Measure solutions to the conservative renewal equation

We prove the existence and uniqueness of measure solutions to the conservative renewal equation and analyze their long time behavior. The solutions are built by using a duality approach. This construction is well suited to apply the Doeblin's argument which ensures the exponential relaxation of the solutions to the equilibrium.


Introduction
We are interested in the conservative renewal equation It is a standard model of population dynamics, sometimes referred to as the McKendrick-von Foerster model. The population is structured by an age variable a 0 which grows at the same speed as time and is reset to zero according to the rate β(a). It is used for instance as a model of cell division, the age of the cells being the time elapsed since the mitosis of their mother. Suppose we follow one cell in a cell line over time and whenever a division occurs we continue to follow only one of the two daughter cells. Equation (1) prescribes the time evolution of the probability distribution n(t, a) of the cell to be at age a at time t, starting with an initial probability distribution n in . Integrating (formally) the equation with respect to age we get the conservation property d dt ∞ 0 n(t, a) da = 0 which ensures that, if n in is a probability distribution (i.e. ∞ 0 n in = 1), then n(t, ·) is a probability distribution for any time t 0. It is also worth noticing that Equation (1) admits stationary solutions which are explicitly given by N (a) = N (0) e − a 0 β(u)du . The problem of asymptotic behavior for Equation (1) consists in investigating the convergence of any solution to a stationary one when time goes to infinity.
Age-structured models have been extendedly studied (existence of solutions and asymptotic behavior) for a long time by many authors in a L 1 setting (see for instance among many others [10,19,12,16,21]). More recently measure solutions to structured population models started to draw attention [9,1,2,4,7,3].
The goal of the mini-course is to define measure solutions to Equation (1), prove existence and uniqueness of such solutions, and demonstrate their exponential convergence to the equilibrium. Considering measure solutions instead of L 1 solutions (i.e. probability density functions) presents the crucial advantage to authorize Dirac masses as initial data. This is very important for the biological problem since it corresponds to the case when the age of the cell at the initial time is known with accuracy.
1 Some recalls about measure theory We first recall some classical results about measure theory and more particularly about its functional point of view. For more details and for proofs of the results we refer to [17].
We endow R + = [0, +∞) with its standard topology and the associated Borel σ-algebra. We denote by M(R + ) the set of finite signed Borel measures on R + .
Theorem (Jordan decomposition). Any µ ∈ M(R + ) admits a unique decomposition of the form µ = µ + −µ − where µ + and µ − are finite positive Borel measures which are mutually singular. The positive measure |µ| = µ + + µ − is called the total variation measure of the measure µ. We call total variation of µ the (finite) quantity We denote by C b (R + ) the vector space of bounded continuous functions on R + . Endowed with the norm f ∞ := sup x 0 |f (x)| it is a Banach space. We also consider the closed subspace C 0 (R + ) of continuous functions which tend to zero at infinity. To any µ ∈ M(R + ) we can associate a continuous linear form The continuity is ensured by the inequality The following theorem ensures that the application µ → T µ is an isometry from M( Theorem (Riesz representation). For any T ∈ C 0 (R + ) ′ there exists a unique µ ∈ M(R + ) such that T = T µ . Additionally we have This theorem ensures that (M(R + ), · T V ) is a Banach space. It also ensures the existence of an isometric inclusion C 0 (R + ) ′ ⊂ C b (R + ) ′ . Notice that this inclusion is strict: there exist nontrivial continuous linear forms on C b (R + ) which are trivial on C 0 (R + ). Such a continuous linear form can be built for instance by using the Hahn-Banach theorem to extend the application which associates, to continuous functions which have a finite limit at infinity, the value of this limit.
More precisely the mapping Ψ : C b (R + ) ′ → C 0 (R + ) ′ defined by Ψ(T ) = T is surjective (due to the Riesz representation theorem). So C 0 (R + ) ′ is isomorphic to Cb(R+) ′ /Ker Ψ, i.e. for all T ∈ C b (R + ) ′ there exists a unique decomposition T = µ + L with µ ∈ C 0 (R + ) ′ = M(R + ) and L ∈ Ker Ψ. Additionally T ∈ C 0 (R + ) ′ if and only if for Abusing notations, we will now denote for f ∈ C b (R + ) and µ ∈ M(R + ) We end by recalling two notions of convergence in M(R + ) which are weaker than the convergence in norm.

Definition of a measure solution
Before giving the definition of a measure solution to Equation (1), we need to state the assumptions on the division rate β. We assume that β is a continuous function on R + which satisfies ∃ a * , β min , β max > 0, ∀a 0, β min 1 [a * ,∞) (a) β(a) β max .
We explain now how we extend the classical sense of Equation (1) to measures. Assume that n(t, a) ∈ C 1 (R + ×R + )∩C(R + ; L 1 (R + )) satisfies (1) in the classical sense, and let f ∈ C 1 c (R + ) the space of continuously differentiable functions with compact support on R + . Then we have after integration of (1) multiplied by f This motivates the definition of a measure solution to Equation (1). From now on we will denote by A the operator defined on . i.e.
Proposition 2. If t → µ t is a solution to Equation (1) in the sense of the definition above, then for all Reciprocally any solution to (4) satisfies (3).
Proof. We start by checking that (3) is also satisfied for , t > 0 and ρ ∈ C 1 (R) a nonincreasing function which satisfies ρ(a) = 1 for a 0 and ρ(a) = 0 for a 1. For all n ∈ N and a 0 (3) and it remains to check that we can pass to the limit n → ∞. By monotone convergence we have We give now another equivalent notion of weak solutions to Equation (1), which will be useful to prove uniqueness. Compared to (3), it uses test functions which depend on both variables t and a.
This is also true by replacing ϕ ∈ C 1 with compact support in time. Proof. Assume that (µ t ) t 0 satisfies Definition 1 and let ϕ ∈ C 1 b (R + × R + ) compactly supported in time. As we have seen in the proof of Proposition 2, we can use ∂ t ϕ(t, ·) ∈ C 1 b (R + ) as a test function in (3). After integration in time we get Reciprocally let T > 0, f ∈ C 1 c (R + ), and assume that t → µ t is narrowly continuous and satisfies (5). We use the function ρ defined in the proof of Proposition 2 to define for n ∈ N and t 0, ρ n (t) = ρ n(t − T ) . It is a decreasing sequence of decreasing C 1 c (R + ) functions which converges pointwise to 1l [0,T ] (t), and ρ ′ n (seen as an element of M(R + )) converges narrowly to −δ T . Using ϕ n (t, a) = ρ n (t)f (a) as a test function in (5) we get

The dual renewal equation
To build a measure solution to Equation (1) we use a duality approach. The idea is to start with the dual problem: find a solution to the dual renewal equation As for the direct problem, we first give a definition of weak solutions for (6) by using the method of characteristics. Assume that f ∈ C 1 (R + × R + ) satisfies (6) in the classical sense. Then easy computations show that for all a 0 the function ψ(t) = f (t, a − t) is solution to the ordinary differential equation ψ ′ (t) + β(a − t)ψ(t) = 0. After integration we get that f satisfies and the change of variable a ← a + t leads to the following definition.

Definition 4.
We say that f ∈ C b (R + × R + ) is a solution to (6) when, for all t, a 0, Formulations (7) and (8) are the same up to changes of variables, but both will be useful in the sequel.
Proof. To prove the existence and uniqueness of a solution we use the Banach fixed point theorem. For , and this solution satisfies f ∞ f 0 ∞ . Since Γ preserves non-negativity if f 0 0, we get similarly that f is nonnegative when f 0 is nonnegative.
More precisely we do it in the closed and Γ-invariant and Finally we get .
We conclude that if f 0 ∈ C 1 b (R + ) then the unique solution belongs to and let f and g the solutions to Equation (6) with initial distributions f 0 and g 0 respectively. Then for all Additionally it is a positive and conservative contraction, i.e for all t 0 we have Finally it satisfies, for all f ∈ C 1 b (R + ) and all t > 0, Proof. Let f ∈ C b (R + ), fix s 0, and define g(t, a) = M t (M s f )(a) and h(t, a) = M t+s f (a). We have g(0, a) = h(0, a) and By uniqueness of the fixed point we deduce that g(t, a) = h(t, a) for all t 0.
The conservativeness M t 1 = 1 is straightforward computations and the positivity and the contraction property follow immediately from Theorem 5.
, and the proof is complete.
. However the restriction of the semigroup to the invariant subspace of bounded and uniformly continuous functions is strongly continuous.

Existence and uniqueness of a measure solution
We are now ready to prove the existence and uniqueness of a measure solution to Equation (1). We define the dual semigroup (M t ) t 0 on M(R + ) = C 0 (R + ) ′ by In other words we have by definition The following lemma ensures that this identity is also satsified for f ∈ C b (R + ). From now on we will denote without ambiguity the quantity (µM t )f = µ(M t f ) by µM t f.
Proof. The identity is true by definition for any f ∈ C 0 (R + ). Let f ∈ C b (R + ) and f n as defined in (2).
The sequence (f n ) n∈N lies in C 0 (R + ) so (µM t )f n = µ(M t f n ) for all n ∈ N. By monotone convergence we clearly have (µM t )f n → (µM t )f. We will prove that µ(M t f n ) → µ(M t f ) by dominated convergence.
Proof. It is a consequence of Proposition 7. If µ 0, then for any f 0 we have Remark. The left semigroup is not strongly continuous, i.e. we do not have lim t→0 µM t − µ T V = 0 for all µ ∈ M(R + ). This is due to the non continuity of the transport semigroup for the total variation distance: for instance for a ∈ R we have δ a+t − δ a T V = 2 for any t > 0. But the left semigroup is weak* continuous. This is an immediate consequence of the strong continuity of the right semigroup on the space of bounded and uniformly continuous functions (which contains C 0 (R + )). The left semigroup is even narrowly continuous as we will see in the proof of the next theorem.
Theorem 10. For any µ in ∈ M(R + ), the orbit map t → µ in M t is the unique measure solution to Equation (1).
Proof. Existence. In this part we use Proposition 7 at different places. Let µ ∈ M(R + ). We start by checking that t → µM t is narrowly continuous. Let f ∈ C b (R + ). Due to the semigroup property, it is sufficient to check that lim t→0 µM t f = µf. But from (7) we have Now we check that (µM t ) t 0 satisfies (3). Let f ∈ C 1 c (R + ). Using the identity ∂ t M t f = M t Af and the Fubini's theorem we have Uniqueness. Because of Proposition (3) we can prove uniqueness on formulation (5). By linearity we can assume that µ in = 0 and we want to prove that the unique family (µ t ) t 0 which satisfies with compact support in time, is the trivial family. If we can prove that for all ψ ∈ C 1 c (R + × R + ) there exists ϕ ∈ C 1 b (R + × R + ) compactly supported in time such that for all t, a 0 ∂ t ϕ(t, a) + ∂ a ϕ(t, a) − β(a)ϕ(t, a) + β(a)ϕ(t, 0) = ψ(t, a), then we get the conclusion. Let ψ ∈ C 1 c (R + × R + ) and let T > 0 such that supp ψ ⊂ [0, T ) × R + . Using the same method as for (6), we can prove the existence of a solution ϕ ∈ C 1 b ([0, T ] × R + ) to (10) with terminal condition ϕ(T, a) = 0. Since ψ ∈ C 1 c ([0, T ) × R + ) we easily check that the extension of ϕ(t, a) by 0 for t > T belongs to C 1 b (R + × R + ), is compactly supported in time, and satisfies (10).

Exponential convergence to the invariant measure
As noticed first by Sharpe and Lotka in [18] the asymptotic behavior of the renewal equation consists in a convergence to a stationary distribution. This property has then been proved by many authors using various methods for L 1 solutions [5,6,11,8,14,15,20]. The generalized entropy method has even been extended to measure solutions in [9]. Here we use a different approach which is based on a coupling argument. More precisely we prove a so-called Doeblin's condition (see [13] for instance) which guarantees exponential convergence of any measure solution to the stationnary distribution. Denote by P(R + ) the set of probability measures, namely the positive measures with mass 1.
Then for α : t 0 > 0 we have for all µ 1 , µ 2 ∈ M(R + ) such that µ 1 (R + ) = µ 2 (R + ) and all t 0 thus satifies µ + , µ − ∈ P(R + ) and the Doeblin's condition ensures that Using the conservativeness of M t 0 we deduce that which gives Now for t 0 we define n = t t 0 and we get by induction This ends the proof since Proposition 12. The renewal semigroup (M t ) t 0 satisfies the Doeblin's condition with t 0 = a * + η, c = η β min e −βmax(a * +η) , and ν the uniform probability measure on [0, η], for any choice of η > 0.
To do so we iterate once the Duhamel formula (7)  As we have already seen in the introduction, the conservative equation (1) admits a unique invariant probability measure, i.e. there exists a unique probability measure µ ∞ such that µ ∞ M t = µ ∞ for all t 0. This probability measure has a density with respect to the Lebesgue measure Corollary 13. For all µ ∈ M(R + ) and all η > 0 we have where t 0 = a * + η, c = η β min e −βmax(a * +η) , and α = − log(1−c) Notice that in the case a * = 0 we obtain by passing to the limit η → 0