Sensitivity Equations for a Size-StructuredPopulation Model

H. T. Banks, Stacey L. Ernstberger and Shuhua Hu

Center for Research in Scientific ComputationNorth Carolina State University

Raleigh, NC 27695-8205

September 28, 2007

Abstract

In this paper we consider the classical Sinko-Streifer size-structured population model andderive sensitivity partial differential equations for the sensitivities of solutions with respect toinitial conditions, growth rate, mortality rate and fecundity rate. Sample numerical resultsto illustrate use of these equations are also presented.

Keywords: Size-structured population models, sensitivity equations, method of character-istics, renewal equations, finite difference schemes.

1 Introduction

Since the seminal work of Sinko and Streifer in [37] (see also [31] and the related efforts ofMcKendrick [33] and von Forester [39]), size-structured population models and their gener-alizations have been widely investigated. Some of these efforts [3, 4, 29, 32] have focusedon establishing well-posedness and stability analysis of models, formulating schemes for nu-merical solutions, and parameter estimation of individual rates. However, to the best of ourknowledge, thus far there is no literature on sensitivity equations and related analysis for size-structured population models. Sensitivity analysis of dynamical systems has drawn the at-tention of numerous researchers [1, 6, 9, 10, 11, 13, 14, 15, 16, 17, 20, 24, 25, 27, 28, 35, 38, 40]for many years because the resulting sensitivity functions can be used in many areas such asoptimization and design [16, 26, 27, 34, 38], computation of standard errors [9, 10, 19, 21, 36],and information theory [12] related quantities (e.g., the Fisher information matrix) as wellas control theory, parameter estimation and inverse problems [5, 8, 9, 10, 11, 40, 41]. Oneof our motivations for investigating sensitivity for size-structured population model derivesfrom our efforts reported in [7], where a shrimp biomass production system and a relatedmedical counter measure production system currently under design and development is mod-eled in part by classical size-structured population models. The growth and mortality ratesof the shrimp population are affected by several environmental factors such as temperatureand salinity (e.g, [30, 42]). Hence, sensitivity of such a model with respect to the growthand mortality rates is one important factor in optimizing the entire production system.

1

In this paper we derive and investigate sensitivity equations for the linear size-structuredpopulation model

ut(x, t) + (g(x)u(x, t))x + m(x)u(x, t) = 0,

g(0)u(0, t) =

∫ x

0

β(x)u(x, t)dx,

u(x, 0) = u0(x).

(1)

Here (x, t) ∈ [0, x] × [0, T ], the function u(x, t) denotes the population density with size xat time t, g is the size-dependent individual growth rate, m represents the size-dependentmortality rate, β is the size-dependent reproduction or fecundity rate, and u0 is the initialpopulation density. The maximum size individuals may obtain in their lifetime is x. Theobjective in this paper is to derive equations for the sensitivities of u with respect to g, m,β, and u0.

For traditional finite dimensional parameter dependent ordinary differential equationsystems

x(t) = f(x(t), θ), x(0) = x0,

one finds that the heuristic differentiation with respect to θ of the system (with an interchangeof derivatives with respect to t and θ and use of the chain rule) to obtain

y(t) =∂f

∂xy(t) +

∂f

∂θ, y(0) = 0,

for y(t) =∂x

∂θ(t) provides the correct sensitivity equations. Moreover, this can be made com-

pletely rigorous, obtaining not only the form of the equations but also their well-posedness.While differentiation of (1) with respect to the functions g, m, β or u0 is somewhat moredelicate, our goal is to establish that a formal, heuristic differentiation also results in thecorrect sensitivity equations and that this heuristic derivation can be made rigorous. Aswe shall see below, the underlying technical details are nontrivial. As we next outline, weconsidered several possible approaches before settling on the one employed here.

A semigroup approach is one of the popular and elegant methods used in the literature(e.g., [3, 4]) to establish the existence and uniqueness of solutions to size-structured popula-tion models. The idea behind this approach is to write the partial differential equation (1)in the abstract form

u(t) = Au(t), u(0) = u0, (2)

in a Banach space with the linear operator A defined by Aϕ = −(gϕ)′ −mϕ. One can thenshow that A is an infinitesimal generator for a C0-semigroup, and thereby establish existenceand uniqueness of solutions to the model. One might then attempt to carry out a sensitivityanalysis for (1) by using the abstract theoretical framework provided in [9], wherein thesensitivity analysis was developed for the general Banach space nonlinear ordinary differentialequation

z(t) = f(t, z(t), µ), z(t0) = z0.

Here the solution z is in a complex Banach space Z, the parameter µ is in a convex subsetM of a topological vector space, and f : R+×Z×M→ Z. However, our efforts in trying to

2

apply this framework to (2) were not productive, as the domain of operator A is dependenton the parameters g and β themselves. For example, the domain of A in [4] is defined as

domA =

{ϕ ∈ H | gϕ ∈ H1(0, x), lim

x→x(gϕ)(x) = 0, (gϕ)(0) =

∫ x

0

β(x)ϕ(x)dx

}, (3)

and A : domA ⊂ H → H for an appropriately defined Hilbert space H.The variational approach in [4] provides another method to establish existence and

uniqueness of solutions to (1), where a Gelfand triple was constructed as V ↪→ H ↪→ V∗with V = domA∗. Here the adjoint operator A∗ of A is given by

A∗ψ = gψ′ −mψ + ψ(0)β, (4)

with domA∗ ={

ψ ∈ H | ψ ∈ H1

loc[0, x), gψ′ ∈ H, limx→x

(gϕψ)(x) = 0 for ϕ ∈ domA}

, and

the inner product on V given by

< ϕ, ψ >V=< (λ0 −A∗)ϕ, (λ0 −A∗)ψ >H . (5)

The variational form is then

d

dt< u(t), ϕ >V∗, V=< u(t),A∗ϕ >H for all ϕ ∈ V , t ≥ 0

u(0) = u0 ∈ H.(6)

At first inspection it might appear more convenient to carry out a sensitivity analysis for(6) instead of (1) since all the model parameters are passed to the operator A∗, which islinearly dependent on those parameters. However, the difficulty in using this approach isthat V is chosen to be the domain of A∗, and the resulting V-norm is dependent on g, m,and β (this can be seen from (4)–(5)). Hence, the sensitivity analysis of u with respect toour model parameters is not readily carried out using either a semigroup or a variationalapproach because the domains of A and A∗ are dependent on the model parameters. Itmight be possible to apply the idea of method of mappings employed in sensitivity analysisfor optimal shape design problems (e.g., [5, 22, 23, 26, 34]), which deals with computation ofderivatives with respect to shape variation. However, it is not a straightforward applicationto our problem since the domain here is an operator domain (functional domain) instead ofa geometric domain as used in optimal shape design problems.

The method of characteristics is another common technique used in the literature (e.g.,see [3, 18, 29, 32]) to establish the existence and uniqueness of solutions to size-structuredpopulation models. Using this approach, one obtains an implicit representation of the solu-tion which can be used to transform the partial differential equation into an integral equation.Then the contraction mapping theorem is applied in order to establish the desired results. Inthis paper we will employ this method to carry out a sensitivity analysis for (1). Althoughcomputationally tedious, this approach is conceptually straightforward and relatively easyto use.

The remainder of the paper is organized as follows. In Section 2, we give some preliminarytheoretical results that are essential to our sensitivity derivations given in Section 3. Finitedifference schemes to obtain numerical solutions of the model as well as the correspondingsensitivity equations are formulated in Section 4, and some sample numerical results are alsopresented.

3

2 Preliminary Results

In order to develop the sensitivity formulations of interest, a number of standing assumptionswill be imposed on the model parameters and initial conditions in (1). We assume

(H1) g ∈ W1,∞(0, x), g > 0 on [0, x), and g(x) = 0;

(H2) m ∈ L∞(0, x), and m ≥ 0 on [0, x];

(H3) β ∈ L∞(0, x), and β ≥ 0 on [0, x];

(H4) u0 ∈ L1(0, x), and u0 ≥ 0 on [0, x].

We will use ‖ · ‖∞ to denote the norm ‖ · ‖L∞(0,x), and ‖ · ‖1 to denote the norm ‖ · ‖L1(0,x)

throughout the paper. Note that assumption (H1) implies that g is in a convex subset ofW1,∞(0, x); assumptions (H2) and (H3) guarantee that both m and β are in a convex subsetof L∞(0, x), and assumption (H4) implies that u0 is in a convex subset of L1(0, x). Hence,we formulate the sensitivity equations for (1) by means of directional derivatives of u withrespect to parameters β, m, u0, and g. The directional derivative is defined by

Definition 2.1 Let Θ be a convex subset in some topological vector space, and f : R+×Θ →R. Given θ and ϑ in Θ, we define the derivative fθ(t; θ, ϑ − θ) of a function f at θ in thedirection ϑ− θ to be

fθ(t; θ, ϑ− θ) = limε→0+

f(t; θ + ε(ϑ− θ))− f(t; θ)

ε, (7)

provided this limit exists.

The method of characteristics can be used to reduce equation (1) as well as each of theassociated sensitivity equations to an equivalent renewal equation (a Volterra equation ofconvolution type), and existence and uniqueness of solutions to these integral equations areestablished via the following result from [2] (Theorem 7.2 on page 220, and Theorem 7.4 onpage 224).

Result 2.1 If Ψ(t) is bounded on [0, T ] and

∫ T

0

|φ(s)|ds < ∞, then the equation

Φ(t) =

∫ t

0

φ(s)Φ(t− s)ds + Ψ(t) (8)

has a unique bounded solution in [0, T ]. If we further assume that Ψ(t) is continuous, thenΦ(t) is also continuous.

Suppose that functions φ and Ψ in equation (8) are both dependent on parameter θ in aconvex subset Θ of some topological space. Then equation (8) becomes

Φ(t; θ) =

∫ t

0

φ(s; θ)Φ(t− s; θ)ds + Ψ(t; θ). (9)

We first give a sensitivity result for (9) because all our sensitivity results for (1) are heavilydependent upon it. We state this formally as a theorem because it is fundamental to thesubsequent results in this paper.

4

Theorem 2.2 Suppose that for a given θ ∈ Θ, φ(t; θ) and Ψ(t; θ) are both bounded on [0, T ].We assume further that for θ, ϑ ∈ Θ, φ has a bounded directional derivative φθ(t; θ, ϑ − θ)on [0, T ] with respect to θ ∈ Θ in the direction ϑ − θ, and Ψ has a bounded directionalderivative Ψθ(t; θ, ϑ − θ) on [0, T ] with respect to θ ∈ Θ in the direction ϑ − θ. Then thedirectional derivative Φθ(t; θ, ϑ− θ) of Φ with respect to θ in the direction ϑ− θ exists. Letz(t) = Φθ(t; θ, ϑ− θ). Then it satisfies the following equation

z(t) =

∫ t

0

φ(s; θ)z(t− s)ds +

∫ t

0

φθ(s; θ, ϑ− θ)Φ(t− s; θ)ds + Ψθ(t; θ, ϑ− θ). (10)

We give arguments to establish these results. Note that the boundedness of φ(t; θ) and Ψ(t; θ)on [0, T ] satisfy the assumptions of Result 2.1. Hence, the boundedness of Φ(t; θ) on [0, T ]is guaranteed. Since φθ(t; θ, ϑ−θ) and Ψθ(t; θ, ϑ−θ) are both bounded on [0, T ], we know that

there exists a positive constant c such that

∣∣∣∣∫ t

0

φθ(s; θ, ϑ− θ)Φ(t− s; θ)ds + Ψθ(t; θ, ϑ− θ)

∣∣∣∣ ≤c on [0, T ]. Thus, Result 2.1 guarantees that there exists a unique bounded solution z(t) to(10). Note that

Φ(t; θ + ε(ϑ− θ))− Φ(t; θ)

=

∫ t

0

φ(s; θ + ε(ϑ− θ))[Φ(t− s; θ + ε(ϑ− θ))− Φ(t− s; θ)]ds

+

∫ t

0

[φ(s; θ + ε(ϑ− θ))− φ(s; θ)]Φ(t− s; θ)ds + Ψ(s; θ + ε(ϑ− θ))−Ψ(s; θ).

Let D(t; θ, ϑ, ε) =Φ(t; θ + ε(ϑ− θ))− Φ(t; θ)

ε− z(t). Then we find that

|D(t; θ, ϑ, ε)| ≤∫ t

0

|φ(t− s; θ + ε(ϑ− θ))||D(s; θ, ϑ, ε)|ds

+

∫ t

0

|φ(s; θ + ε(ϑ− θ))− φ(s; θ)| |z(t− s)|ds

+

∫ t

0

∣∣∣∣φ(s; θ + ε(ϑ− θ))− φ(s; θ)

ε− φθ(s; θ, ϑ− θ)

∣∣∣∣ |Φ(t− s; θ)|ds

+

∣∣∣∣Ψ(t; θ + ε(ϑ− θ))−Ψ(t; θ)

ε−Ψθ(t; θ, ϑ− θ)

∣∣∣∣ .

We observe that the boundedness of φθ(t; θ, ϑ−θ) on [0, T ] implies that limε→0+

φ(t; θ+ε(ϑ−θ)) =

φ(t; θ) uniformly in t. Hence, by the boundedness of z we know that the second term in theright side of the above inequality converges to zero as ε → 0. The boundedness of Φ(t; θ) on[0, T ] and the existence of the directional derivative of φ with respect to θ in the directionϑ − θ imply that the third term of the right side of the above inequality converges to zeroas ε → 0. The fourth term converges to zero as ε → 0 by the existence of the directionalderivative of Ψ with respect to θ in the direction ϑ− θ. Hence, by Gronwall’s inequality wehave

limε→0+

|D(t; θ, ϑ, ε)| = 0,

and thus the theorem follows.

5

3 Sensitivity Equations

In this section, we derive equations for the sensitivity of u with respect to β, m, u0 and g.First, we will use the method of characteristics to obtain the renewal equation for (1) andshow that there exists a unique solution to this equation. We define several functions thatwill be used throughout this paper:

G(x) =

∫ x

0

1

g(ξ)dξ, B(t) =

∫ x

0

β(x)u(x, t)dx,

where u is the unique solution of (1) which is guaranteed to exist–see, e.g., [3, 29]. Byassumption (H1), we find that for ξ ∈ [0, x),

0 < g(ξ) = g(ξ)− g(x) ≤ ‖g′‖∞(x− ξ).

Hence, we have

G(x) ≥ 1

‖g′‖∞

∫ x

0

1

x− ξdξ,

which implies that limx→x

G(x) = ∞. Since g > 0 on [0, x), G is a strictly increasing function

so that G−1 exists and is a strictly increasing map from [0,∞) → [0, x). Therefore, therequirement of g(x) = 0 in (H1) guarantees that the size of an individual always stay less thanx. In addition, to simplify the expressions, the following functions will be used throughoutour derivations:

π(x, t) = G−1(G(x)− t), %(t, ξ) = G−1(t− ξ), ρ(x, t, ξ) = 2t−G(x)− ξ.

Using the method of characteristics, we find that the solution of (1) is given implicitly by:

• If x ≥ G−1(t), then

u(x, t) = u0(G−1(G(x)− t))

g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

). (11)

• If x < G−1(t), then

u(x, t) =B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

). (12)

Hence, we have

B(t) =

∫ G−1(t)

0

β(x)B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

+

∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx.

(13)

6

Let η = G(x). Then x = G−1(η) and dη =1

g(x)dx. Hence, the first term of the right side of

(13) can be rewritten as

∫ G−1(t)

0

β(x)B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

=

∫ t

0

β(G−1(η)) exp

(−

∫ G−1(η)

0

m(ξ)

g(ξ)dξ

)B(t− η)dη

=

∫ t

0

β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)B(t− η)dη.

(14)

Let η = G−1(G(x)− t). Then x = G−1(G(η) + t) and1

g(η)dη =

1

g(x)dx, which implies that

dx =g(G−1(G(η) + t))

g(η)dη. Since lim

τ→∞G−1(τ) = x, we have lim

x→xG−1(G(x) − t) = x for any

t ∈ [0, T ]. Hence, the second term of the right side of (13) can be rewritten as

∫ x

G−1(t)

β(x)u0(G−1(G(x)− t))

g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

)dx

=

∫ x

0

β(G−1(G(η) + t))u0(η) exp

(−

∫ G−1(G(η)+t)

η

m(ξ)

g(ξ)dξ

)dη.

(15)

Let σ = G(ξ)−G(η). Then ξ = G−1(G(η) + σ) and dσ =1

g(ξ)dξ. Hence, we have that

exp

(−

∫ G−1(G(η)+t)

η

m(ξ)

g(ξ)dξ

)= exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

).

Thus, from the above equality and (15) we have

∫ x

G−1(t)

β(x)u0(G−1(G(x)− t))

g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

)dx

=

∫ x

0

u0(η)β(G−1(G(η) + t)) exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

)dη.

(16)

Therefore, by (13), (14) and (16) we obtain the renewal equation

B(t) =

∫ t

0

k(η)B(t− η)dη + F (t), (17)

where

k(η) = β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

),

F (t) =

∫ x

0

u0(η)β(G−1(G(η) + t)) exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

)dη.

(18)

7

Note that 0 ≤ F (t) ≤ ‖β‖∞‖u0‖1, and 0 ≤ k(t) ≤ ‖β‖∞ on [0, T ]. Hence, Result 2.1guarantees that there exists a unique bounded nonnegative solution to (17). Thus, thissolution of (17) must be the same as B used in equations (11) and (12) in representing theunique nonnegative solution to (1).

We observe from equation (12) that the representation form for u(x, t) in the region{(x, t)|0 ≤ x < G−1(t), t > 0} is dependent on B. Hence, in order to find the sensitivityfunctions for u with respect to β, m, u0 and g, we first need to investigate sensitivity forB. To simplify the notation, we use h to denote a given direction in the correspondingparameter space whenever we take a directional derivative with respect to either β, m, u0

or g. In addition, we always suppress the parameter variables (β, m, u0 or g) in a functiondependent on them unless it can cause confusion, such as when taking the derivative withrespect to those parameters.

Since the sensitivity of u with respect to g is much more complicated than those withrespect to β, m or u0, we will consider it subsequently in Section 3.4. At present our goal isto derive the sensitivity equations for B with respect to β, m and u0. First we see that thedirectional derivative of k with respect to β in any given direction h exists. Let kβ(η; β, h)denote the directional derivative of k with respect to β in the direction h. Then it is givenby

kβ(η; β, h) = h(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

), (19)

which implies that |kβ(η; β, h)| ≤ ‖h‖∞ for η ∈ [0, T ]. We also observe that the directionalderivative of F with respect to β in any given direction h exists. Let Fβ(t; β, h) denote thederivative of F with respect to β in the direction h. Then it satisfies

Fβ(t; β, h) =

∫ x

0

h(G−1(G(η) + t))u0(η) exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

)dη, (20)

which implies that |Fβ(t; β, h)| ≤ ‖h‖∞‖u0‖1 for t ∈ [0, T ]. Thus, by Theorem 2.2, thedirectional derivative Bβ(t; β, h) of B with respect to β in the direction h exists and is givenby

Bβ(t; β, h) =

∫ t

0

k(η)Bβ(t− η; β, h)dη +

∫ t

0

kβ(η; β, h)B(t− η)dη + Fβ(t; β, h). (21)

It also follows that there exists a unique bounded solution to equation (21).Next we consider the sensitivity of B with respect to m. By the chain rule we find that

the directional derivative km(η; m,h) of k with respect to m in the direction h satisfies

km(η; m,h) = −β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)(∫ η

0

h(G−1(σ))dσ

)

= −k(η)

(∫ η

0

h(G−1(σ))dσ

),

(22)

and the directional derivative Fm(t; m,h) of F with respect to m in the direction h is givenby

Fm(t; m,h) = −∫ x

0

u0(η)β(π(η,−t)) exp

(−

∫ t

0

m(π(η,−σ))dσ

)(∫ t

0

h(π(η,−σ))dσ

)dη.

(23)

8

Hence, |km(η; m,h)| ≤ T‖h‖∞‖β‖∞ for η ∈ [0, T ], and |Fm(t; m,h)| ≤ T‖h‖∞‖β‖∞‖u0‖1 fort ∈ [0, T ]. Thus, we know from Theorem 2.2 that the directional derivative Bm(t; m,h) of Bwith respect to m in the direction h exists as the unique bounded solution to

Bm(t; m,h) =

∫ t

0

k(η)Bm(t− η; m,h)dη +

∫ t

0

km(η; m,h)B(t− η)dη + Fm(t; m,h). (24)

We observe that k does not depend on u0. Hence, the directional derivative ku0(η; u0, h)of k with respect to u0 in the direction h is given by

ku0(η; u0, h) = 0, 0 ≤ η ≤ T. (25)

We also note that the directional derivative Fu0(t; u0, h) of F with respect to u0 in any givendirection h exists and is readily seen to be

Fu0(t; u0, h) =

∫ x

0

h(η)β(G−1(G(η) + t)) exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

)dη, (26)

which implies that |Fu0(t; u0, h)| ≤ ‖β‖∞‖h‖1 for t ∈ [0, T ]. Let Bu0(t; u0, h) denote thedirectional derivative of B with respect to u0 in the direction h, then by Theorem 2.2 weknow that it exists and is the unique bounded solution to

Bu0(t; u0, h) =

∫ t

0

k(η)Bu0(t− η; u0, h)dη + Fu0(t; u0, h). (27)

Based on the above discussions on the sensitivities for B with respect to β, m and u0, wecan obtain the sensitivities for u with respect to the same functions using the expressionsfor u in (11) and (12). First we see that u is independent of β in the region {(x, t)|x > x ≥G−1(t), t ≥ 0}. Hence, we see that the directional derivative uβ(x, t; β, h) of u with respectto β in the direction h is zero in that region, that is,

uβ(x, t; β, h) = 0, for x ≥ G−1(t). (28)

Note that Bβ(t; β, h) exists for t ∈ [0, T ]. Hence, by (12) we find that

uβ(x, t; β, h) =Bβ(t−G(x); β, h)

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

), for x < G−1(t). (29)

By (11), (12), the existence of Bm(t; m,h) for t ∈ [0, T ] and the chain rule, we have thatthe directional derivative um(x, t; m,h) of u with respect to m in the direction h satisfies

um(x, t; m, h) = −u(x, t)

∫ x

G−1(G(x)−t)

h(τ)

g(τ)dτ, if x ≥ G−1(t), (30)

and

um(x, t; m,h) =Bm(t−G(x); m,h)

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

−u(x, t)

∫ x

0

h(τ)

g(τ)dτ, if x < G−1(t).

(31)

9

By (11), (12) and the existence of Bu0(t; u0, h) for t ∈ [0, T ], we find that if x ≥ G−1(t)then we have

uu0(x, t; u0, h) = h(G−1(G(x)− t))g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

), (32)

and if x < G−1(t) then we have

uu0(x, t; u0, h) =Bu0(t−G(x); u0, h)

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

). (33)

From the computational perspective, it is usually much easier to solve numerically adifferential equation than an integral equation. Hence, in the next three subsections, we willshow that uβ(x, t; β, h), um(x, t; m,h) and uu0(x, t; u0, h) are each respectively the solutionto some first-order hyperbolic partial differential equation. This means that we can use thesepartial differential equations to obtain the numerical solutions for the sensitivity of u withrespect to these functions instead of using integral equations such as (24) coupled with (31).As we shall see, these partial differential equations are in each case precisely those that wouldbe obtained if one simply heuristically differentiated equation (1) with respect to β, m andu0 respectively.

3.1 Sensitivity with Respect to Fecundity Rate

In this subsection, we want to derive the sensitivity partial differential equation for uβ, thesensitivity of u with respect to β. Let v be the unique solution (guaranteed, for example,by results in Section 2 of [3] or a slight modification of the arguments for Proposition 2.2 of[29]) of the initial boundary value problem

vt(x, t) + (g(x)v(x, t))x + m(x)v(x, t) = 0,

g(0)v(0, t) =

∫ x

0

[β(x)v(x, t) + h(x)u(x, t)]dx,

v(x, 0) = 0.

(34)

Our aim here is to characterize the unique solution to (34) and to argue that v =∂u

∂β[h],

the directional derivative of u with respect to β in the direction h. Using the method ofcharacteristics, we find that the implicit representation form for solution to (34) is given asfollows:

• If x ≥ G−1(t), thenv(x, t) = 0. (35)

• If x < G−1(t), then

v(x, t) =V (t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

), (36)

where V (t) =

∫ x

0

[β(x)v(x, t) + h(x)u(x, t)]dx.

10

Hence, we have

V (t) =

∫ G−1(t)

0

β(x)V (t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

+

∫ G−1(t)

0

h(x)B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

+

∫ x

G−1(t)

h(x)u0(G−1(G(x)− t))

g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

)dx.

Using the same transformation as we derived for B(t), we find that

V (t) =

∫ t

0

β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)V (t− η)dη

+

∫ t

0

h(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)B(t− η)dη

+

∫ x

0

h(G−1(G(η) + t))u0(η) exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

)dη,

(37)

which implies, by (19) and (20), that

V (t) =

∫ t

0

k(η)V (t− η)dη +

∫ t

0

kβ(η; β, h)B(t− η)dη + Fβ(t; β, h). (38)

Hence, equations (21), (38) and the uniqueness of solution to (21) imply that

V (t) = Bβ(t; β, h) for any t ∈ [0, T ], (39)

and (35), (36) agree with the unique solution of (34).

Next we argue that∂u

∂β[h] = v, which means that uβ(x, t; β, h) satisfies the initial bound-

ary value problem (34). By (28) and (35) we know that uβ(x, t; β, h) = v(x, t) in the region{(x, t)|x > x ≥ G−1(t), t ≥ 0}. From (36) and (39) we see that

v(x, t) =Bβ(t−G(x); β, h)

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

).

Hence, by the above equality and (29), we have that uβ(x, t; β, h) = v(x, t) in the region{(x, t)|0 ≤ x < G−1(t), t > 0}. Thus, uβ(x, t; β, h) satisfies the initial boundary valueproblem (34) and we can use this system to solve for uβ(x, t; β, h).

3.2 Sensitivity with Respect to Mortality Rate

The sensitivity of u with respect to m is considered in this section. Let w be the uniquesolution to the equation

wt(x, t) + (g(x)w(x, t))x + m(x)w(x, t) + h(x)u(x, t) = 0,

g(0)w(0, t) =

∫ x

0

β(x)w(x, t)dx,

w(x, 0) = 0.

(40)

11

We wish to show that w =∂u

∂m[h], the directional derivative of u with respect to m in the

direction h. Using the method of characteristics, we find that the solution to (40) is givenby

• If x ≥ G−1(t), then

w(x, t) = −∫ t

0

h(π(x, ξ))u(π(x, ξ), t− ξ)g(π(x, ξ))

g(x)exp

(−

∫ x

π(x,ξ)

m(τ)

g(τ)dτ

)dξ.

(41)

• If x < G−1(t), then

w(x, t) =W (t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

−∫ t

t−G(x)

h(%(t, ξ))u(%(t, ξ), ρ(x, t, ξ))g(%(t, ξ))

g(x)exp

(−

∫ x

%(t,ξ)

m(τ)

g(τ)dτ

)dξ,

(42)

where W (t) =

∫ x

0

β(x)w(x, t)dx.

First we simplify the expression in (41). Note that x ≥ G−1(t) implies that

G(π(x, ξ)) = G(x)− ξ ≥ t− ξ.

We further observe that G−1(G(π(x, ξ))− (t− ξ)) = π(x, t). Hence, by (11) we find that

u(π(x, ξ), t− ξ) = u0(π(x, t))g(π(x, t))

g(π(x, ξ))exp

(−

∫ π(x,ξ)

π(x,t)

m(τ)

g(τ)dτ

). (43)

Thus, for the case x ≥ G−1(t) we have that

w(x, t) = −u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

) ∫ t

0

h(π(x, ξ))dξ. (44)

We simplify the expression in (42). Note that x < G−1(t) implies that

G(G−1(t− ξ)) = t− ξ < t− ξ + (t−G(x)) = 2t−G(x)− ξ.

Hence, using equation (12) and the fact that 2t−G(x)− ξ −G(G−1(t− ξ)) = t−G(x), wefind that

u(G−1(t− ξ), 2t−G(x)− ξ) =B(t−G(x))

g(G−1(t− ξ))exp

(−

∫ G−1(t−ξ)

0

m(τ)

g(τ)dτ

). (45)

Using the above equality and (42), for the case x < G−1(t) we have that

w(x, t) =W (t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

−B(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

) ∫ t

t−G(x)

h(G−1(t− ξ))dξ.

(46)

12

By (44) and (46) we find that

W (t) =

∫ G−1(t)

0

β(x)W (t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

−∫ G−1(t)

0

β(x)B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ t

t−G(x)

h(G−1(t− ξ))dξ

)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

) (∫ t

0

h(π(x, ξ))dξ

)dx.

Using the same transformation as we derived for B(t), we obtain that

W (t) =

∫ t

0

β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)W (t− η)dη

−∫ t

0

β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)(∫ η

0

h(G−1(σ))dσ

)B(t− η)dη

−∫ x

0

u0(η)β(π(η,−t)) exp

(−

∫ t

0

m(π(η,−σ))dσ

)(∫ t

0

h(π(η,−σ))dσ

)dη.

From the above equality, equation (22), and equation (23), we see that

W (t) =

∫ t

0

k(η)W (t− η)dη +

∫ t

0

km(η; m, h)B(t− η)dη + Fm(t; m,h). (47)

Hence, equations (24), (47) and the uniqueness of the solution to (24) imply that

W (t) = Bm(t; m,h) for any t ∈ [0, T ]. (48)

Next we argue that∂u

m[h] = w, which means that um(x, t; m,h) satisfies (40). Let

τ = G−1(G(x)− ξ), then dξ = − 1

g(τ)dτ . Hence, we find that

∫ t

0

h(G−1(G(x)− ξ))dξ =

∫ x

G−1(G(x)−t)

h(τ)

g(τ)dτ. (49)

From equations (11), (44) and (49) we observe that

w(x, t) = −u(x, t)

∫ x

G−1(G(x)−t)

h(τ)

g(τ)dτ, if x ≥ G−1(t). (50)

Therefore, by (30) and (50) we have∂u

∂m[h] = w in the region {(x, t)|x > x ≥ G−1(t), t ≥ 0}.

Let τ = G−1(t− ξ), then dξ = − 1

g(τ)dτ . Hence, we find that

∫ t

t−G(x)

h(G−1(t− ξ))dξ =

∫ x

0

h(τ)

g(τ)dτ. (51)

13

From equations (46), (48) and (51) we see that if x < G−1(t) then

w(x, t) =Bm(t−G(x); m,h)

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

−B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

) ∫ x

0

h(τ)

g(τ)dτ.

(52)

By (31) and (52) we have∂u

∂m[h] = w when x < G−1(t). Therefore, um(x, t; m,h) satisfies the

initial boundary value problem (40), and we can use this system to solve for um(x, t; m,h).

3.3 Sensitivity with Respect to Initial Conditions

In this section, we derive the equation for the sensitivity function uu0 . Let r be the uniquesolution of the initial boundary value problem

rt(x, t) + (g(x)r(x, t))x + m(x)r(x, t) = 0,

g(0)r(0, t) =

∫ x

0

β(x)r(x, t)dx,

r(x, 0) = h(x).

(53)

We argue that r =∂u

∂u0

[h], the directional derivative of u with respect to u0 in the direction

h. By the method of characteristics, we find that the solution of (53) is given by

• If x ≥ G−1(t), then

r(x, t) = h(G−1(G(x)− t))g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

). (54)

• If x < G−1(t), then

r(x, t) =R(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

), (55)

where R(t) =

∫ x

0

β(x)r(x, t)dx.

Hence, we have

R(t) =

∫ G−1(t)

0

β(x)R(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

+

∫ x

G−1(t)

β(x)h(G−1(G(x)− t))g(G−1(G(x)− t))

g(x)exp

(−

∫ x

G−1(G(x)−t)

m(ξ)

g(ξ)dξ

)dx.

14

Using the same transformation as we derived for B(t) we find that

R(t) =

∫ t

0

β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

)R(t− η)dη,

+

∫ x

0

h(η)β(G−1(G(η) + t)) exp

(−

∫ t

0

m(G−1(G(η) + σ))dσ

)dη.

(56)

By (25), (26) and the above equality, we have

R(t) =

∫ t

0

k(η)R(t− η)dη + Fu0(t; u0, h), (57)

(recall that ku0 = 0). Hence, by (27), (57) and the uniqueness of solution to (27) we have

R(t) = Bu0(t; u0, h) for any t ∈ [0, T ]. (58)

By (32), (54), (33), (55) and (58), we also know that∂u

∂u0

[h] = r. This means that

uu0(x, t; u0, h) satisfies (53) and we can use this system to solve for uu0(x, t; u0, h).

3.4 Sensitivity with Respect to g

In this section, we investigate the sensitivity of u with respect to g. As we shall see, ourconsiderations follow the ideas in the previous sections for other sensitivities albeit withsubstantially more tedium in the details. Let e satisfy uniquely the initial boundary valuesystem

et(x, t) + (g(x)e(x, t))x + m(x)e(x, t) + (h(x)u(x, t))x = 0,

g(0)e(0, t) + h(0)u(0, t) =

∫ x

0

β(x)e(x, t)dx,

e(x, 0) = 0.

(59)

Our goal is to argue that e =∂u

∂g[h], the directional derivative of u with respect to g in the

direction h. In order to carry out the necessary computations, we strengthen the conditionson g, m, β and u0 by assuming

(A1) g ∈ C1(0, x), g > 0 on [0, x), and g(x) = 0;

(A2) m ∈ C1(0, x), m ≥ 0 on [0, x];

(A3) β ∈ C1(0, x), β ≥ 0 on [0, x];

(A4) u0 ∈ C1(0, x), u0 ≥ 0 on [0, x];

(A5) u0 satisfies the compatibility condition

g(0)u0(0) =

∫ x

0

β(x)u0(x)dx. (60)

15

The notation ‖ · ‖C will be used throughout this section to denote the supremum norm‖ · ‖C(0,x). We first establish that the directional derivative ug(x, t; g, h) of u at g in any givendirection h exists, and then we will derive the solution e(x, t) to (59) via the method of

characteristics and argue that∂u

∂g[h] = e.

By assumption (A1), we know that G(x) is twice continuously differentiable on [0, x),which implies that G−1 is twice continuously differentiable on [0,∞). Hence, π(x, t) istwice continuously differentiable on {(x, t)|x > x ≥ G−1(t), t ≥ 0}. We observe thatG(π(x, t)) = G(x)− t. Hence, we have G′(π(x, t))πx(x, t) = G′(x). Furthermore, we see that

G′(x) =1

g(x). (61)

Hence, we obtain that

πx(x, t) =g(π(x, t))

g(x). (62)

The directional derivative Gg(x; g, h) of G at g in any given direction h exists and is givenby

Gg(x; g, h) = −∫ x

0

h(τ)

g2(τ)dτ. (63)

Moreover, we find that

G′(π(x, t))πg(x, t; g, h) + Gg(π(x, t); g, h) = Gg(x; g, h),

where πg(x, t; g, h) denotes the directional derivative of π at g in the given direction h. Thus,using equations (61) and (63) and the above equality, we obtain

πg(x, t; g, h) = −g(π(x, t))

∫ x

π(x,t)

h(τ)

g2(τ)dτ. (64)

Let $(x, t) = exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

). Then by the chain rule and (64) we know that the

derivative of $g(x, t; g, h) of $ at g in any given direction h exists and is given by

$g(x, t; g, h)

= exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)[m(π(x, t))

g(π(x, t))πg(x, t; g, h) +

∫ x

π(x,t)

m(ξ)h(ξ)

g2(ξ)dξ

]

= exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)[−m(π(x, t))

∫ x

π(x,t)

h(τ)

g2(τ)dτ +

∫ x

π(x,t)

m(ξ)h(ξ)

g2(ξ)dξ

].

(65)

Let χ(x, t) =g(G−1(G(x)− t))

g(x). Then by assumption (A1), the chain rule and (64) we know

that the directional derivative χg(x, t; g, h) of χ at g in any given direction h exists and is

16

given by

χg(x, t; g, h)

=g′(π(x, t))

g(x)πg(x, t; g, h) +

h(π(x, t))

g(x)− g(π(x, t))h(x)

g2(x)

= −g′(π(x, t))

g(x)g(π(x, t))

∫ x

π(x,t)

h(τ)

g2(τ)dτ +

h(π(x, t))

g(x)− g(π(x, t))h(x)

g2(x).

(66)

Thus, assumption (A4) and equations (64)-(66) imply that ug(x, t; g, h) exists in the region{(x, t)|x > x ≥ G−1(t), t ≥ 0} and is given by

ug(x, t; g, h)

= −u′0(π(x, t))g2(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

) ∫ x

π(x,t)

h(τ)

g2(τ)dτ

−u0(π(x, t))g′(π(x, t))

g(x)g(π(x, t)) exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

) ∫ x

π(x,t)

h(τ)

g2(τ)dτ

+u0(π(x, t)) exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)[h(π(x, t))

g(x)− g(π(x, t))h(x)

g2(x)

]

−u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)[m(π(x, t))

∫ x

π(x,t)

h(τ)

g2(τ)dτ

]

+u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)[∫ x

π(x,t)

h(ξ)m(ξ)

g2(ξ)dξ

].

(67)

Before we consider the sensitivity of u with respect to g in the region {(x, t)|0 ≤ x <G−1(t), t > 0}, we need to carry out a sensitivity analysis for the renewal equation (17) withrespect to g. This is because from (12) u is dependent on B in this region. First we needto show that F of (18) has a bounded directional derivative with respect to g. By equation(64), we have

πg(η,−t; g, h) = g(π(η,−t))

∫ π(η,−t)

η

h(τ)

g2(τ)dτ. (68)

Observe that π(η,−t) = G−1(G(η) + t). Hence, using assumptions (A2) and (A3), the chainrule and equation (68), we know that the directional derivative Fg(t; g, h) of F at g in anygiven direction h exists and it is given as follows

Fg(t; g, h)

=

∫ x

0

u0(η)β′(π(η,−t))g(π(η,−t))

(∫ π(η,−t)

η

h(τ)

g2(τ)dτ

)exp

(−

∫ t

0

m(π(η,−σ))dσ

)dη

−∫ x

0

u0(η)β(π(η,−t)) exp

(−

∫ t

0

m(π(η,−σ))dσ

)[∫ t

0

m′(π(η,−σ))πg(η,−σ; g, h)dσ

]dη.

(69)Note that lim

η→xπ(η,−t) = x for any t ∈ [0, T ]. Hence, if we define π(x,−t) = x, then π(η,−t)

is continuous on [0, x]× [0, T ]. Thus, from equation (68) and the continuity of g and h, we

17

know that πg(η,−t; g, h) is continuous on [0, x] × [0, T ]. Therefore, there exists a positiveconstant ν such that |πg(η,−t; g, h)| ≤ ν for (η, t) ∈ [0, x]× [0, T ]. Hence, by equation (69)we have that

|Fg(t; g, h)| ≤ νx‖u0‖C(‖β′‖C + T‖β‖C‖m′‖C) for t ∈ [0, T ].

Next we show that k has a bounded directional derivative with respect to g. Let κ(η) =G−1(η). Then G(κ(η)) = η. Hence, we find that

G′(κ(η))κg(η; g, h) + Gg(κ(η); g, h) = 0,

where κg(η; g, h) denotes the directional derivative of κ at g in the direction h. Using theabove equality and equation (63), we obtain that

κg(η; g, h) = g(κ(η))

∫ κ(η)

0

h(τ)

g2(τ)dτ. (70)

By assumption (A3), the chain rule and equation (70), we have that the directional derivativekg(η; g, h) of k at g in any given direction h exists and is given by

kg(η; g, h) = β′(G−1(η))g(G−1(η))

(∫ G−1(η)

0

h(τ)

g2(τ)dτ

)exp

(−

∫ η

0

m(G−1(σ))dσ

)

−β(G−1(η)) exp

(−

∫ η

0

m(G−1(σ))dσ

) [∫ η

0

m′(G−1(σ))κg(σ; g, h)dσ

].

(71)Observing that G−1(T ) < x, from equation (70) we have

|κg(η; g, h)| ≤ x‖g‖C‖h/g2‖C(0,G−1(T )) for η ∈ [0, T ].

By this bound and (71) we obtain

|kg(η; g, h)| ≤ x(‖β′‖C + T‖β‖C‖m′‖C)‖g‖C‖h/g2‖C(0,G−1(T )) for η ∈ [0, T ].

Therefore, Theorem 2.2 implies that the directional derivative Bg(t; g, h) of B at g in anygiven direction h exists and is the unique bounded solution of

Bg(t; g, h) =

∫ t

0

k(η)Bg(t− η; g, h)dη +

∫ t

0

kg(η; g, h)B(t− η)dη + Fg(t; g, h). (72)

Note that assumptions (A1)-(A4) imply that both k and F are continuously differentiableon [0, T ]. Hence, B is continuously differentiable on [0, T ]. Using the chain rule and equation(63), we have that ug(x, t; g, h) exists in the region {(x, t)|0 ≤ x < G−1(t), t > 0}, and isgiven by

ug(x, t; g, h) =Bg(t−G(x); g, h)

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)

+B′(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)

g2(ξ)dξ

)

−B(t−G(x))

g2(x)h(x) exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)

+B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)m(ξ)

g2(ξ)dξ

).

(73)

18

Finally we show that∂u

∂g[h] = e, which yields that ug(x, t; g, h) satisfies the initial bound-

ary value problem (59). Using the method of characteristics, we find that the solution to(59) is given by

• If x ≥ G−1(t), then

e(x, t) = −∫ t

0

h′(π(x, ξ))u(π(x, ξ), t− ξ)g(π(x, ξ))

g(x)exp

(−

∫ x

π(x,ξ)

m(τ)

g(τ)dτ

)dξ

−∫ t

0

h(π(x, ξ))u1(π(x, ξ), t− ξ)g(π(x, ξ))

g(x)exp

(−

∫ x

π(x,ξ)

m(τ)

g(τ)dτ

)dξ.

(74)

where u1 =∂u

∂xdenotes the first partial derivative of u with respect to x.

• If x < G−1(t), then

e(x, t) =

[E(t−G(x))

g(x)− h(0)

g(0)

B(t−G(x))

g(x)

]exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

−∫ t

t−G(x)

h′(%(t, ξ))u(%(t, ξ), ρ(x, t, ξ))g(%(t, ξ))

g(x)exp

(−

∫ x

%(t,ξ)

m(τ)

g(τ)dτ

)dξ

−∫ t

t−G(x)

h(%(t, ξ))u1(%(t, ξ), ρ(x, t, ξ))g(%(t, ξ))

g(x)exp

(−

∫ x

%(t,ξ)

m(τ)

g(τ)dτ

)dξ,

(75)

where E(t) =

∫ x

0

β(x)e(x, t)dx.

We simplify the expression in (74). Note that x ≥ G−1(t) implies that

G(π(x, ξ)) = G(x)− ξ ≥ t− ξ.

Let τ = G−1(G(x)−ξ), then dξ = − 1

g(τ)dτ . Hence, we have

∫ t

0

h′(π(x, ξ))dξ =

∫ x

π(x,t)

h′(τ)

g(τ)dτ .

Thus, using equation (43) we can rewrite the first term of the right side of (74) as

∫ t

0

h′(π(x, ξ))u(π(x, ξ), t− ξ)g(π(x, ξ))

g(x)exp

(−

∫ x

π(x,ξ)

m(τ)

g(τ)dτ

)dξ

= u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)(∫ x

π(x,t)

h′(τ)

g(τ)dτ

).

(76)

19

Note that assumptions (A1), (A2) and (A4) imply that u is continuously differentiable inthe region {(x, t)|x > x ≥ G−1(t), t ≥ 0}. Thus, by equation (62) we find that

ux(x, t) = u′0(π(x, t))g2(π(x, t))

g2(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)

+u0(π(x, t))g′(π(x, t))

g(x)

g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)

−u0(π(x, t))g(π(x, t))g′(x)

g2(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)

+u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)(m(π(x, t))

g(x)

)

−u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)(m(x)

g(x)

).

(77)

Observe that G−1(G(π(x, ξ))− (t− ξ)) = G−1(G(x)− t) = π(x, t). Hence, we have that

u1(π(x, ξ), t− ξ)

= u′0(π(x, t))g2(π(x, t))

g2(π(x, ξ))exp

(−

∫ π(x,ξ)

π(x,t)

m(τ)

g(τ)dτ

)

+u0(π(x, t))g′(π(x, t))

g(π(x, ξ))

g(π(x, t))

g(π(x, ξ))exp

(−

∫ π(x,ξ)

π(x,t)

m(τ)

g(τ)dτ

)

−u0(π(x, t))g(π(x, t))g′(π(x, ξ))

g2(π(x, ξ))exp

(−

∫ π(x,ξ)

π(x,t)

m(τ)

g(τ)dτ

)

+u0(π(x, t))g(π(x, t))

g(π(x, ξ))exp

(−

∫ π(x,ξ)

π(x,t)

m(τ)

g(τ)dτ

)(m(π(x, t))

g(π(x, ξ))

)

−u0(π(x, t))g(π(x, t))

g(π(x, ξ))exp

(−

∫ π(x,ξ)

π(x,t)

m(τ)

g(τ)dτ

)(m(π(x, ξ))

g(π(x, ξ))

).

(78)

Let τ = G−1(G(x)− ξ). Then we find that the following three equalities hold

∫ t

0

h(π(x, ξ))

g(π(x, ξ))dξ =

∫ x

π(x,t)

h(τ)

g2(τ)dτ,

∫ t

0

h(π(x, ξ))g′(π(x, ξ))

g(π(x, ξ))dξ =

∫ x

π(x,t)

h(τ)g′(τ)

g2(τ)dτ,

∫ t

0

h(π(x, ξ))m(π(x, ξ))

g(π(x, ξ))dξ =

∫ x

π(x,t)

h(τ)m(τ)

g2(τ)dτ.

20

Using the above equalities and equation (78), we can rewrite the second term in the rightside of (74) as

∫ t

0

h(π(x, ξ))u1(π(x, ξ), t− ξ)g(π(x, ξ))

g(x)exp

(−

∫ x

π(x,ξ)

m(τ)

g(τ)dτ

)dξ

= u′0(π(x, t))g2(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)(∫ x

π(x,t)

h(τ)

g2(τ)dτ

)

+u0(π(x, t))g′(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)(∫ x

π(x,t)

h(τ)

g2(τ)dτ

)

−u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)(∫ x

π(x,t)

h(τ)g′(τ)

g2(τ)dτ

)

+u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

) [m(π(x, t))

∫ x

π(x,t)

h(τ)

g2(τ)dτ

]

−u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)[∫ x

π(x,t)

h(τ)m(τ)

g2(τ)dτ

].

(79)

Observe that

∫ x

π(x,t)

h(τ)g′(τ)

g2(τ)dτ −

∫ x

π(x,t)

h′(τ)

g(τ)dτ =

h(π(x, t))

g(π(x, t))− h(x)

g(x). Thus, by equations

(74), (76) and (79), we find that for the case x ≥ G−1(t)

e(x, t) = −u′0(π(x, t))g2(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

) (∫ x

π(x,t)

h(τ)

g2(τ)dτ

)

−u0(π(x, t))g′(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)(∫ x

π(x,t)

h(τ)

g2(τ)dτ

)

+u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)(h(π(x, t))

g(π(x, t))− h(x)

g(x)

)

−u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)[m(π(x, t))

∫ x

π(x,t)

h(τ)

g2(τ)dτ

]

+u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(τ)

g(τ)dτ

)[∫ x

π(x,t)

h(τ)m(τ)

g2(τ)dτ

].

(80)Thus, equations (67) and (80) imply that

ug(x, t; g, h) = e(x, t), if x ≥ G−1(t). (81)

We next simplify the expression in (75). Let τ = G−1(t− ξ). Then we find that∫ t

t−G(x)

h′(G−1(t− ξ))dξ =

∫ x

0

h′(τ)

g(τ)dτ =

h(x)

g(x)− h(0)

g(0)+

∫ x

0

h(τ)

g2(τ)g′(τ)dτ.

Thus, using (45) and this equality, we can rewrite the second term of the right side of (75)as ∫ t

t−G(x)

h′(%(t, ξ))u(%(t, ξ), ρ(x, t, ξ))g(%(t, ξ))

g(x)exp

(−

∫ x

%(t,ξ)

m(τ)

g(τ)dτ

)dξ

=B(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)[h(x)

g(x)− h(0)

g(0)+

∫ x

0

h(τ)

g2(τ)g′(τ)dτ

].

(82)

21

The continuous differentiability of B, and assumptions (A1) and (A2) imply that u is con-tinuously differentiable in the region {(x, t)|0 ≤ x < G−1(t), t > 0}, and ux is given by

ux(x, t) = −B′(t−G(x))

g2(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)

−B(t−G(x))

g2(x)g′(x) exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)

−B(t−G(x))

g(x)

m(x)

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

).

(83)

Observe that ρ(x, t, ξ)−G(%(x, ξ)) = t−G(x). Hence, we find that

u1(%(x, ξ), ρ(x, t, ξ)) = −B′(t−G(x))

g2(%(x, ξ))exp

(−

∫ %(x,ξ)

0

m(τ)

g(τ)dτ

)

−B(t−G(x))

g2(%(x, ξ))g′(%(x, ξ)) exp

(−

∫ %(x,ξ)

0

m(τ)

g(τ)dτ

)

−B(t−G(x))

g(%(x, ξ))

m(%(x, ξ))

g(%(x, ξ))exp

(−

∫ %(x,ξ)

0

m(τ)

g(τ)dτ

).

(84)

Let τ = G−1(t− ξ). Then dξ = − 1

g(τ)dτ . Hence, the following three equalities hold:

∫ t

t−G(x)

h(%(t, ξ))

g(%(t, ξ))dξ =

∫ x

0

h(τ)

g2(τ)dτ,

∫ t

t−G(x)

h(%(t, ξ))g′(%(t, ξ))

g(%(t, ξ))dξ =

∫ x

0

h(τ)

g2(τ)g′(τ)dτ,

∫ t

t−G(x)

h(%(t, ξ))m(%(t, ξ))

g(%(t, ξ))dξ =

∫ x

0

h(τ)m(τ)

g2(τ)dτ.

Thus, by the above three equalities, the third term of the right side of (75) can be rewrittenas ∫ t

t−G(x)

h(%(t, ξ))u1(%(t, ξ), ρ(x, t, ξ))g(%(t, ξ))

g(x)exp

(−

∫ x

%(t,ξ)

m(τ)

g(τ)dτ

)dξ

= −B′(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)(∫ x

0

h(τ)

g2(τ)dτ

)

−B(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)(∫ x

0

h(τ)

g2(τ)g′(τ)dτ

)

−B(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)(∫ x

0

h(τ)m(τ)

g2(τ)dτ

).

(85)

22

Therefore, from equations (75), (82) and (85), we obtain that if x < G−1(t) then

e(x, t) =E(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

+B′(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)(∫ x

0

h(τ)

g2(τ)dτ

)

−B(t−G(x))

g2(x)h(x) exp

(−

∫ x

0

m(τ)

g(τ)dτ

)

+B(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)(∫ x

0

h(τ)m(τ)

g2(τ)dτ

).

(86)

From equations (73) and (86), we see that if we want to show that ug(x, t; g, h) = e(x, t)for the case x < G−1(t), then we need to show that E(t) = Bg(t; g, h) for t ∈ [0, T ]. Let

x = G−1(η). Then η = G(x) and dη =1

g(x)dx. Hence, we have

∫ t

0

kg(η; g, h)B(t− η)dη =

∫ G−1(t)

0

kg(G(x); g, h)B(t−G(x))

g(x)dx. (87)

Let ξ = G−1(σ) so that dσ =1

g(ξ)dξ. Hence, we have

exp

(−

∫ G(x)

0

m(G−1(σ))dσ

)= exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

), (88)

and ∫ G(x)

0

m′(G−1(σ))g(G−1(σ))

(∫ G−1(σ)

0

h(τ)

g2(τ)dτ

)dσ

=

∫ x

0

m′(ξ)(∫ ξ

0

h(τ)

g2(τ)dτ

)dξ

= m(x)

∫ x

0

h(ξ)

g2(ξ)dξ −

∫ x

0

h(ξ)m(ξ)

g2(ξ)dξ.

(89)

Hence, by equations (71), (88) and (89) we find that

kg(G(x); g, h) = β′(x)g(x)

(∫ x

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)

−β(x)m(x)

(∫ x

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)

+β(x) exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

) (∫ x

0

h(ξ)m(ξ)

g2(ξ)dξ

).

(90)

23

From equations (87) and (90), we obtain that

∫ t

0

kg(η; g, h)B(t− η)dη

=

∫ G−1(t)

0

β′(x)B(t−G(x))

(∫ x

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

−∫ G−1(t)

0

β(x)B(t−G(x))

g(x)m(x)

(∫ x

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

+

∫ G−1(t)

0

β(x)B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)m(ξ)

g2(ξ)dξ

)dx.

(91)

Next observe that the first term of the right side of (91) can be rewritten as

∫ G−1(t)

0

β′(x)B(t−G(x))

(∫ x

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

= β(G−1(t))B(0)

(∫ G−1(t)

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ G−1(t)

0

m(ξ)

g(ξ)dξ

)

+

∫ G−1(t)

0

β(x)B′(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)

g2(ξ)dξ

)dx

+

∫ G−1(t)

0

β(x)B(t−G(x))

g(x)m(x) exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)

g2(ξ)dξ

)dx

−∫ G−1(t)

0

β(x)B(t−G(x))

g2(x)h(x) exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx.

(92)

Hence, from equations (91) and (92) we observe that

∫ t

0

kg(η; g, h)B(t− η)dη

= β(G−1(t))B(0)

(∫ G−1(t)

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ G−1(t)

0

m(ξ)

g(ξ)dξ

)

+

∫ G−1(t)

0

β(x)B′(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)

g2(ξ)dξ

)dx

−∫ G−1(t)

0

β(x)B(t−G(x))

g2(x)h(x) exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)dx

+

∫ G−1(t)

0

β(x)B(t−G(x))

g(x)exp

(−

∫ x

0

m(ξ)

g(ξ)dξ

)(∫ x

0

h(ξ)m(ξ)

g2(ξ)dξ

)dx.

(93)

We return to (69) to simplify the expression for Fg(t; g, h). Let ξ = G−1(G(x)− t+σ). Then

dσ =1

g(ξ)dξ. Hence, we find that

exp

(−

∫ t

0

m(G−1(G(x)− t + σ))dσ

)= exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

). (94)

24

Let x = G−1(G(η) + t), then η = π(x, t) and π(η,−σ) = G−1(G(x) − t + σ). Hence, usingthis transformation, equations (94) and (62), and the facts that π(π(x, t),−t) = x andπ(G−1(t), t) = 0, we can rewrite the first term of the right side of (69) as follows:

∫ x

0

u0(η)β′(π(η,−t))g(π(η,−t))

(∫ π(η,−t)

η

h(τ)

g2(τ)dτ

)exp

(−

∫ t

0

m(π(η,−σ))dσ

)dη

=

∫ x

G−1(t)

β′(x)u0(π(x, t))g(π(x, t))

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

= −β(G−1(t))u0(0)g(0)

(∫ G−1(t)

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ G−1(t)

0

m(ξ)

g(ξ)dξ

)

−∫ x

G−1(t)

β(x)u′0(π(x, t))g2(π(x, t))

g(x)

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g′(π(x, t))g(π(x, t))

g(x)

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

[h(x)

g2(x)− h(π(x, t))

g(x)g(π(x, t))

]exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

+

∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)m(x)

g(x)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)m(π(x, t))

g(x)dx.

(95)By the same transformation as we derived for (94), we obtain that

∫ t

0

m′(G−1(G(x)− t + σ))g(G−1(G(x)− t + σ))

(∫ G−1(G(x)−t+σ)

π(x,t)

h(τ)

g2(τ)dτ

)dσ

=

∫ x

π(x,t)

m′(ξ)(∫ ξ

π(x,t)

h(τ)

g2(τ)dτ

)dξ

= m(x)

∫ x

π(x,t)

h(τ)

g2(τ)dτ −

∫ x

π(x,t)

h(ξ)m(ξ)

g2(ξ)dξ.

(96)

Hence, by equations (96), (94) and (68), the second term in the right side of (69) can berewritten as∫ x

0

u0(η)β(π(η,−t)) exp

(−

∫ t

0

m(π(η,−σ))dσ

)[∫ t

0

m′(π(η,−σ))πg(η,−σ; g, h)dσ

]dη

=

∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)[m(x)

∫ x

π(x,t)

h(τ)

g2(τ)dτ

]dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)(∫ x

π(x,t)

h(ξ)m(ξ)

g2(ξ)dξ

)dx.

(97)

25

Thus, equations (69), (97) and (95) imply that

Fg(t; g, h)

= −β(G−1(t))u0(0)g(0)

(∫ G−1(t)

0

h(ξ)

g2(ξ)dξ

)exp

(−

∫ G−1(t)

0

m(ξ)

g(ξ)dξ

)

−∫ x

G−1(t)

β(x)u′0(π(x, t))g2(π(x, t))

g(x)

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g′(π(x, t))g(π(x, t))

g(x)

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

[h(x)

g2(x)− h(π(x, t))

g(x)g(π(x, t))

]exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)dx

−∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

(∫ x

π(x,t)

h(ξ)

g2(ξ)dξ

)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)m(π(x, t))

g(x)dx

+

∫ x

G−1(t)

β(x)u0(π(x, t))g(π(x, t))

g(x)exp

(−

∫ x

π(x,t)

m(ξ)

g(ξ)dξ

)(∫ x

π(x,t)

h(ξ)m(ξ)

g2(ξ)dξ

)dx.

(98)

Let η = G(x). Then dη =1

g(x)dx. Hence we find that

∫ G−1(t)

0

β(x)E(t−G(x))

g(x)exp

(−

∫ x

0

m(τ)

g(τ)dτ

)dx =

∫ t

0

k(η)E(t− η)dη. (99)

Observe that assumption (A5) implies that B(0) = g(0)u0(0). Hence, from equations (80),(86), (93), (98) we see that

E(t) =

∫ t

0

k(η)E(t− η)dη +

∫ t

0

kg(η; g, h)B(t− η)dη + Fg(t; g, h). (100)

Using equations (72) and (100), and the uniqueness of the solution to (72), we have that

E(t) = Bg(t; g, h) for any t ∈ [0, T ]. (101)

Hence, equations (73), (86) and (101) imply that

ug(x, t; g, h) = e(x, t), if x < G−1(t). (102)

Thus, from (81) and (102) we know that ug(x, t; g, h) satisfies the initial boundary valueproblem (59), which means that we can use this system to solve for ug(x, t; g, h).

4 Numerical Results

We illustrate the ease in computation of the sensitivities derived in Section 3 with a simpleexample. In partiular, we consider computation of the sensitivities of u with respect to mand g, and we use an implicit finite difference scheme to approximate the population density

26

u(x, t) as well as w(x, t) = um(x, t; m,h) and e(x, t) = ug(x, t; g, h). Let ∆x = x/n and∆t = T/l be the spatial and time mesh sizes, respectively. The mesh points are given byxj = j∆x, j = 0, 1, 2, . . . , n, and tk = k∆t, k = 0, 1, 2, . . . , l. We denote by uk

j the finite

difference approximation of u(xj, tk), and we let u0j = u0(xj), gj = g(xj), mj = m(xj), and

βj = β(xj), j = 0, 1, 2, . . . , n. The finite difference scheme used to approximate the solutionu(x, t) to (1) is given as

uk+1j − uk

j

∆t+

gjuk+1j − gj−1u

k+1j−1

∆x+ mju

k+1j = 0, 1 ≤ j ≤ n,

g0uk+10 =

n∑i=1

βiuki ∆x.

We denote by wkj the finite difference approximation of w(xj, tk), and we let w0

j = 0 andhj = h(xj). The scheme used to approximate the solution w(x, t) to (40) is given by

wk+1j − wk

j

∆t+

gjwk+1j − gj−1w

k+1j−1

∆x+ mjw

k+1j + hju

k+1j = 0, 1 ≤ j ≤ n,

g0wk+10 =

n∑i=1

βiwki ∆x.

Let ekj represent the finite difference approximation of e(xj, tk), and e0

j = 0, j = 0, 1, 2, . . . , n.Then the scheme used to approximate the solution e(x, t) to (59) is given by

ek+1j − ek

j

∆t+

gjek+1j − gj−1e

k+1j−1

∆x+ mje

k+1j +

hjuk+1j − hj−1u

k+1j−1

∆x= 0, 1 ≤ j ≤ n,

g0ek+10 + h0u

k+10 =

n∑i=1

βieki ∆x.

In this example, we assume that maximum size x = 1 and maximum time period T = 5.The initial condition and model parameters are given as follows

u0(x) = 10x3(1− x), g(x) = 0.2(1− x), m(x) = 0.4, β(x) = 0.

Then by (11) and (12) we can solve analytically for u(x, t):

u(x, t) =

{10(1− x)[1− (1− x)e0.2t]3 x ≥ 1− e−0.2t

0 x < 1− e−0.2t.

Note that m is a constant, and hence we allow the direction h = 1 when we take thedirectional derivative of u with respect to m. Then by (50) and (52) we can solve analyticallyfor w(x, t):

w(x, t) =

{ −10t(1− x)[1− (1− x)e0.2t]3 x ≥ 1− e−0.2t

0 x < 1− e−0.2t.

In our example for ug, the direction h is chosen to be h = 2g in taking the directionalderivative of u with respect to g. By (80) and (86) we can solve analytically for e(x, t):

e(x, t) =

{4t(1− x)[1− (1− x)e0.2t]2[5(1− (1− x)e0.2t)− 3] x ≥ 1− e−0.2t

0 x < 1− e−0.2t.

27

In Figures 1 and 2, the approximate solutions with ∆x = 10−4 and ∆t = 10−3 aredepicted. The analytical solution, the approximate solution and the difference betweenthem for u(x, t) are plotted in Figure 1, those for w(x, t) and e(x, t) plotted in Figure 2.These figures demonstrate that numerical solutions for u(x, t), w(x, t) and e(x, t) providereasonable approximations to the analytical solutions. However, we also observe that theabsolute value of the error becomes larger as time evolves. The reason for this is that thepopulation distribution becomes narrower and steeper (see Figures 3-5) as time progressesbecause g′ = −0.2 < 0 in this example; this provides computational challenges.

Figure 1: (top left): Analytical solution for u(x, t). (top right): The finite difference approx-imation ufd(x, t) of u(x, t). (bottom): Error between the analytical solution and the finitedifference approximation.

28

Figure 2: (left column): Analytical solution, approximate solution and error between themfor w(x, t). (right column): Analytical solution, approximate solution and error betweenthem for e(x, t).

29

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

x

u(x,

t)

u(x,1)u

fd(x,1)

u(x,5)u

fd(x,5)

0 0.2 0.4 0.6 0.8 1−8

−6

−4

−2

0

2

4x 10

−4

x

u(x,

t)−

u fd(x

,t)

u(x,1)−ufd

(x,1)

u(x,5)−ufd

(x,5)

Figure 3: (left): Analytical solution and approximate solution for population density at t = 1and t = 5. (right): Error between analytical solution and approximate solution at t = 1 andt = 5.

0 0.2 0.4 0.6 0.8 1−2

−1.8

−1.6

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

x

w(x

,t)

w(x,1)w

fd(x,1)

w(x,5)w

fd(x,5)

0 0.2 0.4 0.6 0.8 1−2

−1

0

1

2

3

4x 10

−3

x

w(x

,t)−

wfd

(x,t)

w(x,1)−wfd

(x,1)

w(x,5)−wfd

(x,5)

Figure 4: (left): Analytical solution and approximate solution for sensitivity with respect tom at t = 1 and t = 5. (right): Error between analytical solution and approximate solutionat t = 1 and t = 5.

0 0.2 0.4 0.6 0.8 1−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

e(x,

t)

e(x,1)e

fd(x,1)

e(x,5)e

fd(x,5)

0 0.2 0.4 0.6 0.8 1−4

−3

−2

−1

0

1

2

3

4

5

6x 10

−3

x

e(x,

t)−

e fd(x

,t)

e(x,1)−efd

(x,1)

e(x,5)−efd

(x,5)

Figure 5: (left): Analytical solution and approximate solution for sensitivity with respect tog at t = 1 and t = 5. (right): Error between analytical solution and approximate solutionat t = 1 and t = 5.

30

5 Concluding Remarks

In this paper, we have considered the widely used Sinko-Streifer/McKendrick-von Foersterequations for size-structured populations. We addressed the challenging problem of a rigorousderivation of the partial differential equations for the sensitivities of the population densitywith respect to initial conditions u0, growth g, mortality m and fecundity β. We used themethod of characteristics to carry out the sensitivity analysis for the linear size-structuredpopulation model. This approach is tedious but very straightforward, and we believe it can beapplied to a diverse class of more complicated structured population and biomass productionmodels such as the size and class age structured epidemic model in [7]. A reassuring aspectof our investigations is that they reveal that the correct sensitivity equations can be formallyobtained by simply differentiating the population equation of interest with respect to thefunctions u0, g, m and β.

Acknowledgements

This research was supported in part by the National Institute of Allergy and InfectiousDisease under grant 9R01AI071915-05, in part by the U.S. Air Force Office of Scientific Re-search under grant AFOSR-FA9550-04-1-0220 and in part by the Defense Advanced ResearchProjects Agency and Advanced BioNutrition Corp under grant ABNDTRA-0507-03.

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