 µ(v) dv da ≥

∫ ∞

0

m(a)ea−R a0 µ(v) dv da > 1.

This gives us σ0 > −1. Therefore we can state:

Theorem 28 σ0 > −1 ⇔∫∞0

m(a)ea−R a0 µ(v) dv da > 1

33

6.1 Biological analysis of the requirement σ0 > −1

We will here discuss the biological properties of a population that has σ0 > −1.Hopefully it is reasonable to assume that most populations have σ0 > −1.

Consider an individual that lives to be older than a = Am, under its lifetimeit gets

∫∞0

m(a) da children. If∫∞0

m(a) da is less than 1 we surely have apopulation that will die out and biologically its not a possible scenario. Ifinstead

∫∞0

m(a) da is greater than 1 we have:

1 <

∫ ∞

0

m(a) da ≤∫ ∞

0

m(a)ea−R a0 µ(v) dv da

and σ0 > −1 for this population. So we can assume that σ0 > −1 for biologicalpopulations.

34

7 Conclusions

We have proved the existence of a unique solution to the model with age struc-ture. We have found the asymptotics for the time-independent case, and boundsfor some cases of the time-dependent case.

It would be interesting to �nd more bounds and eventually the completeasmptotics for the time-dependent case. We would like to �nd out when theage structure model gives the same asymptotics as the simple model withoutage structure. The next step could be to take in consideration birth and deathrates that depend on the total number of individuals.

The growth of a population probably depend on its spatial structure. There-fore analysis of how migration in�uences the asymptotics is interesting. The goalis to determine under which conditions the approximation with the simple modelwithout age structure is valid.

35

References

[1] J. D. Murray, Mathematical Biology, Springer-Verlag, Berlin Hiedelberg,1989

[2] E. Kreyszig: Introductory Functional Analysis with Applications, Wiley,USA, 1989

[3] E. Zeidler: Functional Analysis and its Applications vol, 1, Springer, NewYork, 1984

[4] L. Råde and B. Westergren, Mathematics Handbook for Science and Engi-neering, 5th edition, Studentlitteratur, Lund, 2004

[5] A. David Wunsch, Comples Variables with Applications, 2th edition,Addison-Wesley Publishing Company, USA, 1999

36

A Banach theory

To examine if integral equation has solutions, the Banach �xed point theoremcan be used. In order to prepare for this theorem, a couple of de�nitions areneeded.

A.1 De�nitions

The following de�nitions are presented in [2].

De�nition 29 (Metric space) A metric space is a pair (X,d), where X isa set and d is a metric on X. The metric d is a function from X ×X to R withthe following properties, for x, y, z ∈ X:

M(1) d is real-valued, �nite and nonnegativeM(2) d(x, y) = 0 ⇔ x = yM(3) d(x, y) = d(y, x)M(4) d(x, y) ≤ d(x, z) + d(z, y).

De�nition 30 (Completeness) A sequence in a metric space is called Cauchyif for every ε > 0 there is a N ∈ N, N = N(ε) such that

d(xm, xn) < ε for m, n > N

The space X is said to be complete if every Cauchy sequence has a limit whichbelongs to X.

De�nition 31 (Norm) A norm on a vector space is a real-valued function‖ · ‖ on X, with the properties:

N(1) ‖x‖ ≥ 0N(2) ‖x‖ = 0 ⇔ x = 0N(3) ‖αx‖ = |α|‖x‖N(4) ‖x + y‖ ≤ ‖x‖+ ‖y‖.

A norm on a vector space X induce a metric d(x, y) = ‖x− y‖, on X.

De�nition 32 (Banach space) A Banach space is a normed vector spacewhich is complete with respect to the metric de�ned by the norm.

De�nition 33 (Contraction) Let X = (X, d) be a metric space. A mappingT : X → X is a contraction on X if there is a α, α ∈ [0, 1) such that, forx, y ∈ X,

d(Tx, Ty) ≤ αd(x, y).

A.2 The Banach �xed point theorem

Theorem 34 (Banach �xed point theorem) Consider a complete metricspace X = (X, d). Let T : X → X be a contraction on X. Then T has preciselyone �xed point, i.e., a point x ∈ X such that T (x) = x.

37

Proof: Take a x0 ∈ X. Construct x1 = T (x0), x2 = T (x1), . . ., xn = T (xn−1).Since T is a contraction,

d(xn, xn+1) = d(Tnx0, Tnx1) ≤ αnd(x0, x1).

So for m ≥ n we have:

d(xm, xn) ≤ d(xm, xm+1) + . . . + d(xn−1, xn)≤ (αm + αm+1 . . . + αn)d(x0, x1)

≤ αm

1− αd(x0, x1) → 0 when m,n →∞.

The sequence (xn) is therefore a Cauchy sequence. By assumption X is completeand therefore the limit x to (xn) exists. Since xn = T (xn−1) and T is continuous,it follows that x = T (x). We have now found a �xed point, x, and the onlything left is to prove that it is unique.

Assume that there is another �xed point y. Then

d(x, y) = d(Tx, Ty) ≤ αd(x, y).

Since α < 1, this gives d(x, y) = 0 and hence x = y. �

If the contraction T , depends on a additional parameter, p ∈ P , we haveTp : X → X and the equation Tpxp = xp, xp ∈ X. In [3] the followingcorollary is presented.

Corollary 35 Suppose that

(i) P is a metric space, called the parameter space,(ii) for each p the operator Tp is a contraction on X

with α independent of p,(iii) for p0 ∈ P and x ∈ X, limp→p0 Tpx = Tp0x.

Then, for each p ∈ P , the equation Tpxp = xp has exactly one solution xp ∈ X,and limp→p0 xp = xp0 .

Proof: Let xp be the solution of the integral equation, Tpxp = xp. We get

d(xp, xp0) = d(Tpxp, Tp0xp0) ≤ d(Tpxp, Tpxp0) + d(Tpxp0 , Tp0xp0)≤ αd(xp, xp0) + d(Tpxp0 , Tp0xp0)

and therefore

d(xp, xp0) ≤1

1− αd(Tpxp0 , Tp0xp0) → 0 when p → p0. �

38

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