Sensitivity Equations for a Size-Structured
Population Model
H. T. Banks, Stacey L. Ernstberger and Shuhua Hu Center for Research in Scientific Computation
North Carolina State University Raleigh, NC 27695-8205
September 28, 2007
Abstract
In this paper we consider the classical Sinko-Streifer size-structured population model and derive sensitivity partial differential equations for the sensitivities of solutions with respect to initial conditions, growth rate, mortality rate and fecundity rate. Sample numerical results to 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
In this paper we derive and investigate sensitivity equations for the linear size-structured population model
ut(x, t) + (g(x)u(x, t))x+m(x)u(x, t) = 0, g(0)u(0, t) =
Z 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 x
at time t, g is the size-dependent individual growth rate, m represents the size-dependent mortality rate, β is the size-dependent reproduction or fecundity rate, and u0 is the initial population density. The maximum size individuals may obtain in their lifetime is ¯x. The objective 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 equation systems
˙
x(t) = f(x(t), θ), x(0) =x0,
one finds that the heuristic differentiation with respect toθof the system (with an interchange of 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,
fory(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 more delicate, our goal is to establish that a formal, heuristic differentiation also results in the correct sensitivity equations and that this heuristic derivation can be made rigorous. As we shall see below, the underlying technical details are nontrivial. As we next outline, we considered 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 equapopula-tion (1) in the abstract form
˙
u(t) = Au(t), u(0) =u0, (2) in a Banach space with the linear operator Adefined by Aϕ=−(gϕ)0−mϕ. One can then show thatAis an infinitesimal generator for aC0-semigroup, and thereby establish existence and uniqueness of solutions to the model. One might then attempt to carry out a sensitivity analysis for (1) by using the abstract theoretical framework provided in [9], wherein the sensitivity analysis was developed for the general Banach space nonlinear ordinary differential equation
˙
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 subset
apply this framework to (2) were not productive, as the domain of operator A isdependent on the parameters g and β themselves. For example, the domain ofA in [4] is defined as
domA = ½
ϕ∈ H |gϕ∈ H1(0,x¯), lim
x→x¯(gϕ)(x) = 0, (gϕ)(0) = Z x¯
0
β(x)ϕ(x)dx
¾
, (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 operatorA∗ of A is given by
A∗ψ =gψ0 −mψ+ψ(0)β, (4)
with domA∗ = nψ ∈ H |ψ ∈ H1
loc[0,x¯), gψ0 ∈ H, xlim→¯x(gϕψ)(x) = 0 for ϕ∈domA o
, 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 is
linearly dependent on those parameters. However, the difficulty in using this approach is that 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 to our model parameters is not readily carried out using either a semigroup or a variational approach because the domains of A and A∗ are dependent on the model parameters. It might be possible to apply the idea of method of mappings employed in sensitivity analysis for optimal shape design problems (e.g., [5, 22, 23, 26, 34]), which deals with computation of derivatives with respect to shape variation. However, it is not a straightforward application to our problem since the domain here is an operator domain (functional domain) instead of a 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-structured population models. Using this approach, one obtains an implicit representation of the solu-tion which can be used to transform the partial differential equasolu-tion into an integral equasolu-tion. Then the contraction mapping theorem is applied in order to establish the desired results. In this paper we will employ this method to carry out a sensitivity analysis for (1). Although computationally tedious, this approach is conceptually straightforward and relatively easy to use.
2
Preliminary Results
In order to develop the sensitivity formulations of interest, a number of standing assumptions will be imposed on the model parameters and initial conditions in (1). We assume
(H1) g ∈ W1,∞(0,x¯),g >0 on [0,x¯), andg(¯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 k · k∞ to denote the norm k · kL∞(0,¯x), andk · k1 to denote the norm k · kL1(0,x¯)
throughout the paper. Note that assumption (H1) implies that g is in a convex subset of
W1,∞(0,x¯); assumptions (H2) and (H3) guarantee that bothmand β are in a convex subset of L∞(0,x¯), and assumption (H4) implies that u
0 is in a convex subset of L1(0,x¯). Hence, we formulate the sensitivity equations for (1) by means of directional derivatives of u with respect to parameters β, m, u0, and g. The directional derivative is defined by
Definition 2.1 LetΘbe a convex subset in some topological vector space, andf :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 the
direction ϑ−θ to be
fθ(t;θ, ϑ−θ) = lim ²→0+
f(t;θ+²(ϑ−θ))−f(t;θ)
² , (7)
provided this limit exists.
The method of characteristics can be used to reduce equation (1) as well as each of the associated sensitivity equations to an equivalent renewal equation (a Volterra equation of convolution type), and existence and uniqueness of solutions to these integral equations are established via the following result from [2] (Theorem 7.2 on page 220, and Theorem 7.4 on page 224).
Result 2.1 If Ψ(t) is bounded on [0, T] and
Z T
0
|φ(s)|ds <∞, then the equation
Φ(t) = Z 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, then
Φ(t) is also continuous.
Suppose that functions φ and Ψ in equation (8) are both dependent on parameter θ in a convex subset Θ of some topological space. Then equation (8) becomes
Φ(t;θ) = Z t
0
Theorem 2.2 Suppose that for a given θ∈Θ, φ(t;θ)andΨ(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 directional derivative Ψθ(t;θ, ϑ−θ) on [0, T] with respect to θ ∈ Θ in the direction ϑ −θ. Then the
directional derivative Φθ(t;θ, ϑ−θ) of Φ with respect to θ in the direction ϑ−θ exists. Let
z(t) = Φθ(t;θ, ϑ−θ). Then it satisfies the following equation
z(t) = Z t
0
φ(s;θ)z(t−s)ds+ Z 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 ofφ(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. Sinceφθ(t;θ, ϑ−θ) and Ψθ(t;θ, ϑ−θ) are both bounded on [0, T], we know that there exists a positive constantcsuch that
¯ ¯ ¯ ¯
Z t
0
φθ(s;θ, ϑ−θ)Φ(t−s;θ)ds+ Ψθ(t;θ, ϑ−θ) ¯ ¯ ¯ ¯≤
con [0, T]. Thus, Result 2.1 guarantees that there exists a unique bounded solution z(t) to (10). Note that
Φ(t;θ+²(ϑ−θ))−Φ(t;θ) =
Z t
0
φ(s;θ+²(ϑ−θ))[Φ(t−s;θ+²(ϑ−θ))−Φ(t−s;θ)]ds
+ Z t
0
[φ(s;θ+²(ϑ−θ))−φ(s;θ)]Φ(t−s;θ)ds+ Ψ(s;θ+²(ϑ−θ))−Ψ(s;θ).
LetD(t;θ, ϑ, ²) = Φ(t;θ+²(ϑ−θ))−Φ(t;θ)
² −z(t). Then we find that
|D(t;θ, ϑ, ²)| ≤
Z t
0
|φ(t−s;θ+²(ϑ−θ))||D(s;θ, ϑ, ²)|ds
+ Z t
0
|φ(s;θ+²(ϑ−θ))−φ(s;θ)| |z(t−s)|ds
+ Z t
0 ¯ ¯ ¯
¯φ(s;θ+²(ϑ−²θ))−φ(s;θ) −φθ(s;θ, ϑ−θ) ¯ ¯ ¯
¯|Φ(t−s;θ)|ds +
¯ ¯ ¯
¯Ψ(t;θ+²(ϑ−²θ))−Ψ(t;θ)−Ψθ(t;θ, ϑ−θ) ¯ ¯ ¯ ¯.
We observe that the boundedness ofφθ(t;θ, ϑ−θ) on [0, T] implies that lim
²→0+φ(t;θ+²(ϑ−θ)) =
φ(t;θ) uniformly in t. Hence, by the boundedness ofz we know that the second term in the right side of the above inequality converges to zero as²→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 direction
ϑ−θ imply that the third term of the right side of the above inequality converges to zero as ² → 0. The fourth term converges to zero as ² → 0 by the existence of the directional derivative of Ψ with respect to θ in the direction ϑ−θ. Hence, by Gronwall’s inequality we have
lim
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) and show that there exists a unique solution to this equation. We define several functions that will be used throughout this paper:
G(x) = Z x
0 1
g(ξ)dξ, B(t) = Z 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]. By assumption (H1), we find that for ξ ∈[0,x¯),
0< g(ξ) =g(ξ)−g(¯x)≤ kg0k∞(¯x−ξ).
Hence, we have
G(x)≥ 1 kg0k
∞
Z x
0 1 ¯
x−ξdξ,
which implies that lim
x→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, the requirement ofg(¯x) = 0 in (H1) guarantees that the size of an individual always stay less than ¯
x. In addition, to simplify the expressions, the following functions will be used throughout our 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:
• Ifx≥G−1(t), then
u(x, t) =u0(G−1(G(x)−t))g(G
−1(G(x)−t))
g(x) exp
µ
−
Z x
G−1(G(x)−t)
m(ξ)
g(ξ)dξ ¶
. (11)
• Ifx < G−1(t), then
u(x, t) = B(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
. (12)
Hence, we have
B(t) =
Z G−1(t)
0
β(x)B(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
dx
+ Z x¯
G−1(t)
β(x)u0(π(x, t))
g(π(x, t))
g(x) exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶
dx.
Letη=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
Z G−1(t)
0
β(x)B(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
dx
= Z t
0
β(G−1(η)) exp Ã
−
Z G−1(η)
0
m(ξ)
g(ξ)dξ !
B(t−η)dη
= Z t
0
β(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ ¶
B(t−η)dη.
(14)
Letη =G−1(G(x)−t). Thenx=G−1(G(η) +t) and 1
g(η)dη= 1
g(x)dx, which implies that
dx = g(G−1(G(η) +t))
g(η) dη. Since limτ→∞G
−1(τ) = ¯x, we have lim x→x¯G
−1(G(x)−t) = ¯x for any
t∈[0, T]. Hence, the second term of the right side of (13) can be rewritten as Z x¯
G−1(t)
β(x)u0(G−1(G(x)−t))g(G
−1(G(x)−t))
g(x) exp
µ
−
Z x
G−1(G(x)−t)
m(ξ)
g(ξ)dξ ¶
dx
= Z x¯
0
β(G−1(G(η) +t))u
0(η) exp Ã
−
Z G−1(G(η)+t)
η
m(ξ)
g(ξ)dξ !
dη.
(15)
Letσ =G(ξ)−G(η). Then ξ =G−1(G(η) +σ) and dσ= 1
g(ξ)dξ. Hence, we have that
exp Ã
−
Z G−1(G(η)+t)
η
m(ξ)
g(ξ)dξ ! = exp µ − Z t 0
m(G−1(G(η) +σ))dσ ¶
.
Thus, from the above equality and (15) we have Z x¯
G−1(t)
β(x)u0(G−1(G(x)−t))g(G
−1(G(x)−t))
g(x) exp
µ
−
Z x
G−1(G(x)−t)
m(ξ)
g(ξ)dξ ¶
dx
= Z x¯
0
u0(η)β(G−1(G(η) +t)) exp µ
−
Z t
0
m(G−1(G(η) +σ))dσ
¶
dη.
(16)
Therefore, by (13), (14) and (16) we obtain the renewal equation
B(t) = Z t
0
k(η)B(t−η)dη+F(t), (17)
where
k(η) = β(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ
¶
,
F(t) = Z x¯
0
u0(η)β(G−1(G(η) +t)) exp µ
−
Z t
0
m(G−1(G(η) +σ))dσ
¶
dη.
Note that 0 ≤ F(t) ≤ kβk∞ku0k1, and 0 ≤ k(t) ≤ kβk∞ on [0, T]. Hence, Result 2.1
guarantees that there exists a unique bounded nonnegative solution to (17). Thus, this solution of (17) must be the same as B used in equations (11) and (12) in representing the unique 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 sensitivity functions for u with respect to β, m, u0 and g, we first need to investigate sensitivity for
B. To simplify the notation, we use h to denote a given direction in the corresponding parameter 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 function dependent on them unless it can cause confusion, such as when taking the derivative with respect to those parameters.
Since the sensitivity of u with respect to g is much more complicated than those with respect to β,m or u0, we will consider it subsequently in Section 3.4. At present our goal is to derive the sensitivity equations for B with respect to β, m and u0. First we see that the directional 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 given by
kβ(η;β, h) = h(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ ¶
, (19)
which implies that |kβ(η;β, h)| ≤ khk∞ for η ∈ [0, T]. We also observe that the directional
derivative of F with respect to β in any given direction h exists. Let Fβ(t;β, h) denote the derivative ofF with respect to β in the direction h. Then it satisfies
Fβ(t;β, h) = Z x¯
0
h(G−1(G(η) +t))u
0(η) exp µ
−
Z t
0
m(G−1(G(η) +σ))dσ ¶
dη, (20)
which implies that |Fβ(t;β, h)| ≤ khk∞ku0k1 for t ∈ [0, T]. Thus, by Theorem 2.2, the directional derivativeBβ(t;β, h) of B with respect toβ in the direction hexists and is given by
Bβ(t;β, h) = Z t
0
k(η)Bβ(t−η;β, h)dη+ Z 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 µ
−
Z η
0
m(G−1(σ))dσ
¶ µZ η
0
h(G−1(σ))dσ ¶
=−k(η) µZ η
0
h(G−1(σ))dσ ¶
,
(22)
and the directional derivative Fm(t;m, h) of F with respect tom in the directionh is given by
Fm(t;m, h) = − Z x¯
0
u0(η)β(π(η,−t)) exp µ
−
Z t
0
m(π(η,−σ))dσ
¶ µZ t
0
h(π(η,−σ))dσ
¶
dη.
Hence, |km(η;m, h)| ≤Tkhk∞kβk∞ forη∈[0, T], and|Fm(t;m, h)| ≤Tkhk∞kβk∞ku0k1 for
t∈[0, T]. Thus, we know from Theorem 2.2 that the directional derivativeBm(t;m, h) of B with respect to m in the direction h exists as the unique bounded solution to
Bm(t;m, h) = Z t
0
k(η)Bm(t−η;m, h)dη+ Z t
0
km(η;m, h)B(t−η)dη+Fm(t;m, h). (24) We observe that k does not depend onu0. Hence, the directional derivative ku0(η;u0, h)
of k with respect tou0 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 tou0 in any given
direction h exists and is readily seen to be
Fu0(t;u0, h) =
Z x¯
0
h(η)β(G−1(G(η) +t)) exp µ
−
Z t
0
m(G−1(G(η) +σ))dσ ¶
dη, (26)
which implies that |Fu0(t;u0, h)| ≤ kβk∞khk1 for t ∈ [0, T]. Let Bu0(t;u0, h) denote the
directional derivative of B with respect to u0 in the direction h, then by Theorem 2.2 we know that it exists and is the unique bounded solution to
Bu0(t;u0, h) =
Z 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,mand u0, we can obtain the sensitivities for u with respect to the same functions using the expressions for 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 respect toβ in the direction h is zero in that region, that is,
uβ(x, t;β, h) = 0, for x≥G−1(t). (28) Note thatBβ(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
µ
−
Z 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 that the directional derivative um(x, t;m, h) ofu with respect to m in the direction h satisfies
um(x, t;m, h) =−u(x, t) Z 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
µ
−
Z x
0
m(τ)
g(τ)dτ ¶
−u(x, t) Z x
0
h(τ)
g(τ)dτ, if x < G
−1(t).
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
µ
−
Z 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
µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
. (33)
From the computational perspective, it is usually much easier to solve numerically a differential equation than an integral equation. Hence, in the next three subsections, we will show that uβ(x, t;β, h), um(x, t;m, h) and uu0(x, t;u0, h) are each respectively the solution
to some first-order hyperbolic partial differential equation. This means that we can use these partial differential equations to obtain the numerical solutions for the sensitivity of u with respect 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 would be obtained if one simply heuristically differentiated equation (1) with respect to β, m and
u0 respectively.
3.1
Sensitivity with Respect to Fecundity Rate
In this subsection, we want to derive the sensitivity partial differential equation for uβ, the sensitivity 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) = Z 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 of characteristics, we find that the implicit representation form for solution to (34) is given as follows:
• Ifx≥G−1(t), then
v(x, t) = 0. (35)
• Ifx < G−1(t), then
v(x, t) = V(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
, (36)
where V(t) = Z x¯
0
Hence, we have
V(t) =
Z G−1(t)
0
β(x)V(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
dx
+
Z G−1(t)
0
h(x)B(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
dx
+ Z x¯
G−1(t)
h(x)u0(G−1(G(x)−t))g(G
−1(G(x)−t))
g(x) exp
µ
−
Z 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) = Z t
0
β(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ ¶
V(t−η)dη
+ Z t
0
h(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ ¶
B(t−η)dη
+ Z x¯
0
h(G−1(G(η) +t))u
0(η) exp µ
−
Z t
0
m(G−1(G(η) +σ))dσ ¶
dη,
(37)
which implies, by (19) and (20), that
V(t) = Z t
0
k(η)V(t−η)dη+ Z 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 thatuβ(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
µ
−
Z 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 value problem (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 unique solution 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) = Z x¯
0
β(x)w(x, t)dx, w(x,0) = 0.
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 given by
• Ifx≥G−1(t), then
w(x, t) =−
Z t
0
h(π(x, ξ))u(π(x, ξ), t−ξ)g(π(x, ξ))
g(x) exp µ
−
Z x
π(x,ξ)
m(τ)
g(τ)dτ ¶
dξ.
(41)
• Ifx < G−1(t), then
w(x, t) = W(t−G(x))
g(x) exp
µ
−
Z x
0
m(τ)
g(τ)dτ ¶
−
Z t
t−G(x)
h(%(t, ξ))u(%(t, ξ), ρ(x, t, ξ))g(%(t, ξ))
g(x) exp µ
−
Z x
%(t,ξ)
m(τ)
g(τ)dτ ¶
dξ,
(42) where W(t) =
Z 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 Ã
−
Z 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 µ
−
Z x
π(x,t)
m(τ)
g(τ)dτ ¶ Z 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), we find that
u(G−1(t−ξ),2t−G(x)−ξ) = B(t−G(x))
g(G−1(t−ξ))exp Ã
−
Z G−1(t−ξ)
0
m(τ)
g(τ)dτ !
. (45)
Using the above equality and (42), for the casex < G−1(t) we have that
w(x, t) = W(t−G(x))
g(x) exp
µ
−
Z x
0
m(τ)
g(τ)dτ ¶
−B(t−G(x))
g(x) exp µ
−
Z x
0
m(τ)
g(τ)dτ ¶ Z t
t−G(x)
h(G−1(t−ξ))dξ.
By (44) and (46) we find that
W(t) =
Z G−1(t)
0
β(x)W(t−G(x))
g(x) exp
µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
dx
−
Z G−1(t)
0
β(x)B(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ
¶ µZ t
t−G(x)
h(G−1(t−ξ))dξ
¶
dx
−
Z x¯
G−1(t)
β(x)u0(π(x, t))
g(π(x, t))
g(x) exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ
¶ µZ t
0
h(π(x, ξ))dξ
¶
dx.
Using the same transformation as we derived for B(t), we obtain that
W(t) = Z t
0
β(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ ¶
W(t−η)dη
−
Z t
0
β(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ
¶ µZ η
0
h(G−1(σ))dσ ¶
B(t−η)dη
−
Z x¯
0
u0(η)β(π(η,−t)) exp µ
−
Z t
0
m(π(η,−σ))dσ
¶ µZ t
0
h(π(η,−σ))dσ
¶
dη.
From the above equality, equation (22), and equation (23), we see that
W(t) = Z t
0
k(η)W(t−η)dη+ Z 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 anyt ∈[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), thendξ =− 1
g(τ)dτ. Hence, we find that Z t
0
h(G−1(G(x)−ξ))dξ = Z 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) Z 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] =win the region{(x, t)|x > x¯ ≥G
−1(t), t≥0}. Letτ =G−1(t−ξ), then dξ=− 1
g(τ)dτ. Hence, we find that Z t
t−G(x)
h(G−1(t−ξ))dξ = Z x
0
h(τ)
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
µ
−
Z x
0
m(τ)
g(τ)dτ ¶
−B(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶ Z x
0
h(τ)
g(τ)dτ.
(52)
By (31) and (52) we have ∂u
∂m[h] = wwhenx < 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 unique
solution 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) = Z x¯
0
β(x)r(x, t)dx, r(x,0) = h(x).
(53)
We argue thatr = ∂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
• Ifx≥G−1(t), then
r(x, t) = h(G−1(G(x)−t))g(G−1(G(x)−t))
g(x) exp
µ
−
Z x
G−1(G(x)−t)
m(ξ)
g(ξ)dξ ¶
. (54)
• Ifx < G−1(t), then
r(x, t) = R(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
, (55)
where R(t) = Z ¯x
0
β(x)r(x, t)dx. Hence, we have
R(t) =
Z G−1(t)
0
β(x)R(t−G(x))
g(x) exp µ
−
Z x
0
m(ξ)
g(ξ)dξ ¶
dx
+ Z x¯
G−1(t)
β(x)h(G−1(G(x)−t))g(G−1(G(x)−t))
g(x) exp
µ
−
Z x
G−1(G(x)−t)
m(ξ)
g(ξ)dξ ¶
Using the same transformation as we derived for B(t) we find that
R(t) = Z t
0
β(G−1(η)) exp µ
−
Z η
0
m(G−1(σ))dσ ¶
R(t−η)dη,
+ Z x¯
0
h(η)β(G−1(G(η) +t)) exp µ
−
Z t
0
m(G−1(G(η) +σ))dσ ¶
dη.
(56)
By (25), (26) and the above equality, we have
R(t) = Z 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, our considerations follow the ideas in the previous sections for other sensitivities albeit with substantially more tedium in the details. Let e satisfy uniquely the initial boundary value system
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) =
Z ¯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 conditions ong, m,β and u0 by assuming
(A1) g ∈ C1(0,x¯),g >0 on [0,x¯), andg(¯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) = Z x¯
0
The notation k · kC will be used throughout this section to denote the supremum norm
k · kC(0,x¯). We first establish that the directional derivativeug(x, t;g, h) of uatg in any given
direction 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) is twice continuously differentiable on {(x, t)|x > x¯ ≥ G−1(t), t ≥ 0}. We observe that
G(π(x, t)) = G(x)−t. Hence, we haveG0(π(x, t))π
x(x, t) = G0(x). Furthermore, we see that
G0(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 given by
Gg(x;g, h) = − Z x
0
h(τ)
g2(τ)dτ. (63)
Moreover, we find that
G0(π(x, t))π
g(x, t;g, h) +Gg(π(x, t);g, h) =Gg(x;g, h),
whereπg(x, t;g, h) denotes the directional derivative ofπatg in the given directionh. Thus, using equations (61) and (63) and the above equality, we obtain
πg(x, t;g, h) = −g(π(x, t)) Z x
π(x,t)
h(τ)
g2(τ)dτ. (64)
Let $(x, t) = exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶
. Then by the chain rule and (64) we know that the derivative of$g(x, t;g, h) of $ atg in any given direction h exists and is given by
$g(x, t;g, h) = exp
µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶ ·
m(π(x, t))
g(π(x, t))πg(x, t;g, h) + Z x
π(x,t)
m(ξ)h(ξ)
g2(ξ) dξ ¸
= exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶ ·
−m(π(x, t)) Z x
π(x,t)
h(τ)
g2(τ)dτ + Z x
π(x,t)
m(ξ)h(ξ)
g2(ξ) dξ ¸
.
(65)
Letχ(x, t) = g(G
−1(G(x)−t))
given by
χg(x, t;g, h) = g0(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
0(π(x, t))
g(x) g(π(x, t)) Z 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) =−u0
0(π(x, t))
g2(π(x, t))
g(x) exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶ Z x
π(x,t)
h(τ)
g2(τ)dτ
−u0(π(x, t))
g0(π(x, t))
g(x) g(π(x, t)) exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶ Z x
π(x,t)
h(τ)
g2(τ)dτ +u0(π(x, t)) exp
µ
−
Z 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 µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ ¶ ·
m(π(x, t)) Z x
π(x,t)
h(τ)
g2(τ)dτ ¸
+u0(π(x, t))
g(π(x, t))
g(x) exp µ
−
Z x
π(x,t)
m(ξ)
g(ξ)dξ
¶ ·Z 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) with respect to g. This is because from (12) u is dependent on B in this region. First we need to show that F of (18) has a bounded directional derivative with respect tog. By equation (64), we have
πg(η,−t;g, h) = g(π(η,−t))
Z 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 chain rule and equation (68), we know that the directional derivative Fg(t;g, h) of F at g in any given direction h exists and it is given as follows
Fg(t;g, h)
= Z x¯
0
u0(η)β0(π(η,−t))g(π(η,−t))
ÃZ π(η,−t)
η
h(τ)
g2(τ)dτ !
exp µ
−
Z t
0
m(π(η,−σ))dσ
¶
dη
−
Z x¯
0
u0(η)β(π(η,−t)) exp µ
−
Z t
0
m(π(η,−σ))dσ
¶ ·Z t
0
m0(π(η,−σ))π
g(η,−σ;g, h)dσ ¸
dη.
(69) Note that lim
