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Renewal age and size structured numerical methods

1 · Prerequisites

2 · Summary

This page develops two related but distinct structured-population models. The age model follows unit-speed age characteristics with mortality and a renewal birth boundary. It supplies the characteristic/Volterra solution, the absorbing finite-age version, positive birth quadrature, and an explicit upwind finite-volume scheme. Under the transport-plus-loss CFL, the scheme preserves nonnegativity and satisfies a discrete mass bound. Compatible smooth data give a first-order cell-mass and reconstructed-density estimate when the true birth quadrature defect is controlled. BV data need no initial boundary compatibility; the separate compactness argument gives (L^1) convergence without a Taylor rate.

The cutoff analysis keeps survival, outflow, and the initial-age tail visible as a separate error. The cohort method advances ages exactly and approximates survival and births; its consistency result is in bounded-Lipschitz measure norm, not (L^1) for atomic measures. A narrow maturation profile approaches a delayed birth term under stated Lipschitz bounds, while a finite-width kernel is not an exact delay. Capped adaptive indicators do not certify true quadrature error.

The size model uses a fixed interval, strictly positive growth, an absorbing upper boundary, parent removal, and a measurable nonlocal daughter kernel. Its variable-width finite-volume update uses positive daughter coefficients and gives exact discrete number and center-weighted-size balances. The first moment is conserved by division only when daughter-size production cancels parent removal; an arbitrary kernel need not satisfy that identity. Moving boundaries, zero growth, and discontinuous coefficients are outside this model.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Linear age-structured renewal transport problem

Definition

Fix a finite horizon T>0. Let mortality and fertility be bounded Lipschitz functions μ,β:[0,∞)→[0,∞), and let n0∈L1(0,∞) be nonnegative. The linear age-structured renewal transport problem is

nt(t,a)+na(t,a)=−μ(a)n(t,a),a>0,0<t≤T,

with initial density n(0,a)=n0(a) and inflow condition

n(t,0)=B(t),B(t):=∫0∞β(a)n(t,a) da.

Here a is age, B is the birth rate, and the characteristic age path solves the first-order initial value problem a′(s)=1 in the sense of First-order systems, initial value problems, and solutions on intervals, hence a(s)=a(0)+s. Define survival from age a over duration s≥0 by

S(a,s):=exp⁡ ⁣(−∫aa+sμ(r) dr).

A weak solution on [0,T] is a nonnegative n∈C([0,T];L1(0,∞)) with continuous B:[0,T]→[0,∞) satisfying B(t)=∫0∞β(a)n(t,a) da and, for every ϕ∈Cc1([0,T]×[0,∞)) with ϕ(T,a)=0,

−∫0T ⁣∫0∞n(t,a)(ϕt+ϕa−μ(a)ϕ) da dt=∫0∞n0(a)ϕ(0,a) da+∫0TB(t)ϕ(t,0) dt.

A mild solution is a weak solution that also satisfies, for almost every a≥0,

n(t,a)={n0(a−t)S(a−t,t),a≥t,B(t−a)S(0,a),0≤a<t,

The formula follows the age characteristics a′(s)=1 and their mortality survival. No delay equation or finite-age cutoff is identified with this model by definition.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Size transport with a nonlocal division kernel

Definition

Fix a finite horizon T>0 and a nondegenerate fixed size interval [0,X] with X>0. Let g,μ,γ:[0,X]→[0,∞) be bounded Lipschitz functions, with g(x)≥gmin⁡>0. Here g is size growth, μ is ordinary mortality, and γ is the rate at which a dividing parent is removed. For each parent size y∈[0,X], let Ky be a finite nonnegative Borel measure of daughter sizes, supported on [0,y]. Assume y↦Ky(E) is measurable for every Borel set E⊆[0,X], and that the number moment k0(y):=Ky([0,y]) is uniformly bounded; the size moment k1(y):=∫[0,y]x Ky(dx) is then finite.

Because Ky may have atoms, represent the population by a finite nonnegative Borel measure νt on [0,X) rather than requiring a density. Given an initial measure ν0, a weak solution is a curve t↦νt that is continuous when tested against continuous functions on [0,X] vanishing at X and satisfies, for every ϕ∈C1([0,T]×[0,X]) with ϕ(t,X)=0,

∫[0,X)ϕ(T,x) νT(dx)−∫[0,X)ϕ(0,x) ν0(dx)=∫0T ⁣∫[0,X)(ϕt+gϕx−(μ+γ)ϕ)(t,x) νt(dx) dt+∫0T ⁣∫[0,X) ⁣(∫[0,y]ϕ(t,x) Ky(dx))νt(dy) dt.

This is the measure-valued transport equation with daughter production ∫[0,X)Ky(dx) νt(dy), no incoming source at x=0, and absorbing outflow at x=X. The test condition ϕ(t,X)=0 encodes removal when a characteristic reaches the upper boundary.

Let Π[0,X](x):=min⁡{X,max⁡{0,x}} and extend the growth field to g~(x):=g(Π[0,X](x)) on R. This is a bounded globally Lipschitz vector field. By Size characteristics, positivity, and discrete number and size-moment balances ↗, its IVP has a unique solution χy on [0,T] from every initial size y∈[0,X]. A model characteristic is χy restricted to [0,min⁡{T,τy}], where τy is its first hitting time of X (taking τy=+∞ if there is no hit on [0,T], and τy=0 if y=X). On this restricted interval, g~=g. For y<X the restricted interval is nondegenerate, and the characteristic is a first-order IVP solution in the sense of First-order systems, initial value problems, and solutions on intervals. For y=X the restriction is only the stopped path on {0}; no derivative on that singleton is asserted, while the extended path χX remains an IVP solution on [0,T]. The interval is fixed; moving boundaries, zero growth and discontinuous coefficients are outside this definition.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Characteristic formula and Volterra solution for age renewal transport

Statement

For Linear age-structured renewal transport problem, put M(a)=∫0aμ(r) dr, K(a)=β(a)e−M(a), and

G(t)=∫0∞β(x+t)exp⁡ ⁣(−∫xx+tμ(r) dr)n0(x) dx.

There is a unique mild solution. Its birth rate is the unique continuous solution of

B(t)=G(t)+∫0tK(a)B(t−a) da,

and its density, for almost every a≥0, is

n(t,a)={n0(a−t)exp⁡ ⁣(−∫a−taμ(r) dr),a≥t,B(t−a)e−M(a),0≤a<t.

Facts & Assumptions

Given: The model Linear age-structured renewal transport problem with bounded nonnegative Lipschitz μ,β and n0∈L+1.

[F1]

Linear age-structured renewal transport problem: characteristic age paths solve a′=1, survival is S(a,s)=exp⁡(−∫aa+sμ), and a mild solution satisfies the renewal boundary equation.

[F2]

First-order systems, initial value problems, and solutions on intervals: a solution of a first-order initial value problem is differentiable along its interval and satisfies its differential equation and initial value.

Proof

1.1F1F2algebra

The old-cohort branch in the Statement follows the characteristic a(s)=x+s and survival S(x,t) from [F1]; the newborn branch starts from the boundary value B(t−a) at time t−a and follows survival S(0,a). These are the characteristic branches in the definition of a mild solution. We use them to construct a candidate after solving the Volterra equation; no pointwise differentiability of a weak solution is assumed.

1.2givenF1algebra

The functions K and G are continuous. Indeed, K is a product of Lipschitz functions; for s,t∈[0,T], the integrand multiplier in G changes by at most (Lip⁡(β)+∥β∥∞∥μ∥∞)∣t−s∣, since ∣e−u−e−v∣≤∣u−v∣ for u,v≥0. Multiplication by n0∈L1 and integration gives continuity of G.

2.1F1step 1.1algebra

Substitute the two branches into B(t)=∫0∞β(a)n(t,a) da and split at a=t. In the old-cohort part set x=a−t; the result is exactly G(t)+∫0tβ(a)e−M(a)B(t−a) da.

2.2step 1.2constructalgebra

On any interval of length τ with τ∥K∥∞<1, the right side of the Volterra equation is a contraction in the sup norm on continuous birth-rate functions: the difference of two outputs is bounded by τ∥K∥∞ times the input difference. If ∥K∥∞=0, the equation is simply B=G. Starting at time zero and then treating the already determined history as a fixed forcing term, finitely many such intervals cover [0,T]; each interval therefore has one continuous solution and the solutions agree at their shared endpoints.

2.3F1step 1.1algebra

Let ϕ∈Cc1([0,T]×[0,∞)) satisfy ϕ(T,a)=0. On an initial cohort with coordinate x=a−t≥0, the chain rule gives ddt[n0(x)S(x,t)ϕ(t,x+t)]=n(t,x+t)(ϕt+ϕa−μϕ)(t,x+t) for almost every t. Integrating from 0 to T makes the terminal term zero and leaves −n0(x)ϕ(0,x); integration over x≥0 gives the initial-data contribution to the weak identity.

2.4F1step 1.1algebra

For a newborn cohort with birth time s=t−a≥0, the chain rule gives dda[B(s)S(0,a)ϕ(s+a,a)]=n(s+a,a)(ϕt+ϕa−μϕ)(s+a,a). Integrating from a=0 to T−s gives −B(s)ϕ(s,0) because ϕ(T,T−s)=0; integrating over s∈[0,T] gives the renewal-boundary contribution.

3.1givenstep 2.2inductionalgebra

When n0≥0, G,K≥0. Picard iteration from the zero function on the first interval preserves nonnegativity and converges to its unique fixed point; the same argument applies on each continuation interval because its known-history forcing is nonnegative. Thus B≥0, and the characteristic formula gives n≥0.

4.1givenstep 1.1step 2.2step 3.1algebra

The formula has finite L1 norm because its old-cohort contribution is at most ∥n0∥1 and its newborn contribution is at most ∫0tB(s) ds. On the old branch, continuity in time follows from translation continuity of n0 in L1 and dominated convergence for the survival multiplier. On the newborn branch, continuity follows from continuity and boundedness of B and S, with the interval of newly entering cohorts having length tending to zero. Thus n∈C([0,T];L1).

5.1F1step 2.1step 4.1step 2.3step 2.4algebra

The old-cohort region a≥t and newborn region a<t partition the time-age domain up to a measure-zero diagonal. Adding their integrals from steps 2.3 and 2.4, then multiplying by −1, gives exactly the weak identity in [F1] with initial term ∫n0(x)ϕ(0,x)dx and boundary term ∫0TB(s)ϕ(s,0)ds. Step 2.1 gives B(t)=∫0∞β(a)n(t,a)da, and step 4.1 gives n∈C([0,T];L1) and nonnegativity. Thus the characteristic formula is a weak, hence mild, solution.

6.1F1step 1.1step 2.1step 2.2∎

Any mild solution has a birth rate satisfying the Volterra equation in step 2.1. Uniqueness from step 2.2 forces that same birth rate, and the characteristic formula in step 1.1 then forces the same density.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Finite-age renewal model with absorbing outflow

Definition

Fix 0<T<A, bounded nonnegative Lipschitz rates μ,β:[0,A]→[0,∞), and initial density n0A∈L+1(0,A). The finite-age renewal model with absorbing outflow is the age transport problem on 0<a<A with the same renewal inflow BA(t)=∫0Aβ(a)nA(t,a) da as in Linear age-structured renewal transport problem. A cohort follows age speed 1, retains its mortality survival S(a,s) until it reaches A, and is then removed; no cohort enters from ages above A. The initial density is n0A on [0,A]. A mild solution is the characteristic formula on 0<a<A, with an initial-cohort branch for a≥t, a newborn branch for a<t, and zero extension to a≥A. Since A>T, no cohort born during [0,T] reaches the outflow boundary.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Finite-volume size transport with positive division quadrature

Definition

Input a partition 0=x0<x1<⋯<xJ=X with J≥1, a time step Δt>0, a nonnegative current mass vector Mk=(Mjk)j=1J, and the coefficients and kernel from Size transport with a nonlocal division kernel. Partition [0,X] into disjoint cells Cj=[xj−1,xj) for j<J and CJ=[xJ−1,X], with widths hj=xj−xj−1>0 and centers xjc=(xj−1+xj)/2. Set ujk=Mjk/hj, the incoming flux F0k=0, the internal and outgoing upwind fluxes Fjk=g(xj)ujk for 1≤j≤J, and dj=μ(xjc)+γ(xjc).

For each parent cell Cℓ, form nonnegative daughter coefficients

Ajℓ=1hℓ QM(y↦Ky(Cj);xℓ−1,xℓ,1),

using Composite midpoint rule on a uniform grid on the parent-size interval. Since the one-panel midpoint rule is QM(f;xℓ−1,xℓ,1)=hℓf(xℓc), this gives Ajℓ=Kxℓc(Cj)≥0. If Δt(g(xj)/hj+dj)>1 for any cell, return invalid-cfl without an update. Otherwise the finite-volume update is

Mjk+1=Mjk−Δt(Fjk−Fj−1k)−Δt djMjk+Δt∑ℓ=1JAjℓMℓk.

The flux FJk is absorbing outflow. The admissible variable-width CFL is Δt(g(xj)/hj+dj)≤1 for every cell. If a grid or any required coefficient/kernel evaluation fails, return failure without a partial update. Otherwise return Mk+1 together with the fluxes and daughter coefficients used in the step.

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Age-cohort quadrature along exact characteristics

Definition

Input a finite horizon T>0, a time step Δt>0, an age cutoff A>0, bounded nonnegative rates μ,β on [0,A], and a finite positive initial quadrature (ai,mi0)i=1I supported in [0,A) for the initial age measure on that interval. Set N=⌊T/Δt⌋; the procedure returns levels at tk=kΔt for 0≤k≤N, with final time NΔt≤T. One option, when a nonnegative real-valued initial density n0 is evaluable at every point of [0,A], is to choose any finite partition 0=b0<⋯<bI=A and apply Composite midpoint rule on a uniform grid with n=1 separately on each cell [bi−1,bi]: set Δai=bi−bi−1, ai=(bi−1+bi)/2, and mi0=QM(n0;bi−1,bi,1)=Δai n0(ai) for 1≤i≤I. The cell widths need not be equal. A cohort at age aik and mass mik follows the exact age characteristic

aik+1=aik+Δt

and updates its survival mass by forward Euler,

mik+1=(1−Δt μ(aik))mik.

If Δt∥μ∥∞>1, return status invalid-step with no updates. Otherwise initialize the birth and outflow records as empty and set νh0=∑imi0δai0. For each k=0,…,N−1, first calculate Bhk=∑iβ(aik)mik. Advance every existing cohort by aik+1=aik+Δt and update its survival mass by mik+1=(1−Δtμ(aik))mik. If an updated age is at least A, record that updated mass as discrete outflow and discard the cohort; otherwise retain it. Then insert a newborn cohort of age 0 and mass ΔtBhk at time tk+1. Record Bhk and the step's outflow, and form νhk+1 from the retained cohorts and this newborn. A failed rate evaluation returns failure with no partial update. The procedure returns the finite sequence (νhk)k=0N and its birth/outflow records. The bounded-Lipschitz quadrature defect of the initial measure and the Euler survival error are separate inputs to any consistency claim.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Positive age-renewal evolution and L1 data stability

Statement

For Linear age-structured renewal transport problem, let β∞=∥β∥∞. The solution map Sμ,β(t):n0↦n(t,⋅) is linear and positive, and

∥Sμ,β(t)n0∥1≤eβ∞t∥n0∥1.

If n,n~ use possibly different nonnegative bounded coefficients (μ,β) and (μ~,β~) and initial data n0,n~0, put δ=∥μ−μ~∥∞+∥β−β~∥∞ and β~∞=∥β~∥∞. Then

∥n(t)−n~(t)∥1≤eβ∞t(∥n0−n~0∥1+δteβ~∞t∥n~0∥1).

Facts & Assumptions

Given: The two age-renewal models and the characteristic mild solutions supplied by Characteristic formula and Volterra solution for age renewal transport.

[F1]

Linear age-structured renewal transport problem: μ,β are nonnegative and bounded, the inflow is B(t)=∫β(a)n(t,a) da, and survival is the exponential along an age characteristic.

[F2]

Characteristic formula and Volterra solution for age renewal transport: each nonnegative initial density has one nonnegative mild solution, given by the characteristic formula and Volterra birth equation.

[F3]

Gronwall's integral inequality with variable and constant coefficients: if u(t)≤a(t)+∫0tb(s)u(s) ds with nonnegative continuous u,b and nondecreasing a, then u(t)≤a(t)e∫0tb.

Proof

1.1F1F2algebra

For nonnegative initial data, sums and nonnegative scalar multiples of the solutions in [F2] satisfy the corresponding characteristic formulas and renewal equations in [F1]; uniqueness in [F2] therefore makes S(t) additive and nonnegatively homogeneous on L+1. For arbitrary real f∈L1, put f+=max⁡{f,0}, f−=max⁡{−f,0} and define S(t)f=S(t)f+−S(t)f−. If also f=p−q with p,q≥0, then p+f−=q+f+, so cone additivity gives S(t)p−S(t)q=S(t)f+−S(t)f−. This independence of decomposition gives additivity on L1, and nonnegative homogeneity together with S(t)(−f)=−S(t)f gives real linearity. The extension obeys the same characteristic formulas and renewal relation by subtraction; it agrees with [F2] on L+1 and is positive there.

1.2F1F2algebra

Integrating the two characteristic branches in [F2] gives N(t)=∫0∞n0(x)S(x,t) dx+∫0tB(t−a)S(0,a) da. Bounded μ and continuous B permit differentiation under these integrals; using ∂tS(x,t)=−μ(x+t)S(x,t) and ∂aS(0,a)=−μ(a)S(0,a) yields N′(t)=B(t)−∫0∞μ(a)n(t,a) da for almost every t. Since 0≤B(t)≤β∞N(t) and μn≥0, we have N′(t)≤β∞N(t).

2.1F3step 1.1step 1.2algebra

Integrating the differential inequality in step 1.2 gives N(t)≤N(0)+∫0tβ∞N(s) ds. Apply [F3] with a(t)=N(0) and b=β∞ to obtain the asserted bound for nonnegative data. For arbitrary real f∈L1, positivity and linearity from step 1.1 give ∣S(t)f∣≤S(t)f++S(t)f−=S(t)∣f∣ almost everywhere. Integrating and applying the nonnegative-data bound proves ∥S(t)f∥1≤eβ∞t∥f∥1.

2.2F1F2step 1.2algebra

Subtract the two transport equations and write w=n−n~. Along characteristics, its interior forcing is (μ~−μ)n~ and its inflow forcing is ∫βw+∫(β−β~)n~. Taking absolute values in the characteristic integral formula, integrating over age, and using nonnegative survival factors gives E(t)≤E(0)+∫0t[β∞E(s)+δN~(s)] ds, where E=∥w∥1 and N~=∥n~∥1.

3.1F3step 2.1step 2.2algebra∎

By step 2.1 applied to n~, N~(s)≤eβ~∞s∥n~0∥1. Thus the nondecreasing function a(t)=E(0)+δteβ~∞t∥n~0∥1 bounds the forcing integral in step 2.2, and [F3] yields the claimed coefficient-and-initial-data stability estimate.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Size characteristics, positivity, and discrete number and size-moment balances

Statement

Under Size transport with a nonlocal division kernel, for every initial size y∈[0,X] the extended IVP specified there has a unique solution χy on [0,T]. The solution is strictly increasing and agrees with the original growth field until it first reaches X, so its restriction through that first hit is the model characteristic.

For Finite-volume size transport with positive division quadrature, assume nonnegative initial masses, nonnegative quadrature coefficients Ajℓ, and the stated CFL. Then all Mjk remain nonnegative. Define the cell number and center-weighted size moment by Nhk=∑jMjk and Whk=∑jxjcMjk. With k0,ℓh=∑jAjℓ and k1,ℓh=∑jxjcAjℓ, the exact balances are

Nhk+1−Nhk=Δt(−FJk−∑jdjMjk+∑ℓk0,ℓhMℓk),

Whk+1−Whk=Δt(∑j=1J−1(xj+1c−xjc)Fjk−xJcFJk−∑jxjcdjMjk+∑ℓk1,ℓhMℓk).

Thus a division kernel preserves the discrete size moment only when its first-moment production cancels the parent-removal term; this identity is not automatic for an arbitrary kernel.

Facts & Assumptions

[F1]

Size transport with a nonlocal division kernel: the total loss rate is d=μ+γ, and each daughter kernel is a nonnegative measure supported on sizes no larger than its parent.

[F2]

Finite-volume size transport with positive division quadrature: the update is Mjk+1=Mjk−Δt(Fj−Fj−1)−ΔtdjMjk+Δt∑ℓAjℓMℓk, with nonnegative Ajℓ and the stated CFL.

[F3]

Picard-Lindelöf local existence and uniqueness for first-order systems: a continuous vector field that is locally Lipschitz in the state variable has a unique local IVP solution; its quantitative clause gives a solution on any centered cylinder where the boundedness and contraction conditions hold, and its uniqueness clause applies on common intervals.

[A1]

The initial cell masses are nonnegative.

Proof

1.1A1F1F2inductionalgebra

In [F2], the coefficient of Mjk outside the division source is 1−Δt(g(xj)/hj+dj), which is nonnegative by the CFL. For j≥2 the incoming transport term has coefficient Δtg(xj−1)/hj−1≥0; for j=1 the incoming flux is F0=0. Every division term has coefficient ΔtAjℓ≥0. Hence each new mass is a sum of nonnegative terms when [A1] holds. Induction on k proves Mjk≥0.

1.2F2algebra

Sum [F2] over j. The finite flux sum telescopes to FJ−F0=FJ because F0=0. Interchanging the finite sums in the kernel source gives ∑ℓ(∑jAjℓ)Mℓ=∑ℓk0,ℓhMℓ. Subtracting Nhk yields the number identity.

2.1F1F2step 1.2algebra

Multiply [F2] by xjc and sum over j. For the flux term, finite summation by parts gives ∑jxjc(Fj−Fj−1)=xJcFJ−∑j=1J−1(xj+1c−xjc)Fj since F0=0. The loss term is −∑jxjcdjMj, and the source term is ∑ℓ(∑jxjcAjℓ)Mℓ=∑ℓk1,ℓhMℓ. Rearranging proves the weighted-moment identity and the stated condition for cancellation.

3.1F1F3constructalgebra∎

Let M=∥g∥∞ and let L be a Lipschitz constant for g on [0,X]. The projection Π[0,X] is 1-Lipschitz, so the extension g~=g∘Π[0,X] is globally bounded by M and globally Lipschitz with constant L on the open domain R. Since g≥gmin⁡>0, choose h0>0 with h0M≤1 and Lh0<1. For every starting time t0∈R and state z∈R, the cylinder [t0−h0,t0+h0]×[z−1,z+1] lies in R×R, and [F3] gives a unique solution through (t0,z) on that interval; the same bounds apply after every restart. Restricting and concatenating at most ⌈T/h0⌉ such forward segments gives a solution on [0,T], and the uniqueness clause of [F3] makes it unique there. Along this solution χy′(s)=g~(χy(s))≥gmin⁡, so it is strictly increasing. Starting from y∈[0,X], it stays in [0,X] through its first hitting of X, where g~=g; that restricted segment is the model characteristic.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Characteristic solution and mass bound for the finite-age renewal model

Statement

For Finite-age renewal model with absorbing outflow, define K(a)=β(a)S(0,a) and

GA(t)=∫0A−tβ(x+t)S(x,t)n0A(x) dx,0≤t≤T.

There is a unique continuous nonnegative birth rate satisfying

BA(t)=GA(t)+∫0tK(a)BA(t−a) da,

and the finite-age density is, for almost every 0<a<A,

nA(t,a)={n0A(a−t)S(a−t,t),t≤a<A,BA(t−a)S(0,a),0≤a<t.

It is the unique mild solution, has BA(t)=∫0Aβ(a)nA(t,a) da, and obeys

∥nA(t,⋅)∥1≤eβ∞t∥n0A∥1,β∞=∥β∥∞.

For t≤T<A, its exact cumulative number outflow is

OA(t)=∫A−tAn0A(x)S(x,A−x) dx.

Facts & Assumptions

Given: Bounded nonnegative Lipschitz rates, n0A∈L+1(0,A), and the characteristic survival law in Finite-age renewal model with absorbing outflow.

[F1]

Finite-age renewal model with absorbing outflow: characteristics have age a+s and survival S(a,s)=exp⁡(−∫aa+sμ); age A is absorbing outflow, and the birth rate integrates βnA only over [0,A].

[F2]

Gronwall's integral inequality with variable and constant coefficients: if u(t)≤a(t)+∫0tb(s)u(s) ds with nonnegative continuous u,b and nondecreasing a, then u(t)≤a(t)e∫0tb.

Proof

1.1givenF1algebra

The kernel K is continuous on [0,T]. The moving-limit forcing GA is also continuous: on the common part of the integration intervals its integrand changes uniformly continuously with t, and the integral over the symmetric difference tends to zero by absolute continuity of the integral of n0A. It is nonnegative.

2.1step 1.1constructalgebra

On any interval of length δ with δ∥K∥∞<1, the map b↦GA(t)+∫0tK(a)b(t−a)da is a contraction on the unknown continuous birth-rate segment, with contraction factor at most δ∥K∥∞ after the already determined history is placed in the forcing. If K=0, the equation reduces to BA=GA. Finitely many such intervals cover [0,T], and at each endpoint the defining equation fixes the same value from either adjacent interval; this constructs a unique continuous solution. Picard iteration from zero on each interval preserves nonnegativity because GA,K≥0, so BA≥0.

3.1F1step 2.1algebra

Substitute the displayed density formula into ∫0Aβ(a)nA(t,a)da. On the initial-cohort part t≤a<A, use x=a−t to obtain GA(t). On the newborn part 0≤a<t, the integral is ∫0tβ(a)S(0,a)BA(t−a)da=∫0tK(a)BA(t−a)da. The Volterra equation therefore gives exactly the renewal boundary relation. Both parts are nonnegative and have finite L1 mass, bounded by ∥n0A∥1+∫0tBA(s)ds.

4.1F1F2step 3.1algebra

Integrating the two characteristic branches gives NA(t)=∫0A−tn0A(x)S(x,t)dx+∫0tBA(t−a)S(0,a)da. This mass function is absolutely continuous; the moving-limit rule and ∂tS(x,t)=−μ(x+t)S(x,t) show for almost every t that NA′(t)=BA(t)−∫0Aμ(a)nA(t,a)da−n0A(A−t)S(A−t,t). The final term is nonnegative outflow, so NA′(t)≤BA(t)≤β∞NA(t). Integrating and applying [F2] yields NA(t)≤eβ∞tNA(0).

5.1F1step 2.1step 3.1step 4.1algebra∎

For t≤T<A, newborn cohorts have age at most T and cannot reach A. An initial cohort with age x exits by time t precisely when A−t≤x<A; it survives until exit by S(x,A−x). Integrating its initial density over those ages gives the stated OA(t). Any mild solution must follow these same characteristics and its birth rate must solve the Volterra equation from step 3.1, whose solution is unique by step 2.1. Thus the density and outflow are unique. Translation continuity in L1, continuity of BA and of S, and the vanishing measure of the newly entering cohort region give continuity of nA in time with values in L1(0,A).

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Positive composite or capped-adaptive quadrature for renewal births

Definition

Input a uniform age partition 0=a0<a1<⋯<aJ=A with Δa=aj−aj−1, nonnegative cell masses Mj, a bounded nonnegative fertility function β:[0,A]→[0,β∞], and one rule setting for each cell. A setting is either midpoint with an integer panel count qj≥1, or capped adaptive with finite depth Dj∈N, positive integer panel-call budget Pj≥1, and indicator tolerance τj>0. If no settings are supplied, use midpoint with qj=1 in every cell. In each cell Cj=[aj−1,aj], let uj=Mj/Δa and approximate ∫Cjβ(a)uj da by its selected positive rule:

  1. Midpoint mode: use ujQM(β;aj−1,aj,qj) from Composite midpoint rule on a uniform grid and record status midpoint.
  2. Capped adaptive mode: call Finite-depth adaptive Simpson with an indicator on β over Cj with inputs (Dj,Pj,τj), then multiply its estimate by uj and retain its status and indicator. The fertility evaluator returns a finite value at every requested node or reports failure.

If an evaluation fails, return failure and no birth rate. Otherwise return

Bh=∑j=1JujQj(β),

the per-cell statuses, and the sum of any reported adaptive indicators. Both rules have nonnegative quadrature weights summing to Δa, so 0≤Bh≤β∞∑jMj. An adaptive accepted status means only that its indicators meet the requested budget; neither that indicator nor a cap-reached estimate bounds the true quadrature error. Any convergence theorem using this procedure must separately assume or prove a true quadrature-defect bound.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Weak consistency of age-cohort characteristic quadrature

Statement

Let μ,β:[0,∞)→[0,∞) be bounded C1 functions with bounded derivatives, let n0∈C1([0,∞))∩L1(0,∞) be nonnegative, and fix T>0, Δt>0, and A>T. Put n^0=n01[0,A−T) and τA=∫A−T∞n0(a) da. Choose the initial positive cohort quadrature for n^0(a) da with support in [0,A−T) and bounded-Lipschitz defect ϵ0 on test functions of norm at most 1. For the procedure Age-cohort quadrature along exact characteristics, set N=⌊T/Δt⌋, consider its levels 0≤k≤N, and write β∞:=∥β∥∞. Assume Δt∥μ∥∞≤1. Define

∥η∥BL∗:=sup⁡∥ϕ∥∞+Lip⁡(ϕ)≤1∣∫ϕ dη∣

for a finite signed measure η. Then, for fixed T, there is CT independent of Δt, initial cohort count and cutoff such that

max⁡kΔt≤T∥νhk−n(tk,a) da∥BL∗≤CT(ϵ0+Δt+τA).

The measure norm is weak and does not assert L1 convergence of atomic cohorts to a density. The constant CT may depend on T, the stated coefficient bounds and ∥n0∥1, but is independent of the time step, cutoff and number of cohorts. For midpoint initial cohorts on [0,A−T] and Lipschitz n0, ϵ0=OA(Δa); the omitted initial tail τA remains separate.

Facts & Assumptions

Given: The renewal model, its characteristic solution, the cohort algorithm and the supplier one-step and discrete-recurrence results.

[F1]

Linear age-structured renewal transport problem: survival is S(a,s)=exp⁡(−∫aa+sμ) and births reset cohort age to zero.

[F2]

Characteristic formula and Volterra solution for age renewal transport: the exact birth rate satisfies B(t)=G(t)+∫0tK(t−s)B(s) ds, with K(s)=β(s)S(0,s), and the density is the corresponding characteristic pushforward.

[F3]

Age-cohort quadrature along exact characteristics: old cohort ages advance by Δt, cohort masses use the forward-Euler survival step, and each discrete birth mass is ΔtBhk at age zero.

[F4]

Forward Euler has uniform local order one and a one-step Lipschitz bound: on a compact tube with a C1 vector field, forward Euler has local defect O(Δt2) and one-step Lipschitz factor 1+LΔt.

[F5]

Positive age-renewal evolution and L1 data stability: for two solutions with the same rates and different initial data, their L1 difference is at most eβ∞t times the initial L1 difference.

[L1]

Finite geometric error bound from a one-step recurrence: a recurrence ek+1≤(1+Lh)ek+Ch2 on a fixed finite horizon has maximum error O(h).

Proof

1.1F1F2F5algebra

Let n^ be the exact solution with initial density n^0. Applying the characteristic formula [F2] and changing variables from age a to birth time s=t−a gives, for every bounded Lipschitz ϕ, ∫ϕ(a)n^(t,a)da=∫0A−Tϕ(x+t)S(x,t)n0(x)dx+∫0tϕ(t−s)S(0,t−s)B^(s)ds. The birth kernel K(s)=β(s)S(0,s) is bounded and Lipschitz, and G^(t)=∫0A−Tβ(x+t)S(x,t)n0(x)dx is bounded and Lipschitz because β,μ have bounded first derivatives and n0 is integrable. The mass estimate [F5] bounds B^ by β∞eβ∞T∥n0∥1. Subtracting its Volterra equation at t and t+h, the old-history kernel difference is at most Lip⁡(K)h∫0T∣B^∣ and the new interval at most h∥K∥∞∥B^∥∞. Thus B^ is Lipschitz on [0,T].

2.1F3F4L1step 1.1algebra

If N=0, there are no time steps and the survival-error claim is zero. Suppose N≥1, and put Th=NΔt>0. A cohort born with mass m0 and initial age a0 has exact survival mass solving m′(s)=−μ(a0+s)m(s); let q=m0 be its reference mass. To apply [F4] at a0=0, define the extension μˉ(r)=μ(r) for r≥0 and μˉ(r)=μ(0)+μ+′(0)rer for r<0. The one-sided derivative μ+′(0) exists; the two formulas and their first derivatives match at zero, and μˉ and μˉ′ are bounded on R. For a cohort whose reference mass is q>0, the vector field f(s,m)=−μˉ(a0+s)m is C1 on the open tube V=(−1,Th+1)×(−q,2q), where Th=NΔt, with all bounds required by [F4] uniform in a0 and proportional to q except for the state-Lipschitz constant, which is bounded by ∥μˉ∥∞. The exact continuation m(s)=mstartexp⁡(−∫0sμˉ(a0+r) dr) exists on an open interval containing [0,Th]; after shrinking that interval at the endpoints its graph lies in V. Its exact forward solution from any m∈[0,q] and its Euler input/output states lie in K=[0,Th]×[0,q]: mortality is nonnegative for ages a0+s≥0, and Δt∥μ∥∞≤1. Thus [F4] gives one-step defect at most CΔt2q and a stability factor 1+LΔt, with constants independent of a0 and proportional to q. Applying [L1] on the aligned horizon Th=NΔt gives survival-weight error at most CTΔtq over the full run; the zero-mass case is exact. For the discrete total cohort mass Nhk, positivity and Bhk≤β∞Nhk give Nhk+1≤(1+Δtβ∞)Nhk, hence Nhk≤eβ∞TNh0. Since Nh0≤∥n0∥1+ϵ0 (test the initial BL defect with ϕ≡1), the sum of the initial cohort masses and all inserted newborn masses ∑j<kΔtBhj is at most CT(∥n0∥1+ϵ0). Thus the sum of all cohort survival-weight errors is at most CTΔt(∥n0∥1+ϵ0).

3.1F2F3step 2.1algebra

Write ek=∣Bhk−B^(tk)∣. The initial-cohort contribution is at most CT(ϵ0+Δt): the test x↦β(x+tk)S(x,tk) has bounded-Lipschitz norm at most CT, and step 2.1 bounds the accumulated Euler survival error. Each cohort born during [tj,tj+1] is inserted at tj+1 with mass ΔtBhj, so at tk its age is (k−j−1)Δt. The exact renewal convolution is the integral of K(tk−s)B^(s) over s∈[0,tk]. Since K and B^ are Lipschitz, replacing this integral by the left-endpoint sum with age (k−j−1)Δt costs at most CTΔt (the age sample differs from tk−tj by one step). Replacing B^(tj) by Bhj contributes at most ∥K∥∞Δt∑j<kej, and replacing exact newborn survival by its Euler product contributes at most CTΔt by step 2.1. Therefore ek≤CT(ϵ0+Δt)+∥K∥∞Δt∑j<kej. Set rk=CT(ϵ0+Δt)+∥K∥∞Δt∑j<kej. Then ek≤rk and rk+1=rk+∥K∥∞Δtek≤(1+∥K∥∞Δt)rk, so ek≤e∥K∥∞TCT(ϵ0+Δt).

4.1F1F2F5step 1.1step 2.1step 3.1algebra∎

For any test function with ∥ϕ∥∞+Lip⁡(ϕ)≤1, the initial cohort pushforward differs from the first integral for n^ in step 1.1 by at most CT(ϵ0+Δt), using step 2.1 and the bounded-Lipschitz norm of x↦ϕ(x+t)S(x,t). The newborn cohort sum differs from the second integral by its left-Riemann error CTΔt plus Δt∑j<kej≤CT(ϵ0+Δt) by step 3.1. Thus ∥νhk−n^(tk,a)da∥BL∗≤CT(ϵ0+Δt). The initial data difference is ∥n0−n^0∥1=τA, so [F5] gives ∥n(tk)−n^(tk)∥1≤eβ∞TτA. The inequality ∥f(a)da∥BL∗≤∥f∥1 and the triangle inequality prove the stated bound.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Finite-volume upwind scheme for age renewal transport

Definition

For Linear age-structured renewal transport problem, input a finite horizon T>0, a cutoff A>T, integers J,N≥1, and use the uniform grid aj=jΔa with Δa=A/J and cells Cj=[aj−1,aj]. Set Δt=T/N and tk=kΔt for 0≤k≤N. Use the model rates μ,β and input nonnegative initial cell masses Mj0. Also input one positive-birth-quadrature setting for each cell as specified in Positive composite or capped-adaptive quadrature for renewal births; if omitted, use its default one-panel midpoint setting. Keep these settings fixed at every time step. Evaluate μj=μ((aj−1+aj)/2). Require the transport-plus-loss CFL

λ+Δt μ∞≤1,λ=Δt/Δa,μ∞=∥μ∥∞.

If the CFL inequality fails, return status invalid-cfl without an update.

Initialize birth, quadrature-status, and outflow records as empty, and set cumulative outflow Oh0=0. For each k=0,…,N−1, put ujk=Mjk/Δa, compute Bhk by calling Positive composite or capped-adaptive quadrature for renewal births with the current masses, model fertility β, and the selected cell settings, and set the upwind fluxes F0k=Bhk and Fjk=ujk for 1≤j≤J. Update, for 1≤j≤J,

Mjk+1=Mjk−Δt(Fjk−Fj−1k)−Δt μjMjk.

The right flux FJk is absorbing outflow: record and discard the step mass ΔtFJk, set Ohk+1=Ohk+ΔtFJk, and allow no inflow from ages above A. Append Bhk, the quadrature status, and the step outflow record, then proceed to step k+1. Return (Mk)k=0N with the birth, quadrature-status, step-outflow, and cumulative-outflow records. An invalid grid or failed rate/quadrature evaluation returns failure with no partial current step. A cap-reached adaptive quadrature estimate may be used for a finite step, but its status is retained and carries no accuracy claim.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Positivity, L1 growth and data stability of the renewal finite-volume step

Statement

For the bounded fertility rate in the age model, write β∞:=∥β∥∞. Whenever Positive composite or capped-adaptive quadrature for renewal births returns an estimate for nonnegative cell masses and a bounded nonnegative fertility rate with upper bound β∞, its birth value satisfies 0≤Bh≤β∞∑jMj.

In the update of Finite-volume upwind scheme for age renewal transport, suppose the initial masses are nonnegative, the CFL holds, and each cell quadrature has nonnegative weights summing to Δa. Then all updated masses remain nonnegative and, with Nhk=∑j=1JMjk,

Nhk+1≤(1+Δt β∞)Nhk.

For two runs with the same grid, rates and quadrature rule but different nonnegative initial mass vectors M0,M~0, the one-step difference satisfies

∥Mk+1−M~k+1∥ℓ1≤(1+Δt β∞)∥Mk−M~k∥ℓ1.

Facts & Assumptions

[F1]

Positive composite or capped-adaptive quadrature for renewal births: each cell quadrature has nonnegative weights with total weight Δa, so its birth contribution is nonnegative and at most β∞Mj.

[F2]

Finite-volume upwind scheme for age renewal transport: Mjk+1=Mjk−Δt(Fjk−Fj−1k)−ΔtμjMjk, with F0k=Bhk, Fjk=Mjk/Δa for j≥1, and Δt/Δa+Δtμ∞≤1.

[A1]

The initial masses in both runs are nonnegative, and both runs use the same rates and positive quadrature rule.

Proof

1.1F1constructalgebra

In midpoint mode, each of the qj samples has weight Δa/qj>0, so the weights sum to the cell width. In capped-adaptive mode, every returned estimate is a finite sum of Simpson panel estimates on its leaf intervals; a two-panel Simpson block has weights proportional to (1,4,1), all positive, and their sum is the block width. The leaf intervals partition the cell, so the combined weights are nonnegative and sum to its width, including when a cap is reached. Hence the cell contribution lies between 0 and ujβ∞Δa=β∞Mj, and summing over cells gives the stated birth bound. A failure returns no estimate.

1.2A1F1F2inductionalgebra

Substituting Fjk=Mjk/Δa in [F2] gives M1k+1=(1−λ−Δtμ1)M1k+ΔtBhk and, for j≥2, Mjk+1=(1−λ−Δtμj)Mjk+λMj−1k. The CFL makes every displayed coefficient nonnegative, while [F1] makes Bhk≥0; induction in k proves nonnegativity.

2.1F1F2step 1.1algebra

Summing [F2] over cells telescopes the fluxes to FJk−F0k, hence Nhk+1=Nhk−ΔtFJk+ΔtBhk−Δt∑jμjMjk. Since FJk,μj,Mjk≥0 and step 1.1 gives Bhk≤β∞Nhk, the asserted mass-growth inequality follows.

3.1A1F1step 1.1step 1.2algebra∎

The update is a linear matrix with nonnegative entries by step 1.2. In each column, transport and mortality leave total coefficient at most 1, because outflow or mortality removes mass; the birth quadrature adds at most Δtβ∞ to that column by step 1.1. Thus its induced ℓ1 norm is at most 1+Δtβ∞. Applying this matrix to Mk−M~k proves the difference estimate.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

First-order finite-volume convergence for compatible smooth renewal data

Statement

Fix T>0 and a cutoff A>T. Let nA be the finite-age solution from Characteristic solution and mass bound for the finite-age renewal model with initial density n0A∈C2([0,A]) and assume μ,β∈C2([0,A]) are nonnegative and bounded. Assume nA extends to C2([0,T]×[0,A]). Its smoothness at the renewal corner includes n0A(0)=BA(0) and −(n0A)′(0)−μ(0)n0A(0)=BA′(0); no boundary value is imposed at the absorbing outflow. Take integers J,N≥1, set Δa=A/J and Δt=T/N, and use the corresponding uniform age and time grids. Initialize with exact cell masses and use the one-panel midpoint rule in each cell from Positive composite or capped-adaptive quadrature for renewal births. Assume Δt/Δa∈[c−,c+] for fixed 0<c−≤c+<∞ and the transport-plus-loss CFL. Define

ϵQ:=max⁡0≤k<N∣Bh(M‾(tk))−∫0Aβ(a)nA(tk,a) da∣,

where Bh(M‾(tk)) is the birth-rate quadrature applied to the exact cell-average masses M‾j(tk)=∫CjnA(tk,a) da. Then the cell-mass error Ek=∑j∣Mjk−∫CjnA(tk,a) da∣ obeys

max⁡0≤k≤NEk≤CT(Δa+Δt+ϵQ).

The piecewise-constant reconstruction uhk=Mjk/Δa on cell Cj and zero above A satisfies the same L1 error order. For the midpoint rule, ϵQ≤CTΔa under the stated regularity, so the method is first order in age mesh and time step. If the capped adaptive rule is used instead, the estimate requires an independently established true ϵQ; its reported indicator is not substituted for that hypothesis.

Facts & Assumptions

Given: The smooth finite-cutoff solution and finite-volume scheme from Finite-age renewal model with absorbing outflow, Characteristic solution and mass bound for the finite-age renewal model and Finite-volume upwind scheme for age renewal transport, under the hypotheses in the Statement.

[F1]

Finite-age renewal model with absorbing outflow: nt+na=−μn on 0<a<A, the renewal inflow is BA(t)=∫0Aβ(a)nA(t,a) da, and the upper boundary is absorbing outflow.

[F2]

Characteristic solution and mass bound for the finite-age renewal model: nA follows characteristics, satisfies the renewal relation, and its exact initial-cohort outflow is absorbing.

[F3]

Finite-volume upwind scheme for age renewal transport: the cell-mass step is Mjk+1=Mjk−Δt(Fjk−Fj−1k)−ΔtμjMjk with upwind face fluxes and absorbing top outflow.

[F4]

Positivity, L1 growth and data stability of the renewal finite-volume step: for the same coefficients and positive quadrature, a one-step mass difference is amplified by at most 1+Δtβ∞.

[F5]

Composite midpoint error bound for a twice continuously differentiable function: for f∈C2[a,b], the composite midpoint error on width h is at most (b−a)h2max⁡∣f′′∣/24.

[F6]

Finite geometric error bound from a one-step recurrence: if e0≤C0h and ek+1≤(1+Lh)ek+Ch2 on a fixed grid Nh=T, then max⁡kek≤CTh.

Proof

1.1F1F2constructalgebra

Let M‾j(t)=∫CjnA(t,a) da and u‾j(t)=M‾j(t)/Δa. The smooth finite-age characteristic solution [F2] satisfies the transport equation and renewal boundary in [F1]. Integrating [F1] over one cell and one time step gives M‾j(tk+1)=M‾j(tk)−∫tktk+1[nA(s,aj)−nA(s,aj−1)]ds−∫tktk+1∫Cjμ(a)nA(s,a) da ds, with nA(s,a0)=BA(s) and the outgoing trace at aJ=A.

2.1givenF5step 1.1algebra

Taylor's formula in age gives u‾j(t)−nA(t,aj)=−Δa2(nA)a(t,aj)+O(Δa2) uniformly. For interior cells, subtracting adjacent face-flux errors cancels the leading term and leaves O((Δa)2) per cell because (nA)aa is bounded. In the first and last cells there is only one uncancelled endpoint face error, of size O(Δa); these two cells contribute O(ΔtΔa) in total. Taylor's formula in time gives a face-integral remainder O(Δt2Δa) per cell after adjacent differences, using bounded (nA)ta, hence O(Δt2) after summing. At the inlet, the smooth trace gives (nA)t(t,a0)=BA′(t), so the time derivative of nA(t,a1)−BA(t) is also bounded by Δa∥(nA)ta∥∞; thus the boundary birth flux has no additional O(\Delta t^2) term beyond this face remainder and ΔtϵQ. For the reaction term, first note that u‾j(t)=nA(t,ajc)+O((Δa)2) by the symmetric cell average. Thus replacing μ(a)nA(t,a) by μ(ajc)u‾j(t) has per-cell error O((Δa)3): the midpoint formula [F5] bounds the error for μnA, and the cell-average replacement changes the midpoint value by O((Δa)2) times the cell width. Summing and multiplying by Δt gives O(Δt(Δa)2). Freezing the reaction term at tk instead of integrating in time contributes O(Δt2) after summing cells, since (μnA)t is bounded. Therefore ∥ρk∥ℓ1≤CΔt(Δa+Δt+ϵQ) for sufficiently fine meshes with Δa≤1.

3.1F3F4step 2.1algebra

Set ek=∑j∣Mjk−M‾j(tk)∣. Apply the one-step stability [F4] to the numerical and exact cell-average states and add the local defect from step 2.1; this yields ek+1≤(1+β∞Δt)ek+CΔt(Δa+Δt+ϵQ), with e0=0.

4.1F5F6step 3.1algebra∎

Put D=Δa+Δt+ϵQ. Iterating the recurrence in step 3.1 gives ek≤CΔtD∑r=0k−1(1+β∞Δt)r≤CTD, where the bound also holds when β∞=0, in which case the sum is k≤T/Δt. For the midpoint birth rule, split each cell defect into u‾j[QM(β;Cj,1)−∫Cjβ] and ∫Cjβ(a)(u‾j−nA(tk,a)) da. The first term is O((Δa)3) per cell by [F5] and bounded nA; in the second, subtract β at the cell midpoint because ∫Cj(u‾j−nA)=0, then use ∣β(a)−β(ajc)∣≤Lip⁡(β)Δa and ∫Cj∣u‾j−nA∣=O((Δa)2) from bounded (nA)a. Summing gives ϵQ=O((Δa)2), hence in particular O(Δa). With the two-sided mesh ratio, D=O(Δt); the recurrence then has additive term C′Δt2, and [F6] gives the stated first-order estimate. Finally, on each cell ∫Cj∣nA(tk,a)−u‾j(tk)∣da≤(Δa)2∥(nA)a∥∞; summing over A/Δa cells adds only O(Δa) to the cell-mass error, proving the reconstruction claim.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

L1 convergence for BV data without renewal-boundary compatibility

Statement

Let n0≥0 belong to BV(0,∞)∩L1(0,∞), where BV means finite total variation, and let μ,β be bounded nonnegative Lipschitz rates. Fix a finite horizon T>0. Initialize the finite-volume scheme with exact cell masses Mj0=∫Cjn0(a) da. Assume every positive birth-quadrature evaluation succeeds and the quadrature rule settings are fixed over time. The compatibility between the one-sided initial trace and the renewal birth rate is not assumed. For any sequence of age cutoffs Am→∞ with Am>T for all sufficiently large m, and admissible uniform grids with integers Jm,Nm≥1, Am=JmΔam and T=NmΔtm, assume Δam,Δtm>0 and Δam,Δtm→0, Δtm/Δam∈[c−,c+] for fixed positive bounds, and the transport-plus-loss CFL; extend the cellwise-constant reconstructions by zero above each cutoff. Linear interpolation between time levels then converges to the mild solution in C([0,T];L1(0,∞)). This is an L1 convergence statement; it asserts no smooth Taylor error rate.

Facts & Assumptions

Given: The BV initial density, bounded Lipschitz rates, and the finite-volume update in Linear age-structured renewal transport problem and Finite-volume upwind scheme for age renewal transport.

[F1]

Linear age-structured renewal transport problem: a weak solution is continuous in time with values in L1 and satisfies the displayed test-function identity, initial trace and renewal birth relation.

[F2]

Characteristic formula and Volterra solution for age renewal transport: the characteristic formula and Volterra equation have a unique mild solution.

[F3]

Finite-volume upwind scheme for age renewal transport: the cell update is the explicit flux balance; birth weights are nonnegative, supported in each cell and sum to the cell width; the right boundary is absorbing.

[F4]

Positivity, L1 growth and data stability of the renewal finite-volume step: discrete masses stay nonnegative and their total is at most 1+Δtβ∞ times the preceding total.

Proof

1.1givenF3construct

Write ujk=Mjk/Δa, Bk=Bhk, λ=Δt/Δa, and Vk=∣u1k−Bk∣+∑j=2J∣ujk−uj−1k∣. The initial cell averages satisfy V0≤TV⁡(n0)+∥n0∥∞+β∞∥n0∥1, which is finite because an integrable function of finite total variation on the half-line has finite essential supremum; the first term bounds differences of adjacent cell averages by averaging the variation measure.

2.1givenF3F4step 1.1algebra

Set u0k=Bk and put d1k=u1k−Bk and djk=ujk−uj−1k for 2≤j≤J. For j≥2, subtract adjacent updates from [F3] to obtain ujk+1−uj−1k+1=(1−λ−Δtμj)(ujk−uj−1k)+λ(uj−1k−uj−2k)+Δt(μj−1−μj)uj−1k. The CFL makes the first two coefficients nonnegative, and ∣μj−μj−1∣≤Lip⁡(μ)Δa. For J≥2, summing their absolute-value bounds for j=2,…,J gives at most ∑j=2J−1∣djk∣+(1−λ)∣dJk∣+λ∣d1k∣+ΔtLip⁡(μ)Nk: for 2≤j<J the two transport coefficients add to at most 1, the last-cell coefficient is at most 1−λ, and the boundary difference contributes λ∣d1k∣. When J=1, omit this adjacent-difference sum and use only the following first-cell estimate. The first-cell jump is d1k+1=(1−λ−Δtμ1)d1k+(1−Δtμ1)Bk−Bk+1. The quadrature weights are fixed and nonnegative, so ∣Bk+1−Bk∣≤β∞∑j∣Mjk+1−Mjk∣. The update and the definition of Vk give ∑j∣Mjk+1−Mjk∣≤Δt(Vk+μ∞Nk). Also ∣(1−Δtμ1)Bk−Bk+1∣≤∣Bk−Bk+1∣+Δtμ∞Bk and Bk≤β∞Nk. Adding the first-cell bound to the preceding sum combines the d1k coefficients as (1−λ−Δtμ1)+λ≤1; all other variation coefficients are at most 1. Therefore Vk+1≤(1+CΔt)Vk+CΔtNk, with C depending only on the rate bounds and Lipschitz constants.

3.1F4step 2.1algebra

Iterating [F4] gives Nk≤(1+Δtβ∞)kN0≤ekΔtβ∞N0≤eβ∞TN0. Iterating the recurrence in step 2.1 gives Vk≤(1+CΔt)kV0+CΔt∑r=0k−1(1+CΔt)k−1−rNr≤eCT(V0+CTeβ∞TN0). The same update gives ∑jΔa∣ujk+1−ujk∣≤Δt(Vk+μ∞Nk); hence the linear time interpolation is uniformly Lipschitz into L1.

4.1F2step 3.1construct

On any fixed interval [0,R], the uniform L1 and variation bounds make the time slices precompact in L1: replacing each slice by averages on blocks of width ρ changes its L1 norm by at most ρVk, while those block averages lie in a bounded finite-dimensional set. A diagonal extraction at rational times and the uniform time-Lipschitz bound from step 3.1 give a subsequence converging uniformly in time in L1 on every bounded age interval. A newborn cohort has age at most T; under the upwind update it can move at most one cell per step, so all newborn mass lies below age T/c−+Δa. For R>T/c−+Δa, births therefore contribute nothing above R. An initial cell moves by at most k cells, so any initial mass reconstructed above R came from initial cells whose lower endpoint exceeds R−kΔa−Δa≥R−T/c−−Δa. Its mass is at most ∫R−T/c−−2Δa∞n0 (extend n0 by zero to negative ages); mortality and outflow only decrease that inherited mass. This tail tends to zero uniformly as R→∞. For the exact solution, [F2] shows that newborns have age at most T, while mass above R>T comes only from initial ages above R−T and is at most ∫R−T∞n0. Hence the subsequence converges in C([0,T];L1(0,∞)).

5.1F1F3step 4.1algebra

Let ϕjk=ϕ(tk,(aj−1+aj)/2) for a compactly supported C1 test function with terminal value zero. Multiplying [F3] by ϕjk+1 and summing by parts gives the exact discrete weak identity −∑k,jMjk(ϕjk+1−ϕjk)−Δt∑k,j<JFjk(ϕj+1k+1−ϕjk+1)+Δt∑k,jμjMjkϕjk+1=∑jMj0ϕj0+Δt∑kBkϕ1k+1; the outflow term vanishes because the test is supported below the cutoff. Uniform L1 convergence, Lipschitz continuity of μ,β, and mesh refinement make the sums converge to the weak transport/loss terms. Since each positive birth rule is supported in its cell and its weights sum to Δa, ∣Bk−∫β(a)uh(tk,a)da∣≤2Lip⁡(β)Δa Nk, so the birth sum converges to the boundary term. The limit therefore satisfies [F1] with the given initial trace.

6.1F1F2step 5.1algebra∎

In characteristic coordinates x=a−t on a>t, the weak identity [F1] becomes ∂tv(t,x)=−μ(x+t)v(t,x) in distributions; multiplication by exp⁡(∫xx+tμ) makes its distributional t-derivative zero, and the initial trace gives v(t,x)=n0(x)S(x,t). On 0<a<t, use s=t−a and obtain ∂aw(s,a)=−μ(a)w(s,a) with trace w(s,0)=B(s); the same integrating-factor argument gives w(s,a)=B(s)S(0,a). Substitution into the renewal relation in [F1] yields the Volterra equation of [F2], whose solution is unique. Thus every subsequential limit equals the mild solution. Since every subsequence has a further subsequence with this same limit, the entire sequence converges.

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Age-cutoff survival, outflow and infinite-age tail bounds

Statement

Let A>T and let n0≥0 be integrable. For an initial cohort of age a, its survival through duration s satisfies

e−∥μ∥∞s≤S(a,s)≤e−μ∗s,μ∗:=ess inf⁡r≥0μ(r).

No cohort born during [0,T] reaches age A. The exact number of initial individuals that exits through age A by time t≤T is

OA(t)=∫A−tAn0(a)S(a,A−a) da≤∫A−T∞n0(a) da.

Let nA be the finite-age solution from Characteristic solution and mass bound for the finite-age renewal model with initial data n0A=n01[0,A], extended by zero above A. Then

sup⁡0≤t≤T∥n(t)−nA(t)∥1≤2eβ∞T∫A−T∞n0(a) da.

Under the smooth discretization hypotheses of First-order finite-volume convergence for compatible smooth renewal data, let uhk be its piecewise-constant reconstruction, extended by zero above A. At every time level,

∥uhk−n(tk,⋅)∥1≤2eβ∞T∫A−T∞n0(a) da+CT(Δa+Δt+ϵQ).

The cutoff term and the mesh/time/quadrature term are separately displayed.

Facts & Assumptions

Given: The infinite-age model and mild solution from Linear age-structured renewal transport problem and Characteristic formula and Volterra solution for age renewal transport, with nonnegative initial density and A>T.

[F1]

Linear age-structured renewal transport problem: a cohort of age a follows a(s)=a+s and survives for duration s by S(a,s).

[F2]

Characteristic formula and Volterra solution for age renewal transport: the infinite-age mild solution is unique and is given by characteristics and its birth-rate Volterra equation.

[F3]

Characteristic solution and mass bound for the finite-age renewal model: the absorbing finite-age solution is unique and has mass at most eβ∞t times its initial mass.

[F4]

Positive age-renewal evolution and L1 data stability: infinite-age solutions with the same rates differ in L1 by at most eβ∞t times the difference in initial data.

[F5]

First-order finite-volume convergence for compatible smooth renewal data: for the exact finite-age solution, the numerical cell-mass error is bounded separately by CT(Δa+Δt+ϵQ).

Proof

1.1givenF1algebra

For an initial cohort at age a, the survival factor is the exponential of minus the integrated mortality along a(s)=a+s by [F1]. Since μ∗≤μ(r)≤∥μ∥∞ almost everywhere, integrating these inequalities over [a,a+s] and exponentiating proves the two survival bounds.

2.1givenstep 1.1algebra

A newborn has age at most t≤T<A by time t, so it cannot contribute to outflow at A. An initial individual of age a exits by time t exactly when A−t≤a≤A, and its survival to the exit age is S(a,A−a). Integrating over those initial ages gives the formula for OA(t); using S≤1 and A−t≥A−T gives the stated upper bound.

3.1givenF2F3F4step 2.1algebra

Split n0=n01[0,A−T]+n01(A−T,∞). By linearity of the Volterra equations in [F2] and [F3], it suffices to compare the evolutions of these two parts. Every initial cohort in the first part remains below age A through time T, and its newborn descendants have age at most T<A; the characteristic formulas therefore make the infinite-age and finite-age evolutions identical for this part. The infinite-age evolution of the second part has mass at most eβ∞T∫A−T∞n0 by [F4]. The finite-age model starts this part only with its restriction to [0,A], whose mass is no larger than the same tail; its mass is at most that same exponential factor by [F3]. The triangle inequality gives the factor-two bound in the Statement.

4.1F5step 3.1algebra∎

Under the smooth hypotheses of [F5], the piecewise-constant reconstruction differs in L1 from the finite-age solution by at most CT(Δa+Δt+ϵQ) at each time level. Add this to the infinite-versus-cutoff error from step 3.1 by the triangle inequality. This is the displayed fully discrete bound, with the mesh/time/quadrature and age-tail contributions separate.

5 · Examples, counterexamples and false statements

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