Christoph S Garbe

Andre Butz

Isabelle Dadou

Boris Dewitte

Véronique Garçon

Serena Illig

Aurélien Paulmier

Joël Sudre

Hussein Yahia

V Garc ¸on

Climatically-Active Gases in the Eastern Boundary Upwelling and Oxygen Minimum Zone (OMZ) Systems

Keywords: Air-Sea Interactions, Fluxes, Green House Gases, Satellite Retrieval, Partial differential equations

published or not. The documents may come

INTRODUCTION

The EBUS (Eastern Boundary Upwelling Systems) and OMZs (Oxygen Minimum Zone) contribute very significantly to the gas exchange between the ocean and the atmosphere [START_REF] Waldron | Carbon export and sequestration in the southern benguela upwelling system: lower and upper estimates[END_REF][START_REF] Hales | Atmospheric co2 uptake by a coastal upwelling system[END_REF], notably with respect to the greenhouse gases (hereafter GHG). Invasion or outgasing fluxes of radiatively-active gases at the air-sea interface result in coupled or decoupled sink and source configurations [START_REF] Paulmier | The oxygen minimum zone (omz) off chile as intense source of co 2 and n 2 o[END_REF]. From the in-situ ocean measurements, for example, the uncertainty of the net global ocean-atmosphere CO 2 fluxes is between 20 and 30%, and could be much higher in the EBUS-OMZ. Off Peru, very few in-situ data are available presently, which justifies alternative approaches for assessing the fluxes, and notably satellite, earth-observation data, and the outputs of modeling systems as well. Here we will focus on the Humboldt system off Peru (see Figure 1).

GHG vertical column densities (VCD) can be extracted from satellite spectrometers. The accuracy of these VCDs need to be highly accurate in order to make extraction of sources feasible. To this accuracy is extremely challenging, This project is funded by ESA through the STSE OceanFlux initiative (http://www.oceanflux-upwelling.eu). Fig. 1. Surface oceanic circulation together with Sea Surface Temperature, annual mean. Following [START_REF] Montes | On the pathways of the equatorial subsurface currents in the eastern equatorial Pacific and their contributions to the Peru-Chile Undercurrent[END_REF].

particularly above water bodies, as water strongly absorbs infra red (IR) radiation. To increase the amount of reflected light, specular reflections (sun glint) can be used on some instruments such as GOSAT [START_REF] Butz | Toward accurate CO 2 and CH 4 observations from GOSAT[END_REF]. Also, denoising techniques from image processing may be used for improving the signalto-noise ratio (SNR).

GHG air-sea fluxes determination can be inferred from inverse modeling applied to VCDs, using state of the art modeling, at low spatial resolution [START_REF] Garbe | Optimal Control in Image Processing[END_REF][START_REF] Garbe | Modeling of atmospheric transport of chemical species in the polar regions[END_REF]. Due to the long life-time of GHGs in the atmosphere, significant contributions of the VCDs may have been transported from far afield. in correctly choosing the boundary conditions of the inverse modeling approach, these effects can be separated. Only through this sep-6150 978-1-4673-1159-5/12/$31.00 ©2012 IEEE IGARSS 2012

aration can the sources be determined accurately. These approaches will be tested on simulated data, which will make an assessment of the accuracy, limitations and requirements of the satellite instrument feasible.

For accurately linking sources of GHGs to EBUS and OMZs, the resolution of the source regions needs to be increased. This task develops on new non-linear and multiscale processing methods for complex signals [START_REF] Turiel | Microcanonical multifractal formalism -a geometrical approach to multifractal systems: Part i. singularity analysis[END_REF] to infer a higher spatial resolution mapping of the fluxes and the associated sinks and sources between the atmosphere and the ocean. Such an inference takes into account the cascading properties of physical variables across the scales in complex signals. The use of coupled satellite data (e.g. SST and/or Ocean color) that carry turbulence information associated to ocean dynamics is taken into account at unprecedented detail level to incorporate turbulence effects in the evaluation of the air-sea fluxes.

Based on a parameterization, the transfer velocities K can be computed from satellite products. This leads to a measurement of p ocean GHG , which in turn will be validated by the output of a coupled physical biogeochemical model and in-situ measurements.

CONSTRAINING SURFACE FLUXES BY EO DATA

In the atmosphere, the temporal evolution of a tracer c is given by the equation :

∂c ∂t = -u∇c + 1 ρ ∇ (ρT d ∇c) + 1 ρ g + F, (1)
with the concentration c, the wind field u, the density of air ρ, the turbulent diffusivity tensor T d , the chemical transformation rate g and the net flux at the air-sea interface F . Using optimal control and inverse problem modeling, in this contribution we will derive the mapping of F using Earth Observation and satellite data [START_REF] Garbe | Optimal Control in Image Processing[END_REF][START_REF] Garbe | Modeling of atmospheric transport of chemical species in the polar regions[END_REF]. It will estimate sources and sinks of gases, based on vertical column densities (VCD), at the spatial resolution of VCD data. For the retrieval of VCD of GHGs, we employ the approach detailed in [START_REF] Butz | Retrievals of atmospheric CO 2 from simulated spaceborne measurements of backscattered near-infrared sunlight: accounting for aerosol effects[END_REF]. This methods yields very accurate results due to its ability to simultaneously retrieve gas concentrations and particle scattering properties of the atmosphere using an efficient radiative transfer (RT) model [START_REF] Otto | Efficient calculation of intensity and polarization spectra in vertically inhomogeneous scattering and absorbing atmospheres[END_REF].

Following [START_REF] Garbe | Modeling of atmospheric transport of chemical species in the polar regions[END_REF], the underlying problem can be formulated as

∂c ∂t = ∇ • (D ∇c) -u • ∇c + F, c(t 0 , x) = c 0 (x), c(t, x) = c in (t, x) on Γ I , D ∂c ∂ν = 0 on Γ O , D ∂c ν -β c = γ on Γ G .
(

) 2
Since the problem is ill-posed, we regularize it by introducing the Tikhonov functional given by

J(q, c) = c -C(c) Z + α 2 F Q . ( 3
)
The first term corresponds to the data term and includes given observations. Here C(c) corresponds to the VCDs. The second term is the regularization term with a small parameter α ≥ 0 . Thus, we aim at the solution of the following optimal control problem: Minimize J(q, c) (

subject to the state equation [START_REF] Hales | Atmospheric co2 uptake by a coastal upwelling system[END_REF]. The inverse problem is discretized using a dG(0)cG(1) Galerkin discretization. Thus, for the temporal integration we employ the backward Euler method which corresponds to the discontinuous Galerkin method. A detailed description of the discretization scheme can be found in [START_REF] Eriksson | Computational differential equations[END_REF]. For the discretization in space we use piecewise trilinear functions. This corresponds to Q 1 elements. To increase computational efficiency, the problem can be solved with a goal oriented adaptive grid refenement [START_REF] Garbe | Modeling of atmospheric transport of chemical species in the polar regions[END_REF].

The relation between the net flux F and the partial pressure p GHG of a GHG is of the form (for typical GHG such as CO 2 , N 2 O, CH 4 ):

F = αK p air GHG -p ocean GHG , (5)
α being the gas solubility (which depends mostly of SST and salinity) and K is a function of wind, salinity, temperature, sea state, which can all be obtained from satellite data. Since p air GHG can be assumed to be constant over the upwelling region under study, the scalar p ocean GHG can then be obtained from inverse modeling and satellite data at VCD resolution.

The quantity p ocean GHG is a complex signal depending, at any spatial resolution, on sea surface temperature, salinity, chlorophyll concentration, dissolved inorganic carbon, alkalinity and nutrients concentrations. Both the biological pump, with chlorophyll a as a proxy, and the physical pump, driven by the temperature and salinity (e.g. solubility, water mass), govern the evolution of pGHG, when dealing with CO 2 for instance, in the surface ocean. For other GHG such as N 2 O, the partial pressure of N 2 O will also depend on the evolution of the various chemical forms of nitrogen and on the activity of the bacteria involved directly or indirectly in the major N transformation processes such as nitrification, denitrification, anammox, etc. It is therefore a competitive research challenge to use high resolution satellite data (at various spatial resolutions) coupled with low spatial resolution model outputs for p ocean GHG to derive a higher spatial resolution mapping of p ocean GHG . Recent advances in non-linear and multiscale processing of complex and turbulent signals [START_REF] Yahia | Motion analysis in oceanographic satellite images using multiscale methods and the energy cascade[END_REF] allow to make use of cascading properties relevant to turbulence to assess high spatial resolution of scalar or vectorial quantities from low resolution satellite data and inference across the scales. The methodology makes use of the framework of reconstructible systems and optimal wavelets, which shows a particular ability to perform fusion of satellite data acquired at different spatial resolutions [START_REF] Pottier | Inferring missing data in satellite cholorophyll maps using turbulent cascading[END_REF], thus making it particularly appropriate for merging different kinds of satellite products. High resolution maps (resolving submesoscale features) will be provided at different timescales of interest, from intraseasonal to seasonal over the upwelling under study.

MULTISCALE PROPERTIES OF THE SIGNAL USED

As described in the previous section, p ocean GHG is a complex signal depending on temperature, salinity, chorophyll, dissolved inorganic carbon, alkalinity and nutrients concentrations. Such a signal is therefore expected to feature cascading, multiscale and other characteristic properties found in turbulent signals. We will present preliminary results which prove that a simulated p ocean GHG signal possesses the key signatures characteristic of complex multiscale signals, notably through a study of log-histograms and singularity spectrums. These characteristics prove that p ocean GHG signals are prone to be studied with the help of multiscale methods [START_REF] Turiel | Microcanonical multifractal formalism -a geometrical approach to multifractal systems: Part i. singularity analysis[END_REF] to infer information along the scales.

The multiscale properties of a signal are well understood though the study of singularity spectrum and singularity exponents which give access to the determination of geometric subsets responsible of cascading properties [START_REF] Arneodo | Multifractales et Turbulence[END_REF][START_REF] Turiel | Microcanonical multifractal formalism -a geometrical approach to multifractal systems: Part i. singularity analysis[END_REF]. Figure 2 shows a computation of the singularity exponents on a simulated p ocean CO2 signal obtained from the ROMS coupled physical/biogeochemical simulation model [START_REF] Penven | A numerical study of the southern Benguela circulation with an application to fish recruitment[END_REF][START_REF] Penven | Generation of cyclonic eddies by the agulhas current in the lee of the agulhas bank[END_REF][START_REF] Veitch | The benguela: A laboratory for comparative modeling studies[END_REF][START_REF] Vu | Physical and biogeochemical processes maintaining the oxygen minimum zone in the benguela upwelling system using an eddy resolving model[END_REF]. The computation is performed using the FluidExponents software developed at INRIA. The spatial distribution of the singularity exponents underlines the complex distribution of fronts and vortices displayed by the p ocean CO2 signal, which possesses turbulent characteristics. For comparison, we show in the same image the singularity exponents computed on the SST signal.

CONCLUSION

In this contribution we present a concerted effort to assess the influence Eastern Boundary Upwelling Systems on gas exchange between ocean and atmosphere. We link a coupled physical-biochemical model to parameters extracted from satellite remote sensing. A multiscale approach based on the singularity exponents is used to increase the resolution of our approach. We will present a framework as described above for determining sources and sinks of GHG from satellite remote sensing with the Peru OMZ as a test bed. First results of this analysis are very promising.

Fig. 2 .

 2 Fig. 2. Left : singularity exponents computed on simulated SST signal from the ROMS simulation model. Right: singularity exponents computed on simulated p ocean CO2 signal from the ROMS simulation model. Note the similarity of coherent structures and the turbulent character of the p ocean CO2 signal. These images are computed using the ROMS model and FluidExponents software. The p ocean CO2 and SST signals used in the computation of the two images are averaged on a 5 days time interval.


Julien Audiffren

Etienne Pardoux

Muller's ratchet clicks in finite time

. We show that for any parameter λ > 0 ( the rate at which mutations occur), for any α > 0 ( the toxicity of the mutations) and for any size N > 0 of the population, the ratchet clicks a.s. in finite time. That is to say the minimum number of deleterious mutations in the population goes to infinity a.s.

Introduction

In natural evolution, deleterious mutations occur much more frequently than beneficial ones. Since the last category is always favored by selection, one may wonder about the advantage of sexual reproduction over the asexual type. The answer has been proposed : in an asexually reproducing population, each individual always inherits all the deleterious mutations of his ancestor (except if another mutation occurs at the same locus on the genome; but this event is rare and we will not consider it) whereas in sexual reproduction, recombinations occur, which allow an individual to take part of a chromosome from each of his parents, therefore giving him a chance to get rid of deleterious mutations. Muller's ratchet can be used as an attempt to translate this phenomenon in a mathematical model, thus explaining the advantage of sexual reproduction [START_REF] Smith | The evolution of Sex[END_REF]. If one considers the best class (the group of fittest individuals) in a given asexual population, Muller's ratchet is said to click when the best class becomes empty. Since beneficial mutations do not occur in this model, it means that all the individuals of the best class have mutated.

The first model for Muller's ratchet due to Haigh [START_REF] Haigh | The accumulation of deleterious genes in a population -Muller's ratchet[END_REF] can be described as follows. Consider an asexual population of fixed sized N which evolves in discrete time, with a multiplicative selection model. Only deleterious mutations occur. Denoting by 0 ≤ α ≤ 1 the deleterious strength of the mutations, and λ > 0 the rate at which 1 1 INTRODUCTION 2 they occur, every generation is constituted as follows : each individual chooses a parent from the previous generation, in such a way that the probability of choosing a specific father with k deleterious mutations is (we denote by N k the number of such individuals in the previous generation) : (

1 -α) k ∞ k=0 N k (1 -α) k .
Next each newborn gains ξ deleterious mutations, where ξ is a Poisson random variable with parameter λ. It is immediate to see that this model clicks a.s. in finite time. Indeed at each generation, with probability (1 -exp(-λ)) N all the individuals mutate, which induces the click.

There are three parameters in our model : N is the size of the population, λ is the mutation rate, α is the fitness decrease due to each mutation.

The Fleming-Viot model for Muller's ratchet proposed by A. Etheridge, P. Pfaffelhuber and A. Wakolbinger in [START_REF] Etheridge | How often does the ratchet click ? facts, heuristics, asymptotics[END_REF] consists of the following infinite set of SDEs for the X k (t)'s, k ≥ 0, where X k (t) denotes the proportion of individuals in the population who carry exactly k deleterious mutations at time t (with X -1 ≡ 0) :

(1.1)

       dX k (t) = [α(M 1 (t)-k)X k (t) + λ(X k-1 (t)-X k (t))]dt+ ≥0, =k X k (t)X (t) N dB k, (t) X k (0) = x k ; k ≥ 0,
where {B k, , k > ≥ 0} are independent Brownian motions, B k, = -B ,k ; and M 1 (t) = k≥0 kX k (t). The first term in the drift models the selective effect of the deleterious mutations. Those individuals who carry less (resp. more) mutations than the average number of mutations in the population have a selective advantage (resp. disadvantage). The second term in the drift reflects the effect of the accumulation of mutations : at rate λ, individuals carrying k -1 mutations gain a k-th mutation, they jump into the kclass, and at the same rate individuals carrying k mutation gain a k + 1-th mutation, they jump out of the k-class. The diffusion term reflects the resampling effect of the birth events, where the factor N -1/2 can be understood as being equivalent to the rescaling of time t → t/N , if N is the "effective population size", which is natural in Kingman's coalescent [START_REF] Kingman | The coalescent, Stoch[END_REF]. For the equivalence between the present model and a more intuitive look-down model à la Donnelly-Kurtz, we refer the reader to [START_REF] Audiffren | Equivalence of a Fleming Viot and look-down models of Muller's ratchet[END_REF].

INTRODUCTION

3

We will show in section 2 that the infinite dimensional system of SDEs (1.1) is well posed provided we choose the initial condition x = (x k , k ≥ 0) ∈ X δ for some δ > 0, where (1.2)

X δ := x ∈ [0, 1] ∞ , ∞ k=0 x k = 1, ∞ k=0 k 2+δ x k < ∞ .
We define T 0 = inf{t > 0, X 0 (t) = 0}. The purpose of the present work is to show that this model of Muller's ratchet is bound to click in finite time, that is to say T 0 < ∞ a.s. We are going to prove the following theorem :

Theorem 1 For any δ > 0, for any choice of initial condition in X δ , let (X k (t)) k∈Z + be the solution of (1.1). Then P(T 0 < ∞) = 1.

We will in fact prove a stronger result, namely Theorem 2 For any δ > 0, for any choice of initial condition in X δ , let (X k (t)) k∈Z + be the solution of (1.1). Then there exists ρ > 0, which depends upon the parameters N , α and λ, such that E [exp(ρT 0 )] < ∞, for all 0 < ρ < ρ.

Clearly, a model for Muller's ratchet must have the property that the ratchet clicks in finite time. In a sense our result says that the Etheridge-Pfaffelhuber-Wakolbinger model for Muller's ratchet is a reasonable model, in the sense that it exhibits a.s. clicking, as the computer simulations had already shown, see [START_REF] Etheridge | How often does the ratchet click ? facts, heuristics, asymptotics[END_REF]. Note that once the zero class is empty, the 1-class takes its place, and some time later a second click happens, at which time both the zero class and the 1-class become empty, and so on. Of course, we would like to know more about the time it takes for the ratchet to click. Here we show that its expectation is finite. We hope to get more precise information in some future work.

There are several difficulties in this model. First, it is an infinite system of SDEs which cannot be reduced to a finite dimensional system. Only X 0 and M 1 enter the coefficients of the equation for X 0 , but the equation for

M 1 (t) brings in the second centered moment M 2 (t) = ∞ k=0 (k -M 1 (t)) 2 X k (t).
The system of SDEs for the centered moments of all orders is infinite as well, the moments of order up to = 2k enter the coefficients of the equation for k-th centered moment, and there is no known solution to it (except in the deterministic case N = +∞, which is solved in [START_REF] Etheridge | How often does the ratchet click ? facts, heuristics, asymptotics[END_REF]). In addition, one has d X 0 , M 1 t = -M 1 (t)X 0 (t) N dt. There is no simple relation between X 0 and M 1 , except that X 0 + M 1 ≥ 1, and (X 0 = 1) ⇒ (M 1 = 0). But we could have X 0 → 0 and M 1 → ∞. Last but not least, the diffusion coefficient in dX k is not a Lipschitz function of X k at 0 and 1, and it vanishes at those two points.

In order to prove the theorem, we will use a three-step proof. First, in section 3 we will show that M 1 cannot grow too fast with a good probability, and we will deduce that for a specific set of initial conditions, the ratchet does click with a strictly positive probability p f in , in a given interval of time.

Next, we show in section 4 that the product X 0 M 2 1 is bound to come back under

2(λ+1) α
after any time, and we use all the previous results to deduce that M 1 is also bound to return under β = λ α after any time, as long as the ratchet does not click. Finally in section 5 we prove that each time M 1 gets below β, the ratchet clicks with a positive probability in a prescribed interval of time. We then conclude with the help of the strong Markov property.

In section 6 we show how the proof of Theorem 1 can be modified into a proof of Theorem 2. The reader may wonder why we do not prove Theorem 2 from the very beginning, and first prove a weaker result. The reason is that the difference between the two proofs is essentially that while proving Theorem 1, we prove that as long as the ratchet has not clicked, M 1 is bound to return below the value β, i.e. the drift of X 0 is bound to become non-positive, which is an interesting result in itself, while the proof of Theorem 2 is based on the same strategy, but with β replaced by a much less explicit quantity.

We shall essentially work with the two dimensional process {X 0 (t), M 1 (t)}, and we shall use the equation for X 1 only in one place, namely in Lemma 5.1 in order to show that X 0 does not get stuck near the value 1. This means that we shall make use only of the three following equations.

(1.3)

                 dX 0 (t) = (αM 1 (t) -λ) X 0 (t)dt + X 0 (t)(1 -X 0 (t)) N dB 0 (t) dX 1 (t) = (α(M 1 (t)-1)X 1 (t)+λ(X 0 (t)-X 1 (t)))dt+ X 1 (t)(1 -X 1 (t)) N dB 1 (t) dM 1 (t) = (λ -αM 2 (t))dt + M 2 (t) N dB(t).
The three Brownian motions B 0 , B 1 and B are standard Brownian motions. But they are not independent, and the three dimensional process (B 0 (t), B 1 (t), B(t)) is not a Gaussian process. But this will play no role in our analysis. This system is not closed, since M 2 enters the coefficients of the last equation. However, the crucial remark is that it will not be necessary to estimate M 2 , in order to estimate M 1 . This is due to the fact that the M 1 -equation takes the form dM 1 (t) = λdt + dZ t , where

Z t = W (A t ) -αN A t , if A t := N -1 t 0 M 2 (s)
ds and {W (t), t ≥ 0} is a standard Brownian motion. The larger M 2 is, the more likely Z t is negative, which produces a smaller M 1 . This means that we should be able to estimate M 1 , without having to estimate M 2 , which is done below in Lemma 3.2 and 4.3. In particular, we show in Lemma 4.4 below that, as long as the ratchet has not clicked, M 1 is bound to return below the level β = λ/α after any time. We believe that this is an interesting qualitative property of the model. Note that Theorem 2 is proved by essentially the same argument as Theorem 1, but with that level β replaced by 2β ∨ (ε/δ), where the constants ε and δ, which are defined in the proof of Theorem 1, have no explicit relation to the constants of the model.

Preliminary results

The aim of this section is to establish a weak existence and uniqueness result for the infinite system of SDEs (1.1), under the condition that the initial condition {X k (0), k ≥ 0} belongs to the set X δ for some δ > 0 (see 1.2 for the definition of this set).

We equip this set with the topology under which a probability x n = (x n k , k ≥ 0) on Z + converges to x = (x k , k ≥ 0) if both it converges weakly, and sup n k≥0 k 2+δ x n k < ∞. More precisely, we will prove in this section Theorem 3 If the initial condition x belongs to X δ , for some δ > 0, then (1.1) has a unique weak solution X(t) = {X k (t), k ≥ 0} which is a. s. continuous with values in X δ .

Remark 2.1 Previous results on this system of SDEs assume that the probability x on Z + possesses exponential moments of arbitrary order, see [START_REF] Cuthbertson | Limits to the rate of adaptation[END_REF], or of some order, see [START_REF] Peter Pfaffelhuber | Muller's ratchet with compensatory mutations[END_REF]. This assumption is naturally requested if one wants to be able to write equations for arbitrary moments of the random measure X(t) on Z + . However, we will need only to make sure that M 1 (t) and M 2 (t) have finite expectation, and for that purpose our weaker condition will be sufficient.

We start with the case α = 0.

2.1

The case α = 0 Proposition 2.2 Suppose that α = 0. Then, for any initial condition x ∈ X δ , (1.1) has a unique weak solution X(t) = {X k (t), k ≥ 0} which is a. s. continuous with values in X δ , and is such that for each λ, δ > 0, there exists a locally bounded function C λ,δ (t) such that

(2.1) E ∞ k=0 k 2+δ X k (t) ≤ C λ,δ (t).
Proof : Let us rewrite our system of SDEs in the particular case α = 0 (again it is written with the convention that X -1 (t) ≡ 0) in the form

(2.2)          dX k (t) = λ(X k-1 (t) -X k (t))dt + dM k (t), k ≥ 0; M k , M t = N -1 t 0 X k (s)(δ k, -X (s))ds, k, ≥ 0; X k (0) = x k , k ≥ 0;
where the M k (t)'s are continuous martingales, and M k , M stands for the joint quadratic variation of the two martingales M k and M . We can apply the result of Theorem 2.1 in [START_REF] Shiga | A certain class of infinite dimensional diffusion processes arising in population genetics[END_REF], which ensures that (2.2) has a unique weak solution. The facts that X k (t) ≥ 0, for all k ≥ 0, t ≥ 0, a.s. and k≥0 X k (t) = 1 for all t ≥ 0 a.s. follow from the results in [START_REF] Shiga | A certain class of infinite dimensional diffusion processes arising in population genetics[END_REF], but we give a proof of the second fact for the convenience of the reader. Summing the k-th equation in the system (2.2)

from k = 0 to k = K yields K k=0 X k (t) = K k=0 x k -λ t 0 X K (s)ds + M 0,K (t),
where M 0,K (t) is a continuous martingale whose quadratic variation is given by

M 0,K t = N -1 t 0 K k=0 X k (s) 1 - K k=0 X k (s) ds.
We first deduce that

E K k=0 X k (t) ≤ K k=0 x k , E ∞ k=0 X k (t) ≤ ∞ k=0 x k ,
where the second inequality follows from the first by monotone convergence. It then follows that t 0 X K (s)ds → 0 as K → ∞, while by dominated convergence

t 0 K k=0 X k (s) 1 - K k=0 X k (s) ds → t 0 ∞ k=0 X k (s) 1 - ∞ k=0 X k (s) ds a.s. as K → ∞. It follows from these considerations that the process Y (t) := ∞ k=0 X k (t) solves the SDE Y (t) = 1 + N -1/2 t 0 Y (s)[1 -Y (s)]d Bs ,
for some standard Brownian motion B. This clearly shows that Y ≡ 1 a.s. We next have

E K k=0 kX k (t) = K k=0 kx k + λE t 0 K k=0 (kX k-1 (s) -kX k (s)) ds, E ∞ k=0 kX k (t) ≤ ∞ k=0 kx k + λt, since K-1 j=0 X j (s) ≤ 1.
Furthermore, using this last inequality in the last step below,

E K k=0 k 2 X k (t) = K k=0 k 2 x k + λE t 0 K k=0 k 2 X k-1 (s) -k 2 X k (s) ds ≤ K k=0 k 2 x k + λE t 0 K-1 j=0 (2j + 1)X j (s)ds, E ∞ k=0 k 2 X k (t) ≤ ∞ k=0 k 2 x k + λt + λ 2 t 2 + 2λt ∞ k=0 kx k .
Let us now suppose that 0 < δ ≤ 1, and we exploit the fact that 2 + δk 1+δ ≤ 3k 2 .

We then deduce that

E K k=0 k 2+δ X k (t) = K k=0 k 2+δ x k + λE t 0 K k=1 k 2+δ X k-1 (s) -k 2+δ X k (s) ds ≤ K k=0 k 2+δ x k + 3λE t 0 k+1 k=1 j 2 X k (s)ds E ∞ k=0 k 2+δ X k (t) ≤ C 2 (λ, t)
from the last estimate. If δ > 1, we need to estimate the third moment in terms of the second, then the fourth in terms of the third, ..., and finally the 2 + δ-th in terms of the 2 + δ -th. So far we have proved that X(t) ∈ X δ a. s. for all t ≥ 0. We now prove that in fact a. s., X(t) ∈ X δ for all t ≥ 0. Our next argument will be very similar to an argument in [START_REF] Peter Pfaffelhuber | Muller's ratchet with compensatory mutations[END_REF]. For any m ≥ 1, t ≥ 0, let

N m,δ (t) := ∞ k=0 inf(k, m) 2+δ X k (t).
It is easy to check that {N m,δ (t), t ≥ 0} is a positive submartingale, to which we can apply Doob's inequality, which, together with the monotone convergence theorem, yields that for any K, T > 0,

P sup 0≤t≤T ∞ k=0 k 2+δ X k (t) > K = lim m→∞ P sup 0≤t≤T N m,δ (t) > K ≤ lim m→∞ K -1 E [N m,δ (T )] = K -1 E ∞ k=0 k 2+δ X k (T ) ≤ K -1 C λ,δ (T ),
where we have used (2.1) for the last inequality. It now follows that for all T > 0,

P sup 0≤t≤T ∞ k=0 k 2+δ X k (t) < ∞ = 1.
The a. s. continuity with values in X δ is now easy to check. ♦

We next want to establish the equation for the first moment M 1 (t) := k≥1 kX k (t). This equation will involve the process

M 2 (t) = k≥1 k 2 X k (t) -[M 1 (t)] 2 .
We know by now that those quantities are well defined and finite.

Proposition 2.3 The first moment solves the SDE

dM 1 (t) = λdt + dM(t),
where {M(t), t ≥ 0} is a continuous martingale satisfying

M, M t = N -1 t 0 M 2 (s)ds,
and for any k ≥ 0,

(2.3) M, M k t = N -1 t 0 (k -M 1 (s))X k (s)ds. Proof : For any K > 1, let M 1,K (t) := K k=1 kX k (t).
We have readily

M 1,K (t) = M 1,K (0) + λ t 0 K-1 j=0 X j (s)ds -λ t 0 KX K (s)ds + M 1,K (t),
where M 1,K (t) is a continuous martingale, with

d M 1,K t = N -1 K k=1 k 2 X k (t) -[M 1,K (t)] 2 dt.
It follows from (2.1) that

E t 0 KX K (s)ds → 0 as K → ∞.
Consequently, all terms in the above equation converge as

K → ∞, yielding that M 1 (t) = M 1 (0) + λt + M(t),
where M(t) is a continuous martingale as follows from the next lemma, which is such that

d M, M t = N -1 M 2 (t)dt. Moreover, if 1 ≤ k ≤ K, d M 1,K , M k t = N -1 [kX k (t)(1 -X k (t)) - =k, ≤K X k (t)X (t)]dt = N -1 X k (t)[k - ≤K X (t)]dt.
The second part of the result follows, by letting K → ∞. ♦

To complete this last proof, we need to establish

Lemma 2.4 The collection of processes {M 1,K (t), t ≥ 0} K≥1 is tight in C([0, +∞)).
Proof : From the Corollary of Theorem 7.4 page 83 in [2], Chebychef's and Doob's inequalities, it suffices to prove that for each T > 0 there exists a constant C(δ, T ) such that for all 0 ≤ s < t ≤ T ,

(2.4)

E |M 1,K (t) -M 1,K (s)| 2+δ ≤ C(δ, T )|t -s| 1+δ/2 .
From the well-known Davis-Burkholder-Gundy inequality (see e.g. p. 160 in [START_REF] Revuz | Continuous Martingales and Brownian Motion[END_REF]), there exists a constant c(δ) such that

E |M 1,K (t) -M 1,K (s)| 2+δ ≤ c(δ)E ( M 1,K t -M 1,K s ) 1+δ/2 .
We have, using Jensen's inequality in two distinct instances,

E   t s k≥0 k 2 X k (r)dr 1+δ/2   ≤ (t -s) δ/2 E t s k≥0 k 2 X k (r) 1+δ/2 dr ≤ (t -s) δ/2 E t s k≥0 k 2+δ X k (r)dr.
(2.4) follows by combining the two last estimates with (2.1). ♦

The general case

We can now prove Theorem 3. We first proceed with the Proof of existence We now introduce a Girsanov transformation. It follows from Proposition 2.3 that there exists a Brownian motion {B(t), t ≥ 0} such that

dM 1 (t) = λdt + M 2 (t) N dB t .
For any α > 0, let

Z α (t) := exp -α √ N t 0 M 2 (s)dB s - α 2 N 2 t 0 M 2 (s)ds .
It is easily seen that

Z α (t) = exp N α M 1 (0) + λt -M 1 (t) - α 2 t 0 M 2 (s)ds ≤ exp (N α [M 1 (0) + λt]) .
It is now clear that {Z α (t), t ≥ 0} is a martingale, and consequently there exists a unique probability measure P α on (Ω, F), such that for all t > 0,

dP α dP Ft = Z α (t).
It now follows from Girsanov's theorem that there exist a P α -standard Brownian motion {B α (t), t ≥ 0} such that

t 0 M 2 (s) N dB(s) = -α t 0 M 2 (s)ds + t 0 M 2 (s) N dB α (s).
Moreover, we deduce from (2.3) and again Girsanov's theorem (see the statement of Theorem VIII.1.4 p. 327 in [START_REF] Revuz | Continuous Martingales and Brownian Motion[END_REF]) that for each k ≥ 0 there exists a Brownian motion

{B α k (t), t ≥ 0} with t 0 X k (s)(1 -X k (s)) N dB k (s) = t 0 α(M 1 (s) -k)X k (s)ds+ t 0 X k (s)(1 -X k (s)) N dB α k (s).
Consequently under P α , we have proved weak existence to our infinite dimensional system (1.1). We can now turn to the Proof of uniqueness We now exploit again Girsanov's theorem to prove weak uniqueness. Consider for some α, δ > 0 > 0 any X δ -valued solution of our SDE, which we rewrite as

X k (t) = x k + t 0 [α(M 1 (s) -k)X k (s) + λ(X k-1 (s) -X k (s))] ds + M k (t), k ≥ 0; M 1 (t) = k≥0 kx k + t 0 [λ -αM 2 (s)] ds + M(t),
where for k, ≥ 0,

M k , M t = N -1 t 0 X k (s)(δ k, -X (s))ds, M k , M t = N -1 t 0 X k (s)((k -M 1 (s))ds, M, M t = N -1 t 0 M 2 (s)ds,
defined on a probability space (Ω, F, P). Let Q α denote the probability law of our solution on the space C([0, +∞); X δ ), and define, for t ≥ 0,

Y α (t) = exp α √ N M t - α 2 N 2 t 0 M 2 (s)ds .
For each n ≥ 1, let

τ n := inf t > 0, t 0 M 2 (s)ds > n .
It is not hard to show that for each n ≥ 1, the probability measure Q defined on C([0, +∞); X δ ) equipped with its Borel σ-field, by

dQ dQ α Fτ n = Y α (τ n )
coincides with the law of the unique weak solution of (2.2) up to time τ n . Hence the restriction of

Q α = (Y α (τ n )) -1
• Q to the σ-algebra F τn coincides with the law of the solution which we have constructed above. Since τ n → ∞ a.s., weak uniqueness is proved.

A comparison theorem for one-dimensional SDEs

We state a result, which will be useful later in this paper. Our processes are defined on a probability space (Ω, F, P), equipped with a filtration (F t , t ≥ 0) which is such that for each k, ≥ 0 {B k, (t), t ≥ 0} is a F t -Brownian motion. We denote by P the corresponding σ-algebra of predictable subsets of R + × Ω.

From the weak existence and uniqueness, we deduce that our system has the strong Markov property, using a very similar proof as in Theorem 6.2.2 from [START_REF] Daniel | Multidimensional Diffusion Processes[END_REF]. Indeed, the proof of that results exploits weak uniqueness of the martingale problem, together with the measurability of the law of the solution, with respect to the starting point. In our case that mapping is easily shown to be continuous.

In the next sections, we will use the following comparison theorem several times. This Lemma can be proved exactly as the comparison Theorem 3.7 from chapter IX of [START_REF] Revuz | Continuous Martingales and Brownian Motion[END_REF].

: Ω × R + × R → R is a P ⊗ B(R) measurable function. Consider the two SDEs (2.5) dY 1 (t) = b 1 (Y 1 (t))dt + σ(Y 1 (t))dB(t), Y 1 (0) = y 1 ; (2.6) dY 2 (t) = b 2 (t, Y 2 (t))dt + σ(Y 2 (t))dB(t), Y 2 (0) = y 2 .
Let Y 1 (resp Y 2 ) be a solution of (2.5) ( resp (2.6)). If y 1 ≤ y 2 (resp y 2 ≤ y 1 ) and outside a measurable subset of Ω of probability zero, ∀t ∈

[0, T ], ∀x ∈ R, b 1 (x) ≤ b 2 (t, x) (resp b 1 (x) ≥ b 2 (t, x)). Then a.s. ∀t ∈ [0, T ] , Y 1 (t) ≤ Y 2 (t) (resp Y 1 (t) ≥ Y 2 (t)).
3 The result for a specific set of initial conditions

From P(E) + P(F ) -P(E ∩ F ) = P(E ∪ F ) ≤ 1,
we deduce the following trivial lemma which will be used several times below :

Lemma 3.1 Let E, F ∈ F. Then P(E ∩ F ) ≥ P(E) + P(F ) -1.
Now first we show that M 1 cannot grow too fast : Lemma 3.2 For all c > 0, t > 0, t > 0,

P sup 0≤r≤t M 1 (t + r) -M 1 (t) ≤ λt + c ≥ 1 -exp(-2αN c) Proof : Define Z t t+s = t+s t M 2 (r) N dB r -α s+t t
M 2 (r)dr. We note that, for any t > 0, {exp(2αN Z t t+u ), u ≥ 0} is both a local martingale and a super-martingale. We also have sup

0≤s≤t M 1 (t + s) -M 1 (t) ≤ sup 0≤s≤t Z t t+s + λt .
But for all c > 0,

P sup 0≤u≤t Z t t+u ≥ c ≤ P sup 0≤u≤t exp 2αN Z t t+u ≥ exp (2αN c) ≤ exp (-2αN c)
where we have taken advantage of the fact that exp 2αN Z t t+u is a local martingale and of Doob's inequality. Then

P sup 0≤r≤t M 1 (t + r) -M 1 (t) ≤ λt + c ≥ 1 -exp (-2αN c) .
♦ Note that we have in fact P sup u≥0 Z t t+u ≥ c ≤ exp (-2αN c). We choose an arbitrary value m > 0 for M 1 (0), which will remain the same throughout this document (for example one could choose m = 1), and we define

(3.1) ε = 1 10N α , t 3 = εN 3λ = 1 30λα , m max = m + λA(t 3 ) + ε 6 ,
where

A(t) = 1 4N t 0 (1 -X 0 (s))ds, (3.2) p 2 = exp(-αN ε 6 ) = exp(- 1 60
),

(3.3) µ = ε 6m max ∧ ε 4 ∧ 1 10 ,
and let δ be a real number, which will be specified below, such that δ ≤ 1 10 ∧ ε m . Now let Y 0 be the solution of the following SDE :

(3.4) dY 0 (t) = dt + 2 Y 0 (t)dW (t) Y 0 (0) = δ
with W a standard Brownian motion.

We will show that starting with X 0 (0) = x 0 ≤ δ , M 1 (0) = m 1 ≤ m, and as long as X 0 M 1 < 2ε and X 0 remains small enough, we can compare X 0 (t) with the solution of (3.4). Lemma 3.3 For any δ > 0, ε given as in (3.1), µ as in (3.3), let

T min = inf{t > 0, X 0 (t)M 1 (t) > 2ε or X 0 (t) > δ + µ}. Then provided that X 0 (0) = x 0 ≤ δ, if A(t) := 1 4 t 0 1-X 0 (s) N
ds, there exists a standard Brownian motion W such that the corresponding solution Y 0 of (3.4) satisfies

X 0 (t) ≤ Y 0 (A(t)), ∀t ∈ [0, T min ] . Proof : We first note that for 0 ≤ t ≤ T min , t 5N ≤ A(t) ≤ t 4N because 4 5 ≤ 1 - X 0 (t) ≤ 1 (
thanks to the choices of µ and δ, and 1-X

0 (t) ≥ 1-δ-µ ≥ 1-1 10 -1 10 ≥ 4 5 ). Define σ(t) = inf{u > 0, A(u) ≥ t} and X0 (t) = X 0 (σ(t)) ( resp M1 (t) = M 1 (σ(t))).
Then there exists a standard Brownian motion W t such that

d X0 (t) = (α M1 (t) -λ) X0 (t) 4N 1 -X0 (t) dt + 2 X0 (t)dW t .
But whenever t ≤ A(T min ),

(α M1 (t) -λ) X0 (t) 4N 1 -X0 (t) ≤ 4αN M1 (t) X0 (t) 1 -X0 (t) ≤ 1,
because the numerator on the right is less than or equal to 4/5, while the denominator is bigger than or equal to the same figure.

The result then follows from Lemma 2.5. ♦

Next we will prove that Y 0 reaches zero with positive probability on a fixed time interval. For any α ∈ R, we define

T α = inf{t > 0, Y 0 (t) = α}.
Lemma 3.4 Let {Y 0 (t), t ≥ 0} be the solution of (3.4). For all p < 1, μ > 0, there exists δ > 0 such that with t 3 defined as in (3.1),

P(T 0 ≤ t 3 ∧ T δ+μ ) ≥ p Proof : Let Ỹ (t) = δ exp (-t + 2W (t)) , D(t) = t 0 Ỹ (s)ds, ρ(t) = inf{s > 0, D(s) > t}.
It is not too hard to show that there exists a Brownian motion W such that

Ỹ (ρ(t)) = δ exp   - t 0 ds Ỹ (ρ(s)) + 2 t 0 d W (s) Ỹ (ρ(s))   .
It now follows from Ito's formula that the process {Y (t) := Ỹ (ρ(t)), t ≥ 0} is the unique strong solution of equation (3.4) driven by W , hence Y 0 (t) = Ỹ (ρ(t)), t ≥ 0. We deduce that T 0 = D(∞) < ∞, and

P(T 0 ≤ t 3 ∧ T δ+μ ) = P ∞ 0 exp(-t + 2W (t))dt ≤ t 3 δ ∩ sup t≥0 exp(-t + 2W (t)) ≤ δ + μ δ → 1,
as δ → 0, since sup t≥0 exp(-t + 2W (t)) < ∞ a.s. ♦

Now we can choose the value of δ which we will be using from now on. Let δ be the largest value of δ such that Lemma 3.4 holds, with p = p 2 defined in (3.2) and μ = µ ( which is a function of m max ) as defined by (3.3). We choose (recall that the value of ε has been defined in (3.1))

(3.5) δ = δ ∧ 1 10 ∧ ε m .
Thanks to Lemma 3.4, when starting at time 0 from δ, Y 0 reaches 0 with probability p 2 before time t 3 ∧ T δ+µ . Then X 0 will do the same before time A(t 3 ) ∧ A(T δ+µ ), provided that X 0 (t)M 1 (t) ≤ 2ε, ∀0 ≤ t ≤ A(t 3 ) ∧ A(T δ+µ ). Hence the fact that T 0 < A(t 3 ) with positive probability, provided x 0 ≤ δ and M 1 (0) ≤ m will follow from the above results and Lemma 3.5 If X 0 (0) ≤ δ and M 1 (0) ≤ m, then we have (again with ε and t 3 given by (3.1))

P sup 0≤t≤A(t 3 ∧T δ+µ ) X 0 (t)M 1 (t) ≤ 2ε = p 3 > 1 -p 2 .
Proof : We use Lemma 3.2. Consider the event

E m,t 3 ,ε = sup 0≤t≤A(t 3 )∧A(T δ+µ ) M 1 (t) ≤ m + λA(t 3 ) + ε 6 .
We have

P(E m,t 3 ,ε ) ≥ P sup 0≤t≤A(t 3 ) M 1 (t) ≤ m 1 + λA(t 3 ) + ε 6 ≥ 1 -exp(-αN ε 3 ) = 1 -exp(- 1 30
).

Since X 0 (t) ≤ δ + µ for t ≤ A(T δ+µ ), on the event E m,t 3 ,ε , sup

0≤t≤A(t 3 )∧A(T δ+µ ) X 0 (t)M 1 (t) ≤ (δ + µ)(m + λA(t 3 ) + ε 6 ) ≤ δm + µm + λA(t 3 ) + ε 6 ≤ ε + ε 6 + ε 12 + ε 6 ≤ 2ε,
where we have used the fact that δ + µ ≤ 1 for the second inequality. ♦ Combining Lemma 3.1, Lemma 3.3, Lemma 3.5 and Lemma 3.4, denoting t 3 = A(t 3 ), we deduce the Corollary 3.6 ∃p fin ≥ p 3 + p 2 -1 > 0 such that

P(T 0 ≤ t 3 |X 0 (0) ≤ δ, M 1 (0) ≤ m) ≥ p f in > 0.
While this Corollary is rather intuitive, we shall need the slightly more general following result, i.e. with a larger set of initial conditions. Given m as above, and δ as in (3.5), let

I = {(x 0 , m 1 ) ∈ [0, 1] × R + , x 0 ≤ δ, x 0 m 1 ≤ δm}
We now prove the (p fin is as defined in Corollary 3.6) Proposition 3.7 For any (x 0 , m 1 ) ∈ I,

P(T 0 ≤ t 3 |X 0 (0) = x 0 , M 1 (0) = m 1 ) ≥ p fin .
Proof : Thanks to the previous Corollary, we only need to consider the case m 1 > m. Let (x 0 , m 1 ) be a point in the set I. First, let us consider the point (δ, m). From the previous section, starting from (δ, m), the process (X 0 , M 1 ) has a strictly positive probability to reach 0 before the time t 3 = A(t 3 ). We will show that the process starting from (x 0 , m 1 ) has a larger probability to reach 0 before time t 3 , which will prove the Proposition.

Let C = m 1 m ≥ 1. Then we have x 0 ≤ δ C . Now we will use the same reasoning as in Lemma 3.4 with a few modifications. Indeed, since the probability that Y 0 (t) reaches 0 before a prescribed time is decreasing in δ, we increase this probability by starting from Y 0 (0) = x 0 = δ ≤ δ C , since C ≥ 1. We will use this new value. Moreover, the starting point satisfies x 0 m 1 ≤ ε. The only thing which is worse than with the starting point (δ, m) is the fact that m 1 is greater than m, hence a greater m max . But this only appears in one place : in the definition of µ.

Note that if we define m max = m 1 + λt 3 + ε 6 , we deduce from Lemma 3.2

P( sup 0≤t≤t 3 M 1 (t) ≤ m max ) ≥ 1 -exp(-αN ε 3 ).
We define µ similarly as µ in (3.3), but with m max replaced by m max , hence since m max ≤ Cm max , µ ≥ µ C . But if we look at the proof of Lemma 3.4, we have, since

t 3 δ ≥ Ct 3 δ ≥ t 3 δ and δ +µ δ = 1 + µ δ ≥ 1 + µ δ , P(T 0 ≤ t 3 ) ≥ P(T 0 ≤ t 3 ∧ T δ +µ ) ≥ P ∞ 0 exp(-t + 2W (t))dt ≤ t 3 δ ∩ sup t≥0 exp(-t + 2W (t)) ≤ δ + µ δ ≥ P ∞ 0 exp(-t + 2W (t))dt ≤ t 3 δ ∩ sup t≥0 exp(-t + 2W (t)) ≤ δ + µ δ
Hence we have a larger probability to reach zero starting from (x 0 , m 1 ) rather than from (δ, m), which concludes the proof. ♦

We sum up in the following Proposition the results obtained in this section, with ε = δm (recall that m has been chosen arbitrarily, δ is prescribed by (3.5), and note that ε ≤ ε). Proposition 3.8 Let X(t) = (X k (t)) k∈Z + be the solution of (1.1) , and M 1 (t) = k≥1 kX k (t). Then ∃p f in > 0 and t 3 such that for any t ≥ 0,

P(T 0 ≤ t + t 3 |X 0 (t) ≤ δ, X 0 (t)M 1 (t) ≤ ε) ≥ p f in > 0.

A recurrence property of M 1

With the help of the results proved in the previous section, we will now prove some results on M 1 . We will show that as long that as the ratchet has not clicked, M 1 is bound to return under some specified value. This particular point will be important in the sequel.

We begin with the following lemma, which is true for any probability on Z + . It will be crucial for establishing one of our first estimates. Lemma 4.1 Let p be a probability on Z + , and let

x k = p(k), m 1 = k≥0 kx k and m 2 = k≥0 (k -m 1 ) 2 x k . Then m 2 ≥ (1 -x 0 )m 2 ≥ x 0 m 2 1 .
Proof : If x 0 = 1, m 1 = m 2 = 0 and the result is true. So it suffices to study the case x 0 < 1. By Jensen's inequality we have

k≥1 x k 1 -x 0 k 2 ≤ k≥1 x k 1 -x 0 k 2
with equality if and only if there exists only one k ≥ 1 such that x k > 0. Then :

k≥1 x k k 2 ≤ (1 -x 0 ) k≥1 x k k 2 , that is m 2 1 ≤ (1 -x 0 ) k≥1 x k k 2 , hence x 0 m 2 1 ≤ (1 -x 0 ) k≥1 k 2 x k -(1 -x 0 )m 2 1 = (1 -x 0 )m 2 .

♦

We now introduce the stopping time

H t λ := inf{s ≥ t, X 0 (s)M 1 (s) 2 ≤ 2 λ + 1 α },
and we define H λ = H 0 λ . Our next claim is Proposition 4.2 For any stopping time T , we have H T λ < +∞ a.s. The Proposition follows from the strong Markov property and

Lemma 4.3 Suppose that X 0 (0)M 1 (0) 2 > 2 λ+1
α . Then H λ < ∞ a.s. Proof : On the interval [0, H λ ],we have from Lemma 4.1

- α 2 M 2 ≤ - α 2 X 0 M 2 1 ≤ -(λ + 1),
hence from the third line of (1.3), M 1 is bounded from above by

(4.1) M 1 (t) = M 1 (0) -t - α 2 t 0 M 2 (s)ds + t 0 M 2 (s) N dB s .
We will show next that

Z t := t 0 M 2 (r) N dB r - α 2 t 0 M 2 (r)dr
is bounded from above a.s. This will imply the result, since on the event {H λ = +∞}, (4.1) holds for all t > 0, which would imply that M 1 eventually becomes negative, and this is absurd.

If we define C(t) = 1 N t 0 M 2 (s)ds, we have Z t = W (C(t)) -αN 2 C(t) where W is a standard Brownian motion. Now, if C(∞) = ∞ then lim t→∞ Z t = -∞, hence Z t is bounded from above. Or else C(∞) < ∞, and we have sup t>0 |Z t | = sup 0<s<C(∞) |W (s) -αN 2 s| < ∞ a.s. ♦
Now we will finally be able to prove that M 1 always returns below β := λ/α, as long as the ratchet does not click. Let S t β = inf{s > t, M 1 (s) ≤ β}. Then we will prove the following lemma : Lemma 4.4 For any stopping time T , P(T 0 ∧ S T β < ∞) = 1

Proof : From the strong Markov property of the solution of (1.1), we may assume that T = 0. First, we let δ inf = δ ∧ ε 2 α 4(λ+1) (recall that ε = δm). Now we introduce the process Y s t , defined ∀s ≥ 0, ∀t ≥ s which is the solution of the following system :

(4.2)    dY s t = Y s t (1 -Y s t ) N dB 0 (t), t ≥ s Y s s = δ inf . We define for any 0 ≤ u ≤ 1 R s u = inf{t ≥ s, Y s t = u}, We have (4.3) R s 0 ∧ R s 1 < +∞ a.s. P(R s 1 < R s 0 ) > 0.
Indeed, for all a ∈ (0, δ inf ), by the non-degeneracy of the diffusion coefficient, Y s t gets out of [a, 1 -a] in finite time. Then if we choose a small enough (using the same reasoning as in Lemma 3.4), we have a chance p f in to reach 0 before a time V > 0 as soon as we start below a (the same with 1 and starting above ≥ 1 -a by symmetry).

Define recursively the stopping times

ξ 1 = inf{t > s; Y s t ∈ (a, 1 -a)}, and for k ≥ 1, ξ k+1 = inf{t > ξ k + V ; Y s t ∈ (a, 1 -a)}. A standard application of the strong Markov property of Y s t yields that P(Y s ξ k+1 ∈ {0, 1}) ≤ (1 -p f in ) k
, hence the first line of (4.3). The second line is essentially obvious. Note that using an argument based upon Green's functions, one can in fact prove that E(R s 0 ∧R s 1 ) < +∞. From this we deduce that there exist K > 0, p > 0 such that

P(R s 1 ≤ K ∧ R s 0 ) ≥ p > 0. In particular P(R s 1 ≤ K) ≥ p > 0.
We define L = K ∨t 3 , where t 3 is as in Proposition 3.8, and the following sequence of stopping times :

U t 0 = inf s > t, X 0 (s)M 2 1 (s) ≤ 2 λ + 1 α ,
and ∀n ≥ 1,

U t n = inf s > U t n-1 + L, X 0 (s)M 2 1 (s) ≤ 2 λ + 1 α .
For all n ≥ 0, U t n is a.s. finite, thanks to Proposition 4.2. Now, at time U t 0 : either we are on the event

A 0 = {X 0 (U t 0 ) ≤ δ inf } (δ inf ≤ δ), in which case X 0 (U t 0 )M 1 (U t 0 ) = X 0 (U t 0 )M 2 1 (U t 0 ) × X 0 (U t 0 ) ≤ 2 λ + 1 α ε 2 α 4(λ + 1)
< ε, and we deduce from Proposition 3.8 that

P(T 0 ≤ U t 0 + L|A 0 ) = p f in > 0.
The other possibility is that we are on the event

A c 0 = B 0 ∪ C 0 , where B 0 = {X 0 (U t 0 ) > δ inf } ∩ { inf U t 0 ≤s≤U t 0 +L M 1 (s) ≥ β}, C 0 = {X 0 (U t 0 ) > δ inf } ∩ { inf U t 0 ≤s≤U t 0 +L M 1 (s) < β}.
On the event C 0 , S β ≤ U t 0 +L. On the event B 0 , inf U t 0 ≤s≤U t 0 +L (αM 1 (s)-λ)X 0 (s) ≥ 0, and then we deduce from Lemma 2.5 that

X 0 (s) ≥ Y U t 0 s . Consequently, if T 1 = inf {t ≥ 0, X 0 (t) = 1}, P(T 1 ≤ U t 0 + L|B 0 ) ≥ p > 0. But if X 0 (s) = 1, then M 1 (s) = 0. Hence P(S β ≤ U t 0 + L|B 0 ) ≥ p > 0.
Finally

P(T 0 ∧ S t β ≤ U t 0 + L) = P(T 0 ∧ S t β ≤ U t 0 + L|A 0 )P(A 0 )+ P(T 0 ∧ S t β ≤ U t 0 + L|B 0 )P(B 0 ) + P(T 0 ∧ S t β ≤ U t 0 + L|C 0 )P(C 0 ) ≥ p f in P(A 0 ) + pP(B 0 ) + P(C 0 ) ≥ p f in ∧ p =: q. P(T 0 ∧ S t β = +∞) ≤ P(T 0 ∧ S t β ≥ U t 0 + L) ≤ 1 -q.
It follows from the strong Markov property of the process X = (X k , k ≥ 0), repeating this argument with U t 0 replaced by U t 1 that

P(T 0 ∧ S t β = +∞) ≤ P(T 0 ∧ S t β ≥ U t 1 + L) ≤ (1 -q) 2 .
Iterating the above argument, we have for all ≥ 0,

P(T 0 ∧ S t β > U t + L) ≤ (1 -q) .
We have proved that P(T 0 ∧ S t β = +∞) = 0. ♦

Reaching the special set from any initial condition

Now we will show that starting from an initial condition ((x k ) k∈Z + , m 1 ) with m 1 ≤ β the process has a probability bounded below by p f in to click before a given time. Since the process is Markovian and this situation repeats itself as long as the ratchet has not clicked, we will conclude that P(T 0 < +∞) = 1.

In this part we denote by (x k ) k≥0 the initial condition of our system, and we suppose that m 1 = k≥0 kx k ≤ β.

One of the difficulties we have to face is that the quadratic variation of X 0 is

X 0 (1-X 0 ) N
, which is not bounded away from 0, near 1 and 0. We need to study three separate cases.

The first case will be described in terms of the constant (5.1)

x max = max 9 10 , 3λ + 5α 5(λ + α) , 1 - 2 λ . 5.1 x 0 ∈ (x max ; 1]
The following lemma will show that if X 0 starts too close to 1, it will quickly go under x max :

Lemma 5.1 Let t 1 = 8 λ 2 . If X 0 (0) > x max , then P( inf s<t 1 X 0 (s) ≤ x max ) ≥ 1 -exp(-N ).
Proof : Let T xmax = inf{s ≥ 0, X 0 (s) ≤ x max }. On the time interval [0, T xmax ), we have X 0 (s) > x max ≥ 3λ + 5α 5(λ + α) .

Since X 1 ≤ 1 -X 0 , on the same interval we have X 1 (s) ≤ 2λ 5(λ+α) while X 0 (s) > 9 10 , hence

αM 1 (s)X 1 (s) + λX 0 (s) -(λ + α)X 1 (s) ≥λX 0 (s) -(λ + α) 2λ 5(λ + α) ≥ λ 2 . Then X 1 (s) ≥ Y 1 (s) for s ∈ [0, T xmax ],
where Y 1 is the solution of the SDE (

   dY 1 (s) = λ 2 ds + Y 1 (1 -Y 1 ) N dB 1 (s), Y 1 (0) = 0, 5.2)
where we stop Y 1 as soon as it reaches 1. We have

P t 1 0 Y 1 (1 -Y 1 ) N dB 1 < -C = P - t 1 0 Y 1 (1 -Y 1 ) N dB 1 > C ≤ P exp -γ t 1 0 Y 1 (1 -Y 1 ) N dB 1 - t 1 0 γ 2 Y 1 (s)(1 -Y 1 (s)) 2N ds > exp γC - γ 2 8N t 1 ≤ exp -γC + γ 2 8N t 1 ,
where the first inequality follows from

Y 1 (s)(1 -Y 1 (s)) ≤ 1/4
, and the second one is Chebychev's inequality. Choosing γ = 4CN/t 1 and C = 2/λ, we deduce that

P t 1 0 Y 1 (1 -Y 1 ) N dB 1 ≥ - 2 λ ≥ 1 -exp (-N ) > 0. Now, since t 1 0 λ 2 ds = 4 λ ,
and on [0, T xmax ), X 0 (s) > 1 -2/λ, hence X 1 (s) < 2/λ, we have the inclusion

t 1 0 Y 1 (1 -Y 1 ) N dB 1 ≥ - 2 λ ⊂ {T xmax < t 1 }, which implies that P (T xmax ≤ t 1 ) ≥ 1 -exp(-N ),
hence the conclusion. ♦

We need to control M 1 on the same time interval of length t 1 . Using Lemma 3.2 we will deduce the following Proposition : Proposition 5.2 Let again x max be given by (5.1),

t 1 = 8 λ 2 , ε 0 = 1 2αN ln 2 1-exp(-N ) and β = β + λt 1 + ε 0 . If X 0 (0) > x max and M 1 (0) < β , then P ({T xmax ≤ t 1 } ∩ {M 1 (T xmax ) ≤ β }) = p init > 0
Proof : It follows from Lemma 5.1 and Lemma 3.2

P(T xmax ≤ t 1 ) ≥ 1 -exp(-N ), P(M 1 (T xmax ) ≤ β ) ≥ 1 -exp(-2αN ε 0 ).
Those two inequalities together with Lemma 3.1 imply

P ({T xmax ≤ t 1 } ∩ {M 1 (T xmax ) ≤ β }) ≥ 1 -exp(-N ) -exp(-2αN ε 0 ) = 1 -exp(-N ) 2 =: p init .

♦

So even if we started with (X 0 (0), M 1 (0)) such that X 0 (0) > x max and M 1 (0) < β, we obtain before time t 1 with probability at least p init > 0 a new initial condition X 0 ≤ x max and M 1 ≤ β , so we can resume with the next case.

X

0 ≤ x max but either X 0 > δ or X 0 M 1 > ε
The idea of this subsection is to show that with a strictly positive probability p trans , both X 0 goes from x max to a δ < δ in finite time, and during the same time interval, M 1 stays small enough so that at the end X 0 M 1 ≤ ε.

We start by showing some inequalities.

Lemma 5.3 Let {V t , t ≥ 0} be a standard Brownian motion, and c > 0 a constant. Then for any t > 0, δ > 0, μ > 0,

P inf 0≤s≤t {cs + V s } ≤ -δ, sup 0≤s≤t {cs + V s } ≤ μ ≥ 1 - 2 π δ √ t + c √ t -2 exp - 1 2 μ √ t -c √ t 2 .
Proof : Using Lemma 3.1, the result follows from the two following computations.

We have, with Z denoting a N (0, 1) random variable,

P inf 0≤s≤t {cs + V s } ≤ -δ ≥ P inf 0≤s≤t V s ≤ -δ -ct = P sup 0≤s≤t V s ≥ δ + ct = 2P(V t ≥ δ + ct) = 1 -P |Z| ≤ δ √ t + c √ t ≥ 1 - 2 π δ √ t + c √ t .
On the other hand,

P sup 0≤s≤t (cs + V s ) ≤ μ ≥ P sup 0≤s≤t V s ≤ μ -ct = 1 -P sup 0≤s≤t V s ≥ μ -ct = 1 -2P Z ≥ μ √ t -c √ t ,
and

P Z ≥ μ √ t -c √ t = P exp(γZ -γ 2 /2) ≥ exp γ μ √ t -c √ t - γ 2 2 ≤ exp -γ μ √ t -c √ t + γ 2 2 .
Choosing γ = μ/ √ t -c √ t, we conclude from the above computations that

P sup 0≤s≤t (cs + V s ) ≤ μ ≥ 1 -2 exp - 1 2 μ √ t -c √ t 2 .

♦

We will choose from now on

(5.3) ε = log(4) 2αN
, so that e -2N αε = 1 4 .

We start from (X 0 , M 1 ) = (x, β ), where 0 < x ≤ x max < 1 (recall the definition (5.1) of x max ) and β < β . Let 0 < μ = 1-xmax 2 . We are going to prove that, with positive probability, X 0 goes down to δ in a finite number of steps, while staying below x + μ (so that 1 -X 0 (t) ≥ a := 1-xmax 2 ), and while M 1 remains under control. Considering the SDE

dX 0 (t) = (αM 1 (t) -λ)X 0 (t)dt + X 0 (t)[1 -X 0 (t)] N dB 0 , let A(t) := t 0 X 0 (s)[1 -X 0 (s)] N ds, and
σ(t) := inf{s > 0, A(s) > t}. Since σ(t) 0 X 0 (s)(1 -X 0 (s)) N ds = t,
we deduce that

dσ(t) dt = N X0 (t)(1 -X0 (t))
, provided we let X0 (t) := X 0 (σ(t)).

Finally

σ(t) = t 0 N X0 (s)(1 -X0 (s))
ds, and if we let M1 (t) := M 1 (σ(t)), we deduce from the above SDE for the process X 0 that

X0 (t) = x + N t 0 α M1 (s) -λ 1 -X0 (s) ds + B(t),
where B(t) is a new standard Brownian motion (we use the same notation as above, which is a slight abuse). At the k-th step of our iterative procedure, k ≥ 1, we let X0 start from xk-1 j=1 δ j , and we stop the process X0 at the first time that it reaches the level xk j=1 δ j . We will choose not only the sequence δ k , but also the sequence s k in such a way that we can deduce from Lemma 5.3 (see (5.11) and (5.12) below) that for each 1 ≤ k ≤ K (K to be defined below), (5.4) P inf

0≤s≤s k {Θ k s + B s } ≤ -δ k , sup 0≤s≤s k {Θ k s + B s } ≤ μ > 1 3 .
We shall make sure that (5.5) Θ 0 = β , and Θ k -Θ k-1 ≥ ε + λs k , with s k := σ(s k ) and ε defined by (5.3), so that we deduce from Lemma 3.2 and our choice of ε that (5.6) P( sup

0≤s≤s k M 1 (s) ≤ Θ k M 1 (0) ≤ Θ k-1 ) ≥ 3/4.
The fact that with positive probability X 0 goes down to δ , while staying below x + μ and M 1 remaining under control, will follow from a combination of (5.4) and (5.6), provided we show that we can choose the two sequences δ k and s k for k ≥ 1 in such a way that not only (5.4) holds, but also that there exists K < ∞ such that

x - K k=1 δ k ≤ δ .
Since during the k-th step we are considering the event that X 0 (t) ≤ x + μ i.e. 1 -X 0 (t) ≥ a, and also X 0 (t) ≥ x -k j=1 δ j , we have that

s k ≤ N a(x -k j=1 δ j )
s k , so that we may, in accordance with (5.5), make the following choice of Θ k in terms of {δ j , s j , 1 ≤ j ≤ k} :

Θ k := β + kε + N λ a k j=1 s j x -j i=1 δ i .
We first want to ensure that (the reason for 0.4 will be made clear below)

δ k √ s k + Θ k √ s k ≤ 0.4,
which we achieve by requesting both that (5.7)

δ k = 0.2 √ s k and (5.8) Θ k √ s k ≤ 0.2 ⇔ s k ≤ 0.2 Θ k 2 .
On the other hand, we shall also request that for each j ≥ 1,

δ j x -j 1 δ i ≤ 1 ⇔ δ j ≤ 1 2 (x - j-1 i=1 δ i ).
This combined with (5.7) implies that

s j x -j i=1 δ i ≤ 25δ j . Consequently β ≤ Θ k ≤ β + kε + 25N λ a sup 1≤j≤k δ j k.
Moreover, a combination of (5.7) and (5.8) yields

δ j = 0.2 √ s j ≤ (0.2) 2 Θ j ≤ (25β ) -1 ,
and from the above inequality follows

Θ k ≤ β + D N k, (5.9) with D N = ε + N λ aβ .
Finally this leads us to choose

δ k = inf κ (β + D N k) , 1 2 (x - k-1 j=1 δ j ) , (5.10) s k = 25δ 2 k .
It still remains to choose κ, which will be done below. Note that (5.7) + (5.8) request us to make sure that κ ≤ 1 25 . We now have Lemma 5.4 ∃K > 0, ∀k > K,

δ k = 1 2 (x - k-1 j=1 δ j ). Proof : We first show that for k ≥ 2, 1 2 (x - k-1 j=1 δ j ) ≤ κ (β + D N k) ⇒ 1 2 (x - k j=1 δ j ) < κ (β + D N (k + 1))
.

Indeed, if the above left inequality holds, then

κ (β +D N (k+1)) 1 2 (x -k-1 j=1 δ j ) ≥ κ (β +D N (k+1)) κ (β +D N k) > 1 2 ,
where the last inequality follows easily from k ≥ 2. Consequently

κ (β + D N (k + 1)) > 1 2 δ k = 1 4 x - k-1 j=1 δ j = 1 2 x - k j=1 δ j .
Finally there exists K ≥ 1 such that

x - K j=1 κ (β + D N j) < 0. Therefore for some k ≤ K , κ (β + D N k) > 1 2 x - k j=1 δ j .

♦

This means that at each k > K, X0 progresses by a step equal to half the remaining distance to zero. Consequently ∃c > 0 such that x k = x -k j=1 δ j ≤ c2 -k . We are looking for the smallest integer k such that c2 -k ≤ δ , δ to be specified below, which implies that

k -1 < log(c) -log(δ ) log(2) ≤ k.
Consequently, since we may as well assume that δ ≤ 1/2,

k ≤ 1 + log(c) log(2) + log(1/δ ) log(2) ≤ [log(2)] -1 2 + log(c) log(2) log 1 δ .
Combining this estimate with (5.9), we deduce that there exists a constant D N such that

Θ k ≤ β + D N log 1 δ .
Hence there exists a δ ≤ δ ∧ 1/2 such that δ Θ k ≤ ε. If we now check that the probability of the previous path is bounded below by a positive constant, we will have that with a positive constant, at the end of the k-th step, both X 0 ≤ δ ≤ δ, and M 1 ≤ Θ k , hence X 0 M 1 ≤ ε, which puts us in a position to apply Proposition 3.8. Given the choice that we have made for ε, see (5.3), it suffices to make sure that

(5.11) 2 π δ k √ s k + Θ k √ s k < 1/3, ∀k ≥ 1,
as well as

(5.12) 2 exp - 1 2 μ √ s k -Θ k √ s k 2 < 1/3, ∀k ≥ 1.
Since 3 -1 π/2 > 0.4, (5.7)+(5.8) implies (5.11).

On the other hand, (5.12) is equivalent to

(5.13) μ √ s k -Θ k √ s k 2 > 2 log 6.
But we have

≥ 25Θ k β + D N k + 10 log 6 κ ≥ 25Θ k β + D N k κ 2 + 10(log 6)κ, μ 5κ/(β + D N k) > Θ k 5κ β + D N k + 2 log 6.
Finally (5.13) follows from the last inequality, (5.10) and (5.7). ♦

We therefore choose κ = 1 25 ∧ β μ 25 + 10 log 6 .

We can now conclude that Proposition 5.6 Suppose that X 0 (0) ≤ x max and M 1 (0) ≤ β . Let T δ = inf{s > 0, X 0 (s) ≤ δ }.

Then

P (T δ ≤ t 2 , X 0 (T δ ) × M 1 (T δ ) ≤ ε) ≥ 1 12 kmax := p trans ,
with t 2 = 25k max , and k max is the number of steps needed to reach δ in the above procedure, while starting from x max .

Proof : It follows from (5.6), (5.11), (5.12), Lemma 5.3 and again Lemma 3.1 that the k-th step in the above procedure happens with probability at least 1/12. It remains to exploit the Markov property, like at the end of the proof of Lemma 4.4. ♦

Conclusion

Proposition 5.2 shows that, if we start with M 1 < β, with probability p init we need to wait at most a length of time t 1 for the pair (X 0 , M 1 ) to reach the set [0, x max ]×[0, β ]. Proposition 5.6 shows that starting from that set, with probability p trans we need to wait at most a length of time t 2 for (X 0 , X 0 M 1 ) to reach the set [0, δ ] × [0, ], with δ < δ. But from Proposition 3.8, starting from this last set, we have a probability p f in to reach 0 during an interval of time of length t 3 . So to sum up, using again the strong Markov property of the system, we have Proposition 5.7 For any finite stopping time T , if M 1 (T ) ≤ β, then P(T 0 < T + t 1 + t 2 + t 3 ) ≥ p f in p trans p init > 0 Moreover Lemma 4.4 implies that this situation will happen infinitely many times as long as the ratchet does not click, which implies Theorem 1, exploiting again the strong Markov property of the solution of (1.1).

Proof of Theorem 2

This final section is devoted to the proof of Theorem 2.

We first note that the reasoning of section 5 can be done with any initial value ρ for M 1 , instead of β. That is to say, with S t ρ = inf {s > t, M 1 (s) ≤ ρ} (and S ρ = S 0 ρ ), Choosing ρ = ε δ ∨ 2λ α , we have : Lemma 6.2 There exist K, p > 0, such that for any initial condition in the set X δ , P(T 0 ∧ S ρ ≤ K) ≥ p

Lemma
Proof : We are going to argue like in the proof of Lemma 4.4. We introduce the process {Y s , s ≥ 0}, which is the solution of the following system : Since αε 2 > 0 we deduce that there exist L > 0, p > 0 such that P(R 1 ≤ L) ≥ p > 0. We choose K = L + t 3 , where t 3 has been defined in Proposition 3.8. Now there are several possibilities : Case 1. inf 0≤s≤L M 1 (s) ≤ ρ, then S ρ < L < K.

Case 2a. inf 0≤s≤L M 1 (s) ≥ ρ and inf 0≤s≤L X 0 (s)M 1 (s) ≤ ε. Then there exists t < L such that X 0 (t)M 1 (t) ≤ ε (which implies X 0 (t) ≤ δ, because M 1 (t) ≥ ρ ≥ ε δ ). In that case we can use Proposition 3.8, and we have P(T 0 ≤ K) ≥ p f in > 0, which implies P(T 0 ∧ S ρ ≤ K) ≥ p f in > 0.

Case 2b. inf 0≤s≤L M 1 (s) ≥ ρ and inf 0≤s≤L X 0 (s)M 1 (s) ≥ ε. In that last case we have (using first X 0 ≥ ε M 1 combined with αM 1 -λ ≥ λ > 0, and next -λ M 1 (s) ≥ -α 2 ) inf 0≤s≤L

(αM 1 (s) -λ)X 0 (s) ≥ inf 0≤s≤L ε(α -λ M 1 (s) )

≥ αε 2 ,
and consequently we can use the comparison theorem (Lemma 2.5), which implies that ∀s ∈ [0, L], X 0 (s) ≥ Y s . Then P(T 1 ≤ L) ≥ p > 0. But when X 0 hits 1, M 1 hits 0. Hence P(S ρ ≤ L) ≥ p > 0.

We may now conclude that there exists p > 0 such that

P(T 0 ∧ S ρ ≤ K) ≥ p ♦
We deduce from the two above Lemmas : Corollary 6.3 There exists K < ∞, and p > 0 such that, for any initial condition in X δ for some δ > 0, P(T 0 ≤ K) ≥ p.

We can now proceed with the Proof of Theorem 2 We deduce from Corollary 6.

Lemma 2 . 5

 25 Let B(t) be a standard F t -Brownian motion, T a stopping time, σ be a 1/2 Hölder function, b 1 : R → R a Lipschitz function and b 2

Lemma 5 . 5 A

 55 sufficient condition for (5.13) is that (5.14) κ ≤ 1 ∧ β μ 25 + 10 log 6 . Proof : It follows from (5.14) μ(β + D N k) > (25 + 10 log 6)κ

For any 0

 0 ≤ u ≤ 1, let R u = inf{s ≥ 0, Y s = u}.

  [START_REF] Cuthbertson | Limits to the rate of adaptation[END_REF] and the strong Markov property that for all n ≥ 0, P(T 0 > nK) ≤ (1 -p) n . ConsequentlyE[e ρT 0 ] ≤ ∞ n=0 e (n+1)ρK P(nK ≤ T 0 ≤ (n + 1)K) ≤ ∞ n=0 e (n+1)ρK (1 -p) n = e ρK ∞ n=0 e ρK (1 -p) n < ∞,provided log(1 -p) + ρK < 0, in other words ρ < ρ := -log(1 -p)/K. ♦

Acknowledgements The authors thank Jean-Stéphane Dhersin, Peter Pfaffelhuber and Anton Wakolbinger for valuable discussions concerning this work.