# Trend to equilibrium for run and tumble equations with non-uniform tumbling
kernels

Josephine Evans111Warwick Mathematics Institute, University of Warwick, Zeeman
Building, Coventry CV4 7AL, United Kingdom<EMAIL_ADDRESS>Havva Yoldaş 222Delft Institute of Applied Mathematics, Faculty of Electrical
Engineering, Mathematics and Computer Science, Delft University of Technology,
Mekelweg 4, 2628CD Delft, The Netherlands<EMAIL_ADDRESS>
###### Abstract

We study the long-time behaviour of a run and tumble model which is a kinetic-
transport equation describing bacterial movement under the effect of a
chemical stimulus. The experiments suggest that the non-uniform tumbling
kernels are physically relevant ones as opposed to the uniform tumbling kernel
which is widely considered in the literature to reduce the complexity of the
mathematical analysis. We consider two cases: (i) the tumbling kernel depends
on the angle between pre- and post-tumbling velocities, (ii) the velocity
space is unbounded and the post-tumbling velocities follow the Maxwellian
velocity distribution. We prove that the probability density distribution of
bacteria converges to an equilibrium distribution with explicit (exponential
for (i) and algebraic for (ii)) convergence rates, for any probability measure
initial data. To the best of our knowledge, our results are the first results
concerning the long-time behaviour of run and tumble equations with non-
uniform tumbling kernels.

†† _Keywords and phrases._ run and tumble equation, hypocoercivity, kinetic
equations, Harris’s theorem.†† _2020 Mathematics Subject Classification._
35B40, 35Q92, 37A25.

###### Contents

  1. 1 Introduction
    1. 1.1 Assumptions and main results
    2. 1.2 Methodology
  2. 2 Angle-dependent tumbling kernel
    1. 2.1 Minorisation condition
    2. 2.2 Foster-Lyapunov condition
  3. 3 Unbounded velocity spaces
    1. 3.1 Minorisation condition
    2. 3.2 Foster-Lyapunov condition
    3. 3.3 Subgeometric convergence rates

##  1 Introduction

We consider a kinetic-transport equation which describes the movement of
biological microorganisms biased towards a chemoattractant. The model is
called the _run and tumble_ equation and introduced in [1, 22] based on some
experimental observations [3] on the bacterium called _Escherichia coli (E.
coli)_. The equation is given by

$\displaystyle\begin{split}\partial_{t}f+v\cdot\nabla_{x}f&=\int_{\mathcal{V}}\lambda(v^{\prime}\cdot\nabla_{x}M(x))\kappa(v,v^{\prime})f(t,x,v^{\prime})\,\mathrm{d}v^{\prime}-\lambda(v\cdot\nabla_{x}M(x))f(t,x,v)\\\
f(0,x,v)&=f_{0}(x,v)\end{split}$ (1)

where $f:=f(t,x,v)\geq 0$ is the density distribution of microorganisms at
time $t\geq 0$ at a position $x\in\mathbb{R}^{d}$, moving with a velocity
$v\in\mathcal{V}\subseteq\mathbb{R}^{d}$.

The term $\lambda(v^{\prime}\cdot\nabla_{x}M(x))\kappa(v,v^{\prime})$ is
called the _tumbling frequency_ where $\lambda:\mathbb{R}\to[0,\infty)$ is the
_tumbling rate_. The tumbling rate $\lambda$ depends on the gradient of the
external signal $M$ along the direction of the velocity $v$, and it is defined
by

$\displaystyle M(x)=m_{0}+\log(S(x)),$

where $m_{0}\in\mathbb{R}^{+}_{0}$ represents the external signal in the
absence of a chemical stimulus and $S(x)$ is a given function for the density
of the chemoattractant. In Eq. (1), the _tumbling or turning kernel_
$\kappa(v,v^{\prime})$ is a probability distribution on the space
$\mathcal{V}$ and gives the probability of moving from velocity $v$ to
velocity $v^{\prime}$, i.e.
$\int_{\mathcal{V}}\kappa(v,v^{\prime})\,\mathrm{d}v^{\prime}=1$.

In the case of peritrichous bacteria such as _E. coli_ and _Salmonella
typhimurium_ , experiments conducted in [3, 17] suggest that $\kappa$ depends
only on the relative angle $\theta$ between the pre- and post-tumbling
velocities $v$ and $v^{\prime}$ respectively. Particularly, for bacterium _E.
coli_ , the tumbling kernel $\kappa$ is given by

$\displaystyle\kappa(v,v^{\prime})=\kappa(\theta)=\frac{g(\theta)}{2\pi\sin\theta},\quad\theta=\arccos\left(\frac{v\cdot
v^{\prime}}{|v||v^{\prime}|}\right),$

where $g(\theta)$ is a sixth order polynomial satisfying $g(0)=g(\pi)=0$ (see
[21, 4] for more details). The exact form of $g$ is provided in [14] by
polynomial fitting to the data of [3]. Here we will work in a bounded velocity
space.

In our previous paper [13] we studied this equation under the assumption that
$\kappa$ was uniformly bounded above and below. However, this assumption is
not realistic as the bacteria are not able to turn a full half circle. We aim
to extend our previous work to the case where the maximum turning angle of the
bacteria may be bounded.

We are also interested in unbounded velocity spaces, i.e.,
$v\in\mathcal{V}=\mathbb{R}^{d}$. In this setting, we consider that the
tumbling kernel is given by the Maxwellian distribution on the post-tumbling
velocities independently from the pre-tumbling velocities, i.e.,

$\displaystyle\kappa(v,v^{\prime})=\mathcal{M}(v)=\frac{1}{(2\pi)^{\frac{d}{2}}}e^{\frac{-|v|^{2}}{2}}.$

To the best of our knowledge, we provide the first results concerning the
long-time behaviour of the run and tumble equation with these non-uniform
tumbling kernels which are more physically relevant in terms of modelling the
chemotactic bacterial motion. We believe the main reason for this is the fact
that the classical hypocoercivity techniques such as [9] cannot be used for
the run and tumble equation even though it is a linear, conservative kinetic
equation. On the other hand, Harris-type theorems (see e.g. [8, 24]) proved
very effective for obtaining quantitative hypocoercivity results, especially
for kinetic equations arising from applied sciences where classical techniques
provide limited results. We elaborated on this fact in our previous paper [13]
in detail and a brief explanation can be found below in the paragraph
_Motivation and novelty_.

#### Summary of previous results.

Previous important works on the linear run and tumble equation include [20,
21, 6, 19, 5]. In [20, 21], the authors study the diffusion approximation to a
linear run and tumble equation and the diffusion limit of this equation to
obtain macroscopic chemotaxis equations, respectively. Using the $L^{2}$
hypocoercivity techniques developed in [9], the authors [6] show the existence
of a unique equilibrium and exponential decay towards it in dimension $d=1$.
This paper works with the assumption that the tumbling rate $\lambda$ can take
two values depending if the bacteria is travelling up or down the gradient of
the chemoattractant density. Though it is expressed differently in [6] the
tumbling rate can be written as

$\displaystyle\lambda(x,v,v^{\prime})=1+\chi\text{sgn}(x\cdot
v),\quad\chi\in(0,1),$

where $\chi$ is called the chemotactic sensitivity. In [19], the authors
extended this results to higher dimensions $d\geq 1$ considering

$\displaystyle\lambda(x,v,v^{\prime})=1-\chi\text{sgn}(\,\partial_{t}S+v\cdot\nabla_{x}S),\quad\chi\in(0,1).$

The result in [19] works under the assumption that the concentration of the
chemoattractant $S(x)$ is radially symmetric and decreasing in $x$ such that
$S(x)\to 0$ as $|x|\to\infty$. In our previous paper, [13] we improve the
result in [19] by proving the exponential convergence to unique equilibrium
without requiring $S(x)$ to be radially symmetric. Our result is valid in an
arbitrary dimension $d\geq 1$ where the tumbling kernel $\lambda$ can take a
much more general form. Most importantly, we show that existence and
convergence to a steady state hold when $\lambda$ is a Lipschitz function. Our
techniques in [13] are based on Harris-type theorems coming from the ergodic
theory of Markov processes. They differ from the techniques used in [19] that
are based on the Krein-Rutman theorem. Moreover in [13], we also consider a
non-linear run and tumble model where we can prove the exponential decay to a
unique equilibrium. In [5], the author studies a non-linear coupled run and
tumble equation in one dimension $d=1$. Even though [5] is mainly concerned
with the travelling wave solutions of the non-linear equation, as an
intermediate step, the author shows the existence of steady states for the
linear equation.

All these previous works deal with situations where the tumbling kernel, which
tells us how the post-tumbling velocities depend on the pre-tumbling
velocities, is essentially uniform, i.e., $\kappa\equiv 1$, and on a bounded
velocity space. The main goal of this work is to look at physically more
realistic cases where the tumbling kernel is not uniform and the bacteria are
not able to turn to every angle in one tumbling event.

#### Motivation and novelty.

In our previous paper [13] we studied Eq. (1) in the case where the velocity
space was a ball and the post-tumbling velocities were uniformly distributed
so that $\kappa\equiv 1$. We showed exponential convergence to equilibrium in
suitable weighted total variation distances. The result was built around
probabilistic techniques called Harris-type theorems coming from the theory of
Markov processes. The goal of this paper is to extend the previous result to a
wider class of tumbling kernels that are non-uniform and physically more
relevant for modelling the motion of chemotactic bacteria.

The typical tools for showing convergence to equilibrium for kinetic equations
come from _hypocoercivity_. It is important to note that "classical"
hypocoercivity techniques such as [23, 9] cannot be applied without strong a
priori knowledge of the steady state which we do not have for the run and
tumble equation. This is the main reason for scarce results on the long-time
behaviour with arbitrary dimensions $d\geq 1$ even though the run and tumble
equation is widely studied. Harris-type theorems (see, e.g., [8, 15, 16] and
references therein) provide the existence and uniqueness of the steady state
as a by-product while simultaneously showing convergence to equilibrium. Using
these tools, one can obtain quantitative hypocoercivity results in weighted
total variation (or in weighted $L^{1}$) distances independently from the
initial data (see, e.g., [7, 24]).

Harris-type theorems are based on verifying two hypotheses: minorisation and
geometric drift conditions. The minorisation condition requires providing a
quantifiable positive lower bound on the process. Therefore, non-uniform
tumbling kernels pose additional challenges compared to the case with a
uniform tumbling kernel when using Harris-type theorems. This is because we
need to prove lower bounds on the law uniform over a large set of initial
conditions. This means that we need to find some possible paths a bacterium
can take when moving from one point in phase space to another. The fact that
the tumbling angle is bounded means that bacteria may only be able to travel
between two points along paths involving many tumbling events. Tracking bounds
on the probabilities of these complex paths (and paths close to them) is
challenging and required us to generate new technical tools.

###  1.1 Assumptions and main results

The two main results of our paper concern angularly dependent tumbling kernels
and unbounded velocity spaces. In both cases, we will make the following
assumptions on the tumbling rate, $\lambda$, and the logarithm of the
chemoattractant concentration, $M$.

  * (H1)

The tumbling rate $\lambda(m):\mathbb{R}\to(0,\infty)$ is a function of the
form

$\displaystyle\lambda(m)=1-\chi\psi(m),\quad\chi\in(0,1)$ ($\lambda$)

where $\psi$ is a bounded (with $\|\psi\|_{\infty}\leq 1$), odd, increasing
function and $m\psi(m)\in W^{1,\infty}_{loc}(\mathbb{R})$.

  * (H2)

There exists a contstant a strictly positive integer $b>0$, such that for
every $B>0$, there exists $c>0$ depending on $B$ so that

$\displaystyle m\psi(m)\geq c|m|^{b}$ ($m$)

for $|m|\leq B$. We note that this holds if $\psi$ is the sign function or if
it is odd and differentiable around zero with strictly positive $k^{th}$
derivative for some $k\geq 1$.

  * (H3)

We suppose that $M(x)\to-\infty$ as $|x|\to\infty$, $|\nabla_{x}M(x)|$ is
bounded and that there exist $R\geq 0$ and $m_{*}>0$ such that whenever
$|x|>R$ we have

$\displaystyle|\nabla_{x}M(x)|\geq m_{*}.$ ($M$)

Moreover, we suppose that $\operatorname{Hess}(M)(x)\to 0$ as $|x|\to\infty$
and $|\operatorname{Hess}(M)(x)|$ is bounded.

The following assumptions concern the tumbling kernels we work with.

  * (H4)

We assume that $\mathcal{V}=V_{0}\mathbb{S}^{d-1}$ and there exist
$\alpha,\beta>0$ such that

$\displaystyle\kappa(v,v^{\prime})=\kappa_{1}(\theta)\geq\beta\mathds{1}_{|\theta|<\alpha}(\theta),$
($\kappa_{1}$)

where $\theta=\arccos\left(\dfrac{v\cdot v^{\prime}}{V_{0}^{2}}\right)$ and
$\kappa_{1}$ is a decreasing function of $|\theta|$ (similar arguments work if
$\kappa_{1}$ is even and bounded below by a decreasing function of
$|\theta|$).

  * (H5)

We assume that $\mathcal{V}=\mathbb{R}^{d}$ and the tumbling kernel is given
by the Maxwellian distribution on the post-tumbling velocities independently
from the pre-tumbling velocities,

$\displaystyle\kappa(v,v^{\prime})=\kappa_{2}(v)=\frac{1}{(2\pi)^{\frac{d}{2}}}e^{-\frac{|v|^{2}}{2}}.$
($\kappa_{2}$)

After the assumptions, we state the main results of the paper below.

######  Theorem 1.1 (Angularly dependent tumbling kernel ($\kappa_{1}$)).

Suppose that $t\mapsto f_{t}$ is the solution of Equation (1) with initial
data $f_{0}\in\mathcal{P}(\mathbb{R}^{2}\times\mathbb{S}^{1})$. We suppose
that hypotheses (H1), (H2), (H3) and (H4) are satisfied. Then there exist
positive constants $C,\sigma$ (independent of $f_{0}$) such that

$\displaystyle\|f_{t}-f_{\infty}\|_{*}\leq Ce^{-\sigma
t}\|f_{0}-f_{\infty}\|_{*},$ (2)

where $f_{\infty}$ is the unique steady state solution to Equation (1). The
norm $\|\cdot\|_{*}$ is the weighted total variation norm defined by

$\displaystyle\|\mu\|_{*}:=\int_{\mathbb{R}^{d}}\int_{\mathbb{S}^{d-1}}\Big{(}1-\frac{\gamma}{1-C_{\kappa}}v\cdot\nabla_{x}M(x)-\gamma
Av\cdot\nabla_{x}M(x)\psi(v\cdot\nabla_{x}M(x))\Big{)}e^{-\gamma
M(x)}|\mu|\,\mathrm{d}v\,\mathrm{d}x,$ (3)

where $\gamma,A,C_{\kappa}>0$ are positive constants that can be computed
explicitly and will be chosen so that $\|\cdot\|_{*}$ is indeed a norm, and
$\mu$ is a finite measure.

######  Remark 1.2.

We believe that Theorem 1.1 works in arbitrary dimension $d\geq 1$. The reason
for stating Theorem 1.1 in dimension $d=2$ is that we provide the proof of
Proposition 2.7 only in $d=2$. We do not believe that there is a major
mathematical obstacle in proving it in higher dimensions. However, even in
$d=2$, the computations become delicate, and the notations get intricate.
Therefore, we decided to provide it in $d=2$ to keep the exposition of our
ideas clear.

######  Theorem 1.3 (Unbounded velocity space with tumbling kernel
($\kappa_{2}$)).

Suppose that $t\mapsto f_{t}$ is the solution of Equation (1) with initial
data $f_{0}\in\mathcal{P}(\mathbb{R}^{d}\times\mathbb{R}^{d})$. We suppose
that hypotheses (H1), (H2), (H3) and (H5) are satisfied and assume further
that $M\operatorname{Hess}(M)$ is bounded. Then there exists a positive
constant $C>0$ such that

$\displaystyle\|f_{t}-f_{\infty}\|_{TV}\leq Ct^{-1}M_{f_{0}}$ (4)

where

$\displaystyle
M_{f_{0}}:=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}f_{0}(x,v)\Big{(}1+M(x)^{2}+2v\cdot\nabla_{x}M(x)M(x)\Big{(}1+\frac{\chi}{1+\chi}\psi\left(v\cdot\nabla_{x}M(x)\Big{)}\Big{)}+Av^{2}\right)\,\mathrm{d}v\,\mathrm{d}x,$
(5)

with $A>0$ is a constant that can be computed explicitly and it is
sufficiently large so that $M_{f_{0}}>0$.

Even though we study the long-time behaviour of the run and tumble equation
(1) with the non-uniform tumbling kernels in this paper, we would like to
briefly comment on the Cauchy theory for these equations.

#### Cauchy theory for Equation(1).

As Eq. (1) is a linear integro-differential equation with bounded
coefficients, showing the existence and uniqueness of global-in-time, measure-
valued solutions is relatively standard. One could either use Picard iteration
arguments to construct short-time solutions and then use the fact that
$\lambda$ is bounded to show these can be glued together globally in time or
one can directly write down a Markov process whose law satisfies Eq. (1). We
briefly explain how to do the latter. Let us generate a Poisson process with
intensity $(1+\chi)$ and call its jump times $J_{1},J_{2},\dots$ and a series
of thinning variables $U_{1},U_{2},\dots$ independent and all having the
uniform law on $[0,1+\chi]$ then we define initial points $(X_{0},V_{0})$
having law $f_{0}$ then set $J_{0}=0$ and for $t\in(J_{i},J_{i+1})$ we write
$X_{t}=X_{J_{i}}+(t-J_{i})V_{J_{i}}$ and $V_{t}=V_{J_{i}}$ then for
$t=J_{i+1}$ we set $X_{t}=X_{J_{i}}+(J_{i+1}-J_{i})V_{J_{i}}$ then if
$U_{i+1}\leq\lambda(V_{J_{i}}\nabla_{x}M(X_{t}))$ we generate $V_{J_{i+1}}$ as
a new random variable having law $\kappa(V_{J_{i}},\cdot)$ and if
$U_{i+1}>\lambda(V_{J_{i}}\nabla_{x}M(X_{t}))$ we set $V_{J_{i+1}}=V_{J_{i}}$.

#### Plan of the paper.

After listing our assumptions and stating our main results in Section 1.1, in
the following section (Section 1.2), we describe our methodology, particularly
we state Harris’s theorem in geometric and subgeometric settings, Theorems 1.4
and 1.6 respectively. Then, Sections 2 and 3 are dedicated to proving Theorems
1.1 and 1.3 for the cases (i) angularly dependent tumbling kernels and (ii)
unbounded velocity spaces respectively.

###  1.2 Methodology

This paper is an extension of the work in [13] and as such is built on
Harris’s theorem from Markov process theory. More precisely, Harris-type
theorems are used in showing geometric (exponential) or sub-geometric
(algebraic) rates of convergence to a unique equilibrium solution for
equations that can be described as Markov processes. Harris-type theorems rely
on verifying two hypotheses: a Foster-Lyapunov condition and a uniform
minorisation condition.

We briefly introduce the functional setting and some notations below and then
we provide the statements of the theorems both in geometric and sub-geometric
settings in the spirit of [16, 18, 8] and the references therein. We skip the
proofs of these theorems.

#### Notations.

We consider a measurable space $(\Omega,\Sigma)$ where $\Omega$ is a Polish
space endowed with a probability measure. We denote the space of probability
measures on $\Omega$ by $\mathcal{P}(\Omega)$. Note that in our setting
$\Omega=\mathbb{R}^{d}\times\mathcal{V}$ so that
$(x,v)\in\mathbb{R}^{d}\times\mathcal{V}=\Omega$. We sometimes use the
notation $z:=(x,v)$.

We define the weighted total variation (or weighted $L^{1}$) distance by

$\displaystyle\|\mu\|_{\phi}:=\int_{\Omega}\phi(z)|\mu|(\,\mathrm{d}z),$

where $\mu$ is a finite measure or a measurable function and
$\phi:\Omega\to[1,+\infty)$ is a measurable weight function.

We call $(S_{t})_{t\geq 0}$ a Markov (or stochastic) semigroup if it is a
linear semigroup conserving mass and positivity. Remark that if $f$ solves Eq.
(1), then $f(t,x,v)=S_{t}f_{0}(x,v)$ and $(S_{t})_{t\geq 0}$ is a Markov
semigroup since Eq. (1) is positivity and mass-preserving. Moreover, we also
use the notation $\,\partial_{t}f=\mathcal{L}[f]$ equivalently to Eq. (1). If
Eq. (1) has a stationary soluton $f_{\infty}$, this means that
$\,\partial_{t}f_{\infty}=\mathcal{L}[f_{\infty}]=0$ and $f_{\infty}$ is an
invariant measure for the semigroup $S_{t}$, i.e.,
$S_{t}f_{\infty}=f_{\infty}$.

######  Theorem 1.4 (Harris’s Theorem).

Let $(S_{t})_{t\geq 0}$ be a Markov semigroup satisfying the following
hypotheses:

######  Hypothesis 1 (Foster-Lyapunov condition).

There exist positive constants $\zeta,D$ and a continuous, measurable function
$\phi:\Omega\to[1,+\infty)$ such that

$\displaystyle\mathcal{L}^{*}[\phi](z)\leq D-\zeta\phi(z).$ (FL1)

######  Hypothesis 2 (Minorisation condition).

There exist a probability measure $\nu$, a constant $\beta\in(0,1)$ and some
time $\tau>0$ such that

$\displaystyle S_{\tau}\delta_{z}\geq\beta\nu,\quad\mbox{for all
}z\in\mathcal{C},$ (M1)

where $\mathcal{C}=\\{z\colon\phi(z)\leq R\\}$ for some
$R>\frac{2D(1-e^{-\zeta\tau})}{\zeta(1-\alpha)}$.

Then $(S_{t})_{t\geq 0}$ has a unique invariant measure $\mu_{\infty}$ and for
any $\mu\in\mathcal{P}(\Omega)$ there exist some constants $C>1,\sigma>0$ such
that for all $t\geq 0$ we have

$\displaystyle\|S_{t}\mu-\mu_{\infty}\|_{\phi}\leq Ce^{-\sigma
t}\|\mu-\mu_{\infty}\|_{\phi}.$ (6)

######  Remark 1.5.

The constants $C,\sigma$ can be computed explicitly in terms of
$D,\tau,\alpha,\beta,\zeta$ (see Remark 3.10 in [15] or Remark 2 in [24]).

There are versions of Harris’s Theorem adapted to weaker Lyapunov conditions,
providing subgeometric convergence results, see, e.g., [2, 10, 11, 12]. Here,
we state and use a version which is found in [15, 8, 24]. We refer the reader,
e.g., to [10] (Theorems 3.10 and 3.12) or to [2] (Theorem 1.2) for different
versions of this theorem.

######  Theorem 1.6 (Subgeometric Harris’s Theorem).

Let $(S_{t})_{t\geq 0}$ be a Markov semigroup satisfying the following
hypotheses:

######  Hypothesis 3 (Weaker Foster-Lyapunov condition).

There exist constants $\zeta>0,\,D\geq 0$ and a continuous function
$\phi:\Omega\to[1,+\infty)$ with pre-compact sub-level sets such that

$\displaystyle\mathcal{L}^{*}[\phi](z)\leq D-\zeta h(\phi),$ (FL2)

where $h:\mathbb{R}_{+}\to\mathbb{R}$ is a strictly concave, positive,
increasing function and $\lim_{u\to+\infty}h^{\prime}(u)=0$.

######  Hypothesis 4 (Minorisation condition).

For every $R>0$, there exist a probability measure $\nu$, a constant
$\beta\in(0,1)$ and some time $\tau>0$ such that

$\displaystyle S_{\tau}\delta_{z}\geq\beta\nu\quad\mbox{for all
}z\in\mathcal{C},$ (M2)

where $\mathcal{C}=\\{z\colon\phi(z)\leq R\\}$.

Then $(S_{t})_{t\geq 0}$ has a unique invariant measure $\mu_{\infty}$
satisfying

$\displaystyle\int h(\phi(z))\mu_{\infty}(\,\mathrm{d}z)\leq D,$

and there exists a constant $C$ such that

$\displaystyle\|S_{t}\mu-\mu_{\infty}\|_{\mathrm{TV}}\leq\frac{C\mu(\phi)}{H_{h}^{-1}(t)}+\frac{C}{(h\circ
H_{h}^{-1})(t)}$

holds for every $\mu(\phi)=\int\phi(z)\mu(\,\mathrm{d}z)$ where the function
$H_{h}$ is defined by

$\displaystyle H_{h}:=\int_{1}^{u}\frac{\,\mathrm{d}s}{h(s)}.$

The proofs of these theorems can be found in [8, 10, 15]. In [10, 15], the
authors make a weaker assumption, namely $h(u)\leq u$ for any $u\geq 1$,
instead of the one which is stated above $\lim_{u\to+\infty}h^{\prime}(u)=0$.
Their assumption allows for linear growth at infinity, whereas
$\lim_{u\to+\infty}h^{\prime}(u)=0$ essentially requires $h$ to be flat at
infinity.

For the case of angularly dependent jump kernels, we are able to recover
exponential convergence using Harris’s theorem (Theorem 1.4). The Foster-
Lyapunov condition is proven in a similar way though more intricate than the
work in our previous paper [13]. The minorisation condition is considerably
more challenging due to the fact that we need to track the dynamics over many
more jumps to produce a lower bound. Therefore, we present the proof that the
minorsiation condition holds in dimension $d=2$ to simplify the computations
of the lower bound. Thus, the convergence result is stated in $d=2$ in Theorem
1.4. However, we believe that there is no serious obstacle to generalising it
to higher dimensions.

For the case of unbounded velocity spaces, we are only able to show
subgeometric rates of convergence. We do this via the subgeometric version of
Harris’s theorem (Theorem 1.6). Here, we are able to use exactly the same
uniform minorisation condition as in our previous paper [13] in the linear
case. The Foster-Lyapunov condition is substantially different and this is
reflected in the subgeometric rate of convergence.

Before verifying these hypotheses for Eq. (1) we would like to add some
comments on the Foster-Lyapunov condition. In order to show that (FL1) holds
true for Eq. (1) we would like to find some function $\phi(z)$ where
$\phi(z)\rightarrow\infty$ as $|z|\rightarrow\infty$ and the existence of some
$\tau>0$, $C>0$ and $\alpha\in(0,1)$ such that

$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathcal{V}}\phi(x,v)f(\tau,x,v)\,\mathrm{d}x\,\mathrm{d}v\leq\alpha\int_{\mathbb{R}^{d}}\int_{\mathcal{V}}\phi(x,v)f_{0}(x,v)\,\mathrm{d}x\,\mathrm{d}v+C,$
(7)

for any initial data
$f_{0}(x,v)\in\mathcal{P}(\mathbb{R}^{d}\times\mathcal{V})$. This is because
for $f$ satisfying $\partial_{t}f=\mathcal{L}[f]$, (7) is equivalent to
showing that

$\displaystyle\mathcal{L}^{*}[\phi]\leq D-\zeta\phi,$ (8)

where $\mathcal{L}^{*}$ is the formal adjoint of $\mathcal{L}$ and
$\zeta=\frac{\log\alpha}{\tau}$ and $D=C\frac{\log\alpha}{\tau(1+\alpha)}$. In
our case we have

$\displaystyle\mathcal{L}[f](x,v)=-v\cdot\nabla_{x}f+\int_{\mathcal{V}}\lambda(v^{\prime}\cdot\nabla_{x}M(x))\kappa(v^{\prime},v)f(t,x,v^{\prime})\,\mathrm{d}v^{\prime}-\lambda(v\cdot\nabla_{x}M(x))f(t,x,v),$
(9)

and therefore,

$\displaystyle\mathcal{L}^{*}[\phi](x,v)=v\cdot\nabla_{x}\phi(x,v)+\lambda(v\cdot\nabla_{x}M(x))\left(\int_{\mathcal{V}}\kappa(v,v^{\prime})\phi(x,v^{\prime})\,\mathrm{d}v^{\prime}-\phi(x,v)\right).$
(10)

##  2 Angle-dependent tumbling kernel

This section is dedicated to the long-time behaviour of the linear run and
tumble equation, Eq. (1) with the angularly dependent tumbling kernel
($\kappa_{1}$). The following two sections are dedicated to verifying the two
hypotheses of Harris’s theorem (Theorem 1.4). At the end of this section, we
provide a proof of Theorem 1.1.

###  2.1 Minorisation condition

In this section, we prove that Hypothesis 2, (M1), is satisfied for Eq. (1)
with the angular-dependent tumbling kernel ($\kappa_{1}$). Our overall
strategy is as follows: we show that the solution $f\equiv f_{t}$ fo Eq. (1)
attains a lower bound using a Duhamel’s formula and bounding below by the part
of $f_{t}$ where there are a particular number of jumps.

To keep the exposition of our strategy clear, we provide the following
computations for only $d=2$. The very same approach can be used to extend the
results to higher dimensions.

Denoting $v_{\theta}:=(\cos\theta,\sin\theta)$ and
$\nabla_{x,y}g:=(\partial_{x}g,\partial_{y}g)$ for any differentiable function
$g$, Eq. (1) becomes

$\displaystyle\partial_{t}f+v_{\theta}\cdot\nabla_{x,y}f=\int
b(\theta^{\prime}-\theta)\lambda(v_{\theta}\cdot\nabla_{x,y}M)f(t,x,y,\theta^{\prime})\mathrm{d}\theta^{\prime}-\lambda(v_{\theta}\cdot\nabla_{x,y}M)f.$
(11)

Next, we define the semigroup $(T_{t})_{t\geq 0}$ accounting for the transport
part,

$\displaystyle(T_{t}f)(t,x,y,\theta):=f(t,x-t\cos\theta,y-t\sin\theta,\theta),$

and the operator $\tilde{L}$,

$\displaystyle(\tilde{L}f)(t,x,y,\theta):=\int\mathds{1}_{|\theta^{\prime}-\theta|<\alpha}(\theta^{\prime})f(t,x,y,\theta^{\prime})\,\mathrm{d}\theta^{\prime}=\int\mathds{1}_{|\theta^{\prime}|<\alpha}(\theta^{\prime})f(t,x,y,\theta+\theta^{\prime})\,\mathrm{d}\theta^{\prime}.$

Then we have the following lemma

######  Lemma 2.1.

Suppose that $b(\theta)\geq\beta\mathds{1}_{|\theta|\leq\alpha}(\theta)$. Then
for any $n\geq 1$ we have

$f(t,x,y,\theta)\geq\\\
\beta^{n}(1-\chi)^{n}e^{-(1+\chi)t}\int_{0}^{t}\int_{0}^{t_{n}}\dots\int_{0}^{t_{2}}\int_{0}^{t_{1}}\left(T_{t-t_{n}}\tilde{L}T_{t_{n}-t_{n-1}}\dots\tilde{L}T_{t_{2}-t_{1}}\tilde{L}T_{t_{1}}f_{0}\right)(x,y,\theta)\mathrm{d}t_{1}\,\mathrm{d}t_{2}\dots\,\mathrm{d}t_{n}.$

###### Proof.

We have that

$\,\partial_{t}(f(t,x-t\cos\theta,y-t\sin\theta,\theta)=\lambda(v_{\theta}\cdot\nabla_{x,y}M(x-t\cos\theta,y-t\cos\theta))\times\\\
\left(\int
b(\theta^{\prime})f(t,x-t\cos\theta,y-t\sin\theta,\theta+\theta^{\prime})\,\mathrm{d}\theta^{\prime}-f(t,x-t\cos\theta,y-t\sin\theta,\theta)\right).$

Therefore writing
$\Lambda(t,x,y,\theta)=\int_{0}^{t}\lambda(v_{\theta}\cdot\nabla_{x}M(x-s\cos\theta,y-s\cos\theta))\,\mathrm{d}s$
we obtain

$f(t,x-t\cos\theta,y-t\sin\theta,\theta)=e^{-\Lambda(t)}f_{0}(x,y,\theta)\\\
+e^{-\Lambda(t)}\int_{0}^{t}\lambda(v_{\theta}\cdot\nabla_{x}M(x-s\cos\theta,y-s\cos\theta))\int
b(\theta^{\prime})f(t,x-s\cos\theta,y-s\sin\theta,\theta+\theta^{\prime})\,\mathrm{d}\theta\,\mathrm{d}s.$

Changing variables and using the fact that $1-\chi\leq\lambda\leq 1+\chi$, we
obtain

$\displaystyle f(t,x,y,\theta)\geq
e^{-(1+\chi)t}T_{t}f_{0}+(1-\chi)\int_{0}^{t}e^{-(1+\chi)(t-s)}\int
b(\theta^{\prime})(T_{t-s}f_{s})(x,y,\theta+\theta^{\prime})\,\mathrm{d}\theta^{\prime}\,\mathrm{d}s.$

Then using the assumption on $b$ we have

$\displaystyle f(t,x,y,\theta)\geq
e^{-(1+\chi)t}T_{t}f_{0}+\beta(1-\chi)\int_{0}^{t}e^{-(1+\chi)(t-s)}T_{t-s}\tilde{L}f_{s}\,\mathrm{d}s.$

We note that the second term above is positive, so we have

$\displaystyle f(t,x,y,\theta)\geq e^{-(1+\chi)t}T_{t}f_{0}.$

Then, we can plug this into the integral term to obtain

$\displaystyle
f(t,x,y,\theta)\geq\beta(1-\chi)e^{-(1+\chi)t}\int_{0}^{t}T_{t-s}\tilde{L}T_{s}f_{0}\,\mathrm{d}s.$

We then continue substituting this into the integral term $n$ times to obtain
the result for any $n\geq 1$. This finishes the proof. ∎

Next, we want to show that we can find an $n\geq 1$ and a range of admissible
times for which we can provide a lower bound on the term
$T_{t-t_{n}}\tilde{L}T_{t_{n}-t_{n-1}}\dots\tilde{L}T_{t_{3}-t_{2}}\tilde{L}T_{t_{2}-t_{1}}\tilde{L}T_{t_{1}}f_{0}$.

Let us define $P_{t}g:=\tilde{L}T_{t}{g}$ for any probability measure $g$.
Then we write $P_{t_{1},t_{2},t_{3}}^{3}:=P_{t_{1}}P_{t_{2}}P_{t_{3}}$ and
note that for any measure $g$, $P_{t_{1},t_{2},t_{3}}^{3}g$ has a density. In
particular,
$P_{t_{1},t_{2},t_{3}}^{3}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)$
is bounded below by a uniform measure of a set with non-empty interior. Hence,
we can bound it below by a constant times the indicator function of red a ball
in the position variable $x$ with a centre at a point which we can compute. We
can then use this computation to show that $P^{3}_{t_{1},t_{2},t_{3}}$ acting
on the indicator function of a ball is bounded below by a constant times the
indicator function of another ball with a slightly larger radius and a
different centre. By tracking how these centres move and how the radii of the
balls grow, we can then get a bound below in $(x,y)$-space by the indicator
function of a ball whose centre is at the origin (rather than depending on the
initial point).

We then need to prove that we can subsequently reach all possible angles
(i.e., velocities) while maintaining a uniform lower bound below in position.
We show this by looking at repeated jumps in a small time period so that we
reach all angles without moving too far. This then allows us to reach all
possible angles but slightly shrinks the ball we found in the lower bound for
the spatial variables.

Therefore, in the following lemma, we look at
$P_{t_{1},t_{2},t_{3}}^{3}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)$
and show that it is bounded below by a constant times the indicator function
of a set which contains a ball whose radius and centre we can compute.

######  Lemma 2.2.

Let $\alpha<\dfrac{\pi}{2}$ and $r_{1},r_{2},r_{3}>0$ are a set of times,
$\varepsilon>0$ and $|s_{i}-r_{i}|<\varepsilon$ for each $i\in\\{1,2,3\\}$.
Then if $\varepsilon$ is sufficiently small we have

$\displaystyle
P^{3}_{s_{1},s_{2},s_{3}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)\geq\frac{1}{s_{2}s_{3}}\mathds{1}_{B((x_{*},y_{*});r)}(x,y)\mathds{1}_{|\theta-\theta_{0}\pm\delta\theta|\leq\frac{\alpha}{2}}(\theta),$

where

$\displaystyle x_{*}=x_{0}$ $\displaystyle+(r_{1}+R)\cos(\theta_{0}),\quad
y_{*}=y_{0}+(r_{1}+R)\sin(\theta_{0}),\quad r=r_{2}r_{3}(1-\cos(\alpha/2)),$
$\displaystyle\delta\theta=\pm\arctan\left(\frac{r_{2}\sin(\alpha/2)}{r_{3}+r_{2}\cos(\alpha/2)}\right),\quad
R=\sqrt{r_{2}^{2}+r_{3}^{2}+2r_{2}r_{3}\cos(\alpha/2)}.$

###### Proof.

Since we have

$T_{s_{1}}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta)=\delta_{x_{0}+s_{1}\cos(\theta_{0})}(x)\delta_{y_{0}+s_{1}\sin(\theta_{0})}(y)\delta_{\theta_{0}}(\theta)\\\
=\delta_{x_{0}}(x-s_{1}\cos(\theta_{0}))\delta_{y_{0}}(y-s_{1}\sin(\theta_{0}))\delta_{\theta_{0}}(\theta),$

and

$\displaystyle\tilde{L}T_{s_{1}}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta)=\int_{-\alpha}^{\alpha}\delta_{x_{0}+s_{1}\cos(\theta_{0})}(x)\delta_{y_{0}+s_{1}\sin(\theta_{0})}(y)\delta_{\theta_{0}}(\theta-\theta^{\prime})\mathrm{d}\theta^{\prime},$

applying $T_{s_{2}}$ once more we obtain,

$T_{s_{2}}\tilde{L}T_{s_{1}}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta)\\\
=\int_{-\alpha}^{\alpha}\delta_{x_{0}}(x-s_{1}\cos(\theta_{0})-s_{2}\cos(\theta_{0}+\theta_{1}))\delta_{y_{0}}(y-s_{1}\sin(\theta_{0})-s_{2}\sin(\theta_{0}+\theta_{1}))\delta_{\theta_{0}}(\theta-\theta_{1})\mathrm{d}\theta_{1}.$

Iterating this yields

$\tilde{L}T_{s_{3}}\tilde{L}T_{s_{2}}\tilde{L}T_{s_{1}}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta)\\\
=\int_{-\alpha}^{\alpha}\int_{-\alpha}^{\alpha}\int_{-\alpha}^{\alpha}\delta_{x_{0}}(x-s_{1}\cos(\theta_{0})-s_{2}\cos(\theta_{0}+\theta_{1})-s_{3}\cos(\theta_{0}+\theta_{1}+\theta_{2}))\\\
\times\delta_{y_{0}}(y-s_{1}\sin(\theta_{0})-s_{2}\sin(\theta_{0}+\theta_{1})-s_{3}\sin(\theta_{0}+\theta_{1}+\theta_{2}))\delta_{\theta_{0}}(\theta-\theta_{1}-\theta_{2}-\theta_{3})\,\mathrm{d}\theta_{1}\,\mathrm{d}\theta_{2}\,\mathrm{d}\theta_{3}.$

We perform a change of variables,

$\displaystyle\tilde{x}=s_{2}\cos(\theta_{0}+\theta_{1})+s_{3}\cos(\theta_{0}+\theta_{1}+\theta_{2}),\quad\tilde{y}=s_{2}\sin(\theta_{0}+\theta_{1})+s_{3}\sin(\theta_{0}+\theta_{1}+\theta_{2}),$

then we have

$\displaystyle\frac{\partial\tilde{x}}{\partial\theta_{1}}=-s_{2}\sin(\theta_{0}+\theta_{1})-s_{3}\sin(\theta_{0}+\theta_{1}+\theta_{2}),\quad\frac{\partial\tilde{y}}{\partial\theta_{1}}=s_{2}\cos(\theta_{0}+\theta_{1})+s_{3}\cos(\theta_{0}+\theta_{1}+\theta_{2})$

and

$\displaystyle\frac{\partial\tilde{x}}{\partial\theta_{2}}=-s_{3}\sin(\theta_{0}+\theta_{1}+\theta_{2}),\quad\frac{\partial\tilde{y}}{\partial\theta_{2}}=s_{3}\cos(\theta_{0}+\theta_{1}+\theta_{2}).$

The Jacobian of this change of variables is
$\,\mathrm{d}\tilde{x}\,\mathrm{d}\tilde{y}=s_{2}s_{3}\sin(\theta_{2})\,\mathrm{d}\theta_{1}\,\mathrm{d}\theta_{2}$.

Moreover, defining $\tilde{\theta}:=\theta_{1}+\theta_{2}+\theta_{3}$ and then
$\mathrm{d}\tilde{\theta}=\mathrm{d}\theta$, we obtain

$\tilde{L}T_{s_{3}}\tilde{L}T_{s_{2}}\tilde{L}T_{s_{1}}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta)=\int_{S(x_{0},y_{0},\theta_{0},s_{1},s_{2},s_{3})}\delta_{x_{0}}(x-s_{1}\cos(\theta_{0})-\tilde{x}))\\\
\times\delta_{y_{0}}(y-s_{1}\sin(\theta_{0})-\tilde{y})\delta_{\theta_{0}}(\theta-\tilde{\theta})\frac{1}{s_{2}s_{3}|\sin(\theta_{2}(\tilde{x},\tilde{y}))|}\,\mathrm{d}\tilde{x}\,\mathrm{d}\tilde{y}\,\mathrm{d}\tilde{\theta},$

where $S(x_{0},y_{0},\theta_{0},s_{1},s_{2},s_{3})$ is the set of possible
values of
$x_{0}+s_{1}\cos(\theta_{0})+\tilde{x},y_{0}+s_{1}\sin(\theta_{0})+\tilde{y}$,
and $\theta_{0}+\tilde{\theta}$, (see Figure 1 below). Then, as
$|\sin(\theta_{2})|\leq 1$ we have

$\displaystyle\tilde{L}T_{s_{3}}\tilde{L}T_{s_{2}}\tilde{L}T_{s_{1}}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta)\geq\frac{1}{s_{2}s_{3}}\mathds{1}_{S(x_{0},y_{0},\theta_{0},s_{1},s_{2},s_{3})}(x,y,\theta).$

Next, we want to show that $S(x_{0},y_{0},\theta_{0},s_{1},s_{2},s_{3})$
contains the ball mentioned in the statement of the lemma. We notice that

$\displaystyle\tilde{x}^{2}+\tilde{y}^{2}=s_{1}^{2}+s_{2}^{2}+2s_{1}s_{2}\cos(\theta_{2})=R(\theta_{2})^{2}.$

We choose $\theta_{2}=\pm\frac{\alpha}{2}$ and
$\theta_{1}=\pm\arctan\left(\frac{s_{3}\sin(\alpha/2)}{s_{2}+s_{3}\cos(\alpha/2)}\right)$
and this will give the centre of the ball.

Now for a given $\theta_{2}$, we choose
$\beta=\arctan\left(\frac{s_{3}\sin(\theta_{2})}{s_{2}+s_{3}\cos(\theta_{2})}\right)$
then we can write

$\displaystyle\tilde{x}$
$\displaystyle=(s_{2}\cos(\beta)+s_{3}\cos(\theta_{2}-\beta))\cos(\theta_{0}+\theta_{1}+\beta)=R(\theta_{2})\cos(\theta_{0}+\theta_{1}+\beta)$
$\displaystyle\tilde{y}$
$\displaystyle=(s_{2}\cos(\beta)+s_{3}\sin(\theta_{2}-\beta))\sin(\theta_{0}+\theta_{1}+\beta)=R(\theta_{2})\sin(\theta_{0}+\theta_{1}+\beta).$

So we can set $\tilde{x}=R\cos(\omega)$ and $\tilde{y}=R\sin(\omega)$ as long
as we can choose

$\displaystyle\theta_{2}=\arccos\left(\frac{R^{2}-s_{2}^{2}-s_{3}^{2}}{s_{2}s_{3}}\right)\,\mbox{
and
}\,\theta_{1}=\omega-\arctan\left(\frac{s_{3}\sin(\theta_{2})}{s_{2}+s_{3}\cos(\theta_{2})}\right).$

We can see that for $\theta_{1}$ and $\theta_{2}$ small we have

$\displaystyle\theta_{1}\approx\omega-\frac{s_{3}\theta_{2}}{s_{2}+s_{3}}.$

Thus, we expect to be able to choose $|\beta|$ up to
$\frac{s_{3}\alpha}{s_{2}+s_{3}}$ which is at least as large as it is needed
to cover the ball given that $\alpha$ is small. Then since all the
computations depend continuously on $s_{1},s_{2},s_{3}$ and the ball is
strictly contained inside the set $S$, we can replace $s_{1},s_{2},s_{3}$ with
$r_{1},r_{2},r_{3}$ provided that $\varepsilon$ is small enough. This
concludes the proof. ∎

Figure 1. This image is an illustration of the set $S$ (dotted region)
mentioned in the proof of Lemma 2.2 for fixed values of $r_{2},r_{3},\alpha$.
The ball found inside the set $S$ is grey coloured.

Using Lemma 2.2 we then prove the following:

######  Lemma 2.3.

Let $r_{1},r_{2},r_{3}>0$ be a set of times, $R,\alpha$ are as in Lemma 2.2.
Then we have

$\displaystyle
P^{3}_{r_{1},r_{2},r_{3}}\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x,y)\mathds{1}_{|\theta-\theta_{0}|<\frac{\alpha}{2}}(\theta)\geq\gamma\mathds{1}_{B(x_{**},y_{**};\tilde{r}+\frac{r}{2})}(x,y)\mathds{1}_{|\theta_{**}-\theta|\leq\alpha}(\theta),$

where

$\displaystyle x_{**}=x_{0}-(r_{1}+R)\cos(\theta_{0}+\delta\theta),\quad
y_{**}=y_{0}-(r_{1}+R)\sin(\theta_{0}+\delta\theta),\quad\theta_{**}=\theta_{0}+\delta\theta$

and $\gamma$ a constant that can be computed explicitly. Let us define the map
$F$,

$\displaystyle F(x_{0},y_{0},\theta_{0}):=(x_{**},y_{**},\theta_{**})$ (12)

which tracks how the centres of the balls move.

###### Proof.

We have

$\displaystyle\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x,y)\mathds{1}_{|\theta-\theta_{0}|<\frac{\alpha}{2}}(\theta)=\int\delta_{x^{\prime}}(x)\delta_{y^{\prime}}(y)\delta_{\theta^{\prime}}(\theta)\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x^{\prime},y^{\prime})\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\frac{\alpha}{2}}(\theta)\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\theta^{\prime},$

where $\int$ represents the triple integral,
$\int_{\mathbb{R}^{2}\times\mathbb{R}^{2}}\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\int_{-\alpha}^{\alpha}\,\mathrm{d}\theta^{\prime}$.
Now applying three times $T_{t}$ and $\tilde{L}$ yields,

$\displaystyle
P^{3}_{r_{1},r_{2},r_{3}}\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x,y)\mathds{1}_{|\theta-\theta_{0}|<\frac{\alpha}{2}}(\theta)$
$\displaystyle\geq$
$\displaystyle\int\left(P^{3}_{r_{1},r_{2},r_{3}}\delta_{x^{\prime}}\delta_{y^{\prime}}\delta_{\theta^{\prime}}\right)(x,y,\theta)\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x^{\prime},y^{\prime})\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\frac{\alpha}{2}}\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\theta^{\prime}$
$\displaystyle\geq$
$\displaystyle\frac{1}{r_{2}r_{3}}\int\mathds{1}_{B((x^{\prime}+(r_{1}+R)\cos(\theta^{\prime}),y^{\prime}+(r_{1}+R)\sin(\theta^{\prime}));r)}(x,y)\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x^{\prime},y^{\prime})\mathds{1}_{|\theta-\theta^{\prime}-\delta\theta|<\frac{\alpha}{2}}\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\frac{\alpha}{2}}\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\theta^{\prime}$
$\displaystyle=$
$\displaystyle\frac{1}{r_{2}r_{3}}\int\mathds{1}_{B((x-(r_{1}+R)\cos(\theta^{\prime}),y-(r_{1}+R)\sin(\theta^{\prime}));r)}(x^{\prime},y^{\prime})\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x^{\prime},y^{\prime})\mathds{1}_{|\theta-\theta^{\prime}-\delta\theta|<\frac{\alpha}{2}}\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\frac{\alpha}{2}}\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\theta^{\prime}$
$\displaystyle=$
$\displaystyle\frac{1}{r_{2}r_{3}}\int\left|B\left((x-(r_{1}+R)\cos(\theta^{\prime}),y-(r_{1}+R)\sin(\theta^{\prime}));r\right)\cap
B\left((x_{0},y_{0});\tilde{r}\right)\right|\mathds{1}_{|\theta-\theta^{\prime}-\delta\theta|<\frac{\alpha}{2}}\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\frac{\alpha}{2}}\,\mathrm{d}\theta^{\prime}$

In the last line above, we fixed $\theta^{\prime}$ and considered the integral
in $x^{\prime}$ and $y^{\prime}$ which will measure the size of the overlap
between the balls
$B((x-(r_{1}+R)\cos\theta^{\prime},y-(r_{1}+R)\sin\theta^{\prime});r)$ and
$B((x_{0},y_{0});\tilde{r})$.

If $(x,y)\in
B((x_{0}+(r_{1}+R)\cos\theta^{\prime},y_{0}+(r_{1}+R)\sin\theta^{\prime});\tilde{r}+\frac{r}{2})$
then we can bound the size of the overlap below by $\frac{\pi r^{2}}{4}$. We
recall from Lemma 2.2 that
$r=r_{2}r_{3}\left(1-\cos(\frac{\alpha}{2})\right)$, so we have

$\left|B\left((x-(r_{1}+R)\cos(\theta^{\prime}),y-(r_{1}+R)\sin(\theta^{\prime}));r\right)\cap
B\left((x_{0},y_{0});\tilde{r}\right)\right|\geq\\\ \frac{\pi
r_{2}r_{3}}{4}\left(1-\cos\left(\frac{\alpha}{2}\right)\right)^{2}\mathds{1}_{B((x_{0}+(r_{1}+R)\cos\theta^{\prime},y_{0}+(r_{1}+R)\sin\theta^{\prime});\tilde{r}+\frac{r}{2})}(x,y).$

Therefore the lower bound becomes, denoting $C(r_{2},r_{3},\alpha):=\frac{\pi
r_{2}r_{3}}{4}\left(1-\cos\left(\frac{\alpha}{2}\right)\right)^{2}$ (a
constant depending on $r_{2},r_{3},\alpha$),

$\displaystyle P^{3}_{r_{1},r_{2},r_{3}}$
$\displaystyle\mathds{1}_{B((x_{0},y_{0});\tilde{r})}(x,y)\mathds{1}_{|\theta-\theta_{0}|<\frac{\alpha}{2}}(\theta)$
$\displaystyle\geq
C(r_{2},r_{3},\alpha)\int\mathds{1}_{B\left((x_{0}+(r_{1}+R)\cos\theta^{\prime},y_{0}+(r_{1}+R)\sin\theta^{\prime});\tilde{r}+\frac{r}{2}\right)}(x,y)\mathds{1}_{|\theta-\theta^{\prime}-\delta\theta|<\frac{\alpha}{2}}\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\alpha}\,\mathrm{d}\theta^{\prime},$
$\displaystyle=C(r_{2},r_{3},\alpha)\int\mathds{1}_{B\left((x_{0}-x,y_{0}-y);\tilde{r}+\frac{r}{2}\right)}((r_{1}+R)\cos(\theta^{\prime}),(r_{1}+R)\sin(\theta^{\prime}))\mathds{1}_{|\theta-\theta^{\prime}-\delta\theta|<\frac{\alpha}{2}}\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\alpha}\,\mathrm{d}\theta^{\prime}$
$\displaystyle=C(r_{2},r_{3},\alpha)\int\mathds{1}_{\left|\theta^{\prime}-\arccos\left(\frac{y_{0}-y}{x_{0}-x}\right)\right|\leq\arccos\left(\frac{\tilde{r}+r/2}{r_{1}+R}\right)}\mathds{1}_{|\theta-\theta^{\prime}-\delta\theta|<\frac{\alpha}{2}}\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\alpha}\,\mathrm{d}\theta^{\prime}$

Thus, if we impose that $(x_{0}-x,y_{0}-y)\in
B\left(\left((R+r_{1})\cos(\theta_{0}+\delta\theta),(R+r_{1})\sin(\theta_{0}+\delta\theta)\right);\tilde{r}+\frac{r}{4}\right)$
then we can bound the last integral above by

$\displaystyle\gamma\mathds{1}_{B\left((x_{0}-(r_{1}+R)\cos(\theta_{0}+\delta\theta),y_{0}-(r_{1}+R)\sin(\theta_{0}+\delta\theta));\tilde{r}+\frac{r}{4}\right)}(x,y)\mathds{1}_{|\theta-\theta_{0}-\delta\theta|<\frac{\alpha}{2}}(\theta),$

where $\gamma$ is a constant depending on $r_{1},r_{2},r_{3}$ and is bounded
uniformly in terms of upper and lower bounds on $r_{1},r_{2},r_{3}$. Roughly
we can compute this constant $\gamma\approx\frac{\pi
r_{2}^{2}r_{3}^{2}\alpha^{2}}{32(r_{1}+R)}$. ∎

As we iterate the process that moves the centre of the balls, we can find a
radius $\hat{R}>0$ so that the path of centres stays inside
$B((x_{0},y_{0});\hat{R})$ forever. We prove this in the following lemma.

######  Lemma 2.4.

Let $k\geq 1$ and consider a sequence of maps
$(x_{k},y_{k},\theta_{k})=F^{k}(x_{0},y_{0},\theta_{0})$ where $F$ is defined
in Lemma 2.3. Then there exists a constant $\hat{R}>0$ depending on
$r_{1},r_{2},r_{3}$ such that $(x_{k},y_{k})\in B((x_{0},y_{0});\hat{R})$ for
every $k\geq 1$.

###### Proof.

We have

$\displaystyle
x_{k}=x_{0}-(r_{1}+R)\cos(\theta_{0}+\delta\theta)-(r_{1}+R)\cos(\theta_{0}+2\delta\theta)-\dots-(r_{1}+R)\cos(\theta_{0}+k\delta\theta),$

and $y_{k}$ is defined similarly. The points $(x_{k},y_{k})$ are illustrated
in Figure 2 below.

We can then compute the ball that all the points lie on, which is a circle
whose radius $\hat{R}$ and centre $C$ are given by

$\displaystyle\hat{R}=\frac{r_{1}+R}{\sin(\delta\theta/2)},\quad
C=(x_{0},y_{0})-\hat{R}\left(\cos(\pi/2-(\theta_{0}+\delta\theta/2)),\sin(\pi/2-(\theta_{0}+\delta\theta/2))\right).$

This finishes the proof. ∎

$\hat{R}$$(x_{1},y_{1})$$\delta\theta$$C$$(x_{0},y_{0})$$(x_{2},y_{2})$$\delta\theta$$\delta\theta$
Figure 2. This image shows the iterates of the map $F$ in $(x,y)$-space. We
notice that each triangle formed by the points
$C,(x_{k},y_{k}),(x_{k+1},y_{k+1})$ is an isosceles triangle and a rotation of
the triangle formed by the points $C,(x_{0},y_{0}),(x_{1},y_{1})$. We can then
compute the circle that they all lie on and the point at the centre $C$ using
standard tools in trigonometry.

In the next lemma, we prove that after $\frac{24\hat{R}}{r}$ steps, the $x,y$
marginal is bounded by a constant times the indicator function of a ball of
radius $\hat{R}$.

######  Lemma 2.5.

If we let $\tilde{n}=\big{\lceil}\frac{24\hat{R}}{r}\big{\rceil}$, then there
exists some constant $\tilde{\gamma}>0$ such that

$\displaystyle
P^{\tilde{n}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)\geq\tilde{\gamma}\mathds{1}_{B((x_{0},y_{0});\hat{R})}(x,y)\
\mathds{1}_{|\theta-\theta_{k}|\leq\alpha}(\theta).$ (13)

###### Proof.

By Lemma 2.4 we can choose a path such that $(x_{k},y_{k})\in
B((x_{0},y_{0});\hat{R})$ and writing $k=\frac{\tilde{n}}{3}$ Lemma 2.3 yields

$\displaystyle
P^{\tilde{n}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)\geq\gamma^{k}\mathds{1}_{B\left((x_{k},y_{k});\left(1+\frac{k-1}{4}\right)r\right)}(x,y)\mathds{1}_{|\theta-\theta_{k}|\leq\alpha}(\theta).$

Since we also have that $\left(1+\frac{k-1}{4}\right)r\geq 2\hat{R}$,
therefore we obtain (13) with $\tilde{\gamma}=\gamma^{k}$. ∎

Next, we look at the angles and prove the following lemma:

######  Lemma 2.6.

Let $n_{*}=\left\lceil\frac{4\pi}{\alpha}\right\rceil$ and suppose that
$t_{1}+t_{2}+\dots+t_{n_{*}}\leq l$. Then we have

$\displaystyle
P^{n_{*}}_{t_{1},t_{2},\dots,t_{n_{*}}}\big{(}\mathds{1}_{B\big{(}(0,0);\frac{\hat{R}}{2}\big{)}}(x,y)\mathds{1}_{|\theta-\theta_{0}|<\alpha}(\theta)\big{)}\geq\gamma\mathds{1}_{B\big{(}(0,0);\frac{\hat{R}}{2}-2l\big{)}},$

uniform in $\theta$.

###### Proof.

First, let us look at

$\displaystyle\int P^{n_{*}}\delta_{x_{0}}\delta_{y_{0}}$
$\displaystyle\delta_{\theta_{0}}(x,y,\theta)\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\int\int
P^{n_{*}-1}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x-t_{n_{*}}\cos(\theta),y-t_{n_{*}}\sin\theta,\theta+\theta^{\prime})\mathds{1}_{|\theta^{\prime}|\leq\alpha}(\theta^{\prime})\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta^{\prime}$
$\displaystyle=\int\int\int
P^{n_{*}-1}(\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}})(x,y,\theta+\theta^{\prime})\mathds{1}_{|\theta^{\prime}|\leq\alpha}(\theta^{\prime})\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta^{\prime}\mathrm{d}\theta.$

We can keep iterating this to obtain

$\displaystyle\int P^{n_{*}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}$
$\displaystyle(x,y,\theta)\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\int\int\cdots\int\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(\theta+\theta_{1}+\theta_{2}+\dots+\theta_{n_{*}})\mathds{1}_{|\theta_{1}|\leq\alpha}\dots\mathds{1}_{|\theta_{n_{*}}|\leq\alpha}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta_{1}\dots\,\mathrm{d}\theta_{n_{*}}$
$\displaystyle=\int\delta_{\theta_{0}}(\theta+\theta_{1}+\theta_{2}+\dots+\theta_{n_{*}})\mathds{1}_{|\theta_{1}|\leq\alpha}\dots\mathds{1}_{|\theta_{n_{*}}|\leq\alpha}\,\mathrm{d}\theta_{1}\dots\,\mathrm{d}\theta_{n_{*}}$
$\displaystyle\geq C.$

For some constant $C$ that doesn’t depend on $\theta_{0},\theta$. Therefore,

$\displaystyle\int
P^{n_{*}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)\,\mathrm{d}x\,\mathrm{d}y\geq
C.$

Hence we can write

$\displaystyle
P^{n_{*}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)=\mu(\theta\mid\theta_{0})\mu((x,y)\mid\theta,(x_{0},y_{0},\theta_{0}))\geq
C\mu((x,y)\mid\theta,(x_{0},y_{0},\theta_{0}))$

where $\mu((x,y)\mid\theta,(x_{0},y_{0},\theta_{0}))$ is the conditional law
of the position variables given $\theta$ and the initial point $(x_{0},y_{0})$
so we have that

$\displaystyle\int\mu((x,y)\mid\theta,(x_{0},y_{0},\theta_{0}))\,\mathrm{d}x\,\mathrm{d}y=1.$

We can also see that

$\displaystyle\operatorname{supp}(\mu((x,y)\mid\theta,(x_{0},y_{0},\theta_{0}))\subseteq\mathds{1}_{B((x_{0},y_{0});l)}$

as the $x$ and $y$ variables can have travelled a distance of at most $l$.
Then

$\displaystyle
P^{n_{*}}\big{(}\mathds{1}_{B((0,0);\hat{R})}\mathds{1}_{|\theta-\theta_{*}|<\alpha}\big{)}\geq\int
c\mu((x,y)\mid\theta,(x^{\prime},y^{\prime},\theta^{\prime}))\mathds{1}_{B((0,0);\hat{R})}(x^{\prime},y^{\prime})\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\alpha}\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\theta^{\prime}.$

Then using the translation invariance of the transport map we can write

$\displaystyle\mu((x,y)\mid\theta,(x^{\prime},y^{\prime},\theta^{\prime}))=\mu((-x^{\prime},-y^{\prime})\mid\theta,(-x,-y,\theta^{\prime})),$

so we have

$\displaystyle
P^{n_{*}}\big{(}\mathds{1}_{B((0,0);\hat{R})}1_{|\theta-\theta_{*}|<\alpha}\big{)}$
$\displaystyle\geq\int\mu((-x^{\prime},-y^{\prime})\mid\theta,(-x,-y,\theta^{\prime}))\mathds{1}_{B((0,0);\hat{R})}(x^{\prime},y^{\prime})\mathds{1}_{|\theta^{\prime}-\theta_{0}|<\alpha}\,\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\theta^{\prime}$
$\displaystyle\geq C\mathds{1}_{B((0,0);\hat{R}-2l)}(x,y).$

Above, we use the fact that if $(x,y)\in B((0,0);\hat{R}-2l)$ then we will be
integrating over the full support of
$\mu((-x^{\prime},-y^{\prime})\mid\theta,(-x,-y,\theta^{\prime}))$. ∎

Using the previous lemmas, we can prove that the minorisation condition (M1)
for Eq. (1) with the angularly dependent tumbling kernel is satisfied.

######  Proposition 2.7.

Let
$f_{0}(x,y,\theta)=\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)$
with $(x_{0},y_{0})\in B\big{(}(0,0);\frac{\hat{R}}{2}\big{)}$ then, after
$\big{\lceil}\frac{24\hat{R}}{r}\big{\rceil}+n_{*}$ steps, provided that the
times are chosen suitably, we have

$\displaystyle f(t,x,y,\theta)\geq
C\mathds{1}_{B\big{(}(0,0);\frac{\hat{R}}{4}\big{)}},$

for some $C$ and $\hat{R}$ that can be computed explicitly.

###### Proof.

Suppose that the first $n_{**}$ inter-jump times are within $\varepsilon$ of
$r_{1},r_{2},r_{3}$ with $\varepsilon$ small enough so that Lemma 2.2 applies
and $\gamma$ is chosen such that Lemma 2.5 holds for any interjump times in
this range with constant $\gamma$. Suppose further that the sum of the last
$n_{*}$ jump times is less than $l$. Then, using the previous result we have

$\displaystyle
P^{n_{**}+n_{*}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}(x,y,\theta)$
$\displaystyle=P^{n_{*}}\left(P^{n_{**}}\delta_{x_{0}}\delta_{y_{0}}\delta_{\theta_{0}}\right)(x,y,\theta)$
using Lemma 2.5 $\displaystyle\geq
P^{n_{*}}\big{(}\gamma^{\frac{n_{**}}{3}}\mathds{1}_{B((x_{0},y_{0});\hat{R})}\mathds{1}_{\left|\theta-\theta_{0}-n_{**}\frac{\delta\theta}{3}\right|\leq\frac{\alpha}{2}}\big{)}$
using that $(x_{0},y_{0})\in B\big{(}(0,0);\frac{\hat{R}}{2}\big{)}$
$\displaystyle\geq
P^{n_{*}}\big{(}\gamma^{\frac{n_{**}}{3}}\mathds{1}_{B((0,0);\frac{\hat{R}}{2})}\mathds{1}_{\left|\theta-\theta_{0}-n_{**}\frac{\delta\theta}{3}\right|\leq\frac{\alpha}{2}}\big{)}$
using Lemma 2.6
$\displaystyle\geq\gamma\mathds{1}_{B\left((0,0);\frac{\hat{R}}{2}-2l\right)}.$

We then take $m=\frac{\hat{R}}{8}$ and substitute this into Lemma 2.1 and
integrate over the admissible possible jump times and obtain

$\displaystyle
f(t,x,v)\geq\hat{\gamma}\mathds{1}_{B\big{(}(0,0);\frac{\hat{R}}{4}\big{)}}$

where $\hat{\gamma}$ is a constant we could, in principle, compute. This
verifies (M1) for (11) with the angularly dependent kernel ($\kappa_{1}$). ∎

###  2.2 Foster-Lyapunov condition

In this section, we verify the Foster-Lyapunov condition (FL1) for Eq. (1) for
the case ($\kappa_{1}$). We remark that the minorisation condition in the
previous section was given in dimension $d=2$ to keep the exposition clear.
The Foster-Lyapunov condition we prove in this section is valid for arbitrary
dimensions $d\geq 1$. We start by proving the following lemma.

######  Lemma 2.8.

Suppose that $\kappa$ is a collision kernel ($\kappa_{1}$) satisfying the
hypothesis (H4) then

$\displaystyle\int_{\mathbb{S}^{d-1}}\kappa(v^{\prime},v)v^{\prime}\cdot\nabla_{x}M(x)\,\mathrm{d}v^{\prime}=C_{\kappa}v\cdot\nabla_{x}M(x),$
(14)

where $C_{\kappa}\leq 1$ is a constant that only depends on the form of the
collision kernel.

###### Proof.

We perform a change of variables and write write
$v^{\prime}=\cos(\theta)v+\sin(\theta)w$ where $w\in S_{v}$ ranges over the
sphere of dimension $d-2$ lying in the hyperplane perpendicular to $v$. This
gives us

$\displaystyle\int_{\mathbb{S}^{d-1}}\kappa(v^{\prime},v)v^{\prime}\cdot\nabla_{x}M(x)\,\mathrm{d}v^{\prime}=\int_{-\pi}^{\pi}\int_{S_{v}}\kappa_{1}(\theta)(\cos(\theta)v\cdot\nabla_{x}M(x)+\sin(\theta)w\cdot\nabla
M(x))\,\mathrm{d}w\,\mathrm{d}\theta.$

Since $\kappa_{1}$ is an even function, integrating first in $\theta$ yields

$\displaystyle|S_{v}|\Big{(}\int\kappa_{1}(\theta)\cos(\theta)\,\mathrm{d}\theta\Big{)}v\cdot\nabla_{x}M(x)=|\mathbb{S}^{d-2}|\Big{(}\int\kappa_{1}(\theta)\cos(\theta)\,\mathrm{d}\theta\Big{)}v\cdot\nabla_{x}M(x).$

where $|S_{v}|$ is the size of $S_{v}$, similarly for $|\mathbb{S}^{d-2}|$.
Therefore we obtain (14) with

$\displaystyle
C_{\kappa}=|\mathbb{S}^{d-2}|\Big{(}\int\kappa_{1}(\theta)\cos(\theta)\,\mathrm{d}\theta\Big{)}.$

∎

######  Lemma 2.9.

Suppose that $\kappa$ is a collision kernel ($\kappa_{1}$) and that hypotheses
(H2) and (H4) are satisfied. Then we have

$\displaystyle\int_{\mathbb{S}^{d-1}}\kappa(v^{\prime},v)v^{\prime}\cdot\nabla_{x}M(x)\psi(v^{\prime}\cdot\nabla_{x}M(x))\,\mathrm{d}v^{\prime}\geq\tilde{\lambda}(\|\nabla_{x}M\|_{\infty},\kappa)|\nabla_{x}M(x)|^{b}.$

###### Proof.

Let us first prove the lemma when $d=2$ which is the application of Theorem
1.1. Performing the same change of variables as in Lemma 2.8, we have

$\int_{\mathbb{S}^{1}}\kappa(v^{\prime},v)v^{\prime}\cdot\nabla_{x}M(x)\psi(v^{\prime}\cdot\nabla_{x}M(x))\,\mathrm{d}v^{\prime}\\\
=\int_{-\pi}^{\pi}\kappa_{1}(\theta)(\cos(\theta)v\cdot\nabla_{x}M+\sin(\theta)v^{\perp}\cdot\nabla_{x}M)\psi(\cos(\theta)v\cdot\nabla_{x}M+\sin(\theta)v^{\perp}\cdot\nabla_{x}M)\,\mathrm{d}\theta.$

Then, using the assumption ($m$) we can bound this integral below by

$\displaystyle
c\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\cos(\theta)v\cdot\nabla_{x}M+\sin(\theta)v^{\perp}\cdot\nabla_{x}M|^{b}\,\mathrm{d}\theta=c|\nabla_{x}M|^{b}\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\cos(\theta-\alpha)|^{b}\,\mathrm{d}\theta,$

where $\alpha$ is the angle between $v$ and $\nabla_{x}M$. If we write

$\displaystyle
F(\alpha)=\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\cos(\theta-\alpha)|^{b}\,\mathrm{d}\theta,$

then we can compute that

$\displaystyle
F^{\prime}(\alpha)=b\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\cos^{2}(\theta-\alpha)|^{\frac{b}{2}-1}\cos(\theta-\alpha)\sin(\theta-\alpha)\,\mathrm{d}\theta$

and by changing variables to get
$F^{\prime}(\alpha)=b\int_{-\pi}^{\pi}\kappa_{1}(\theta+\alpha)|\cos^{2}(\theta)|^{\frac{b}{2}-1}\cos(\theta)\sin(\theta)\,\mathrm{d}\theta$
and then changing variables back we have

$\displaystyle
F^{\prime\prime}(\alpha)=b\int_{-\pi}^{\pi}\kappa_{1}^{\prime}(\theta)|\cos^{2}(\theta-\alpha)|^{\frac{b}{2}-1}\cos(\theta-\alpha)\sin(\theta-\alpha)\,\mathrm{d}\theta.$

Therefore, $F(\alpha)=0$ when
$\alpha=\left\\{0,\pm\frac{\pi}{2},\pm\pi\right\\}$ and in the case
$\alpha=\pm\frac{\pi}{2}$ we have $F^{\prime\prime}(\alpha)<0$ as

$\displaystyle
F^{\prime\prime}\left(\pm\frac{\pi}{2}\right)=-b\int_{-\pi}^{\pi}\kappa_{1}^{\prime}(\theta)|\sin^{2}(\theta)|^{b/2-1}\sin(\theta)\cos(\theta)\,\mathrm{d}\theta,$

which is negative when $\kappa_{1}^{\prime}(\theta)\sin(\theta)\geq 0$ for all
$\theta$ which will be the case if $\kappa_{1}$ is a decreasing function of
$|\theta|$. Therefore, for all $\alpha$ we have

$F(\alpha)\geq\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\sin(\theta)|^{b}\mathrm{d}\theta.$

In the case where we have $d>2$, let us make the change of variables

$\displaystyle
v^{\prime}=\cos(\theta)v+\sin\theta\cos(\psi)u+\sin(\theta)\sin(\psi)p,$

where $u$ is the unit vector in the direction
$\nabla_{x}{M}-\left(\nabla_{x}M\cdot v\right)v$ and $p$ is a variable vector
that ranges over the sphere of vectors of norm $1$ perpendicular to both $v$
and $u$ (which we call $S_{v,u}$. The Jacobian of this change of variable is
$|\sin(\theta)|^{d-2}$. Therefore, we want to find a lower bound for the
integral

$\displaystyle\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{S_{v,u}}\kappa_{1}(\theta)|\sin(\theta)|^{d-2}|\cos(\theta)v\cdot\nabla_{x}M+\sin(\theta)\cos(\psi)u\cdot\nabla_{x}M|^{b}\,\mathrm{d}p\,\mathrm{d}\psi\,\mathrm{d}\theta.$

The integral $\,\mathrm{d}p$ just gives us a constant factor. To evaluate the
rest, we write $\phi_{v}$ to be the angle between $\nabla_{x}M$ and $v$. Then
we have

$\displaystyle|\nabla_{x}M|^{b}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}$
$\displaystyle\kappa_{1}(\theta)|\sin(\theta)|^{d-2}|\cos(\theta)\cos(\phi_{v})+\sin(\theta)\sin(\phi_{v})\cos(\psi)|^{b}\,\mathrm{d}\psi\,\mathrm{d}\theta$
$\displaystyle=|\nabla_{x}M|^{b}\int_{-\pi}^{\pi}(1-\sin^{2}(\phi_{v})\sin^{2}(\psi))^{b/2}\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\sin(\theta)|^{d-2}|\cos(\theta-\alpha(\phi_{v},\psi))|^{b}\,\mathrm{d}\theta\,\mathrm{d}\psi$
$\displaystyle\geq|\nabla_{x}M|^{b}\left(\int_{-\pi}^{\pi}|\cos(\psi)|^{b}\mathrm{d}\psi\right)\inf_{\alpha}\left(\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\sin(\theta)|^{d-2}|\cos(\theta-\alpha)|^{b}\mathrm{d}\theta\right)$
$\displaystyle=|\nabla_{x}M|^{b}\left(\int_{-\pi}^{\pi}|\cos(\psi)|^{b}\mathrm{d}\psi\right)\left(\int_{-\pi}^{\pi}\kappa_{1}(\theta)|\sin(\theta)|^{d-2}|\sin(\theta)|^{b}\mathrm{d}\theta\right).$

In the last line above, we have used similar considerations to the case $d=2$.
∎

Now, we can move on to the proof of the Foster-Lyapunov condition (FL1).

######  Proposition 2.10.

If hypotheses (H1), (H2), (H3) and (H4) are satisfied then we can choose a
constant $A$ so that the function

$\displaystyle\phi(x,v)=\Big{(}1-\frac{\gamma}{1-C_{\kappa}}v\cdot\nabla_{x}M(x)-\gamma
Av\cdot\nabla_{x}M(x)\psi(v\cdot\nabla_{x}M)\Big{)}e^{-\gamma M(x)}$

verifies a Foster-Lyapunov condition (FL1) for Eq. (1).

###### Proof.

First, we notice that if $A$ is sufficiently small then

$\displaystyle\phi(x,v)\geq\Big{(}1-2\gamma\Big{(}\frac{1}{1-C_{\kappa}}+\frac{\chi}{1+\chi}\Big{)}V_{0}\|\nabla_{x}M\|_{\infty}\Big{)}e^{-\gamma
M(x)},$

so we can choose

$\displaystyle\gamma\leq\frac{1}{4V_{0}\|\nabla_{x}M\|_{\infty}}\Big{(}\frac{1}{1-C_{k}}+\frac{\chi}{1+\chi}\Big{)}^{-1}$

and this ensures

$\displaystyle\phi(x,v)\geq\frac{1}{2}e^{-\gamma M(x)}.$ (15)

Now, we differentiate the separate parts of the Lyapunov function, remembering
(10) and using Lemmas 2.8 and 2.9,

$\displaystyle\mathcal{L}^{*}(e^{-\gamma M})=-\gamma
v\cdot\nabla_{x}Me^{-\gamma M},$ $\mathcal{L}^{*}(v\cdot\nabla_{x}Me^{-\gamma
M})=v^{T}\operatorname{Hess}(M)ve^{-\gamma
M}+\gamma(v\cdot\nabla_{x}M)^{2}e^{-\gamma M}\\\
\quad-(1-C_{\kappa})v\cdot\nabla_{x}Me^{-\gamma
M}+\chi(1-C_{\kappa})\psi(v\cdot\nabla_{x}M)v\cdot\nabla_{x}Me^{-\gamma M},$

and

$\displaystyle\mathcal{L}^{*}(v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)$
$\displaystyle e^{-\lambda M})\geq
v^{T}\operatorname{Hess}(M)v\left(\psi(v\cdot\nabla_{x}M)+\psi^{\prime}(v\cdot\nabla_{x}M)v\cdot\nabla_{x}M\right)e^{-\gamma
M}$
$\displaystyle\quad+\gamma(v\cdot\nabla_{x}M)^{2}\psi(v\cdot\nabla_{x}M)e^{-\gamma
M}-v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)(1+\chi)e^{-\gamma M}$
$\displaystyle\quad+\tilde{\lambda}(\|\nabla_{x}M\|_{\infty},\kappa_{1})(1-\chi)|\nabla_{x}M|^{b}.$

Now, we can put this all together in a Lyapunov functional for positive $A$
sufficiently small

$\displaystyle\mathcal{L}^{*}\Big{(}\Big{(}1-\frac{\gamma}{1-C_{\kappa}}v\cdot$
$\displaystyle\nabla_{x}M-\gamma
Av\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)\Big{)}e^{-\gamma M}\Big{)}$
$\displaystyle\leq$ $\displaystyle\,\gamma
v^{T}\operatorname{Hess}(M)ve^{-\gamma
M}\Big{(}\frac{1}{1-C_{\kappa}}+A\Big{(}1+\sup_{z}(z\psi^{\prime}(z)))\Big{)}+\gamma^{2}V_{0}^{2}e^{-\gamma
M}\Big{(}\frac{1}{1-C_{\kappa}}+A\Big{)}$
$\displaystyle+v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)e^{-\gamma
M}\gamma(-\chi+A(1+\chi))-\gamma A(1-\chi)|\nabla_{x}M|^{b}e^{-\gamma M}$

Then, choosing $A\leq\frac{\chi}{1+\chi}$ we have

$\mathcal{L}^{*}\Big{(}\Big{(}1-\frac{\gamma}{1-C_{\kappa}}v\cdot\nabla_{x}M-\gamma
Av\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M\Big{)}e^{-\gamma M}\Big{)}\\\
\leq\gamma\Big{(}C_{1}V_{0}^{2}|\operatorname{Hess}(M)|e^{-\gamma M}+\gamma
C_{2}V_{0}^{2}e^{-\gamma M}-A(1-\chi)|\nabla_{x}M|^{b}\Big{)}e^{-\gamma M}.$

Now, we know that we can choose $R_{*}$ such that if $|x|>R_{*}$ then,

$\displaystyle|\nabla_{x}M(x)|\geq m_{*}\quad\mbox{and }\quad
C_{1}V_{0}^{2}|\operatorname{Hess}(M)(x)|\leq A(1-\chi)\frac{m_{*}^{b}}{4},$

then if we choose $\gamma$ small enough so that $\gamma C_{2}V_{0}^{2}\leq
A(1-\chi)\dfrac{m_{*}^{b}}{4}$, then we will have

$\mathcal{L}^{*}\Big{(}\Big{(}1-\frac{\gamma}{1-C_{\kappa}}v\cdot\nabla_{x}M-\gamma
Av\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M\Big{)}e^{-\gamma M}\Big{)}\\\
\leq\gamma\Big{(}-\frac{1}{2}A(1-\chi)m_{*}^{b}e^{-\gamma
M}\mathds{1}_{|x|>R_{*}}+C_{3}\mathds{1}_{|x|\leq R_{*}}\Big{)},$ (16)

for some constant $C_{3}$. Finally, using (15) we have

$\displaystyle\mathcal{L}^{*}[\phi]\leq-\gamma A(1-\chi)m_{*}^{b}\phi+C_{4}.$

This verifies (FL1) with $\zeta=\gamma A(1-\chi)m_{*}^{b}$ and $D=C_{4}$. ∎

###### Proof of Theorem 1.1.

We verify the two hypotheses of Harris’s theorem in Propositions 2.7 and 2.10.
The contraction in the $\phi-$weighted total variation norm and the existence
of a unique steady state follow by Harris’s theorem (Theorem 1.4). ∎

##  3 Unbounded velocity spaces

This section is dedicated to the long-time behaviour of the linear run and
tumble equation, Eq. (1) posed in an unbounded velocity space
$\mathcal{V}=\mathbb{R}^{d}$ with a tumbling kernel ($\kappa_{2}$). The
following two sections are dedicated to verifying the two hypotheses of the
subgeometric version of Harris’s theorem (Theorem (1.6)). At the end of this
section, we provide a proof of Theorem 1.3.

###  3.1 Minorisation condition

For unbounded velocities, our minorisation part is essentially identical to
that in our previous paper [13]. For the sake of completeness, we include a
proof here. Let us again write $(T_{t}f)(t,x,v)=f(t,x-vt,v)$ for the transport
semigroup and define

$\displaystyle(\hat{L}f)(t,x,v):=\int_{\mathbb{R}^{d}}f(t,x,u)\,\mathrm{d}u\mathds{1}_{|v|\leq
V_{0}}(v).$

Then we have the following lemmas.

######  Lemma 3.1.

There exists a constant $C$ such that

$f(t,x,v)\geq
Ce^{-(1+\chi)t}\int_{0}^{t}\int_{0}^{s}T_{t-s}\hat{L}T_{s-r}\hat{L}T_{r}f_{0}(x,v)\,\mathrm{d}r\,\mathrm{d}s.$

###### Proof.

The proof is exactly the same as in [13] after observing that there exists
some $\tilde{C}>0$ such that

$\displaystyle\kappa_{2}(v)\geq\tilde{C}\mathds{1}_{|v|\leq V_{0}}(v).$

∎

######  Lemma 3.2.

For every $R_{*}>0$, we can take $t=3+\frac{R_{*}}{V_{0}}$ such that any
solution of Eq. (1) with initial data
$f_{0}\in\mathcal{P}(\mathbb{R}^{d}\times\mathbb{R}^{d})$ with $\int_{|x|\leq
R_{*}}\int_{B(0;V_{0})}f_{0}(x,v)\,\mathrm{d}x\,\mathrm{d}v=1$ satisfies

$\displaystyle
f(t,x,v)\geq(1-\chi^{2})e^{-(1+\chi)t}\frac{1}{t^{d}|B(V_{0})|}\mathds{1}_{|x|\leq
V_{0}}(x)\mathds{1}_{|v|\leq V_{0}}(v).$ (17)

###### Proof.

We take $f_{0}(x,v):=\delta_{(x_{0}}\delta_{v_{0})}$ where
$(x_{0},v_{0})\in\mathbb{R}^{d}\times B(0,V_{0})$, is an arbitrary point with
an arbitrary velocity. We only need to consider $x_{0}\in B(0,R_{*})$, then
the bound we obtain depends on $R_{*}$. First, we have that

$\displaystyle T_{r}f_{0}\geq\delta_{x_{0}+rv_{0}}(x)\delta_{v_{0}}(v).$

Applying $\hat{L}$ to this we obtain

$\displaystyle\hat{L}T_{r}f_{0}\geq\delta_{x_{0}+rv_{0}}(x)\mathds{1}_{|v|\leq
V_{0}}(v).$

Performing a change of variables we have

$\displaystyle\int_{\mathbb{R}^{d}}(T_{s-r}\hat{L}\mathcal{T}_{r}f_{0})(x,v)\,\mathrm{d}v\geq\frac{1}{(s-r)^{d}|B(V_{0})|}\mathds{1}_{|x-x_{0}-rv_{0}|\leq
V_{0}(s-r)}(x).$

Therefore we have

$\displaystyle\hat{L}T_{s-r}\hat{L}T_{r}f_{0}\geq\frac{1}{(s-r)^{d}|B(V_{0})|}\mathds{1}_{|x-x_{0}-rv_{0}|\leq
V_{0}(s-r)}(x)\mathds{1}_{|v|\leq V_{0}}(v).$

Applying the transport operator once more we obtain

$\displaystyle
T_{t-s}\hat{L}T_{s-r}\hat{L}T_{r}f_{0}\geq\frac{1}{(s-r)^{d}|B(V_{0})|}\mathds{1}_{|x-(t-s)v-x_{0}-rv_{0}|\leq
V_{0}(s-r)}(x)\mathds{1}_{|v|\leq V_{0}}(v).$

Moreover, we have

$\displaystyle|x|\leq(s-r)V_{0}-(t-s)V_{0}-rV_{0}-R_{*}$

which is implied by $|x-v(t-s)-x_{0}-rv_{0}|\leq(s-r)V_{0}$ since all the
velocities are smaller than $V_{0}$. Then, if we ensure that $(s-r)\geq
2+\frac{R_{*}}{V_{0}}$, $r\leq\frac{1}{2}$ and that $(t-s)\leq\frac{1}{2}$, we
will obtain

$\displaystyle
T_{t-s}\hat{L}T_{s-r}\hat{L}T_{r}f_{0}\geq\frac{1}{(s-r)^{d}|B(V_{0})|}\mathds{1}_{|x|\leq
V_{0}}(x)\mathds{1}_{|v|\leq V_{0}}(v).$

Therefore, setting $t=3+\frac{R_{*}}{V_{0}}$ and restricting the time
integrals to $r\in\big{(}0,\frac{1}{2}\big{)}$,
$s\in\big{(}\frac{5}{2}+\frac{R_{*}}{V_{0}},3+\frac{R_{*}}{V_{0}}\big{)}$, we
obtain

$\displaystyle f(t,x,v)$ $\displaystyle\geq
C\int_{0}^{t}\int_{0}^{s}T_{t-s}\hat{L}T_{s-r}\hat{L}T_{r}f_{0}(x,v)\,\mathrm{d}r\,\mathrm{d}s$
$\displaystyle\geq
C(1-\chi)^{2}e^{-(1+\chi)t}\int_{\frac{5}{2}+\frac{R_{*}}{V_{0}}}^{3+\frac{R_{*}}{V_{0}}}\int_{0}^{\frac{1}{2}}\frac{1}{(s-r)^{d}|B(V_{0})|}\mathds{1}_{|x|\leq
V_{0}}(x)\mathds{1}_{|v|\leq V_{0}}(v)\,\mathrm{d}r\,\mathrm{d}s$
$\displaystyle\geq
C(1-\chi)^{2}e^{-(1+\chi)t}\frac{1}{t^{d}|B(V_{0})|}\mathds{1}_{|x|\leq
V_{0}}(x)\mathds{1}_{|v|\leq V_{0}}(v).$

This gives the uniform lower bound and verifies the minorisation condition
(M2). We can extend this result from the Dirac delta function initial data to
general initial data by using the fact that the associated semigroup is
Markov. ∎

###  3.2 Foster-Lyapunov condition

In this section, we verify the Foster-Lyapunov condition (FL2) for Eq. (1) for
the case ($\kappa_{2}$). Thus, we prove the following lemma.

######  Lemma 3.3.

If hypotheses (H1), (H2), (H3), (H5) are satisfied and assuming that
$M\operatorname{Hess}(M)$ is bounded, then the function

$\displaystyle\phi(v,M)=M^{2}+2v\cdot\nabla_{x}MM\Big{(}1+\frac{\chi}{1+\chi}\psi(v\cdot\nabla_{x}M)\Big{)}+Av^{2}$

verifies a weaker Foster-Lyapunov function (FL2) for a constant $A$
sufficiently large and with computable constants $C>0,\Lambda>0$ so that

$\displaystyle\mathcal{L}^{*}[\phi](v,M)\leq C-\Lambda\sqrt{\phi(v,M)}.$

###### Proof.

In this proof, it is useful to remember that $M$ is negative for $|x|$
sufficiently large. Since it is only defined up to a constant, let us choose
$M<0$. Similar to the previous case, we look at how the adjoint
$\mathcal{L}^{*}$ (defined by (10)) acts on different terms. Precisely, we
have

$\displaystyle\mathcal{L}^{*}(M^{2})$ $\displaystyle=2v\cdot\nabla_{x}MM,$
$\displaystyle\mathcal{L}^{*}(2v\cdot\nabla_{x}MM)$
$\displaystyle=2v^{T}\mbox{Hess}(M)vM+2(v\cdot\nabla_{x}M)^{2}-2v\cdot\nabla_{x}MM-2\chi
v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)|M|$

Then we have, for any $c>0$

$\displaystyle\mathcal{L}^{*}\big{(}cv\cdot\nabla_{x}M$
$\displaystyle\psi(v\cdot\nabla_{x}M)M\big{)}$ $\displaystyle=\,$
$\displaystyle
cv^{T}\operatorname{Hess}(M)vM\left(\psi^{\prime}(v\cdot\nabla_{x}M)v\cdot\nabla_{x}M+\psi(v\cdot\nabla_{x}M)\right)+c(v\cdot\nabla_{x}M)^{2}\psi(v\cdot\nabla_{x}M)$
$\displaystyle+c(1-\chi\psi(v\cdot\nabla_{x}M))M\left(\int\kappa_{2}(v^{\prime})v^{\prime}\cdot\nabla_{x}M\psi(v^{\prime}\cdot\nabla_{x}M)\,\mathrm{d}v^{\prime}-v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)\right)$
$\displaystyle\leq$
$\displaystyle\,c\left(\|\operatorname{Hess}(M)M\|_{\infty}\|\psi(z)(z)\|_{\text{Lip}}+\|\nabla_{x}M\|^{2}_{\infty}\right)|v|^{2}$
$\displaystyle-c(1-\chi)\tilde{\lambda}(\|\nabla_{x}M\|_{\infty},\kappa_{2})|M|+c(1+\chi)v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)|M|.$

Summing these up and choosing $c=\frac{2\chi}{1+\chi}$ we have

$\mathcal{L}^{*}\big{(}M^{2}+2v\cdot\nabla_{x}MM+\frac{2\chi}{1+\chi}v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)M\big{)}\\\
\leq\Big{(}\Big{(}2+\frac{2\chi}{1+\chi}\|\psi(z)(z)\|_{\text{Lip}}\Big{)}\|\operatorname{Hess}(M)M\|_{\infty}+\Big{(}2+\frac{2\chi}{1+\chi}\Big{)}\|\nabla_{x}M\|_{\infty}^{2}\Big{)}|v|^{2}-\frac{2\chi(1-\chi)}{1+\chi}\tilde{\lambda}|M|$

Now we also have

$\displaystyle\mathcal{L}^{*}(|v|^{2})\leq(1+\chi)-(1-\chi)|v|^{2}.$

Therefore choosing

$\displaystyle A\geq
1+\frac{1}{1-\chi}\Big{(}\Big{(}2+\frac{2\chi}{1+\chi}\|\psi(z)(z)\|_{\text{Lip}}\Big{)}\|\operatorname{Hess}(M)M\|_{\infty}+\Big{(}2+\frac{2\chi}{1+\chi}\Big{)}|\nabla_{x}M\|^{2}_{\infty}\Big{)},$

we have

$\displaystyle\mathcal{L}^{*}\Big{(}M^{2}+2v\cdot\nabla_{x}MM+\frac{2\chi}{1+\chi}v\cdot\nabla_{x}M\psi(v\cdot\nabla_{x}M)$
$\displaystyle M+A|v|^{2}\Big{)}$ $\displaystyle\leq
A(1+\chi)-(1-\chi)|v|^{2}-\frac{2\chi(1-\chi)\tilde{\lambda}}{1+\chi}|M|$
$\displaystyle\leq C-\Lambda\sqrt{\phi(v,M)}.$

Lastly, choosing $A$ sufficiently large ensures that $\phi>0$, and for $A$
sufficiently large, $\phi$ is comparable to $M^{2}+v^{2}$. ∎

###  3.3 Subgeometric convergence rates

We can now combine the results of the two previous sections to get a proof of
Theorem 1.3.

###### Proof of Theorem 1.3.

We have verified the hypotheses of the subgeometric Harris’s theorem with the
Foster-Laypunov function being

$\displaystyle\phi(v,M)=M^{2}+2v\cdot\nabla_{x}MM\Big{(}1+\frac{\chi}{1+\chi}\psi(v\cdot\nabla_{x}M)\Big{)}+Av^{2},$

and the function $h(t)=\sqrt{t}$. We can, therefore, compute that the function
$H^{-1}(t)=\big{(}\frac{t}{2}+1\big{)}^{2}$ and $h\circ
H_{h}^{-1}(t)=\big{(}\frac{t}{2}+1\big{)}$. Hence, applying the conclusion of
Theorem 1.6 gives the existence of a steady state $f_{\infty}$ and that

$\|f_{t}-f_{\infty}\|_{TV}\\\ \leq C\Big{(}\frac{t}{2}+1\Big{)}^{-2}\int
f_{0}\left(M^{2}+2v\cdot\nabla_{x}MM\Big{(}1+\frac{\chi}{1+\chi}\psi(v\cdot\nabla_{x}M)\Big{)}+Av^{2}\right)\,\mathrm{d}x\,\mathrm{d}v+C\Big{(}\frac{t}{2}+1\Big{)}^{-1}.$

This proves the result. ∎

## Acknowledgements

The authors would like to thank Emeric Bouin for helpful discussions and for
suggesting to look at unbounded velocity spaces.

The authors would like to thank the Isaac Newton Institute for Mathematical
Sciences for support and hospitality during the programme “Frontiers in
kinetic theory: connecting microscopic to macroscopic scales - KineCon 2022”
when work on this paper was undertaken. This work was supported by EPSRC Grant
Number EP/R014604/1.

J. Evans acknowledges partial support from the Leverhulme Trust, Grant
ECF-2021-134. H. Yoldaş was partially supported by the Vienna Science and
Technology Fund (WWTF) with a Vienna Research Groups for Young Investigators
project, grant VRG17-014 (until October 2022).

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