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Well-Posedness of the Model

CHAPTER 4. SINGULARLY PERTURBED MODEL 28

CHAPTER 4. SINGULARLY PERTURBED MODEL 29 defineL(b,a) asL(b)L1(a). The (i, j)th elementli j(b,a) ofL(b,a) denotes the probability that a person in the jth state at ageawill survive to agebin theith state. We can prove the following.

Lemma 4.6 ForL(a,b), the following holds:

1. L(b,a)is nonnegative, 2. kL(b,a)k ≤eµ(ba),

where the operator norm is related to the l1-type norm inRn. Proof. 1. Clearly,L(b,a) satisfies

d

dbL(b,a)=Q(b)L(b,a), L(a,a)=I.

Letη:=sup

i,a

|qii(a)|. ThenQ(a)+ηIis a nonnegative matrix, and we have d

db

nL(b,a)eη(ba)o

=

Q(b)+ηIL(b,a)eη(ba). By Picard’s iteration method, we have the following representation

L(b,a)eη(ba) =I+ Z b

a

Q(ρ)+ηI dρ +

Z b a

Q1)+ηI Z ρ1

a

Q2)+ηI21+· · · .

Since the right-hand side is nonnegative, we can conclude thatL(b,a) is nonnegative.

2. Next, letli j(b,a) denote the (i, j)th element ofL(b,a). Then d

dbli j(b,a)= Xn

k=1

qik(b)lk j(b,a), where li j(a,a)=δi j, and summing over indexi, we have

d db

Xn

i=1

li j(b,a)= Xn

k=1

Xn

i=1

qik(b)lk j(b,a)= Xn

k=1

−µk(b)

lk j(b,a)

≤(−µ) Xn

i=1

li j(b,a).

Thus we obtain

Xn

i=1

li j(b,a)≤eµ(ba).

CHAPTER 4. SINGULARLY PERTURBED MODEL 30 This shows thatkL(b,a)k ≤eµ(ba). This completes the proof.

In the following series of lemmas, we are going to prove the second condition of Theorem 2.18, and that will subsequently prove our main Theorem 4.5.

Lemma 4.7 Ais a closed linear operator in L1(0,w; Rn).

Proof. Linearity ofA follows immediately from linear properties of differentiation and matrix multiplication.

Next, in order to show that A is a closed operator, we prove that if φn ∈ D(A),

nlim→∞φn =: φ ∈ L1([0,w],Rn) and lim

n→∞Aφn =: v ∈ L1([0,w],Rn), then φ ∈ D(A) and Aφ=v.

Since

Aφn

(a)=− d

daφn(a)+Q(a)φn(a), it follows that

Z a 0

Aφn

(σ)dσ=− Z a

0

d

dσφn(σ)dσ+ Z a

0

Q(σ)φn(σ)dσ

=−φn(a)+φn(0)L(a)+ Z a

0

Q(σ)φn(σ)dσ or, φn(0)=φn(a)+

Z a 0

Aφn

(σ)dσ+ Z a

0

Q(a)φn(σ)dσ.

Then we find that

φn(0)−φm(0) Rn

kLk

L1

φn−φm

L1 +w

AφnAφm

L1 +w

QφnQφm L1. Hence{φn(0)}n

Nis a Cauchy sequence. Let lim

n→∞φn(0)=α. If we define ω(a) :=αL(a)−

Z a

0

v(σ)L(a, σ)dσ,

then we can see thatωis absolutely continuous, because the right hand side is differ- entiable almost everywhere and its derivative is integrable on [0,w] (we assumed that Q(a) is bounded in [0,w]). Also we have

φn(a)−ω(a)=φn(0)L(a)− Z a

0

Aφn

(σ)L(a, σ)dσ−αL(a)+ Z a

0

v(σ)L(a, σ)dσ,

CHAPTER 4. SINGULARLY PERTURBED MODEL 31 which shows that limn→∞

φn−ω

L1 = 0, so φ = ωalmost everywhere. Moreover, from φn(0) = R w

0 B(s)φn(s)ds we haveα = Rw

0 B(s)φ(s)dswhen n → ∞. Since φ = ω almost everywhere, we obtain

α=ω(0)= Z w

0

B(s)ω(s)ds.

Since we can identifyφasω(as an element ofL1) by modifying the values on the null set, so we can say thatφ ∈D(A) andAφ =v. This completes the proof.

Lemma 4.8 ρ(A)⊃ {λ: Reλ >β¯−µ}.

Proof. [28] Define the characteristic matrix ˜Γ: C→B(Rn,Rn) by Γ(λ) :=˜

Z w 0

Γ(a)eλada,

with Γ(a) :=B(a)L(a).

Let us first prove the following result which we are going to use in this lemma.

σ(A)=n

λ:λ ∈C, det

I−Γ(λ)˜

=0o

. (4.6)

To prove (4.6), let λ ∈ C be such that det

I−Γ(λ)˜

= 0. Then there must exist x∈Rn,x,0 such that

I−Γ(λ)˜

x=0. Define

φ(0) :=x, (4.7)

φ(a) :=eλaL(a)φ(0), a≥0. (4.8) Thenφ,0and using

I−Γ(λ)˜

x=0we get φ(0)=

Z w 0

eλaB(a)L(a)φ(0)da

= Z w

0

B(a)φ(a)da,

i.e.,φ∈D(A). Differentiating (4.8) with respect toa, we get φ0(a)=−λeλaL(a)φ(0)+eλa d

daL(a)φ(0)

=−λeλaL(a)φ(0)+eλaQ(a)L(a)φ(0) (by (4.4))

=−λφ(a)+Q(a)φ(a) (by (4.8)

or, λφ(a)=−φ0(a)+Q(a)φ(a)=Aφ(a),

CHAPTER 4. SINGULARLY PERTURBED MODEL 32 and this implies λ ∈ σ(A). Now to prove our lemma, let F(λ), λ ∈ R denote the Frobenius root of the characteristic matrix ˜Γ(λ), let ˜Γi j(λ) be the (i, j)th element of Γ(λ). Let˜ Sp(A) denote the spectral radius of the operator A. Then it can be shown that Sp( ˜Γ(λ))≤F(λ) [24], p.57

. From

Γ(Reλ)˜ ≤

Z w 0

kΓ(v)k evReλdv

≤ Z w

0

kB(v)k kL(v)k evReλdv

≤β¯ Z w

0

e

µ+Reλ

vdv by Lemma 4.6 and assumption (A3)

≤ β¯ Reλ+µ

1−e

Reλ+µ w

, we obtain

F(Reλ)≤max

j n

X

i=1

Γ˜ij(Reλ)=

Γ(Reλ)˜

(4.9)

≤ β¯ Reλ+µ

h1−e(Reλ+µ)wi ,

where for the proof of the inequality in (4.9), we refer to [24, p.63]. Therefore if Reλ > β¯ − µ then Sp( ˜Γ(λ))≤F(Reλ)<1. Thus det(I−Γ(λ))˜ ,0 for Reλ >β¯−µ.

Hence using (4.6), we getρ(A)⊃ {λ:Reλ >β¯−µ}. Lemma 4.9 Ifλ∈ρ(A)withλ >β¯−µ, then R(λ,A)is given by

R(λ,A)ψ(a)=eλaL(a) I− Z w

0

B(b)L(b)eλbdb

!1

× Z w

0

B(a)L(a)eλa Z a

0

eλbL1(b)ψ(b)db da+eλaL(a) Z a

0

eλbL1(b)ψ(b)db.

Proof. Following [65], define

∆(λ)x:=x− Z w

0

eλaB(a)L(a)xda,

wherex∈Rnand Reλ > β−µ. Letψ∈L1. There existsφ∈L1satisfying (λIA)φ=ψ if and only if

λφ(a)+φ0(a)−Q(a)φ(a)=ψ(a), (4.10)

φ(0)= Z w

0

B(a)φ(a)da. (4.11)

CHAPTER 4. SINGULARLY PERTURBED MODEL 33

The solution of (4.10) is given by (see [42, Proposition 5.2, p. 242]) φ(a)=eλaL(a)φ(0)+

Z a 0

eλ(ab)L(a,b)ψ(b)db. (4.12) Substituting this formula forφinto the boundary equation (4.11), we obtain

φ(0)= Z w

0

eλaB(a)L(a)φ(0)da+ Z w

0

B(a)

"Z a 0

eλ(ab)L(a,b)ψ(b)db

# da.

Sinceλ > β−µ,λ<σ(A), so that∆(λ)1exists and thus we get φ(0)= I

Z w 0

eλaB(a)L(a)da

!1Z w 0

B(a)

"Z a 0

eλ(ab)L(a)L1(b)ψ(b)db

#

da. (4.13) Hence, putting the expression ofφ(0) from (4.13) into the equation (4.12), we get

φ(a)=R(λ,A)ψ(a)=eλaL(a) I− Z w

0

B(b)L(b)eλbdb

!1

× Z w

0

B(a)L(a)eλa Z a

0

eλbL1(b)ψ(b)db da+ eλaL(a) Z a

0

eλbL1(b)ψ(b)db.

Lemma 4.10 D(A)=L1(0,w; Cn).

Proof. To prove the result, we follow [28] but we will go through the detailed calcu- lations. If λ >β¯−µ, we can defineφλ = λ(λIA)1ψfor allψ ∈ L1(0,w; Cn). Since φλ ∈ D(A), it is sufficient to show thatφλ → ψasλ → ∞inL1(0,w;Cn). Now using the expression of the resolvent from Lemma 4.9 we have

φλ(a)=λeλaL(a) I− Z w

0

B(b)L(b)eλbdb

!1

× Z w

0

B(u)L(u)eλu Z u

0

eλvL1(v)ψ(v)dv du+ λeλaL(a) Z a

0

eλvL1(v)ψ(v)dv.

We can write

φλ−ψ L

1

≤J1+J2, where

J1=λ Z w

0

eλaL(a) I− Z w

0

B(b)L(b)eλbdb

!1

× Z w

0

B(u)L(u)eλu Z u

0

eλvL1(v)ψ(v)dv du da,

CHAPTER 4. SINGULARLY PERTURBED MODEL 34 and

J2= Z w

0

λeλaL(a) Z a

0

eλvL1(v)ψ(v)dv−ψ(a)

da. (4.14)

Now

I− Z w

0

eλaB(a)L(a)da

≥ 1−

Z w 0

eλaL(a)B(a)da

≥1−β Z w

0

e(λ+µ)ada





∵β Z w

0

e(λ+µ)ada< β¯ λ+µ<1





or,

I− Z w

0

eλaB(a)L(a)da

!1

≤ λ+µ

λ−(β−µ)+βe(λ+µ)w ≤ λ+µ λ−(β−µ). So

J1 ≤λ Z w

0





e(λ+µ)a

I− Z w

0

eλaB(a)L(a)da

!1 Z w

0

B(u)L(u)eλu ×

Z u 0

eλv

L1(v)ψ(v) dv du

# da

≤λβ λ+µ λ−(β−µ)

Z w 0

"

e(λ+µ)a Z w

0

e(λ+µ)u Z u

0

e(λ+µ)v ψ(v)

dv du

# da

=λβ λ+µ λ−(β−µ)

Z w 0

"

e(λ+µ)a Z w

0

e(λ+µ)v ψ(v)

Z w v

e(λ+µ)udu dv

# da

≤λβ 1 λ−(β−µ)

Z w 0

"

e(λ+µ)a Z w

0

e(λ+µ)v.e(λ+µ)v ψ(v)

dv

# da

=λβ 1 λ−(β−µ).

1−e(λ+µ)w

λ+µ .

ψ L

1

or J1 ≤λ1−e(λ+µ)w λ−( ¯β−µ). β¯

λ+µ ψ

L

1 , (4.15)

and hence lim

λ→∞J1 =0.

CHAPTER 4. SINGULARLY PERTURBED MODEL 35 To show limλ→∞J2 =0, we notice that the expressionJ2in (4.14) can be written as J2 =

Z w 0

λL(a)

Z a 0

eλ(av)L1(v)ψ(v)dv−ψ(a) da,

≤ Z w

0

kL(a)k

Z a

0

λeλ(av)L1(v)ψ(v)dv−L1(a)ψ(a)

da

≤ Z w

0

λ

Z a 0

λeµaeλ(av)L1(v)ψ(v)dv−eµaL1(a)ψ(a)

da

∵kL(a)k ≤eµa

= Z w

0

Z a 0

λeµteλteµ(at)L1(a−t)ψ(a−t)dt−eµaL1(a)ψ(a)

da (wheret:=a−v)

= Z w

0

Z a 0

λeλteµtψ(a¯ −t)dt−ψ(a)¯

da, (4.16)

where ¯ψ(a) :=eµaL1(a)ψ(a) and since

ψ(a)¯

=eµa

L1(a) ψ(a)

≤eµaeµa ψ(a)

=

ψ(a) ,

we have ¯ψ∈L1. Now, following [65], to prove that the expression (4.16) tends to zero asλ→ ∞, lett>0 and defineψt ∈L1by

ψt(a) :=





0, a<t;

eµtψ(a¯ −t), a>t.

Since

ψt−ψ¯ L1 =

Z

0

ψt(a)−ψ(a)¯ da

= Z t

0

ψ(a)¯

da+

Z

t

eµtψ(a¯ −t)−ψ(a)¯ da

≤ Z t

0

ψ(a)¯

da+

Z

t

eµt−1

ψ(a¯ −t) da+

Z

t

ψ(a¯ −t)−ψ(a)¯ da

= Z t

0

ψ(a)¯

da+

Z

0

eµt−1 ψ(z)¯

dz+

Z

0

ψ(z)¯ −ψ(t¯ +z) dz,

(4.17) we claim that

limt0

ψt−ψ¯

L1 =0. (4.18)

To prove (4.18), the last term from (4.17) needs some attention and we show that limt0

Z

0

ψ(t¯ +z)−ψ(z)¯

dz=0

or, lim

t0

ψ¯t−ψ¯

L1 =0, (4.19)

where ¯ψt(z) :=ψ(t¯ +z).

CHAPTER 4. SINGULARLY PERTURBED MODEL 36

Now as the continuous functions with compact support are dense inL1([59, Theorem 2.4]), the proof of the fact (4.19) is a simple consequence of the approximation of integrable functions by continuous functions with compact support. In fact for any >0, we can find such a functiongsuch thatkψ¯ −gk

L1 < . Now ψ¯t−ψ¯ = gtg+

ψ¯tgt

− ψ¯ −g,

wheregt(x) :=g(t+x). However,kψ¯tgtkL1 =kψ¯ −gkL1 < , while sincegis continuous and has compact support, we have

gtg L1 =

Z

0

g(t+z)−g(z)

dz→0 as t→0.

Therefore, if |t| < δ, whereδ is sufficiently small, then kgtgkL1 < , and as a result kψ¯t−ψ¯k

L1 <3, whenever|t|< δ. This proves (4.19) and subsequently we have (4.18).

Now from (4.16), we have J2

Z 0

Z a

0

λeλteµtψ(a¯ −t)dt−ψ(a)¯

da

= Z

0

Z a 0

λeλteµtψ(a¯ −t)dt− Z

0

λeλtψ(a)¯ dt

da ∵ Z

0

λeλtdt=1

!

= Z

0

Z

0

λeλth

ψt(a)−ψ(a)¯ i dt

da

≤ Z

0

( λeλt

Z

0

ψt(a)−ψ(a)¯ da

) dt

≤ sup

0t

ψt−ψ¯ L1

Z

0

λeλtdt+2 ψ¯

L1

Z

λeλtdt

≤ sup

0t

ψt−ψ¯

L1 +2 ψ¯

L1eλ. Therefore, using (4.18), we find lim

λ→∞J2 =0.Hence we can conclude that

λlim→∞

φλ−ψ L1 =0,

and this completes the proof.

Lemma 4.11 k(λIA)1k ≤ 1 λ−( ¯β−µ).

Proof. From the proof of Lemma (4.10) we have

IA)1ψ ≤ 1

λJ1+ Z w

0

eλa Z a

0

eλvkL(a,v)k ψ(v)

dv da,

CHAPTER 4. SINGULARLY PERTURBED MODEL 37

where, from (4.15), we have

J1 ≤ λβ¯

λ+µ. 1

λ−( ¯β−µ). ψ

L1 ,

and Z w

0

eλa Z a

0

eλv kL(a,v)k ψ(v)

dv da≤ 1 λ+µ.

ψ L1. So, we have

IA)1 ≤ 1

λ+µ





 β¯

λ−( ¯β−µ) +1







= 1

λ−( ¯β−µ).

Proof of Theorem 4.5. Considering above lemmas, we can see that (A,D(A)) satisfies condition (2) of Hille–Yosida Theorem 2.18. Hence, the proof of Theorem 4.5 follows immediately due to equivalence conditions (1) and (2) in theorem 2.18.

The specific spectral properties of the migration matrix what has been discussed in this chapter is going to be used in the asymptotic analysis of the model in next chapter. Also the existence of the C0-semigroup (Theorem 4.5) is one of the crucial result needed for all of our further analysis.

CHAPTER 5

Asymptotic Analysis of the Perturbed Model

5.1 Aggregated Model

Biological heuristics suggests that no geographical structure should persist for very large interstate migration rates; that is, for →0. Here we also note that both biological and mathematical analysis rely onλ=0 being the dominant simple eigenvalue ofC(a) for eacha∈R+with the corresponding positive eigenvector, denoted byk(a), and the left eigenvector 1 = (1,1, . . . ,1). Vector k(a) is normalized to satisfy 1 ·k = 1 and k(a) = (k1(a), . . . ,kn(a)) is the so-called stable patch structure; that is, the asymptotic (ast → ∞ and disregarding demographic processes) distribution of the population among the patches for a given agea. Thus, in population theory, the components ofk are approximated aski ≈ui/ufori=1, . . . ,n, where

u:=u·1= Xn

i=1

ui.

Adding together equations in (4.1) and using the above approximation we obtain

ut ≈ −ua−µu, (5.1)

where µ := 1 ·Mk is the ‘aggregated’ mortality. This model, supplemented with boundary condition

u(0,t)≈ Z

0

β(a)u(a,t)da, (5.2)

38

CHAPTER 5. ASYMPTOTIC ANALYSIS OF THE PERTURBED MODEL 39 whereβ :=1·Bkis the ‘aggregated fertility’, is expected to provide an approximate description of the averaged population. Thus, (5.1) is the macroscopic and (4.1) the mesoscopic description of the population.

The main goal of this chapter is a rigorous validation of the above heuristics;

that is, that the true total populationucan be approximated by the solution ¯uof the aggregated problem (5.1)–(5.2) (where ‘≈’ is replaced by ‘=’) with an -order error.

The analysis is involved due to the initial and boundary conditions which are not consistent with those of the aggregated model. This makes the problem singularly perturbed and thus necessitates a careful analysis of the boundary, corner and initial layer phenomena.

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