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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.

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