# Quasi-free (p,2p) reactions in inverse kinematics for studying the fission
yield dependence on temperature

A. Graña-González  J. L. Rodríguez-Sánchez  J. Benlliure  G. García-Jiménez
H. Alvarez-Pol  D. Cortina-Gil  L. Atar  L. Audouin  G. Authelet  A. Besteiro
G. Blanchon  K. Boretzky  P. Cabanelas  E. Casarejos  J. Cederkall  A.
Chatillon  A. Corsi  M. Feijoo  D. Galaviz  I. Gasparic  R. Gernhäuser  M.
Heil  A. Heinz  M. Holl  T. Jenegger  L. Ji  H.T. Johansson  O.A. Kiselev  P.
Klenze  A. Knyazev  D. Körper  T. Kröll  I. Lihtar  Y. A. Litvinov  B. Löher
P. Morfouace  D. Mücher  S. Murillo Morales  A. Obertelli  V. Panin  J. Park
S. Paschalis  A. Perea  M. Petri  S. Pirrone  L. Ponnath  A. Revel  H.-B. Rhee
L. Rose  D.M. Rossi  P. Russotto  H. Simon  A. Stott  Y. Sun  C. Sürder  J.
Taieb  R. Taniuchi  O. Tengblad  H.T. Törnqvist  S. Velardita  J. Vesic  B.
Voss  the R³B collaboration

###

Despite the recent experimental and theoretical progress in the investigation
of the nuclear fission process, a complete description still represents a
challenge in nuclear physics because it is a very complex dynamical process,
whose description involves the coupling between intrinsic and collective
degrees of freedom, as well as different quantum-mechanical phenomena. To
improve on the existing data on nuclear fission, we produce fission reactions
of heavy nuclei in inverse kinematics by using quasi-free (p,2p) scattering,
which induce fission through particle-hole excitations that can range from few
to ten’s of MeV. The measurement of the four-momenta of the two outgoing
protons allows to reconstruct the excitation energy of the fissioning compound
nucleus and therefore to study the evolution of the fission yields with
temperature. The realization of this kind of experiment requires a complex
experimental setup, providing full isotopic identification of both fission
fragments and an accurate measurement of the momenta of the two outgoing
protons. This was realized recently at the GSI/FAIR facility and here some
preliminary results are presented.

##  1 Introduction

Nuclear fission plays an important role in the nucleosynthesis r-process,
which was indicated as the mechanism responsible for the production of the
heaviest elements in the universe. Therefore, fission studies in the neutron-
rich region are required in order to measure fission yields and fission
barrier heights to improve r-process calculations [1, 2].
In the last decade, unprecedented fission experiments have been carried out at
the GSI/FAIR facility using the inverse-kinematics technique in combination
with state-of-the-art detectors. For the first time in the history of fission,
it was possible to simultaneously measure and identify both fission fragments
in terms of their mass and atomic numbers [3, 4] and to extract correlations
between fission observables sensitive to the dynamics of the fission process
[5] and the nuclear structure at the scission point [6, 7]. However, a
limitation of those previous experiments was the impossibility of measuring
the excitation energy of the compound nucleus undergoing fission, an
observable that plays a fundamental role in the description of the process.
Depending on the amount of available energy, different regions of the
potential landscape can be populated, leading to different fission paths and
consequently to different fission yields distributions. The gradual weakening
of the influence of shell structure on the fission yields with the increasing
of excitation energies can be quantified with the so-called suppression
function [8], which suppresses the microscopic term of the energy-dependent
effective potential. However, the model parameters used in such an approach
are still not constrained due to the lack of data at different excitation
energies. This lack of data is caused by the use of Coulomb excitation to
induce fission of exotic nuclei and the use of transfer- and fusion-fission
reactions only with stable nuclei. We went further by combining fission with
quasi-free (p,2p) reactions, which allows for the first time to measure many
excitation energies with stable and exotic nuclei providing therefore very
valuable data to constrain the suppression function parameters. Another
interesting question is the possible dependence of the energy sorting of the
fission fragments on the excitation energies of the compound nucleus, since in
the studies in direct kinematics with neutrons, the well known saw-tooth shape
shows higher neutron multiplicities for higher incident neutron energies [9].

In this work, the main scientific goal is to investigate the energy dependence
of the fission yields. Such data can be obtained by using quasi-free (p,2p)
reactions and combining the previous experimental setup with a silicon tracker
based on AMS-type detectors [10] and the calorimeter CALIFA (CALorimeter for
In-Flight detection of gamma-rays and high energy charged pArticles) [11]
developed by the R³B collaboration. Here we report on the analysis of the
first fission experiment performed at the GSI/FAIR accelerator facilities in
Darmstadt (Germany) using the quasi-free (p,2p) reactions as mechanism to
induce fission of 238U projectiles.

##  2 Experiment

The experiment was performed in March 2021 by using primary beams of 238U at
560$A$ MeV delivered by the SIS18 synchrotron, which were guided to the
experimental area to impinge on a liquid hydrogen target in order to produce
quasi-free (p,2p) reactions. We use the inverse-kinematics technique in which
the heavy nucleus undergoing fission is the projectile. This methodology was
developed at GSI and represents a major breakthrough in the characterization
of the fission fragments. The major advantage is that the fission fragments
are emitted in a narrow cone in the forward direction with very high
velocities, permitting to detect both fragments simultaneously and to measure
the isotopic yields.

Figure 1: Schematic view of the experimental setup used in the present work to
study quasi-free (p,2p) induced fission reactions.

Using this methodology, previous fission experiments at GSI already provided
for the first time the complete isotopic identification of both fission
fragments. In the (p,2p)-fission experiment the fragments are identified by
using the same detectors. The experimental setup (see the Fig 1) consists of
several multi-wire proportional chambers (MWPCs) to track the trajectories
before and after the superconducting dipole GLAD, a Twin-MUSIC (Multi-Sampling
Ionization Chamber) divided into two sections which are segmented into 16
anodes to measure the charges of the fission fragments and a ToF-Wall composed
by 28 plastic scintillators to measure the time of flight. With all these
measurements, employing the $\Delta E-B\rho-ToF$ technique, the mass of the
fission fragments can be obtained, allowing therefore for full isotopic
identification. Neutrons emitted from the fission fragments are detected in
NeuLAND (New Large-Area Neutron Detector) [12], located 15m downstream from
the target.
Since our goal is to study the energy dependence of the fission yields, we
access the excitation energy via the aforementioned quasi-free (p, 2p)
reactions which are knockout reactions between protons of the liquid hydrogen
target and protons bound in the heavy projectile nucleus. The products of the
reaction are the two outgoing protons and the excited compound nucleus
characterized by the particle-hole excitations created by the removed proton.
This excited compound nucleus can then de-excite via fission or neutron
evaporation. From the measurement of the momenta of the two outgoing protons
the excitation energy of the compound nucleus can be reconstructed using the
missing energy method [13]. In order to obtain the four-momenta, the energy
and the tracking of the protons are required. To reconstruct the protons
trajectory a silicon tracker is located in front of the target, which was
equipped with two double-plane arms consisting of an array of three AMS-type
[10] 0.3 mm thick double-sided silicon-strip detectors. Then to measure the
energy of the protons the CALIFA calorimeter [11] is surrounding the whole,
which consists of 1504 CsI crystal scintillators covering a polar angle range
between 22 and 90 degrees.

##  3 Results

Figure 2: Left: Two-dimensional cluster plot of the charges $Z_{1}$ and
$Z_{2}$ of both fission fragments for all fissioning systems. The inset shows
the fission fragments distribution for $Z_{1}+Z_{2}=91$. Right: Fissioning
systems distribution.

The first step of the analysis is to obtain the charge identification of both
fission fragments given by the Twin-MUSIC, which is designed with two
independent ionization chambers divided into two sections in order to measure
the atomic number of both fission fragments simultaneously. The correlation of
the derived charge for both fission fragments $Z_{1}$ and $Z_{2}$ is shown in
the Fig. 2-left in a two-dimensional cluster plot. Each diagonal of clusters
in this figure corresponds to a different fissioning system, and the
fissioning system distribution is displayed in the Fig. 2-right. Since the
beam is ${}^{238}_{92}$U, when the desired (p,2p) takes place the resulting
nucleus is an excited ${}^{237}_{91}$Pa, which is the $Z_{1}+Z_{2}=91$ peak,
the one with the highest statistics of this plot. Nevertheless, it is
important to notice that this peak might not have only ${}^{237}_{91}$Pa
systems, it will have the contribution for other lighter 91Pa isotopes as
well, since the ${}^{237}_{91}$Pa can also de-excite via neutron emission and
undergo fission after. It is also possible to obtain different excited 91Pa
isotopes when the reaction is not quasi-free. A selection in the opening angle
between the protons serves to select the quasi-free reactions in which there
is no evaporation of neutrons before fission.
The inset of the Fig. 2-left shows the fission fragment distribution for the
production of 91Pa, with a charge resolution of 0.38 units (FWHM). The
distribution fits well to an overall Gaussian shape, although two small humps
around Z=40 and Z=50 are slightly noticeable. These two peaks are the
asymmetric fission contribution and they appear because of the shell effects
[14] which favour the formation of nuclei with magic numbers around Z=50 in
this case. Without any constrain on the energy (like in this plot) the
symmetric contribution from the high energies will partially shadow the
asymmetric humps and the total distribution will be the convolution of the
three Gaussian distributions. Nevertheless, when this plot is constrained to
low energies, the symmetric component should gradually vanish, since the
fission barrier in ${}_{91}Pa$ is supposed to be higher for symmetric fission.
Therefore, the asymmetric distribution would start to be more and more
noticeable and at some point it would be the only visible one. Once the
excitation energy is reconstructed, it will be possible to plot this same
graph for different excitation energies and to study the evolution of the
fission yields from asymmetric to symmetric while going from low to high
energies.

##  4 Conclusions

The first (p,2p)-fission experiment has been carried out at GSI/FAIR with
projectiles of 238U at energies of $560A$ MeV impinging on a liquid hydrogen
target and using the R³B experimental setup to measure all the reaction
products. These complete kinematics measurements will allow to correlate the
fission yields with the excitation energy of the fissioning compound nucleus
and to study the energy sharing between the two fission fragments.
Additionally, this experiment permits to develop the data analysis and
validate the methodology to use this approach in future experiments with
exotic nuclei. The current state of the analysis is that the charge
identification is completed and the next steps will be the identification in
mass number and the reconstruction of the excitation energy spectrum. In the
near future further (p,2p)-fission experiments are planned to be carried out
at GSI/FAIR with exotic neutron-rich projectiles close to the neutron shell
N=152 to obtain their fission yields and fission barrier heights for
constraining r-process calculations.

##  5 Acknowledgments

This work was partially supported by the Spanish Ministry for Science and
Innovation under grants PGC2018-099746-B-C21 and PGC2018-099746-B-C22, by the
Regional Government of Galicia under the program “Grupos de Referencia
Competitiva” ED431C-2021-38 and by the “María de Maeztu” Units of Excellence
program MDM-2016-0692. J.L.R.S. thanks the support from Xunta de Galicia under
the program of postdoctoral fellowship ED481D-2021-018.

## References

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  * [2] Samuel A. Giuliani et al. 2018 Phys. Rev. C 97 034323
  * [3] J. L. Rodríguez-Sánchez et al. 2015 Phys. Rev. C 91 064616
  * [4] E. Pellereau et al. 2017 Phys. Rev. C 95 054603
  * [5] J. L. Rodríguez-Sánchez et al. 2016 Phys. Rev. C 94 061601(R)
  * [6] A. Chatillon et al. 2019 Phys. Rev. C 99 054628
  * [7] A. Chatillon et al. 2020 Phys. Rev. Lett. 102 202502
  * [8] J. Randrup et al. 2013 Phys. Rev. C 88 064606
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# On the large time behaviour of the solution of an SDE driven by a Poisson
Point Process

Elma Nassar and Etienne Pardoux

###### Abstract

We study a stochastic differential equation driven by a Poisson point process,
which models continuous changes in a population’s environment, as well as the
stochastic fixation of beneficial mutations that might compensate for this
change. The fixation probability of a given mutation increases as the
phenotypic lag $X_{t}$ between the population and the optimum grows larger,
and successful mutations are assumed to fix instantaneously (leading to an
adaptive jump). Our main result is that the process is transient (i.e.,
continued adaptation is impossible) if the rate of environmental change $v$
exceeds a parameter $m$, which can be interpreted as the rate of adaptation in
case every beneficial mutation gets fixed with probability $1$. If $v<m$, the
process is positive recurrent, while in the limiting case $m=v$, null
recurrence or transience depends upon additional technical conditions. We show
how our results can be extended to the case of a time varying rate of
environmental change.

##  1 Introduction

We study the large time behaviour of the solution of a scalar stochastic
differential equation of the type

$X_{t}=X_{0}-v(t)+\int_{[0,t]\times\mathbb{R}\times[0,1]}\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi),$
(1)

where $M$ is a Poisson point process on
$\mathbb{R}_{+}\times\mathbb{R}\times[0,1]$ with mean measure $ds\
\nu(d\alpha)\ d\xi$ and $\varphi(x,\alpha,\xi)={\bf 1}_{\\{\xi\leq
g(x,\alpha)\\}}$. The goal of our work is to understand how a population can
adapt to a deterioration of its fitness, due for instance to continuous change
in the climatic conditions, thanks to mutations which improve its adaptation
to the new environment. $ds\ \nu(d\alpha)$ represents the rate of appearance
of new mutations, while $g(x,\alpha)$ is the probability that a mutation
$\alpha$, which is proposed while the population’s phenotypic lag is given by
$x$, gets fixed. We assume that $g(x,\alpha)\rightarrow 1$ when
$x\rightarrow\pm\infty$ provided that $x\alpha<0$.

We start with the simple case $v(t)=vt$, with $v>0$. With the notation
$m=\int_{0}^{\infty}\alpha\nu(d\alpha)$, in other words $m$ is the mean
movement to the right per time unit produced by the positive mutations if all
of them get fixed, our first result says that the Markov process $X_{t}$ is
positive recurrent if $m>v$, transient if $m<v$, with a speed of escape to
infinity equal to $v-m$. The most interesting case is the limit situation
$m=v$. We show that, depending upon the speed at which
$m(x)=\int_{0}^{\infty}\alpha g(x,\alpha)\nu(d\alpha)$ converges to $m$ as
$x\to-\infty$, the process can be either null recurrent or else transient with
zero speed.

We then generalize our results to the case where $v(t)$ is a more general (and
even possibly random) function of time.

Note that Kersting (1986) has studied similar questions in discrete time.
Similar resuts for a SDE driven by Brownian motion with coefficients that do
not depend upon the time variable would be easy to obtain. Here we use
stochastic calculus and several ad hoc Lyapounov functions. Note that the Itô
formula for processes with jumps leads to less explicit computations than in
the Brownian case. To circumvent this difficulty, for the treatment of the
delicate case $m=v$, we establish a stochastic inequality for $C^{2}$
functions whose second derivative is either increasing or decreasing,
exploiting the fact that all jumps have the same sign, see Lemma 4 in
subsection 4.3 below.

The paper is organized as follows. We define our model in detail in section 2,
refering to models already studied in the biological literature. We establish
existence and uniqueness of a solution to our equation in section 3 (the
result is not immediate since we do not assume that the measure $\nu$ is
finite). Section 4 is devoted to the large time behaviour of $X_{t}$ when
$v(t)=vt$, successively with $m<v$, $m>v$, and $m=v$. Finally section 5 is
devoted to the large time behaviour of $X_{t}$ when $v(t)$ takes a more
general form, but
$\overline{v}=\lim\limits_{t\to\infty}t^{-1}\int_{0}^{t}v(s)ds$ exists.

##  2 The Model

Our starting point is the model by Kopp and Hermisson (2009) of a population
of constant size $N$ that is subject to Gaussian stabilizing selection, with a
moving optimum that increases linearly at rate $v$. The population is assumed
to be monomorphic at all times (i.e., its state is completely characterized by
$x$). Mutations arise according to a Poisson point process with intensity
$ds\nu(d\alpha)$. Note that, in the model considered in Kopp and Hermisson
(2009),

$\nu(d\alpha)=\frac{\Theta}{2}p(\alpha)d\alpha,$ (2)

which translates as follows: mutations appear at rate $\Theta/2=N\mu$ (where
$\mu$ is the _per-capita_ mutation rate and $\Theta=2N\mu$ is a standard
population-genetic parameter), and their phenotypic effects $\alpha$ are drawn
from a distribution with density $p(\alpha)$. Whereas in our model, we do not
impose that $\nu$ has a density, nor that it is a finite measure. We neglect
the possibility of fixation of deleterious mutations. Yet even beneficial
mutations have a significant probability of being lost due to the effects of
genetic drift. A mutation with effect $\alpha$ that arises in a population
with phenotypic lag $x$ has a probability of fixation $g(x,\alpha)$ that
satisfies

  1. 1.

$0\leq g(x,\alpha)\leq{\bf 1}_{\\{\alpha x<0\\}}\times{\bf 1}_{\\{|\alpha|\leq
2|x|\\}}$,

  2. 2.

For all $\alpha\in\mathbb{R}$, $g(x,\alpha)\uparrow{\bf 1}_{\\{\alpha
x<0\\}}$, as $|x|\rightarrow\infty$,

  3. 3.

For any compact set $K\subset\mathbb{R}$, there exists $c_{K}>0$ such that for
all $x,y\in\mathbb{R}$

$\int_{K}|\alpha|\times|g(x,\alpha)-g(y,\alpha)|\leq c_{K}|x-y|.$ (3)

One popular model among theoretical biologists for the fixation probability is

$g(x,\alpha)=\begin{cases}1-\exp(-2s(x,\alpha))\quad\text{ if
}s(x,\alpha)>0,\\\ 0\text{ otherwise}\end{cases}$ (4)

where

$s(x,\alpha)=-\sigma[|\alpha|(2|x|-|\alpha|)]^{+}\times\mathbf{1}_{\\{x\alpha<0\\}}\approx\frac{\mathcal{W}(x+\alpha)}{\mathcal{W}(x)}-1$
(5)

is the selection coefficient; $\mathcal{W}(x)=e^{-\sigma x^{2}}$ being the
fitness of the population when the phenotypic lag is $x$ and $\sigma$
determining the strength of selection. The expression of $g$ defined in (4)
and (5) satisfies the three conditions listed above (see Lemma 1 for the proof
of (3)), and is a good approximation of the fixation probability derived under
a diffusion approximation (Malécot 1952; Kimura 1962), which is valid when the
population size $N$ is large enough. Note that Kopp and Hermisson (2009) used
the even simpler approximation $g(x,\alpha)\approx 2s(x,\alpha)$ (Haldane
1927; for more exact approximations for the fixation probability in changing
environments, see Uecker and Hermisson 2011; Peischl and Kirkpatrick 2012).
Once a mutation gets fixed, it is assumed to do so instantaneously, and the
phenotypic lag $x$ of the population is updated accordingly.

In the particular case $v(t)=vt$, the evolution of the phenotypic lag of the
population can be described by the following particular case of equation (1):

$X_{t}=X_{0}-vt+\int_{[0,t]\times\mathbb{R}\times[0,1]}\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi).$
(6)

Here, $M$ is a Poisson point process over
$\mathbb{R_{+}}\times\mathbb{R}\times[0,1]$ with intensity $ds\ \nu(d\alpha)\
d\xi$. $\nu(d\alpha)$ is a $\sigma$-finite measure on $\mathbb{R}$ describing
the distribution of new mutations up to a multiplicative constant, which
satisfies

$\int_{\mathbb{R}}|\alpha|\wedge 1\nu(d\alpha)<\infty,$ (7)

and

$\varphi(x,\alpha,\xi)=\mathbf{1}_{\\{\xi\leq g(x,\alpha)\\}},$

where the fixation probability $g(x,\alpha)$ has been defined above. The
points of this Poisson point process $(T_{i},A_{i},\Xi_{i})$ are such that the
$(T_{i},A_{i})$ form a Poisson point process over
$\mathbb{R}_{+}\times\mathbb{R}$ of the proposed mutations with intensity
$ds\nu(d\alpha)$, and the $\Xi_{i}$ are i.i.d. $\mathcal{U}[0,1]$, globally
independent of the Poisson point process of the $(T_{i},A_{i})$. $T_{i}$’s are
the times when mutations are proposed and $A_{i}$’s are the effect sizes of
those mutations. The $\Xi_{i}$ are auxiliary variables determining fixation: a
mutation gets instantaneously fixed if $\Xi_{i}\leq g(X_{T_{i}},A_{i})$, and
is lost otherwise.

##  3 Existence and uniqueness

Define for all $x$

$\displaystyle m(x)$ $\displaystyle=\int_{\mathbb{R}}\alpha
g(x,\alpha)\nu(d\alpha),$ $\displaystyle m$
$\displaystyle=\int_{\mathbb{R}_{+}}\alpha\nu(d\alpha),$ (8)
$\displaystyle\psi(x)$ $\displaystyle=m(x)-v,$ (9) $\displaystyle V(x)$
$\displaystyle=\int_{\mathbb{R}}\alpha^{2}g(x,\alpha)\nu(d\alpha),$
$\displaystyle V$ $\displaystyle=\int_{\mathbb{R}_{+}}\alpha^{2}\nu(d\alpha).$
(10)

$m(x)$ is the mean speed towards zero induced by the fixation of random
mutations while $X_{t}=x<0$. $V(x)$ is related to the second moment of the
distribution of these mutations. $m$ and $V$ are the limits of $m(x)$ and
$V(x)$, respectively, in the case that all mutations with $\alpha>0$ go to
fixation (as we will show later, this is the case if $x\rightarrow-\infty$).
Note that our assumptions do not exclude cases where $m=\infty$ and/or
$V=\infty$, unless stated otherwise. However, since $g(x,\cdot)$ has compact
support, for each $x$, $m(x)<\infty$ and $V(x)<\infty$. The cases $m=\infty$
and $V=\infty$ correspond to a heavy tailed $\nu$. It would be quite
acceptable on biological grounds to assume that $m<\infty$ and/or $V<\infty$.
However, we refrain from adding unnecessary assumptions. We rewrite the SDE
(6) as follows

$X_{t}=X_{0}+\int_{0}^{t}\psi(X_{s})ds+\mathcal{M}_{t}$ (11)

where the martingale

$\mathcal{M}_{t}=\int_{0}^{t}\int_{\mathbb{R}_{+}}\int_{0}^{1}\alpha\varphi(X_{s^{-}},\alpha,\xi)\bar{M}(ds,d\alpha,d\xi),$
(12)

with $\bar{M}(ds,d\alpha,\xi)$ being the compensated Poisson measure
$M(ds,d\alpha,d\xi)-ds\nu(d\alpha)d\xi$.

######  Proposition 1.

Equation (11) has a unique solution.

###### Proof.

If $\nu$ is a finite measure, then $M$ has a.s. finitely many points in
$[0,t]\times\mathbb{R}$ for any $t>0$. In that case, the unique solution is
constructed explicitly by adding the successive jumps. In the general case, we
choose an arbitrary compact set $K=[-k,k]$ (with $k>0$). There are finitely
many jumps $(t_{i},\alpha_{i})$ of $M$ with $\alpha_{i}\notin K$. It suffices
to prove existence and uniqueness between two such consecutive jumps. In other
words , it suffices to prove existence and uniqueness under the assumption
$\nu(K^{c})=0$, and hence from (7), we deduce that
$\int\left(|\alpha|+\alpha^{2}\right)\nu(d\alpha)<\infty$, which we assume
from now on. Define for each $t>0$

$\Gamma_{t}(U)=x-vt+\int_{[0,t]\times
K\times[0,1]}\alpha\varphi(U_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi).$ (13)

A solution of equation (11) is a fixed point of the mapping $\Gamma$. Hence it
suffices to prove that $\Gamma$ admits a unique fixed point. For $\lambda>0$,

$\displaystyle\mathbb{E}e^{-\lambda
t}\left|\Gamma_{t}(U)-\Gamma_{t}(V)\right|=$
$\displaystyle-\lambda\mathbb{E}\int_{0}^{t}e^{-\lambda
s}\left|\Gamma_{s}(U)-\Gamma_{s}(V)\right|ds+\mathbb{E}\int_{0}^{t}e^{-\lambda
s}d\left|\Gamma_{s}(U)-\Gamma_{s}(V)\right|$ $\displaystyle\leq$
$\displaystyle-\lambda\mathbb{E}\int_{0}^{t}e^{-\lambda
s}\left|\Gamma_{s}(U)-\Gamma_{s}(V)\right|ds$
$\displaystyle+\mathbb{E}\int_{[0,t]\times K\times[0,1]}|\alpha|e^{-\lambda
s}\left|\varphi(U_{s^{-}},\alpha,\xi)-\varphi(V_{s^{-}},\alpha,\xi)\right|M(ds,d\alpha,d\xi).$

The above inequality follows readily from the fact that, for all $0<s<t$,

$\begin{split}\left|\Gamma_{t}(U)-\Gamma_{t}(V)\right|&-\left|\Gamma_{s}(U)-\Gamma_{s}(V)\right|\\\
&\leq\int_{(s,t)\times
K\times[0,1]}|\alpha|\times|\varphi(U_{r^{-}},\alpha,\xi)-\varphi(V_{r^{-}},\alpha,\xi)|M(dr,d\alpha,d\xi).\end{split}$

Thus,

$\begin{split}\lambda\mathbb{E}\int_{0}^{t}e^{-\lambda
s}\left|\Gamma_{s}(U)-\Gamma_{s}(V)\right|ds&\leq\mathbb{E}\int_{0}^{t}\int_{K}|\alpha|e^{-\lambda
t}\left|g(U_{s},\alpha)-g(V_{s},\alpha)\right|\nu(d\alpha)ds\\\ &\leq
c_{K}\mathbb{E}\int_{0}^{t}e^{-\lambda s}|U_{s}-V_{s}|ds.\\\ \end{split}$ (14)

The last inequality is due to the assumption (3). Let $T$ be arbitrary. Define
for all $\lambda>0$ the norm on the Banach space $L^{1}(\Omega\times[0,T])$,

$\|Z\|_{T,\lambda}=\mathbb{E}\int_{0}^{T}e^{-\lambda t}|Z_{t}|dt.$

We choose $\lambda_{0}>c_{K}$. We deduce from (14) that

$\mathbb{E}\|\Gamma(U)-\Gamma(V)\|_{T,\lambda_{0}}\leq\frac{c_{K}}{\lambda_{0}}\mathbb{E}\|U-V\|_{T,\lambda_{0}}.$

Since $c_{K}/\lambda_{0}<1$, $\Gamma$ has a unique fixed point such that
$\Gamma_{t}(U)=U_{t}$ for all $0\leq t\leq T$. Since $T$ is arbitrary, the
result is proved. ∎

We now prove that $g$ given by (4) and (5) satisfies the assumption (3).

######  Lemma 1.

For any compact set $K\subset\mathbb{R}$ and for all $u,v\in\mathbb{R}$,

$\int_{K}|\alpha(g(u,\alpha)-g(v,\alpha))|\nu(d\alpha)\leq c_{K}|u-v|,$

where $c_{K}=4\sigma\left(\int_{K}\alpha^{2}\nu(d\alpha)\right)$.

###### Proof.

For $0<u<v$ we have that

$\begin{split}\int_{K}|\alpha(g(u,\alpha)-g(v,\alpha))|\nu(d\alpha)=\
&\int_{\mathbb{R_{-}}\bigcap
K}|\alpha\left(e^{-2\sigma|\alpha|(2|v|-|\alpha|)^{+}}-e^{-2\sigma|\alpha|(2|u|-|\alpha|)^{+}}\right)|\nu(d\alpha)\\\
=&\int_{[-2u,0]\bigcap
K}|\alpha\left(e^{-2\sigma|\alpha|(2|v|-|\alpha|)}-e^{-2\sigma|\alpha|(2|u|-|\alpha|)}\right)|\nu(d\alpha)\\\
&+\int_{[-2v,-2u]\bigcap
K}|\alpha|\left|e^{-2\sigma|\alpha|(2|v|-|\alpha|))}-1\right|\nu(d\alpha)\\\
\leq\ &4\sigma\left(\int_{[-2u,0]\bigcap
K}\alpha^{2}\nu(d\alpha)\right)\times|u-v|\\\ &+2\sigma\int_{[-2v,-2u]\bigcap
K}\alpha^{2}(2v+\alpha)\nu(d\alpha)\\\ \leq\
&4\sigma\left(\int_{K}\alpha^{2}\nu(d\alpha)\right)\times|u-v|.\end{split}$

A similar estimate can easily be obtained for $v<u<0$. For $u<0<v$ , we have
that

$\begin{split}\int_{\mathbb{R}}|\alpha(g(u,\alpha)-g(v,\alpha))|\nu(d\alpha)\leq\
&\int_{\mathbb{R_{-}}\bigcap K}|\alpha
g(v,\alpha)|\nu(d\alpha)+\int_{\mathbb{R_{+}}\bigcap K}|\alpha
g(u,\alpha)|\nu(d\alpha)\\\ \leq\ &2\sigma\int_{\mathbb{R}_{-}\bigcap
K}\alpha^{2}(2v+\alpha)^{+}\nu(d\alpha)+2\sigma\int_{\mathbb{R}_{+}\bigcap
K}\alpha^{2}(-2u-\alpha)^{+}\nu(d\alpha)\\\ \leq\
&4\sigma\int_{\mathbb{R}_{-}\bigcap
K}\alpha^{2}|v|\nu(d\alpha)+4\sigma\int_{\mathbb{R}_{+}\bigcap
K}\alpha^{2}|u|\nu(d\alpha)\\\ \leq\
&4\sigma\left(\int_{K}\alpha^{2}\nu(d\alpha)\right)\times|u-v|.\\\
\end{split}$

∎

##  4 Classification of the large-time behaviour

######  Proposition 2.

If $X_{0}>0$, then $X_{t}$ becomes negative after a finite time a.s.

###### Proof.

Let $T_{0}=\inf(t>0,X_{t}<0)$. Since $g(x,\alpha)=0$ for $x\alpha>0$,

$\begin{split}X_{t\wedge T_{0}^{-}}&=X_{0}-v(t\wedge T_{0})+\int_{0}^{t\wedge
T_{0}^{-}}\int_{\mathbb{R}}\int_{0}^{1}\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi)\\\
&\leq X_{0}-v(t\wedge T_{0}),\end{split}$

hence

$t\wedge T_{0}\leq\frac{X_{0}-X_{t\wedge T_{0}^{-}}}{v}<\frac{X_{0}}{v}.$

Let $t$ tend to $\infty$.

$T_{0}\leq\frac{X_{0}}{v}<\infty$

∎

Whether the process $X_{t}$ is positive recurrent, nul recurrent or transient
depends only upon its behavior while $X_{t}<0$. Hence, we only need to
consider in detail the case when $X_{t}$ is negative, in which case only
positive mutations ($\alpha>0$) have a positive probability of fixation.

######  Proposition 3.

The functions $x\longmapsto m(x)$ and $x\longmapsto V(x)$ are continuous and
decreasing on $\mathbb{R_{-}}$ and

$\begin{split}m(x)&\xrightarrow[x\rightarrow-\infty]{}m,\\\
V(x)&\xrightarrow[x\rightarrow-\infty]{}V.\\\ \end{split}$ (15)

###### Proof.

We prove this result for the function $x\longmapsto m(x)$. A similar argument
applies to $V(x)$. Let

$\displaystyle h:\mathbb{R}_{-}\times\mathbb{R}_{+}$
$\displaystyle\to\mathbb{R}_{+}$ $\displaystyle(x,\alpha)$
$\displaystyle\mapsto h(x,\alpha)=\alpha g(x,\alpha).$

We have that $h(x,\cdot)\in L^{1}(\nu)$, and $x\mapsto h(x,\alpha)$ is
decreasing. For each fixed $\alpha>0$, $0\leq h(x,\alpha)\uparrow\alpha$, as
$x\rightarrow-\infty$. By the monotone convergence theorem, it follows that

$m(x)=\int_{\mathbb{R}}h(x,\alpha)\nu(d\alpha)\xrightarrow[x\rightarrow-\infty]{}m.$

Continuity is proved similarly. ∎

To determine the large-time behavior of the process, we now consider
successively, the three cases $v>m$, $v<m$ and $v=m$.

###  4.1 The case $v>m$

In particular, here $m=\int_{0}^{\infty}\alpha\nu(d\alpha)$ is finite. Let

$\mathcal{N}_{t}=\int_{[0,t]\times\mathbb{R}\times[0,1]}\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi)$

be the sum of all the jumps on the time interval $[0,t]$. We have that

$\mathcal{N}_{t}=\mathcal{N}_{t}^{(+)}+\mathcal{N}_{t}^{(-)}\leq\mathcal{N}_{t}^{(-)},$

where

$\begin{split}\mathcal{N}_{t}^{(+)}&=\mathbf{1}_{\\{X_{s^{-}}>0\\}}d\mathcal{N}_{s}\\\
\mathcal{N}_{t}^{(-)}&=\mathbf{1}_{\\{X_{s^{-}}<0\\}}d\mathcal{N}_{s}\\\
\end{split}$

Let $m^{(-)}(x)=\mathbf{1}_{\\{x<0\\}}m(x)$, hence

$\mathcal{M}_{t}^{(-)}=\mathcal{N}^{(-)}_{t}-\int_{0}^{t}m^{(-)}(X_{s})ds.$

Thus,

$X_{t}\leq X_{0}+\int_{0}^{t}(m^{-}(X_{s})-v)ds+\mathcal{M}_{t}^{(-)}$

######  Lemma 2.

If $m<\infty$, then

$\lim_{t\rightarrow\infty}\frac{\mathcal{M}^{(-)}_{t}}{t}=0.$ (16)

###### Proof.

$\mathcal{M}^{(-)}_{t}$ is a square-integrable martingale, such that
$\mathbb{E}{\mathcal{M}^{(-)}_{t}}=0$. Define $M^{0}$ as the Poisson random
measure for new mutations i.e. a Poisson point process on
$\mathbb{R}_{+}\times\mathbb{R}$ with intensity $ds\nu(d\alpha)$. For all
$i\in\mathbb{N}^{*}$ and $n\in\mathbb{N}^{*}$, define

$\begin{split}\xi_{i}&=\int_{i-1}^{i}\int_{0}^{\infty}\int_{0}^{1}\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi),\\\
\omega_{i}&=\int_{i-1}^{i}m^{-}(X_{s})ds,\\\
\eta_{i}&=\int_{i-1}^{i}\int_{0}^{\infty}\alpha M^{0}(ds,d\alpha),\\\
Y_{i}&=\xi_{i}-\omega_{i},\\\ \mathcal{M}_{n}&=\sum_{i=1}^{n}Y_{i},\\\
\end{split}$

Note that for all $i\in\mathbb{N}^{*}$, $0\leq\xi_{i}\leq\eta_{i}$ and
$0\leq\omega_{i}\leq m$. We first establish

######  Lemma 3.

$\mbox{If }\quad\frac{\sum_{i=1}^{n}Y_{i}}{n}\xrightarrow[n]{}0,\quad\mbox{
then}\quad\frac{\mathcal{M}^{-}_{t}}{t}\xrightarrow[t\rightarrow\infty]{}0.$

###### Proof.

$\frac{\mathcal{M}^{-}_{t}}{t}=\frac{\mathcal{M}^{-}_{\lfloor
t\rfloor}}{\lfloor t\rfloor}\times\frac{\lfloor
t\rfloor}{t}+\frac{\tilde{\mathcal{M}^{-}}_{t}}{t},$

where

$\begin{split}\frac{\tilde{\mathcal{M}}^{-}_{t}}{t}=&\frac{1}{t}\left(\int_{\lfloor
t\rfloor}^{t}\int\int\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi)-\int_{\lfloor
t\rfloor}^{t}m^{-}(X_{s})ds\right)\\\ &\leq\frac{1}{t}\left(\int_{\lfloor
t\rfloor}^{\lceil
t\rceil}\int\int\alpha\varphi(X_{s^{-}},\alpha,\xi)M(ds,d\alpha,d\xi)+\int_{\lfloor
t\rfloor}^{\lceil t\rceil}m^{-}(X_{s})ds\right)\\\
&=\frac{1}{t}\left(\xi_{\lceil t\rceil}+\omega_{\lceil
t\rceil}\right)=\frac{\lceil t\rceil}{t}\times\frac{1}{\lceil
t\rceil}\left(Y_{\lceil t\rceil}+2\omega_{\lceil
t\rceil}\right)\xrightarrow[t\rightarrow\infty]{}0,\\\ \end{split}$

since for all $n>0$,

$\frac{Y_{n+1}}{n+1}=\frac{\sum_{i=1}^{n+1}Y_{i}}{n+1}-\frac{\sum_{i=1}^{n}Y_{i}}{n}\times\frac{n}{n+1}\xrightarrow[n\rightarrow\infty]{}0$

and

$0\leq\frac{|\omega_{n}|}{n}\leq\frac{m}{n},$

hence

$\frac{\omega_{n}}{n}\xrightarrow[n\rightarrow\infty]{}0.$

∎

Back to the proof of Lemma 1. We now define

$\begin{split}A_{i}&=\\{\eta_{i}>i\\},\\\
\tilde{Y}_{i}&=Y_{i}\mathbf{1}_{\\{\eta_{i}\leq i\\}}.\\\ \end{split}$

Since the $\left(\eta_{i},i\in\mathbb{N}^{*}\right)$ are i.i.d, integrable and

$\mathbb{P}(\eta_{i}>i)=\sum_{i\geq
1}\mathbb{P}(\eta_{1}>i)\leq\mathbb{E}\eta_{1}<\infty,$

it follows from Borel Cantelli’s Lemma that $\mathbb{P}(\limsup A_{i})=0$.
Hence, a.s. there exists $N(\alpha)$ such that for all $n>N(\alpha)$, we have
$\tilde{Y}_{n}=Y_{n}$. But since
$\mathbb{E}(\tilde{Y}_{n})\rightarrow\mathbb{E}(Y_{1})$ due to the dominated
convergence theorem, it is sufficient to prove that

$\frac{\sum_{i=1}^{n}\left(\tilde{Y}_{i}-\mathbb{E}(\tilde{Y}_{i})\right)}{n}\xrightarrow[n]{}0.$

Due to corollary 3.22 in Breiman (1968)111In the proof of this theorem, we
replace Kolmogorov’s inegality by Doob’s inegality for martingales, and the
result holds in our case., it is again sufficient to prove that

$\sum_{i=1}^{\infty}\frac{\mathbb{E}(\tilde{Y_{i}}^{2})}{i^{2}}<\infty.$

Indeed, we have that

$\sum_{i=1}^{\infty}\frac{\mathbb{E}(\tilde{Y_{i}}^{2})}{i^{2}}<2m.$

The underlying calculation can be found in the proof of theorem 3.30 in
Breiman (1968). ∎

######  Remark 1.

In the case $m<\infty$ and $X_{t}\rightarrow-\infty$, we have that

$\frac{1}{t}\mathcal{M}_{t}^{(+)}\rightarrow 0,$

since eventually $X_{t}$ becomes negative. Furthermore, if we assume that
$\int_{-\infty}^{0}\alpha\nu(d\alpha)>-\infty$ then the previous Lemma implies
that

$\frac{\mathcal{M}_{t}}{t}\rightarrow 0,$

whether $X_{t}\rightarrow-\infty$ or not. But we refrain from adding any
supplementary assumption on $\nu$.

######  Proposition 4.

In the case $v>m$, $X_{t}\rightarrow-\infty$ with speed $v-m$ in the sense
that

$\frac{X_{t}}{t}\xrightarrow[]{a.s.}m-v\text{ as }t\rightarrow\infty.$

###### Proof.

$\frac{X_{t}}{t}=\frac{X_{0}}{t}-v+\frac{1}{t}\int_{0}^{t}m(X_{s})ds+\frac{\mathcal{M}_{t}}{t}\leq\frac{X_{0}}{t}-v+m+\frac{\mathcal{M}_{t}}{t}.$

Hence

$\limsup_{t\rightarrow\infty}\frac{X_{t}}{t}\leq-v+m.$ (17)

On the other hand, it follows from (15) that

$\forall\epsilon>0\quad\exists K_{\epsilon}>0\mbox{ such that }x\leq-
K_{\epsilon}\Rightarrow m(x)>m-\epsilon,$ (18)

and since $X_{t}\xrightarrow[t\rightarrow\infty]{}-\infty$ by (17), we have

$\forall\epsilon>0\quad\exists t_{\epsilon}>0\mbox{ such that }\forall s\geq
t_{\epsilon}\Rightarrow X_{s}\leq-K_{\epsilon}.$ (19)

Statements (18) and (19) combined give

$\forall\epsilon>0\quad\exists t_{\epsilon}\mbox{ such that }\forall s\geq
t_{\epsilon}\Rightarrow m(X_{s})>m-\epsilon.$

Then, $\forall\epsilon>0$ and $t>t_{\epsilon}$

$\begin{split}\frac{X_{t}}{t}&\geq\frac{X_{t_{\epsilon}}}{t}+\frac{1}{t}\int_{t_{\epsilon}}^{t}(m(X_{s})-v)ds+\frac{\mathcal{M}_{t}-\mathcal{M}_{t_{\epsilon}}}{t}\\\
&\geq\frac{X_{t_{\epsilon}}}{t}+(m-\epsilon-v)\times\frac{t-t_{\epsilon}}{t}+\frac{\mathcal{M}_{t}-\mathcal{M}_{t_{\epsilon}}}{t}.\end{split}$

Hence,

$\liminf_{t\rightarrow\infty}\frac{X_{t}}{t}\geq-v+m$

We conclude that $X_{t}\rightarrow-\infty$ with speed $v-m$. ∎

###  4.2 The case $v<m$

The assumption is satisfied, in particular, when $m=\infty$.

######  Proposition 5.

In the case $v<m$, $X_{t}$ is positive recurrent.

###### Proof.

Since $x\longmapsto m(x)$ is continuous, decreasing from $\mathbb{R}_{-}$ to
$\mathbb{R}_{+}$ and $m(0)=0<v<m$, $\exists$ $N>0$ such that $m(-N)=v$. We
choose an arbitrary $K>N$, so that for all $x<-K$

$\psi(x)>m(-K)-v>0.$

Assume that $X_{0}<-K$, and define the stopping time

$T_{K}=\inf\\{t>0,X_{t}\geq-K\\}$

We have

$\begin{split}X_{t\wedge T_{K}}&=X_{0}+\int_{0}^{t\wedge
T_{K}}\psi(X_{s})ds+\int_{[0,t\wedge
T_{K}]\times\mathbb{R}_{+}\times[0,1]}\alpha\varphi(X_{s^{-}},\alpha,\xi)\bar{M}(ds,d\alpha,d\xi)\\\
&>X_{0}+(m(-K)-v)(t\wedge T_{K})+\mathcal{M}_{t\wedge T_{K}}.\end{split}$

Thus,

$0>\mathbb{E}(X_{t\wedge T_{K}})>X_{0}+(m(-K)-v)\mathbb{E}(t\wedge T_{K}).$

Now let $t$ tend to $\infty$. It follows by monotone convergence that

$\mathbb{E}(T_{K})<\frac{-X_{0}}{m(-K)-v}<\infty.$ (20)

Given any fixed $T>0$, let $p$ denote the lower bound of the probability that,
starting from any given point $x\in[-K,0)$ at time $t_{0}$, $X$ hits
$[0,+\infty)$ before time $T$. Clearly $p>0$. We now define a geometric random
variable $\beta$ with success probability $p$. Let us restart our process $X$
at time $t_{0}=T_{K}$ from $x_{0}\in[-K,0)$. If $X$ hits zero before time $T$,
then $\beta=1$. If not, we look at the position $X_{T}$ of $X$ at time $T$.
Two cases are possible:

  * •

If $X_{T}<-K$, we wait until $X$ goes back above $-K$. Since
$X_{T}\geq-(K+vT)$, the time $\alpha_{2}$ needed to do so satisfies

$\mathbb{E}(\alpha_{2})\leq\frac{K+vT}{m(-K)-v}.$

This calculation is similar to (20).

  * •

If $X_{T}\geq-K$, we start afresh from there, since the probability to reach
zero in less than $T$ is greater than or equal to $p$.

So either at time $T$ or at time $T+\alpha_{2}$, we start again from a level
which is above $-K$. If $[0,+\infty)$ is reached during the next time interval
of length $T$, then $\beta=2$. If not, we repeat the procedure. A.s. one of
the mutually independent trials is successful. We have that

$T_{0}<T_{K}+\sum_{i=1}^{\beta}\left(T+\alpha_{i}\right),$

where the random variables $\left(\alpha_{i}\right)_{i}$ are i.i.d, globally
independent of $\beta$. Hence

$\mathbb{E}T_{0}<\mathbb{E}T_{K}+\frac{1}{p}\left(T+\frac{K+vT}{m(-K)-v})\right),$

and the process is positive recurrent. ∎

###  4.3 The case $v=m$

We first state a lemma that we will apply several times in this section.

######  Lemma 4.

Let $X_{t}$ be a FV càdlàg process.

  1. 1.

If $\Phi\in C^{1}$, then

$\Phi(X_{t})=\Phi(X_{0})+\int_{0}^{t}\Phi^{\prime}(X_{s^{-}})dX_{s}+\sum_{s\leq
t,\Delta X_{s}\neq 0}\Phi(X_{s^{-}}+\Delta
X_{s})-\Phi(X_{s^{-}})-\Phi^{\prime}(X_{s^{-}})\Delta X_{s},$

where $\Delta X_{s}=X_{s}-X_{s^{-}}$, $\forall s$.

  2. 2.

Moreover, if $\Phi\in C^{2}$ such that $\Phi^{\prime\prime}$ is an increasing
function and $\Delta X_{s}\geq 0$ for all $s$, then

$\Phi(X_{t})-\Phi(X_{0})-\int_{0}^{t}\Phi^{\prime}(X_{s^{-}})dX_{s}\geq\frac{1}{2}\sum_{s\leq
t,\Delta X_{s}\neq 0}\Phi^{\prime\prime}(X_{s})(\Delta X_{s})^{2}.$

If $\Phi\in C^{2}$ such that $\Phi^{\prime\prime}$ is a decreasing function
and $\Delta X_{s}\geq 0$ for all $s$, then

$\Phi(X_{t})-\Phi(X_{0})-\int_{0}^{t}\Phi^{\prime}(X_{s^{-}})dX_{s}\leq\frac{1}{2}\sum_{s\leq
t,\Delta X_{s}\neq 0}\Phi^{\prime\prime}(X_{s^{-}})(\Delta X_{s})^{2}.$

In particular, choosing $\Phi(x)=x^{2}$, we deduce that

$X_{t}^{2}=X_{0}^{2}+2\int_{0}^{t}X_{s^{-}}dX_{s}+\sum_{s\leq t}\left(\Delta
X_{s}\right)^{2}.$ (21)

###### Proof.

The first part of this lemma is a well known result (see Protter 2005). We
will only prove part 2 of the lemma. If $\Phi\in C^{2}$ then it follows from
Taylor’s formula that there exists a random function $\beta$ taking its values
in $[0,1]$ such that for all $s$

$\Phi(X_{s})-\Phi(X_{s^{-}})-\Phi^{\prime}(X_{s^{-}})\Delta
X_{s}=\frac{1}{2}\Phi^{\prime\prime}(X_{s^{-}}+\beta_{s}\Delta
X_{s})\left(\Delta X_{s}\right)^{2}.$

If $\Phi^{\prime\prime}$ is an increasing function and $y\geq 0$ then

$\Phi^{\prime\prime}(x)\leq\Phi^{\prime\prime}(x+\beta_{s}y)\leq\Phi^{\prime\prime}(x+y).$

If $\Phi^{\prime\prime}$ is a decreasing function and $y\geq 0$ then

$\Phi^{\prime\prime}(x+y)\leq\Phi^{\prime\prime}(x+\beta_{s}y)\leq\Phi^{\prime\prime}(x).$

∎

Note that $V\leq\infty$ and at this stage we do not assume that $V$ is finite.
In the case $m=v$, the asymptotic behavior of the process $X_{t}$ depends on
the asymptotic behavior of the mean net rate of adaptation $\psi(x)$ defined
in (9) as $x\rightarrow-\infty$.

######  Proposition 6.

We assume that $m=v$. If moreover

$\limsup_{x\rightarrow-\infty}\left|x\psi(x)\right|<\frac{V}{2},$ (22)

then the process $X_{t}$ is null recurrent.

We first establish

######  Lemma 5.

Under the condition $m<\infty$, we have that

$\lim_{x\to-\infty}\frac{V(x)}{|x|}=0.$

###### Proof.

Consider, for $x<x_{0}<0$,

$\displaystyle\frac{V(x)}{|x|}$
$\displaystyle\leq\frac{1}{|x|}\int_{0}^{2|x|}\alpha^{2}\nu(d\alpha)$
$\displaystyle=\frac{1}{|x|}\int_{0}^{2|x_{0}|}\alpha^{2}\nu(d\alpha)+\frac{1}{|x|}\int_{2|x_{0}|}^{2|x|}\alpha^{2}\nu(d\alpha)$
$\displaystyle\leq\frac{1}{|x|}\int_{0}^{2|x_{0}|}\alpha^{2}\nu(d\alpha)+2\int_{2|x_{0}|}^{\infty}\alpha\nu(d\alpha),$

hence

$\limsup_{x\to-\infty}\frac{V(x)}{|x|}\leq
2\int_{2|x_{0}|}^{\infty}\alpha\nu(d\alpha),$

and our condition implies that the last right hand side tends to 0, as
$x_{0}\to-\infty$. The results follows. ∎

We can now return to the

###### Proof of Proposition 6.

First note that, since $m=v$ implies $\psi(x)\leq 0$ for all $x\leq 0$,
condition (22) is equivalent to

$\liminf_{x\rightarrow-\infty}|x|\psi(x)>-\frac{V}{2}.$

To prove recurrence under condition (22), we recall that

$X_{t}=X_{0}+\int_{0}^{t}\psi(X_{s})ds+\mathcal{M}_{t}.$ (23)

We will apply Lemma 3 with $f(x)=\log|x|$, with $x<0$. Here $f^{\prime\prime}$
is decreasing. Hence as long as $X_{t}$ remains negative,

$\begin{split}\log|X_{t}|&\leq\log|X_{0}|+\int_{0}^{t}\frac{\psi(X_{s})}{X_{s}}ds+\int_{0}^{t}\frac{1}{X_{s^{-}}}d\mathcal{M}_{s}-\frac{1}{2}\sum_{s\leq
t}\frac{(\Delta X_{s})^{2}}{X^{2}_{s^{-}}}\\\
&\leq\log|X_{0}|+\int_{0}^{t}\frac{\psi(X_{s})}{X_{s}}ds+\int_{0}^{t}\frac{1}{X_{s^{-}}}d\mathcal{M}_{s}\\\
&-\frac{1}{2}\int_{0}^{t}\int_{\mathbb{R}_{+}}\int_{0}^{1}\frac{\alpha^{2}\varphi(X_{s^{-}},\alpha,\xi)}{X_{s^{-}}^{2}}\bar{M}(ds,d\alpha,d\xi)-\frac{1}{2}\int_{0}^{t}\frac{V(X_{s})}{X_{s}^{2}}ds\\\
&=\log|X_{0}|+\int_{0}^{t}\left(\frac{\psi(X_{s})}{X_{s}}-\frac{V(X_{s})}{2X_{s}^{2}}\right)ds+\hat{\mathcal{M}}_{t},\\\
\end{split}$

where $\hat{\mathcal{M}}$ is a martingale. For all $a<b<0$, define the
stopping time

$S_{a,b}=\inf(t>0,X_{t}\leq a\mbox{ or }X_{t}\geq b).$

It follows from our assumption that there exists $L>0$ such that

$\inf_{x\leq-L}\left(|x|\psi(x)+\frac{V(x)}{2}\right)>0.$ (24)

For any $N>L$, from Doob’s optional sampling theorem, if $-N<X_{0}<L$,

$\begin{split}\mathbb{E}\log|X_{t\wedge
S_{-N,-L}}|&\leq\log|X_{0}|+\mathbb{E}\int_{0}^{t\wedge
S_{-N,-L}}\left(\frac{\psi(X_{s})}{X_{s}}-\frac{V(X_{s})}{2X_{s}^{2}}\right)ds.\\\
\end{split}$

Letting $t$ tend to $\infty$,

$\mathbb{E}\log|X_{S_{-N,-L}}|\leq\log|X_{0}|.$

Define the stopping times

$\begin{split}T_{-L}^{\uparrow}&=\inf(t>0,X_{t}\geq-L),\\\
T_{-N}^{\downarrow}&=\inf(t>0,X_{t}\leq-N).\end{split}$

It follows from the previous estimate that

$\log N\times\mathbb{P}(T_{-N}^{\downarrow}<T_{-L}^{\uparrow})<\log|X_{0}|.$

We deduce that $\mathbb{P}(T_{-N}^{\downarrow}<T_{-L}^{\uparrow})\rightarrow
0$ as $N$ tend to $\infty$. We conclude that the process returns a.s. an
infinite number of times above $-L$, hence also above $0$ by a classical
argument (see the proof of Proposition 5). Therefore, the process $X$ is
recurrent.

Let now $X_{0}<-(L+1)$. For all $N>L$, multiplying (23) by $-1$, we have

$|X_{t\wedge S_{-N,-L}}|=|X_{0}|-\int_{0}^{t\wedge
S_{-N,-L}}\psi(X_{s})ds-\int_{0}^{t\wedge S_{-N,-L}}d\mathcal{M}_{s},$

By Doob’s theorem and letting $t$ tend to $\infty$, since again $\psi(x)\leq
0$ for $x\leq 0$

$\begin{split}\mathbb{E}|X_{S_{-N,-L}}|=|X_{0}|-\mathbb{E}\int_{0}^{S_{-N,-L}}\psi(X_{s})ds&\geq|X_{0}|\mbox{,
hence}\\\
L\mathbb{P}(T_{-L}^{\uparrow}<T_{-N}^{\downarrow})+N\mathbb{P}(T_{-N}^{\downarrow}<T_{-L}^{\uparrow})&\geq|X_{0}|.\\\
\end{split}$

We have

$\liminf_{N\rightarrow\infty}N\mathbb{P}(T_{-N}^{\downarrow}<T_{-L}^{\uparrow})\geq|X_{0}|-L>0.$
(25)

It follows from Lemma 3 that

$X_{t}^{2}=X_{0}^{2}-\int_{0}^{t}2|X_{s}|\psi(X_{s})ds+\int_{0}^{t}2X_{s^{-}}d\mathcal{M}_{s}+\sum_{s\leq
t}(\Delta X_{s})^{2}.$

On the other hand,

$\begin{split}\sum_{s\leq t}(\Delta
X_{s})^{2}&=\int_{0}^{t}\int_{\mathbb{R_{+}}}\int_{0}^{1}\alpha^{2}\varphi(X_{s^{-}},\alpha,\xi)\bar{M}(ds,d\alpha,d\xi)\\\
&+\int_{0}^{t}\int_{\mathbb{R_{+}}}\alpha^{2}g(X_{s^{-}},\alpha)\nu(d\alpha)ds.\\\
\end{split}$

Thus, from (24) and the monotonicity of $V(x)$

$X_{t\wedge S_{-N,-L}}^{2}\leq X_{0}^{2}+\int_{0}^{t\wedge
S_{-N,-L}}2V(-N)ds+\tilde{\mathcal{M}}_{t\wedge S_{-N,-L}},$

where $\tilde{\mathcal{M}}_{\cdot\wedge S_{-N,-L}}$ is a martingale. Letting
$t$ tend to $\infty$, we have for all $\epsilon>0$

$\begin{split}\mathbb{E}X_{S_{-N,-L}}^{2}&\leq
X_{0}^{2}+2V(-N)\mathbb{E}S_{-N,-L}\text{, hence}\\\
\mathbb{E}S_{-N,-L}&\geq\frac{L^{2}\mathbb{P}(T_{-L}^{\uparrow}<T_{-N}^{\downarrow})+N^{2}\mathbb{P}(T_{-N}^{\downarrow}<T_{-L}^{\uparrow})-X_{0}^{2}}{2V(-N)}.\\\
\end{split}$

It follows by monotone convergence that

$\mathbb{E}(T_{-L}^{\uparrow})=\lim_{N\rightarrow\infty}\mathbb{E}S_{-N,-L}\geq\liminf_{N\rightarrow\infty}\left\\{N\mathbb{P}(T_{-N}^{\downarrow}<T_{-L}^{\uparrow})\times\frac{N}{2V(-N)}-\frac{X_{0}^{2}}{2V(-N)}\right\\}.$

Combining this with Lemma 5 and (25), we deduce that
$\mathbb{E}T_{-L}^{\uparrow}=\infty$ and the process is null recurrent. ∎

######  Remark 2.

Condition (22) is rather weak. It is satisfied as soon as both the measure
$\nu$ and $V$ are finite. We give the proof below. It is also satisfied for
some measures that don’t have a second moment such as

$\nu(d\alpha)\approx\frac{d\alpha}{\alpha^{2+\delta}}\mathbf{1}_{\\{\alpha\geq
1\\}},\quad\frac{1}{2}<\delta\leq 1.$

######  Proposition 7.

If $\nu$ is a finite measure, $V<\infty$ and the fixation probability is given
by (4) and (5), then (22) is satisfied.

###### Proof.

Let for all $x$,

$\begin{split}D(x)=|x\psi(x)|-\frac{V(x)}{2}&=|x|\int_{2|x|}^{\infty}\alpha\nu(d\alpha)-\int_{0}^{2|x|}\frac{\alpha^{2}}{2}\nu(d\alpha)\\\
&+\int_{0}^{2|x|}\left(|x|\alpha+\frac{\alpha^{2}}{2}\right)e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha).\\\
\end{split}$

It follows by dominated convergence that

$\frac{1}{2}\int_{0}^{\infty}\alpha^{2}e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\mathbf{1}_{[0,2|x|]}\nu(d\alpha)\xrightarrow[x\rightarrow-\infty]{}0,$

since ,

$\begin{split}&\alpha^{2}e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\mathbf{1}_{[0,2|x|]}\xrightarrow[x\rightarrow-\infty]{a.s.}0\text{
for all }\alpha>0,\\\ \text{and
}&\alpha^{2}e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\mathbf{1}_{[0,2|x|]}\leq\alpha^{2}\in
L^{1}(\nu).\\\ \end{split}$

On the other hand,

$\int_{0}^{2|x|}|x|\alpha
e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha)=\int_{0}^{|x|}|x|\alpha
e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha)+\int_{|x|}^{2|x|}|x|\alpha
e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha).$

Note that if $0\leq\alpha\leq|x|$ then $2|x|-\alpha\geq|x|$, thus

$e^{-2\sigma\alpha(2|x|-\alpha)}\leq e^{-2\sigma\alpha|x|}.$

In addition the function

$\begin{array}[]{cccc}f_{\sigma}:&\mathbb{R}_{+}&\rightarrow&\mathbb{R}_{+}\\\
&z&\rightarrow&f_{\sigma}(z)=ze^{-\sigma z}\end{array}$

has a maximum for $z=\frac{1}{\sigma}$. It follows that

$\int_{0}^{|x|}|x|\alpha
e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha)\leq\frac{2}{\sigma
e}\int_{0}^{\infty}e^{-\sigma|x|\alpha}\nu(d\alpha)\xrightarrow[x\rightarrow-\infty]{}0,$

again by dominated convergence. The second term also goes to 0 when
$x\rightarrow-\infty$. In fact,

$\int_{|x|}^{2|x|}|x|\alpha
e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha)\leq\int_{0}^{\infty}\alpha^{2}e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\nu(d\alpha),$

and by the same argument as before the result follows since

$\begin{split}&\alpha^{2}e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\mathbf{1}_{[|x|,2|x|]}\xrightarrow[x\rightarrow-\infty]{a.s.}0\text{
for all }\alpha>0,\\\ \text{and
}&\alpha^{2}e^{-2\sigma\alpha\left(2|x|-\alpha\right)}\mathbf{1}_{[|x|,2|x|]}\leq\alpha^{2}\in
L^{1}(\nu).\end{split}$

Furthermore, it follows from the fact that $V$ is finite that

$|x|\int_{2|x|}^{\infty}\alpha\nu(d\alpha)\leq\int_{2|x|}^{\infty}\frac{\alpha^{2}}{2}\nu(d\alpha)\xrightarrow[x\rightarrow-\infty]{}0.$

Hence,

$\limsup_{x\rightarrow-\infty}D(x)=-\frac{V}{2}<0.$

∎

We now consider the case $m=v$ and
$\liminf\limits_{x\rightarrow-\infty}\left|x\psi(x)\right|>\frac{V}{2}$, which
implies in particular that $V<\infty$.

######  Proposition 8.

Assume that $m=v$ and

$\liminf_{x\rightarrow-\infty}\left|x\psi(x)\right|>\frac{V}{2}.$ (26)

If, moreover, there exist $0<p_{0}<1$ and $0<\beta_{0}<1$ such that for all
$0<\beta<\beta_{0}$

$|x|^{p_{0}+2}\int_{-\beta
x}^{\infty}\alpha^{2}g(x,\alpha)\nu(d\alpha)\xrightarrow[x\rightarrow-\infty]{}0,$
(27)

then $X_{t}$ is transient, that is, $X_{t}\rightarrow-\infty$, and moreover
$\frac{X_{t}}{t}\rightarrow 0$.

######  Remark 3.

The conditions of Proposition 8 are satisfied in the case where both $\nu$ is
infinite and its tail is thin enough, while $g$ is given by (4) and (5). For
example, if

$\nu(d\alpha)=\left(\frac{1}{\alpha^{1+\delta}}\mathbf{1}_{|\alpha|<1}+\rho(\alpha)\mathbf{1}_{|\alpha|>1}\right)d\alpha,$

where $\rho(\alpha)\leq C|\alpha|^{-(5+\delta^{\prime})},\>|\alpha|>1$ for
some $\delta,\delta^{\prime}>0$. Condition (26) follows from the fact that
$V<\infty$ while $|x\psi(x)|\rightarrow\infty$ as $|x|\rightarrow\infty$,
since, cf. proof of Proposition 7,

$\begin{split}|x|\int_{0}^{1}\alpha
e^{-2\sigma\alpha(2|x|-\alpha)}\nu(d\alpha)&\geq|x|\int_{0}^{1}\alpha^{-\delta}e^{-4\sigma\alpha|x|}d\alpha\\\
&=|x|^{\delta}\int_{0}^{|x|}e^{-4\sigma z}\frac{dz}{z^{\delta}}.\\\
\end{split}$

Condition (27) is easy to check.

###### Proof.

First note that condition (26) is equivalent to

$\limsup_{x\rightarrow-\infty}|x|\psi(x)<-\frac{V}{2}.$

Hence there exist $K>0$ and $0<p\leq p_{0}$ such that

$\sup_{x\leq-K}\left(|x|\psi(x)+(2p+1)\frac{V(x)}{2}\right)<0.$ (28)

Let $f$ be the $C^{2}(\mathbb{R})$-function such that $f(-1)=1$,
$f^{\prime}(-1)=p$, and

$f^{\prime\prime}(x)=\frac{p(p+1)}{|x|^{p+2}}\mathbf{1}_{\\{x\leq-1\\}}+p(p+1)\mathbf{1}_{\\{x\geq-1\\}},$

with $p$ being a real number in $(0,1)$ for which (28) holds. Then it follows
from Lemma 3 applied to $f$, since $f^{\prime\prime}$ is an increasing
function,

$f(X_{t})\leq
f(X_{0})+\int_{0}^{t}\psi(X_{s})f^{\prime}(X_{s})ds+\frac{1}{2}\int_{0}^{t}\int_{0}^{\infty}f^{\prime\prime}(X_{s}+\alpha)\alpha^{2}g(X_{s},\alpha)\nu(d\alpha)ds+\mathcal{N}_{t},$

where the martingale $\mathcal{N}$ is defined by

$\mathcal{N}_{t}=\frac{1}{2}\int_{0}^{t}\int_{0}^{\infty}\int_{0}^{1}\left[f^{\prime}(X_{s^{-}})+f^{\prime\prime}(X_{s^{-}}+\alpha)\alpha^{2}\right]\varphi(X_{s^{-}},\alpha,\xi)\bar{M}(ds,d\alpha,d\xi).$

Let us admit for the moment:

######  Lemma 6.

If (27) holds, then

$\lim_{x\rightarrow-\infty}|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)=p(p+1)V.$

This implies that

$\lim_{x\rightarrow-\infty}|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)<\lim_{x\rightarrow-\infty}p(2p+1)V(x).$

Hence, there exists $N\geq K$ such that for all $x\leq-N$,

$\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)<p(2p+1)\frac{V(x)}{|x|^{p+2}}.$

Thus, for all $k>0$ satisfying $-kN<X_{0}<-N$,

$\begin{split}f(X_{t\wedge S_{-kN,-N}})\ \leq\ &f(X_{0})+\int_{0}^{t\wedge
S_{-kN,-N}}\frac{p}{|X_{s}|^{p+1}}\left[\psi(X_{s})+(2p+1)\frac{V(X_{s})}{2|X_{s}|}\right]ds\\\
&+\mathcal{N}_{t\wedge S_{-kN,-N}}.\end{split}$

Now if $k\geq 3$, letting $X_{0}=-2N$, it follows from (28) that

$\mathbb{E}(f(X_{t\wedge S_{-kN,-N}}))\leq\frac{1}{(2N)^{p}}.$

Thus, if we let $t$ tend to $\infty$,

$\frac{1}{N^{p}}\mathbb{P}(S_{-kN,-N}=T_{-N}^{\uparrow}))\leq\mathbb{E}\frac{1}{|X_{S_{-kN,-N}}|^{p}}\leq\frac{1}{(2N)^{p}}.$

Now letting $k$ tend to $\infty$,

$\mathbb{P}(T_{-N}^{\uparrow}<\infty)\leq\frac{1}{2^{p}}.$

Thus, the process is transient, which means

$X_{t}\xrightarrow[t\rightarrow\infty]{}-\infty.$

And since $m=v<\infty$, it follows from Lemma 1 that
$\frac{\mathcal{M}_{t}}{t}\rightarrow 0$, hence

$\frac{X_{t}}{t}\xrightarrow[t\rightarrow\infty]{}0.$

∎

###### Proof of Lemma 4.

For any $0<\beta<\beta_{0}<1$, if $x<-(1-\beta)^{-1}$,

$\begin{split}|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)g(x,\alpha)\alpha^{2}\nu(d\alpha)&=|x|^{p+2}\int_{0}^{-\beta
x}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)\\\
&+|x|^{p+2}\int_{-\beta
x}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)\\\
&\leq\int_{0}^{-\beta
x}\frac{p(p+1)}{(1-\beta)^{p+2}}\alpha^{2}g(x,\alpha)\nu(d\alpha)\\\
&+|x|^{p+2}p(p+1)\int_{-\beta x}^{\infty}\alpha^{2}g(x,\alpha)\nu(d\alpha).\\\
\end{split}$

On the other hand,

$\begin{split}|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)&\geq
p(p+1)\int_{0}^{-\beta
x}\frac{|x|^{p+2}}{|x+\alpha|^{p+2}}\alpha^{2}g(x,\alpha)\nu(d\alpha)\\\
&>p(p+1)\int_{0}^{-\beta x}\alpha^{2}g(x,\alpha)\nu(d\alpha).\\\ \end{split}$

Letting $x\rightarrow-\infty$ in the two above inequalities, we deduce from
(27), which holds with $p_{0}$ replaced by $p\leq p_{0}$,

$\begin{split}p(p+1)V&\leq\liminf_{x\rightarrow-\infty}|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)\\\
&\leq\limsup_{x\rightarrow-\infty}|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)\\\
&\leq\frac{p(p+1)}{(1-\beta)^{p+2}}V.\\\ \end{split}$

Thus, letting $\beta\rightarrow 0$, it follows that

$|x|^{p+2}\int_{0}^{\infty}f^{\prime\prime}(x+\alpha)\alpha^{2}g(x,\alpha)\nu(d\alpha)\xrightarrow[x\rightarrow-\infty]{}p(p+1)V.$

∎

######  Remark 4.

We have not been able to precise the large time behavior of the process
$X_{t}$ when the measure $\nu$ is of the type

$\nu(d\alpha)\approx\frac{d\alpha}{\alpha^{2+\delta}}\mathbf{1}_{\\{\alpha\geq
1\\}},\quad 0<\delta\leq\frac{1}{2},$

which still satisfies $m<\infty$. In this case, $V=\infty$,
$|x\psi(x)|\rightarrow\infty$ as $|x|\rightarrow\infty$, and (27) also fails.

##  5 Generalization to the case of a time-variable speed

In the following, we treat the case where the speed of environmental change is
a random function of time

$v(t)=\int_{0}^{t}v_{1}(s)ds+\mathcal{R}_{t}$ (29)

where $v_{1}$ is a random function such that

$\frac{1}{t}\int_{0}^{t}v_{1}(s)ds\xrightarrow[t\rightarrow\infty]{}\bar{v},$

and $\mathcal{R}$ is a stochastic process. The stochastic equation describing
the evolution of phenotypic lag becomes

$X_{t}=X_{0}-v(t)+\int_{0}^{t}m(X_{s})ds+\mathcal{M}_{t}.$ (30)

As above, we study three cases:

###  5.1 The case $\bar{v}>m$

Here we assume that $\mathcal{R}$ satisfies the condition

$\frac{\mathcal{R}_{t}}{t}\xrightarrow[t\rightarrow\infty]{}0.$

This condition is verified by a Brownian motion for example. It is easy to see
that results (15) and (16) hold in the new context of equation (30). Following
the steps of the proof in section 1, we can see that $X_{t}\rightarrow-\infty$
with speed $\bar{v}-m$.

###  5.2 The case $\bar{v}<m$

Define $\mathcal{T}$ as the set of bounded stopping times. Now we assume that
there exists $0<c<\infty$ such that $\mathbb{E}\mathcal{R}_{T}\leq c$ for all
$T\in\mathcal{T}$. This condition is verified for example by a process sum of
a martingale and a bounded process. In this case, we will prove that the
process $X_{t}$ is positive recurrent. We can see from (15) and (29) that
there exist $M,N>0$ such that for $y<-M$ and $t>N$,

$m(x)-\frac{1}{t}\int_{0}^{t}v_{1}(s)ds>\frac{m-\bar{v}}{2}.$

We remind that

$T_{-M}^{\uparrow}=\inf\\{t>0,X_{t}\geq-M\\}.$

For the purpose of notation and without loss of generality, we denote
$T_{-M}^{\uparrow}$ by $T$. Assume that $X_{0}<-M$. It follows that for all
$t>N$,

$\mathbb{E}\int_{0}^{t\wedge
T}\left[m(X_{s})-v_{1}(s)\right]ds<-X_{0}+\mathbb{E}\mathcal{R}_{t\wedge T},$

since $X_{s}<0$ for $s<T$. We have

$\begin{split}\mathbb{E}\int_{0}^{t\wedge
T}\left[m(X_{s})-v_{1}(s)\right]ds&=\mathbb{E}\mathbf{1}_{T\geq
N}\int_{0}^{t\wedge T}\left[m(X_{s})-v_{1}(s)\right]ds\\\
&+\mathbb{E}\mathbf{1}_{T<N}\int_{0}^{t\wedge
T}\left[m(X_{s})-v_{1}(s)\right]ds\\\ &<-X_{0}+\mathbb{E}\mathcal{R}_{t\wedge
T},\\\ \end{split}$

and hence,

$\begin{split}\frac{m-\bar{v}}{2}\mathbb{E}\left(\mathbf{1}_{T\geq N}(t\wedge
T)\right)&\leq-X_{0}+\mathbb{E}\mathcal{R}_{t\wedge
T}-\mathbb{E}\mathbf{1}_{T<N}\int_{0}^{t\wedge
T}\left[m(X_{s})-v_{1}(s)\right]ds\\\ &\leq-
X_{0}+\mathbb{E}\mathcal{R}_{t\wedge
T}+\int_{0}^{N}v_{1}^{+}(s)ds.\end{split}$

Now let $t$ tend to $\infty$, yielding

$\mathbb{E}(\mathbf{1}_{T\geq
N}T)<\frac{2}{m-\bar{v}}\left(-X_{0}+\int_{0}^{N}v_{1}^{+}(s)ds+c\right)<\infty.$

Thus, $\mathbb{E}(T)<N+\mathbb{E}(\mathbf{1}_{T\geq N}T)<\infty$. From here,
it is not hard to prove that $\mathbb{E}X_{T^{\uparrow}_{0}}<\infty$. Thus,
$X_{t}$ is positive recurrent.

###  5.3 The case $\bar{v}=m$

Here we assume that $\mathcal{R}_{t}\equiv 0$. Even in this case stronger
assumptions need to be made. Define

$\begin{split}v_{\sup}&=\sup\limits_{\begin{subarray}{c}s\end{subarray}}v_{1}(s),\\\
v_{\inf}&=\inf\limits_{\begin{subarray}{c}s\end{subarray}}v_{1}(s),\\\
\psi_{\sup}(x)&=m(x)-v_{\sup},\\\ \psi_{\inf}(x)&=m(x)-v_{\inf},\\\
\end{split}$

We define two sets of assumptions:

Assumptions A

  * •

$v_{\sup}<\infty$,

  * •

$\liminf\limits_{x\rightarrow-\infty}|x|\psi_{\sup}(x)>-\frac{V}{2}$.

Assumptions B

  * •

$v_{\inf}<\infty$,

  * •

$\limsup\limits_{x\rightarrow-\infty}|x|\psi_{\inf}(x)<-\frac{V}{2}$.

Under the set of assumptions A, we can prove that the process is recurrent. We
have, however, not been able to prove null recurrence in the case of non-
constant $v$.

###### Ideas of Proof.

Apply Lemma 3 to the process in equation (30) with $f(x)=\log|x|$, with $x<0$.
Here $f^{\prime\prime}$ is decreasing. Hence, as long as $X_{t}$ remains
negative,

$\begin{split}\log|X_{t}|&\leq\log|X_{0}|+\int_{0}^{t}\left(\frac{\psi_{\sup}(X_{s})}{X_{s}}-\frac{V(X_{s})}{2X_{s}^{2}}\right)ds+\int_{0}^{t}\frac{v_{\sup}-v_{1}(s)}{X_{s}}ds+\mathcal{M^{\prime}}_{t}\\\
&<\log|X_{0}|+\int_{0}^{t}\left(\frac{\psi_{\sup}(X_{s})}{X_{s}}-\frac{V(X_{s})}{2X_{s}^{2}}\right)ds+\mathcal{M^{\prime}}_{t},\\\
\end{split}$

where $\mathcal{M^{\prime}}$ is a martingale. Then we continue the proof as
for the case of constant speed. ∎

Under the set of assumptions B and hypothesis (27), we can prove that

$X_{t}\xrightarrow[t\rightarrow\infty]{}-\infty\quad\text{ and
}\quad\frac{X_{t}}{t}\xrightarrow[t\rightarrow\infty]{}0.$

###### Ideas of Proof.

We take the same function $f$ we constructed in the case of constant speed. We
have $f^{\prime}>0$, and

$\begin{split}f(X_{t})&\leq
f(X_{0})+\int_{0}^{t}\psi_{\inf}(X_{s})f^{\prime}(X_{s})ds+\int_{0}^{t}(v_{\inf}-v_{1}(s))f^{\prime}(X_{s})ds\\\
&+\frac{1}{2}\int_{0}^{t}\int_{0}^{\infty}f^{\prime\prime}(X_{s}+\alpha)\alpha^{2}g(X_{s},\alpha)\nu(d\alpha)ds+\mathcal{N^{\prime}}_{t}\\\
&\leq
f(X_{0})+\int_{0}^{t}\psi_{\inf}(X_{s})f^{\prime}(X_{s})ds+\frac{1}{2}\int_{0}^{t}\int_{0}^{\infty}f^{\prime\prime}(X_{s}+\alpha)\alpha^{2}g(X_{s},\alpha)\nu(d\alpha)ds+\mathcal{N^{\prime}}_{t},\\\
\end{split}$

where $\mathcal{N^{\prime}}$ is a martingale. Then we continue the proof as
for the case of constant speed. ∎

#### Acknowledgement

The authors want to thank Michael Kopp for formulating the biological question
which led us to this research, and for many stimulating discussions in the
course of our work.

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