Population Growth Analysis of an Age Structure Population Model

Population growth analysis of an age structure population model

Nina Håkansson

LITH-MAT-EX05/16SE

Examensarbete: 20p

Level: D

Examiner: Vladimir Kozlov Department of mathematics Applied mathematics Linköpings universitet

Supervisors: Vladimir Kozlov and Bengt Ove Turesson Department of mathematics Applied mathematics Linköpings universitet

Uno Wennergren Department of biology Theory and Modeling Linköpings universitet

Linköping May 2005

Datum Avdelning, Institution Date Division, Department

Matematiska Institutionen may 2005 581 83 LINKOPING¨ SWEDEN

Spr˚ak Rapporttyp ISBN Language Report category ISRN Svenska/Swedish Licentiatavhandling LiTH - MAT - EX - - 05/16 - - SE x Engelska/English x Examensarbete C-uppsats Serietitel och serienummer ISSN D-uppsats Title of series, numbering 0348-2960 Ovrig¨ rapport

URL f¨orelektronisk version http://www.ep.liu.se/exjobb/mai/2005/tm/016/

Titel Population growth — analysis of an age structure population model Title

F¨orfattare Nina H˚akansson Author

Sammanfattning Abstract This report presents an analysis of a partial diﬀerential equation, resulting from population model with age structure. The existence and uniqueness of a solution to the equation are proved. We look at stability of the solution. The asymptotic behaviour of the solution is treated. The report also contains a section about the connection between the solution to the age structure population model and a simple model without age structure.

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

Abstract

This report presents an analysis of a partial dierential equation, resulting from population model with age structure. The existence and uniqueness of a solution to the equation are proved. We look at stability of the solution. The asymptotic behaviour of the solution is treated. The report also contains a section about the connection between the solution to the age structure population model and a simple model without age structure.

Contents

1 Introduction 9 1.1 Population models ...... 9 1.1.1 Population models without age structure ...... 9 1.1.2 Population models with age structure time-independent case ...... 9 1.1.3 Population models with age structure time-dependent case ...... 10 1.1.4 Other population models with age structure ...... 10 1.2 Presentation of the model to study ...... 11 1.2.1 What to study ...... 12

2 Existence and uniqueness of solutions 13 2.1 The integral equation ...... 13 2.2 Existence and uniqueness of the solution ...... 15

3 Stability of the model 17 3.1 Stability, considering m ...... 17

4 Asymptotics of N the time-independent case 18 4.1 Representation of n(0, · ) ...... 19 4.1.1 Domain of convergence ...... 19 4.1.2 The Laplace transform of n(0, · ) ...... 20 4.1.3 Asymptotic behaviour of Ln(0, · ) ...... 22 4.1.4 The representation of n(0, · ) ...... 22 4.2 Asymptotics of N ...... 24

5 Asymptotics of Nthe time-dependent case 26 5.1 Upper bounds for n(0, · ) ...... 26 5.2 Lower bounds for n(0, · ) ...... 29 5.3 Upper and lower bounds for N ...... 31 5.4 Comparison with the time-independent case ...... 32

6 Comparison with the model without age structure 33

6.1 Biological analysis of the requirement σ0 > −1 ...... 34 7 Conclusions 35

A Banach theory 37 A.1 Denitions ...... 37 A.2 The Banach xed point theorem ...... 37

1 Introduction

This section presents population models of dierent types. In the following sections 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 time t = 0, can be described with dierential equations of varying complexity and accuracy. 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) = births − deaths + migration. dt The simplest model occurs when there is no migration and the death and birth rates are proportional to N. This model is frequently used in physics, chemistry and biology:

dN(t) = bN(t) − dN(t) with solution N(t) = N e(b−d)t, (1) dt 0 where N0 = N(0) is the initial population, b the constant birth rate and d the constant 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; the environment is probably limiting N. At a certain population size there will, for example, be insucient resources to support a larger population. The logistic growth (suggested by Verhulst 1836) includes such a limiting part:

dN(t) N(t) = rN(t) 1 − . dt K

Assume r > 0. If N0 < K the population will increase, but it will never exceed N = K. If N0 > K the population will instead decrease. The constant K is called the carrying capacity of the environment and is the largest N that the environment can support. This model predicts that N → K as t → ∞. A delay model compensates for the time it takes for new individuals to reach maturity. It has the form:

dN(t) = f(N(t),N(t − T )), dt where T is a constant called the delay. This model describes a population with a maturation period and then a birth rate independent of the age distribution of the population.

1.1.2 Population models with age structure time-independent case

In a model with age structure, the death rate µ and birth rate m depends on the age 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 of age a at time t. If we look at an age group at age a, from time t to t + dt the change 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) ∂n(a, t) = dt + da. ∂t ∂a Our a represents chronological age, therefore da = dt. If the equation is divided with dt the result is a partial dierential equation: ∂n(a, t) ∂n(a, t) + = −µ(a)n(a, t), a, t ≥ 0. ∂t ∂a The newborn part of the population contributes only to n(0, t), which gives the boundary condition at a = 0: Z ∞ n(0, t) = m(a)n(a, t) dt, t ≥ 0. 0 For t = 0, we have a start population with a certain age distribution f(a) and the 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 for example pollution, the death and birth rates for the population will probably change over time too. The model

∂n(a, t) ∂n(a, t) + = −µ(a, t)n(a, t), a, t ≥ 0, ∂t ∂a describes this. Its boundary conditions are Z ∞ n(0, t) = m(a, t)n(a, t) dt, t ≥ 0 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 the same species and may also depend on R ∞ , the total number of µ N = 0 n(a, t) da individuals. The model ∂n(a, t) ∂n(a, t) + = −µ(a, t, N)n(a, t) ∂t ∂a

10 describes this. Its boundary conditions are:

Z ∞ n(0, t) = m(a, t, N)n(a, t) dt, t ≥ 0 0 n(a, 0) = f(a), a ≥ 0.

A population can be inuenced by other populations and the birth rate for , , may also depend on R ∞ , the total number of individuals n mn P = 0 p(a, t) da in another population. The model

 ∂n(a, t) ∂n(a, t)  + = −µ (a, t, P )n(a, t)  ∂t ∂a n ∂p(a, t) ∂p(a, t)  + = −µ (a, t, N)p(a, t) ∂t ∂a p describes this. Its boundary conditions are

 Z ∞  n(0, t) = mn(a, t)n(a, t) dt, n(a, 0) = fn(a),  0 Z ∞   p(0, t) = mp(a, t)p(a, t) dt, p(a, 0) = fp(a). 0 1.2 Presentation of the model to study We will for the most part study the dierential equation from section 1.1.3:

∂n(a, t) ∂n(a, t) + = −µ(a, t)n(a, t), (2) ∂t ∂a with the boundary conditions

Z ∞ n(0, t) = m(a, t)n(a, t) dt, n(a, 0) = f(a). (3) 0 The function n(a, t), the number of individuals of age a at time t, is dened for 0 ≤ t < ∞. It is positive and assumed to be continuous. The growth of n(a, t) is assumed to be at most exponential. The death rate µ(a, t), dened for a, t ≥ 0, is at the lowest no deaths and at most all individuals dead, so µ(a, t) ∈ [0, 1] for alla a, t. Since no individual can became innitely old, the death rate becomes equal to 1 for large a. That is, µ(a, t) = 1, a > Aµ for some constant Aµ. The birthrate m(a, t), dened for a, t ≥ 0, is a positive function. It has compact 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 some constant M. Since f describes the start population, it is positive, bounded and has compact support. We will assume that the start population contains some individuals young enough to eventually have children, otherwise we get n(0, t) = 0 for all t. From a biological perspective it is interesting to study the increase and decrease of the total population. Therefore we will also study R ∞ , 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 dierential equation (2) has exactly one solution with at most exponential growth. This is proved with help from the Banach xed point theorem. Therefore we will have to transform the dierential equation to an integral equation. We will also consider the stability of the model in section 3, using a corollary to Banach's theorem. The asymptotics of the solution can be studied in two ways. In section 4 we will use the Laplace transform. This method requires extra assumptions on the functions and is too dicult for the time-dependent case, so we will use it on the time-independent case. For the time-dependent case, in section 5, we will use the second method. The asymptotics will be estimated using upper and lower bounds. Section 6 compares the solutionof the model with agestructure with 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 most exponential growth. The dierential 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) = (T n)(t), t ≥ 0, so that we can apply the Banach xed point theorem (Appendix A.2) to show that 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), satises

− R a µ(v,v+t−a) dv f(a − t)e a−t , a ≥ t n(a, t) = (4) R a µ(v,v+t−a) dv n(0, t − a)e 0 , a < t where n(0, t) solves the integral equation:

Z t − R a µ(v,v+t−a) dv n(0, t) = m(a, t)n(0, t − a)e 0 da 0 Z ∞ − R a µ(v,v+t−a) dv + m(a, t)f(a − t)e a−t da. (5) t If m and f are dierentiable, then n(a, t) described by (4),(5) solves problem (2),(3).

Proof: To obtain this result, we start by transforming the equation into an ordinary dierential equation. Let a and t be functions of x and y:

a = x + y, t = x − y.

Then the partial dierential equation can be rewritten as an ordinary dierential equation since

dn(x + y, x − y) ∂n(a, t) dt ∂n(a, t) da ∂n(a, t) ∂n(a, t) = + = + . dx ∂t dx ∂a dx ∂t ∂a This dierential equation has the solution

− R x µ(s+y,s−y)ds n(x + y, x − y) = C(2y)e y , where is to be determined. With and as variables and a−t , we get C a t v = s + 2

− R a µ(v,v+t−a) dv n(a, t) = C(a − t)e a−t . (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

− R a µ(v,v+t−a) dv n(a, t) = f(a − t)e a−t .

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

R 0 µ(v,v+t) dv C(−t) = n(0, t)e −t .

If we now replace t with t − a, we get

R 0 µ(v,v+t−a) dv C(a − t) = n(0, t − a)e a−t .

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

R a µ(v,v+t−a) dv n(a, t) = n(0, t − a)e a−t 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,

Z ∞ n(0, t) = m(a)n(a, t) da. (7) 0

If we insert the expression for n(a, t) from equation (4) into (7), we obtain the integral equation (5). Now let us consider the last part of Theorem 1. If m and f are dierentiable functions, then substituting t − a = x we have

Z t − R t−x µ(v,v+x) dv n(0, t) = m(t − x, t)n(0, x)e 0 dx 0 Z Am − R x+t µ(v,v−x) dv + m(x + t, t)f(x)e x dx 0 is clearly dierentiable since the right side is dierentiable. From (4) it follows that n is dierentiable.

To make the equations more readable, we will sometimes write only µ when what we really mean is µ(v, v+t−a). The integral equation (5) can be rewritten as n(0, t) = (T n)(t), if T is dened as

(T n)(t) = (Kn)(t) + F (t), t ≥ 0, where Z t − R a µ(v,v+t−a) dv (Kn)(t) = m(a, t)n(0, t − a)e 0 da 0 and Z ∞ − R a µ(v,v+t−a) dv F (t) = m(a, t)f(a − t)e a−t da. t

14 2.2 Existence and uniqueness of the solution In this section, we will prove that the integral equation (5) has exactly one solution of at most exponential growth.

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

−Λt kukΛ = sup |u(t)|e . t≥0

Theorem 2 The equation n(0, t) = (T n)(t), t ≥ 0 has exactly one solution in BΛ for a suciently 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 contraction on 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 k · kΛ is in fact a norm and that the space is complete with respect to the metric dened by the norm.

We start by checking that k · kΛ is a norm. The norm is obviously non- negative. Only zero has the norm equal to zero, since if kukΛ = 0, then u must be identically zero. Conversely k0kΛ = 0. The norm is homogeneous, since

−Λt −Λt −Λt kaukΛ = sup |au(t)|e = sup |a||u(t)|e = |a| sup |u(t)|e = |a|kukΛ. t≥0 t≥0 t≥0 The triangle inequality holds, since

−Λt −Λt ku + vkΛ = sup |u(t) + v(t)|e ≤ sup(|u(t)| + |v(t)|)e t≥0 t≥0 −Λt −Λt ≤ sup |u(t)|e + sup |v(t)|e = kukΛ + kvkΛ. t≥0 t≥0

So k · kΛ is a norm. −Λt To prove that BΛ is a complete space, set vi(t) = e ui(t), where (u)i ⊂ BΛ is a Cauchy sequence. Then (v)i is a Cauchy sequence in B0. Indeed,

−Λt kum − unkΛ = sup |um(t) − un(t)|e = sup |vm(t) − vn(t)| = kvm − vnk0. t≥0 t≥0

So BΛ with the norm k · k is complete since B0 = C[0, ∞) with the ordinary supremum norm is complete.

Proposition 4 The operator T is a contraction on BΛ for Λ suciently large.

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

− R a µ(v,v+t−a) dv |m(a, t)e 0 | ≤ M. The following inequality comes in handy:

|n(0, t)|e−Λ(t) ≤ sup |n(0, t)|e−Λ(t) = kn(0, · )k. t≥0

15 For the mapping T we observe:

Z t −Λt − R a µ dv |(T n)(t)|e = m(a, t)e 0 n(0, t − a) da 0 Z ∞ − R a µ dv −Λt + m(a, t)f(a − t)e a−t da e t Z t Z ∞ ≤ M|n(0, t − a)|e−Λ(t−a)e−Λa da + M|f(a − t)|e−Λt da 0 t Z t −Λa ≤ Mk(n(0, · )kΛe da + MN(0). 0 This gives us

−Λt −Λt 1 e k(T n)kΛ = sup |(T n)(t)|e = sup Mkn(0, · )kΛ − + MN(0) t≥0 t≥0 Λ Λ

≤ αkn(0, · )kΛ + MN(0), where α = M/Λ. We have proved that T : BΛ → BΛ. If we choose Λ > M then α < 1. We have Z t − R a µ dv k(T n1 − T n2)kΛ = sup m(a, t)e 0 (n1(0, t − a) − n2(0, t − a)) da t≥0 0

≤ αkn1(0, · ) − n2(0, · )kΛ. This shows that that T is a contraction.

Finally, since BΛ is a Banach space and T is a contraction for Λ > M Theorem 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 support dened on , with the norm . We want to [0, ∞) × [0, ∞) kmkP = supa,t≥0 m(a, t) prove the following theorem:

Theorem 5 The integral equation nm(0, t) = (Tmnm)(t) has exactly one solution and tends to in as tends to in . nm nm nm0 BΛ m m0 P Proof: We have to check conditions (i)-(iii) described in the corollary to the Banach xed point theorem.

(i) P is a metric space.

(ii) The operator Tm is a contraction on BΛ with bound depedent on M and independent of m. (iii) Let ε > 0. Then

Z t −Λ(t−a) −Λa kTmn − Tm0 nkΛ ≤ sup |m(a, t) − m0(a, t)||n(0, t − a)e |e da t≥0 0 Z ∞ −Λ(t−a) −Λa + sup |m(a, t) − m0(a, t)||f(a − t)e |e da t≥0 t Z t −Λa ≤ sup km − m0kP kn(0, · )kΛe da t≥0 0 Z ∞ −Λa + sup km − m0kP kfk−Λe da t≥0 t km − m k ≤ 0 P (kn(0, · )k + kfk ) < ε Λ Λ −Λ if Λε km − m0kP < δ = . kn(0, · )kΛ + kfk−Λ This proves the theorem.

17 4 Asymptotics of N the time-independent case To determine the asymptotic behaviour of , R ∞ , the number N N(t) = 0 n(a, t) da of individuals in a population at time t, in the time-independent case, we will use the Laplace transform. This method will require the extra assumption of bounded derivatives on µ, m and f. If m and µ are independent of time we have the dierential equation: ∂n(a, t) ∂n(a, t) + = −µ(a)n(a, t) ∂t ∂a with the solution

− R a µ(v) dv f(a − t)e a−t , a ≥ t n(a, t) = − R a µ(v) dv n(0, t − a)e 0 , a < t where n(0, t) satises the integral equation

Z t Z ∞ − R a µ(v) dv − R a µ(v) dv n(0, t) = m(a)n(0, t − a)e 0 da + m(a)f(a − t)e a−t da. 0 t

We write A(t) ∼ B(t) if limt→∞ A(t)/B(t) = 1. Theorem 6 The function N has the following asymptotic behaviour:

σ0t N(t) ∼ Ce if σ0 > −1 −t N(t) ∼ Cte if σ0 = −1 −t N(t) ∼ Ce if σ0 < −1 where σ0 satises Z ∞ −σ a−R a µ(v) dv m(a)e 0 0 da = 1. 0 We rst prove a special case of Riemann-Lebesgues lemma that will be useful later.

Lemma 7 Let R(a) be a continuous function on R with compact support and σ ∈ R. Then Z ∞ R(a)e−σae−iωa da → 0, when |ω| → ∞ 0 and Z 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, let ε . Choose so that for −σa1 −σa2 when α = A+2π ω0 ω > ω0 |R(a1)e − R(a2)e | < α 2π . Choose so that Aω Aω0 . |a1 − a2| ≤ ω M ∈ N 2π < M ≤ 2π + 1 We divide the integral into M integrals. Suppose that ω ≥ ω0. Then

∞ M−1 2π(k+1) Z X Z ω R(a)e−σa−iωa da = R(a)e−σa(cos(aω) − sin(aω)) da. 2πk 0 k=0 ω

18 For an arbitrary integral in the sum, we have

2π(k+1) 2π(k+1) Z ω Z ω −σa 2πk −σ 2πk R(a)e cos(aω) da ≤ R( )e ω cos(aω) da 2πk 2πk ω ω ω 2π(k+1) Z ω −σa 2πk −σ 2πk α + (R(a)e − R( )e ω ) cos(aω) da ≤ 2π . 2πk ω ω ω The integral is bounded by ε:

Z ∞ M−1 −σa X α (Aω + 2π)α R(a)e cos(aω) da ≤ 2π < < (A + 2π)α = ε. ω ω 0 0

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

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

− R t µ(v) dt g(t) = m(t)e 0 , t ≥ 0, (8) and Z ∞ − R a µ(v) dv F (t) = m(a)f(a − t)e a−t da, t ≥ 0. (9) t

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

n(0, t) = Keσ0t + O(e(σ0−δ)t) as t → ∞ where

LF (σ) K = , ∂ (1 − Lg(σ)) ∂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. We know that n is a function with at most exponential growth:

−Λt kn(a, · )kΛ = sup |n(a, t)|e = B < ∞, t≥0 which gives us an upper bound for n:

n(a, t) ≤ BeΛt.

The Laplace transform of , R ∞ −st converges if Re n Ln(a, s) = 0 n(a, t)e dt, (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 equation for n(0, · ):

Z t Z ∞ − R a µ(v) dv − R a µ(v) dv n(0, t) = m(a)n(0, t − a)e 0 da + m(a)f(a − t)e a−t da. 0 t 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 enough to eventually have children. Therefore LF is not identically zero. Using the Laplace transform, we get

LF (s) Ln(0, s) = (10) 1 − Lg(s) where Z ∞ −st−R t µ(v) dv Lg(s) = m(t)e 0 dt 0 and Z ∞ Z ∞ −st−R a µ(v) dv LF (s) = m(a)f(a − t)e a−t da dt. 0 t The term Ln(0, s) is dened where Re(s) > Λ but the right hand side of (10) has an extension to the whole complex plane. In [5], page 423, the following sucient condition for inversion of the Laplace transformation is found. If k there exist positive constants h, R0 and k such that |Ln(0, s)| ≤ h/|s| when |s| > R0, we can nd n(0, · ) by using the inversion formula for the Laplace transformation: 1 Z σ+i∞ n(0, t) = Ln(0, s)estds, 2πi σ−i∞ where σ > Λ. In Lemma 12 we will see that the condition is satised. The integral can be calculated by integration along a rectangle and calculations of the residues of the poles. Therefore we need to determine the poles of the right hand 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

Z Am Z Am − R a µ(v) dv −st LF (s) = m(a)f(a − t)e t−a e da dt. 0 t By dierenting under the integral signs using the fact that e−st is entire, it follows that LF is entire. The same argument applies to Lg:

Z Am − R t µ(v) dv −st Lg(s) = m(t)e 0 e dt. 0 Now we examine where Lg(s)=1.

20 Lemma 10 The function 1 − Lg has exactly one real zero, σ0, and the zero is of order one. Proof: Let s = σ + iω. Recall that m is not allowed to be 0 everywhere. For s = σ, Lg is a real-valued, continuous and decreasing function of σ: ∞ Z R t − µ(v,) dv −σ1t −σ2t Lg(σ1) − Lg(σ2) = m(t)e 0 (e − e ) dt > 0 if σ1 < σ2. 0 If , then : σ > M = supa≥0 m(a) Lg(s) < 1 Z ∞ M Lg(σ) ≤ Me−σt dt = < 1. (11) 0 σ ∞ R a There exists so that R −σt− 0 µ(v) dv , since is not σ Lg(σ) = 0 m(a)e dt > 1 m 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

Z Am d −σt−R t µ(v) dv (1 − Lg(σ)) = tm(t)e 0 dt dσ 0 and the last integral is not 0 since the integrand is non-negative and not identically zero.

Notice that (11) gives us the information σ0 ≤ M. We need to prove that there are no complex zeroes in the area of integration. Lemma 11 There exists a δ such that the function 1 − Lg has no complex zero in the half plane Re(s) > σ0 − δ. Proof: We consider ve cases.

For s = σ + iω and σ > σ0, we have Z ∞ − R a µ(v) dv −σa Re(Lg(s)) = m(a)e 0 e cos(aω) da 0 Z ∞ − R a µ(v) dv −σ a < m(a)e 0 e 0 da = 1 0

For s = σ0 + iω and ω 6= 0, we have, since cos(aw) < 1 almost everywhere, Z ∞ − R a µ(v) dv −σ a Re(Lg(s)) = m(a)e 0 e 0 cos(aω) da 0 Z ∞ − R a µ(v) dv −σ a < m(a)e 0 e 0 da = 1. 0

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 fact −σt that Lg(s) is a integral over e 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) is analytic, around each point on the line segment. It follows from Heine-Borel therorem 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 will make it possible to calculate the integral in the inversion formula. To be able to integrate by parts we must make the extra assumptions on f, m and µ.

Lemma 12 Let m and µ be dierentiable with bounded derivatives and f two times dierentiable with bounded derivatives. Then

n(0, 0) 1 Ln(0, s) = + O as |Re(s)| → ∞. s s2

Proof: We start with Lg(s):

R t Z ∞ − µ(v) dv Am −st−R t µ(v,) dv m(t)e 0 −st Lg(s) = m(t)e 0 dt = − e 0 s 0 Z Am 1 0 − R t µ(v) dv −st 1 + (m (t) − m(t)µ(t))e 0 e dt = O . s 0 s This gives us 1 1 = 1 + O . 1 − Lg(s) s Now look at LF (s):

Z Am Z Am Z Am −st−R a µ(v) dv −st LF (s) = m(a)f(a − t)e a−t da dt = F (t)e dt, 0 t 0 where F (t) has bounded derivatives. So we have

0 Am Z Am −F (t) −st F (t) −st 1 00 −st LF (s) = e + 2 e + 2 F (t)e dt s s 0 s 0 1 Z Am 1 n(0, 0) 1 = m(a)f(a) da + O 2 = + O 2 . s 0 s s s

The result for Ln(0, s) is

n(0, 0) 1 1 n(0, 0) 1 Ln(0, s) = + O 1 + O = + O . s s2 s s s2

4.1.4 The representation of n(0, · ) Proof of Theorem 8: To nd n(0, t) we will use Laplace inversion formula and integrate along a line in the half plane σ > Λ:

1 Z σ+i∞ n(0, t) = Ln(0, s)estds. 2πi σ−i∞

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 ω 6 L2 6 L1

σ0 - σ r

? L4 -

We have 1 Z 1 Z n(0, t) = − lim Ln(0, s)estds − lim Ln(0, s)estds ω→∞ ω→∞ 2πi L2 2πi L3 1 Z st σ0t − lim Ln(0, s)e ds + Res Ln(0, σ0)e . ω→∞ 2πi L4

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

n(0, 0) Z σ0−δ+i∞ e−st 1 Z σ0−δ+i∞ e−st ds + 2 O(1)ds. 2πi σ0−δ−i∞ s 2πi σ0−δ−i∞ s If we change the integration variable to ω and use the fact that the integral of an odd function is zero,we see that these integrals equal

n(0, 0) Z ∞ (σ − δ) cos(ωt) Z ∞ ω sin(ωt) (σ0−δ)t 0 e 2 2 dω + 2 2 dω π 0 ω + (σ0 − δ) 0 ω + (σ0 − δ) Z ∞ e−iωt (σ0−δ)t +ie 2 O(1) dω. −∞ (σ0 − δ + iω) Observe that the last integral is bounded since

Z ∞ 1 2 2 dω < ∞. −∞ (σ0 − δ) + ω So with help from [4], page 181, we get

1 Z σ0−δ+i∞ st (σ0−δ)t (σ0−δ)t Ln(0, s)e ds = 2n(0, 0)H(σ0 − δ) + O(e ) = O(e ), 2πi σ0−δ−i∞ where H denotes the heaviside-function. The residue of the pole at σ0 is

σ0t LF (σ0)e Res σ0t σ0t Ln(0, σ0)e = ∂ = Ke . ∂σ (1 − Lg(σ))σ=σ0

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

−σt−R t µ(v) dv m(t)e 0

23 and Z ∞ −σt−R a µ(v) dv m(a)f(a − t)e a−t da t 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 dened by:

Z ∞ Z t Z ∞ − R a µ(v) dv − R a µ(v) dv N(t) = n(a, t) da = n(0, t−a)e 0 da+ f(a−t)e a−t da. 0 0 t 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):

Z t Z t σ (t−a)−R a µ(v) dv (σ −δ)(t−a)−R a µ(v) dv N(t) = K e 0 0 da + e 0 0 O(1) da 0 0 Z ∞ − R a µ(v) dv + f(a − t)e a−t da = I1 + I2 + I3. t

To estimate these integrals, we will use the fact that µ(a) = 1 for a ≥ Aµ; no individual can be innitely old. Consider N(t) for t > Aµ. We will use ci to denote constants. We start with I1:

t Z R a σ0(t−a)− µ(v) dv I1 = K e 0 da 0

Z Aµ Z t A σ t −σ a−R a µ(v) dv σ t −σ a−(a−A )−R µ µ(v) dv = e 0 K e 0 0 da + e 0 K e 0 µ 0 da. 0 Aµ

So for σ0 6= −1 we get

σ0t −t I1 = c1(σ0)e − c2(σ0)e and for σ0 = −1 −t −t I1 = c3e t + c4(σ0)e .

For I2 we can use the same calculations as for I1. We have chosen δ so that σ0 − δ 6= −1. We have

t t Z R a Z R a (σ0−δ)(t−a)− µ(v) dv (σ0−δ)t −(σ0−δ)a− µ(v) dv I2 = e 0 O(1) da = O(1)e e 0 da 0 0

Z t A Z Aµ (σ −δ)t −(σ −δ+1)a−R µ µ(v) dv+A −(σ −δ)a−R a µ(v) dv = O(1)e 0 e 0 0 µ da + e 0 0 Aµ 0 = O(1)e(σ0−δ)t − O(1)e−t, where O(1) denotes a bounded function.

24 Finally, we calculate the integral I3:

∞ ∞ Z R a Z R Aµ R x+t − µ(v) dv − x µ(v) dv− A µ(v) dv I3 = f(a − t)e a−t da = f(x)e µ dx t 0 ∞ Z R Aµ −t −x− µ(v) dv+Aµ −t = e f(x)e x dx = c5e . 0

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

σ0t −t (σ0−δ)t −t −t N(t) = c1(σ0)e − c2(σ0)e + O(1)e − O(1)e + c5e and for σ0 = −1:

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

Notice that the population is increasing if σ0 > 0.

25 5 Asymptotics of Nthe 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 work when m or µ depend on time, and it requires that m, f and µ are dierentiable functions. In this section we will instead try to nd upper and lower bounds for N to get information about the asymptotic behaviour. We need rst to consider bounds n+(t) and n−(t) to n(0, t). Recall that n(0, t) satises the integral equation (5):

Z t Z ∞ − R a µ dv − R a µ dv n(0, t) = m(a, t)n(0, t − a)e 0 da + m(a, t)f(a − t)e a−t da 0 t = (Kn)(t) + F (t).

For a function n we use the notation E(n)(t) = n(0, t) − (Kn)(t) − F (t). We will rst prove two lemma that describe sucient conditions for such bounds n+ and n−.

Lemma 13 If n+ satises E(n+)(t) ≥ 0 for all t ≥ 0, and n, n+ is of at most exponential growth, then n+ ≥ n(0, · ), where n(0, · ) is the solution to the integral equation (5).

Proof: Let n1, n2 ... denote the sequence of approximations to n obtained from the Picard iteration scheme with n+ as the initial approximation. Then n1 = (Kn+) + F ≤ n+. Suppose that nk−1 ≤ n+ for some k ≥ 2. Then nk = (Knk−1) + F ≤ (Kn+) + F ≤ n+. By induction, nk ≤ n+ for all k, and thus by letting , . k → ∞ n ≤ n+ In an analog way, we can prove the following lemma.

Lemma 14 If n− satises E(n−)(t) ≤ 0 for all t ≥ 0, and n+ is of at most exponential growth, then n− ≤ n(0, · ) where n(0, · ) is the solution to the integral equation (5).

Let σ0(t1) = −∞ if m(a, t1) ≡ 0 and otherwise, for a given t, let σ0(t) be the function that satises the following equation:

Z ∞ −σ (t)a−R a µ(v,v+t−a) dv m(a, t)e 0 0 da = 1. 0

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

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

∞ Z R a − a−t µ dv Ff1 (t) = m(a, t)f1(a − t)e da. t We will need a lemma.

26 Lemma 15 Suppose that solve the integral equation (5), with as nf1 (0, · ) f1 start 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 for all nf (0, t) ≤ Cnf1 (0, t) t ≥ 0. Proof: We have

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

Since f1(a) > 0 for x ∈ [0,Am] and f(a) is bounded, there exists a constant C so that, for another f, ∞ Z R a − a−t µ(v,v+t−a) dv CFf1 (t) − Ff (t) = m(a, t)(Cf1(a − t) − f(a − t))e da ≥ 0. t 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 . Since

nf1 (0, · ) − K(nf1 ) − Ff1 = 0, we have . Thus is an upper bound for . This is E(Cnf1 ) ≥ 0 Cnf1 (0, · ) nf (0, · ) true for an arbitrary and therefore f Cnf1 (0, · ) ≥ n(0, · ). Now we can seek upper bounds for n under some conditions on σ0(t). We start 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, c and σ, such that

σt ct n(0, t) ≤ n+(t) = D1e H(T − t) + D2e H(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 , for nf1 ≤ n+

−σa−R a µ(v,v−a) dv f1(a) = D1e 0

ct σt for a ∈ [0,Am]. Choose the constants so that D2e ≥ D1e for t ∈ [0,T ]. Then Z t σt σt −σa − R a µ dv E(n+) = D1e H(T − t) − D1e m(a, t)e e 0 H(T − t + a) da 0 Z t ct ct −ca−R a µ dv +D2e H(t − T ) − D2e m(a, t)e 0 H(t − T − a) da 0 Z ∞ − R a µ dv + m(a, t)f1(a − t)e a−t da. t For t > T , we get Z ∞ ct σ(t−a) − R a µ dv E(n+) = D2e − m(a, t)D1e e 0 da t−T Z t−T ct −ca − R a µ dv −D2e m(a, t)e e 0 da. 0

27 σ(t−a) c(t−a) Using the fact that D1e ≤ D2e , we get

Z t ct ct −ca − R a µ dv E(n+) ≥ D2e − D2e m(a, t)e e 0 da ≥ 0. 0 Since, by assumtion, the last integral is not greater than 1. For t ≤ T , we get

t Z R a σt σt −σa − 0 µ dv E(n+) = D1e − D1e m(a, t)e e da − Ff1 (t). 0

There exists a σ such that E(n+) ≥ 0, σ ≥ M will always do. Since Z ∞ Mt σt −Ma − R a µ(v,v+t−a) dv E(n+) ≥ D1e − D1e Me e 0 da ≥ 0. 0 | {z } ≤1 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 ct n(0, t) ≤ n+(t) = D2e where n is the solution to (5) and H denotes the Heaviside function.

ct σt σt Proof: Since we chose D2e ≥ D1e for t ∈ [0,T ], we have D1e H(T − t) + ct ct ct D2e H(t − T ) ≤ D2e and n+(t) = D2e 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 that σ0(t) 6= −∞. Under this condition we can nd an upper bound for n(0, · ) for a σ0 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 σ such that

R t σt σ0(τ)d τ n(0, t) ≤ n+(t) = D1e H((T + Am) − t) + D2e 0 H(t − (T + Am)).

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

−σa−R a µ(v,v−a) dv f1(a) = D1e 0 for a ∈ [0,Am].

R t σ0(τ)d τ σt Choose D1, D2 and σ such that D2e 0 ≥ D1e for t ∈ [0,T + Am]. For t ≤ T + Am we get E(n+) ≥ 0 using the same calculations as in the previous case for t ≤ T . Consider E(n+) for t ≥ T + Am:

∞ R t Z R a σ0(τ)d τ σ(t−a) − µ dv E(n+) = D2e 0 − m(a, t)D1e e 0 da t−(T +Am)

t−(T +Am) Z R t−a R a σ0(τ)d τ − µ dv −D2 m(a, t)e 0 e 0 da. 0

28 R t σ0(τ)d τ σt If we now use the fact that D2e 0 ≥ D1e for t ∈ [0,T + Am], we get

∞ R t Z R t−a R a σ0(τ)d τ σ0(τ)d τ − µ dv E(n+) ≥ D2e 0 − D2 m(a, t)e 0 e 0 da 0 ∞ R t Z R t R a σ0(τ)d τ − σ0(τ)d τ − µ dv = D2e 0 1 − m(a, t)e t−a e 0 da . 0

R t − σ0(τ)d τ −aσ0(t) Since σ0 is a decreasing function, we know that e t−a ≤ e . Using this, we obtain

∞ R t Z R a σ0(τ)d τ −aσ0(t) − µ dv E(n+) ≥ D2e 0 1 − m(a, t)e e 0 da = 0. 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

R t σ0(τ)d τ n(0, t) ≤ n+(t) = D2e 0 .

R t σ0(τ)d τ σt Proof: We chose D1, D2 and σ such that D2e 0 ≥ D1e 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 condition that 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. Also suppose that both g(u, x) and f(x) ≥ 0 for all x. Then Z x g(u, x)f(x − u)du > 0 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 for u ∈ [u2, u2 + ε]. Let x ∈ (x1 + u2, x1 + u2 + δ + ε). For x ≥ u2 + ε we know 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 the interval [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 integrate over the interval u ∈ [u2, u2 + x1 + ρ]. When we integrate over the interval u ∈ [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 that F (t) > 0 for an interval of length δ otherwise we get the solution n(0, t) = 0. − R a µ dv We also know that g(a, t) = m(a, t)e 0 > 0 for an interval of length ε. Therefore, using the lemma above, we can choose p such that (KpF )(t) > 0 for an interval of length Am. We will start to look for a lower bound for n, for σ0(t) that eventually is bounded from below.

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

ct p−1 n(0, t) ≥ n−(t) = H(t − TF )Ce + F (t) + (KF )(t) + ... + (K F )(t)

p Proof: Choose TF > T such that (K F )(t) > 0 for t ∈ [TF ,TF + Am] this is possible according to Lemma 20. Consider rst E(n−)(t) for t < TF :

ct ct p E(n−) = H(t − TF )Ce −(KH(t − TF )Ce )(t) − (K F )(t) ≤ 0. | {z } =0

For t ≥ TF , Z t ct ct −ca − R a µ(v) dv P E(n−) = Ce − Ce m(a)e e 0 H(t − TF − a) da − (K )F (t), 0 which can be rewritten: Z ∞ ct −ca−R a µ dv E(n−) = Ce 1 − m(a, t)e 0 da (12) 0 Z ∞ ct −ca−R a µ dv p +Ce m(a, t)e 0 da − (K )F (t). (13) t−TF

Notice that the term in (12) is negative since σ0(t) ≥ c for t ≥ T . The rst term in (13) equals 0 for t ≥ TF + Am since m(a, t) = 0 for a ≥ Am. Notice p that (K F )(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 constants C and TF such that

ct n(0, t) ≥ n−(t) = H(t − TF )Ce . Proof: We know that

σ0t ct p−1 H(t − TF )Ce ≤ H(t − TF )Ce + F (t) + (KF )(t) + ... + (K F )(t).

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

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

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

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

∞ R t Z R t R a − σ0(τ)dτ − σ0(τ)dτ− µ dv E(n−) = Ce 0 1 − m(a, t)e t−a 0 da 0 Z ∞ − R t σ (τ)dτ−R a µ dv p+1 +C m(a, t)e t−a 0 0 da − (K F )(t). t−TF

30 R t − σ0(τ)dτ −aσ0(t) Since σ0(t) is increasing for t ≥ T we know that e t−a ≥ e . For t ≥ TF , this implies that ∞ R t Z R a − σ0(τ)dτ −σ0(t)a− µ dv E(n−) ≤ Ce 0 1 − m(a, t)e 0 da 0 Z ∞ − R t σ (τ)dτ−R a µ dv p+1 +C m(a, t)e t−a 0 0 da − (K )F (t) t−TF Z ∞ − R t σ (τ)dτ−R a µ dv p+1 = C m(a, t)e t−a 0 0 da −(K )F (t). t−TF | {z } bounded As in the proof for Lemma 21 we can choose and so that . C p E(n−) ≤ 0 Clearly the following theorem is true.

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

R t − σ0(τ)d τ n(0, t) ≥ H(t − TF )Ce 0 .

5.3 Upper and lower bounds for N Now we are ready to look at N, which has the following equation: Z ∞ Z t Z ∞ − R a µ dv − R a µ dv N(t) = n(a, t) da = n(0, t − a)e 0 da + f(a − t)e a−t da. 0 0 t

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 constants C1 and C2 such that, for t > T + Aµ, we have

ct −t N(t) ≤ C1e + C2e , if c 6= −1, and −t −t N(t) ≤ C1e + C2te , if c = −1.

Proof: Insert our rst upperbound n+(t), from Theorem 17, which was calculated under the given condition on σ0, into the equation for N. Then Z t Z ∞ c(t−a) − R a µ(v,v+t−a) dv − R a µ(v,v+t−a) dv N(t) ≤ D2e e 0 da + f(a − t)e a−t da. 0 t

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

Aµ t Z R a Z R Aµ c(t−a) − µ dv c(t−a) −a+Aµ− µ(v,v+t−a) dv N(t) ≤ D2e e 0 da + D2e e 0 da 0 Aµ ∞ Z − R Aµ µ(v,v+x) dv−R x+t µ(v,v+x) dv + f(x)e x Aµ dx 0 Z Aµ Z t Z ∞ c(t−a) c(t−a) −a+Aµ −(x+t−Aµ) ≤ D2e da + D2e e da + F e dx. 0 Aµ 0

This proves the theorem.

31 Theorem 26 Suppose that σ0(t) ≥ c for all t ≥ T . Then there exists constants C1, C2 and TF such that, for t > TF + Aµ, we have ct −t N(t) ≥ C1e + C2e , if c 6= −1 and −t −t N(t) ≥ C1e + C3te , if c = −1.

Proof: Insert n−(0, t) from Theorem 22 into the equation for N. Z t Z ∞ c(t−a) − R a µ dv − R a µ dv N(t) ≥ H(t − a − TF )Ce e 0 da + f(a − t)e a−t da. 0 t − R a µds −a Since e 0 ≥ e , we have

Z t−TF Z ∞ c(t−a) −a −t N(t) ≥ Ce e da + f(a − t)e da. 0 t

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

R t R t σ0(τ)d τ σ0(τ)d τ −(c+1)t −t N(t) ≤ C1e 0 + C2e 0 e + C3e , if c 6= −1, and

R t R t σ0(τ)d τ σ0(τ)d τ −t N(t) ≤ C1e 0 + C2e 0 (t − Aµ) + C3e , if c = −1.

R t−a σ (τ)d τ −σ (t)a −ca Proof: We use the fact that e t 0 ≤ e 0 ≤ e . We obtain

Aµ ∞ Z R t−a R a Z R a σ0(τ)d τ − µ(v,v+t−a) dv − µ dv N(t) = D2e 0 e 0 da + f(a − t)e a−t da 0 t t Z R (t−a) R Aµ σ0(τ)d τ +Aµ−a− µ(v,v+t−a) dv + D2e 0 e 0 da Aµ

Aµ t R t Z R t Z R t σ0(τ)d τ − σ0(τ)dτ − σ0(τ)d τ −a ≤ D2e 0 e t−a da + e (t−a) e da 0 Aµ −t +C3e .

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

Aµ t R t Z R t Z σ0(τ)d τ −aσ0(t) σ0(τ)d τ −(σ0(t)+1)a −t D2e 0 e da + D2e 0 e da + C4e 0 Aµ

Aµ t R t Z R t Z σ0(τ)d τ −ac σ0(τ)d τ −(c+1)a −t ≤ D2e 0 e da + D2e 0 e da + C4e . 0 Aµ 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 that we have the following estimate:

σ0t σ0t C1e ≤ n(0, t) ≤ C2e . This gives the same asymptotic behaviour for N as we found in Section 4.

32 6 Comparison with the model without age structure

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

Z ∞ Z ∞ −σ a−R a d dv −(σ +d)a b 1 = be 0 0 da = be 0 da = . 0 0 σ0 + d

Nicely enough σ0 = b − d. So for m and µ constant and σ0 > −1 the solution to the age structure population model, for the time-independent case, has the σ t same asymptotics, N(t) v e 0 , as the most simple model. For non constant m and µ σ0 is the asymptotic birthrate minus death rate for the total population:

R ∞ m(a)n(a, t) da − R ∞ µ(a)n(a, t) da b − d = lim 0 0 t→∞ N(t)

t R a t R a R σ0(t−a) − 0 µ(v) dv R σ0(t−a) − 0 µ(v) dv 0 m(a)Ke e da − 0 µ(a)Ke e da = lim t R a t ∞ t→∞ R σ0(t−a) − 0 µ(v) dv R (σ0−δ)(t−a) R 0 Ke e da + 0 O(1)e da + t n(a, t) da

t ∞ R a R (σ0−δ)(t−a) R − a−t µ(v) dv 0 O(1)e da − t f(a − t))e da + t R a t ∞ R σ0(t−a) − 0 µ(v) dv R (σ0−δ)(t−a) R 0 Ke e da + 0 O(1)e da + t n(a, t) da

t R a R t R −σ0a 0 µ(v) dv 0 −σ0t− µ(v) dv 1 − 0 e (e ) da e 0 = lim t R a = lim t R a + σ0 = σ0 t→∞ R −σ0a − 0 µ(v) dv t→∞ R −σ0a− 0 µ(v) dv 0 e e da 0 e da

It is interesting to nd out when the simplest model has the same asymptotic solution as the solution to the age structure population model. That is, what conditions on m and µ correspond to σ0 > −1? If σ0 > −1 we have:

Z ∞ Z ∞ −σ a−a+a−R a µ(v) dv a−R a µ(v) dv 1 = m(a)e 0 0 da < m(a)e 0 da, 0 0

∞ R a and if R a− 0 µ(v) dv , we have for : 0 m(a)e da > 1 σ ≤ −1

Z ∞ Z ∞ −(1+σ )a+a−R a µ(v) dv a−R a µ(v) dv Lg(σ) = m(a)e 0 0 da ≥ m(a)e 0 da > 1. 0 0

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

∞ R a Theorem 28 R a− 0 µ(v) dv σ0 > −1 ⇔ 0 m(a)e 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 lifetime it gets R ∞ children. If R ∞ is less than 1 we surely have a 0 m(a) da 0 m(a) da population that will die out and biologically its not a possible scenario. If instead R ∞ is greater than 1 we have: 0 m(a) da Z ∞ Z ∞ a−R a µ(v) dv 1 < m(a) da ≤ m(a)e 0 da 0 0 and σ0 > −1 for this population. So we can assume that σ0 > −1 for biological populations.

34 7 Conclusions

We have proved the existence of a unique solution to the model with age structure. We have found the asymptotics for the time-independent case, and bounds for some cases of the time-dependent case. It would be interesting to nd more bounds and eventually the complete asmptotics for the time-dependent case. We would like to nd out when the age structure model gives the same asymptotics as the simple model without age structure. The next step could be to take in consideration birth and death rates 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 inuences the asymptotics is interesting. The goal is to determine under which conditions the approximation with the simple model without 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, New York, 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 theorem can be used. In order to prepare for this theorem, a couple of denitions are needed.

A.1 Denitions The following denitions are presented in [2].

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

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

Denition 30 (Completeness) A sequence in a metric space is called Cauchy if 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 which belongs to X.

Denition 31 (Norm) A norm on a vector space is a real-valued function k · k on X, with the properties:

N(1) kxk ≥ 0 N(2) kxk = 0 ⇔ x = 0 N(3) kαxk = |α|kxk N(4) kx + yk ≤ kxk + kyk.

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

Denition 32 (Banach space) A Banach space is a normed vector space which is complete with respect to the metric dened by the norm.

Denition 33 (Contraction) Let X = (X, d) be a metric space. A mapping T : X → X is a contraction on X if there is a α, α ∈ [0, 1) such that, for x, y ∈ X, d(T x, T y) ≤ αd(x, y).

A.2 The Banach xed point theorem Theorem 34 (Banach xed point theorem) Consider a complete metric space X = (X, d). Let T : X → X be a contraction on X. Then T has precisely one 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,

n n n d(xn, xn+1) = d(T x0,T x1) ≤ α d(x0, x1).

So for m ≥ n we have:

d(xm, xn) ≤ d(xm, xm+1) + ... + d(xn−1, xn) m m+1 n ≤ (α + α ... + α )d(x0, x1) αm ≤ d(x , x ) → 0 when m, n → ∞. 1 − α 0 1

The sequence (xn) is therefore a Cauchy sequence. By assumption X is complete and 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 only thing left is to prove that it is unique. Assume that there is another xed point y. Then

d(x, y) = d(T x, T y) ≤ 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 have Tp : X → X and the equation Tpxp = xp, xp ∈ X. In [3] the following corollary 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, for and , (iii) p0 ∈ P x ∈ X limp→p0 Tpx = Tp0 x.

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,Tp0 xp0 ) ≤ d(Tpxp,Tpxp0 ) + d(Tpxp0 ,Tp0 xp0 )

≤ αd(xp, xp0 ) + d(Tpxp0 ,Tp0 xp0 ) and therefore 1 d(x , x ) ≤ d(T x ,T x ) → 0 when p → p . p p0 1 − α p p0 p0 p0 0

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