# Exponential stability and hypoelliptic regularization for the kinetic
Fokker-Planck equation with confining potential

Anton Arnold, Gayrat Toshpulatov

###### Abstract

This paper is concerned with a modified entropy method to establish the large-
time convergence towards the (unique) steady state, for kinetic Fokker-Planck
equations with non-quadratic confinement potentials in whole space. We extend
previous approaches by analyzing Lyapunov functionals with non-constant weight
matrices in the dissipation functional (a generalized Fisher information). We
establish exponential convergence in a weighted $H^{1}$-norm with rates that
become sharp in the case of quadratic potentials. In the defective case for
quadratic potentials, i.e. when the drift matrix has non-trivial Jordan
blocks, the weighted $L^{2}$-distance between a Fokker-Planck-solution and the
steady state has always a sharp decay estimate of the order
$\mathcal{O}\big{(}(1+t)e^{-t\nu/2}\big{)}$, with $\nu$ the friction
parameter. The presented method also gives new hypoelliptic regularization
results for kinetic Fokker-Planck equations (from a weighted $L^{2}$-space to
a weighted $H^{1}$-space).

††Authors addresses: Institute for Analysis and Scientific Computing, TU Wien,
Wiedner Hauptstraße 8–10, 1040 Vienna, Austria<EMAIL_ADDRESS><EMAIL_ADDRESS>
Keywords: Kinetic theory, Fokker-Planck equation, confinement potential,
degenerate evolution, long time behavior, convergence to equilibrium,
hypocoercivity, hypoellictic regularity, Lypunov functional.
2020 Mathematics Subject Classification: 35Q84, 35B40, 35Q82, 82C40.

###### Contents

  1. 1 Introduction
  2. 2 Main results
  3. 3 Modified entropy methods for degenerate Fokker-Planck equations
  4. 4 The choice of the matrix P
  5. 5 Proofs
    1. 5.1 Proof of Theorem 2.3
    2. 5.2 Proof of Proposition 2.5
    3. 5.3 Proof of Theorem 2.7 and Corollary 2.8
  6. 6 Appendix
    1. 6.1  Proof that Assumption 2.2’ implies Assumption 2.2
    2. 6.2  Matrix inequalities for Section 5.1
    3. 6.3  Proof of inequality (80)

##  1 Introduction

This paper is devoted to the study of the long time behavior of the kinetic
Fokker-Planck equation

$\begin{cases}\partial_{t}f+v\cdot\nabla_{x}f-\nabla_{x}V\cdot\nabla_{v}f=\nu\text{div}_{v}(vf)+\sigma\Delta_{v}f,\,\,\,x,\,v\in\mathbb{R}^{n},\,t>0\\\
f(t=0)=f_{0}\in L^{1}(\mathbb{R}^{2n})\end{cases}$ (1)

describing the time evolution of the phase space probability density
$f(t,x,v),$ e.g. in a plasma [28]. Applications range from plasma physics [26,
11] to stellar dynamics [15, 16]. Here $V=V(x)$ is a given smooth, bounded
below confinement potential for the system, and $\nu>0,\,\sigma>0$ denote the
friction and diffusion parameters, respectively. This equation is associated
with the Langevin stochastic differential equation

$\begin{cases}dx_{t}=v_{t}dt\\\ dv_{t}=-\nu v_{t}dt-\nabla
V(x_{t})dt+\sqrt{2\sigma}dB_{t},\end{cases}$

where $\\{B_{t}\\}_{t\geq 0}$ is a Brownian motion in $\mathbb{R}^{n}$ with
covariance $\langle B_{t},B_{t^{\prime}}\rangle=\delta_{t-t^{\prime}}.$
Since the equation conserves mass, i.e.,

$\int_{\mathbb{R}^{2n}}f(t,x,v)dxdv=\int_{\mathbb{R}^{2n}}f_{0}(x,v)dxdv,\,\,\,\,\,t\geq
0,$

we shall always assume (without restriction of generality) that
$\displaystyle\int_{\mathbb{R}^{2n}}f_{0}(x,v)dxdv=1.$ The unique normalized
steady state of (1) is given by

$f_{\infty}(x,v)=c_{V}e^{-\frac{\nu}{\sigma}[V(x)+\frac{|v|^{2}}{2}]},\,\,\,x,v\in\mathbb{R}^{n},$
(2)

where $c_{V}$ is a positive constant such that
$\int_{\mathbb{R}^{2n}}f_{\infty}(x,v)dxdv=1.$ The following equation is also
considered as the kinetic Fokker-Planck equation:

$\partial_{t}h+v\cdot\nabla_{x}h-\nabla_{x}V\cdot\nabla_{v}h=\sigma\Delta_{v}h-\nu
v\cdot\nabla_{v}h,\,\,\,x,\,v\in\mathbb{R}^{n},\,t>0,$ (3)

and to switch from (1) to (3) it suffices to set $h:=f/f_{\infty}.$

It was shown in [19] that, if $V\in C^{\infty}(\mathbb{R}^{n}),$ (3) generates
a $C^{\infty}$ regularizing contraction semigroup in
$L^{2}(\mathbb{R}^{d},f_{\infty}):=\\{g\colon\mathbb{R}^{d}\to\mathbb{R}:g\text{
is measurable and
}\int_{\mathbb{R}^{d}}g^{2}f_{\infty}dxdv<\infty\\},\,\,\,d=2n.$ For well-
posedness with non-smooth potentials, we refer to [29, Theorem 6, Theorem 7].

The long time behavior and exponential convergence of the solution to the
steady state has been studied and there are various results: in [17],
algebraic decay was proved for potentials that are asymptotically quadratic
(as $|x|\to\infty$) and for initial conditions that are bounded below and
above by Gaussians. The authors used logarithmic Sobolev inequalities and
entropy methods. In [21], exponential decay was obtained also for faster
growing potentials and more general initial conditions. That proof is based on
hypoellipticity techniques. In [18] exponential decay in $L^{2}$ was proved,
allowing for potentials with linear or super-linear growth. By using
hypoelliptic methods, Villani proved exponential convergence results in
$H^{1}(\mathbb{R}^{d},f_{\infty}):=\\{g\in
L^{2}(\mathbb{R}^{d},f_{\infty}):|\nabla g|\in
L^{2}(\mathbb{R}^{d},f_{\infty})\\}$ [29, Theorem 35] and in
$L^{2}(\mathbb{R}^{d},f_{\infty})$ [29, Theorem 37]. The main conditions in
Villani’s theorems above, as well as in [18, 9, 10, 30, 12, 13], are the
validity of the Poincaré inequality (5) and the criterion

$\exists\,\,C\geq
0\,\,:\,\,\,\,\,\,\,\,\small\left|\left|\frac{\partial^{2}V(x)}{\partial
x^{2}}\right|\right|\leq C(1+|\nabla V(x)|),\,\,\,\,\forall
x\in\mathbb{R}^{n},$ (4)

where $\displaystyle\left|\left|\frac{\partial^{2}V(x)}{\partial
x^{2}}\right|\right|$ denotes the Frobenius norm of
$\displaystyle\frac{\partial^{2}V(x)}{\partial x^{2}}.$
When $\frac{\partial^{2}V}{\partial x^{2}}$ is bounded, Villani also proved
that the solution converges to the steady state exponentially in the
logarithmic entropy [29, Theorem 39]. This result was extended in [14] to
potentials $V$ satisfying a weighted log-Sobolev inequality and the condition
that $V^{-2\eta}\frac{\partial^{2}V}{\partial x^{2}}$ is bounded for some
$\eta\geq 0.$ Even though Villani’s result allows for a general class of
potentials, the growth condition (4) is not satisfied by potentials with
singularities. This type of potentials, such as Lennard-Jones type
interactions with confinement, are considered in [10] and their method relies
on an explicit construction of a Lypunov function and Gamma calculus.

The associated semigroup of the kinetic Fokker-Planck equation has
instantaneous regularizing properties which is called hypoellipticity [23].
This hypoelliptic regularization is obvious when the confining potential $V$
is zero or quadratic as the fundamental solution can be explicitly computed
(see [25], [23]). For potentials such that $\frac{\partial^{2}V}{\partial
x^{2}}$ is bounded, Hérau [20] obtained short time estimates for a
$L^{2}(\mathbb{R}^{d},f_{\infty})\to H^{1}(\mathbb{R}^{d},f_{\infty})$
regularization by constructing a suitable Lyapunov functional. Based on
interpolation inequalities and a system of differential inequalities, Villani
[29, Appendix A.21] extended Hérau’s result for potentials satisfying (4).

We provide a new method to establish exponential decay of the solution to the
steady state in $H^{1}(\mathbb{R}^{d},f_{\infty})$ for a wide class of
potentials: Our method extends [29, 1, 3] by allowing for more general
Lyapunov functionals. Generalizing the previous approaches, the weight matrix
in the dissipation functional (a generalized Fisher information) may now
depend on $x$ and $v.$ This leads to a new criterion on the potential $V.$ We
provide a formula to estimate easily the exponential decay rate depending on
the parameters of the equation, the constants appearing in the Poincaré
inequality (5) and the growth condition on the potential (see (6) below). As a
test of the effectiveness of our method, we show that our estimate on the
decay rate is sharp when the potential is a quadratic polynomial. Moreover,
our method lets us obtain estimates on the hypoelliptic regularization for
potentials that are more general than in [20].

The organization of this paper is as follows. In Section 2, we define the
assumptions on the potential, state the main results, and present concrete
examples of such potentials. In Section 3, we present the intuition and
explain our method. Section 4 contains important lemmas about matrix
inequalities which are important to construct suitable Lyapunov functionals.
The final section presents the proof of the main results.

##  2 Main results

We make the following assumptions.

######  Assumption 2.1.

There exists a constant $C_{PI}>0$ such that the Poincaré inequality

$\int_{\mathbb{R}^{2n}}h^{2}f_{\infty}dxdv-\left(\int_{\mathbb{R}^{2n}}hf_{\infty}dxdv\right)^{2}\leq\frac{1}{C_{PI}}\int_{\mathbb{R}^{2n}}(|\nabla_{x}h|^{2}+|\nabla_{v}h|^{2})f_{\infty}dxdv\,\,$
(5)

holds for all $h\in H^{1}(\mathbb{R}^{d},f_{\infty}).$

Sufficient conditions on the potential appearing in $f_{\infty}$ so that the
Poincaré inequality holds, e.g. the Bakry-Emery criterion, are presented in
[8, Chapter 4].

######  Assumption 2.2.

There are constants $c\in\mathbb{R}$ and $\tau\in[0,\nu)$ such that the
following $\mathbb{R}^{m\times m}$ matrix, $m\colonequals n(n+1),$

$\begin{pmatrix}\nu\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)&0&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V(x))}{\partial
x^{2}}\\\ 0&\nu\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V(x))}{\partial
x^{2}}\\\ ...&...&...&...&...\\\
0&0&...&\nu\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V(x))}{\partial
x^{2}}\\\ -\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V(x))}{\partial
x^{2}}&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V(x))}{\partial
x^{2}}&...&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V(x))}{\partial
x^{2}}&\frac{\tau\nu}{2\sigma}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)\end{pmatrix}$ (6)

is positive semi-definite for all $x\in\mathbb{R}^{n},$ where
$I\in\mathbb{R}^{n\times n}$ denotes the identity matrix.

Roughly speaking, Assumption 2.2 essentially means that the second order
derivatives of $V$ control the third order ones. It implies that
$\frac{\partial^{2}V(x)}{\partial x^{2}}+cI$ is positive semi-definite for all
$x\in\mathbb{R}^{n},$ and hence the eigenvalues of
$\frac{\partial^{2}V(x)}{\partial x^{2}}$ are uniformly bounded from below. We
note that, in contrast to the Bakry-Emery strategy [7], the eigenvalues here
may take negative values.

Let $\displaystyle\alpha(x)\in\mathbb{R}$ denote the smallest eigenvalue of
$\frac{\partial^{2}V(x)}{\partial x^{2}}$ at $x\in\mathbb{R}^{n}.$ Then the
following condition implies Assumption 2.2. For its proof see Appendix 6.1.
Assumption 2.2’. There are constants $c\in\mathbb{R}$ and $\tau\in[0,\nu)$
such that  $\frac{\partial^{2}V(x)}{\partial x^{2}}+cI$ is positive semi-
definite and

$-\sqrt{\frac{2\tau\nu^{2}}{n\sigma}}(\alpha(x)+c)I\leq\frac{\partial^{2}(\partial_{x_{i}}V(x))}{\partial
x^{2}}\leq\sqrt{\frac{2\tau\nu^{2}}{n\sigma}}(\alpha(x)+c)I$ (7)

for all  $x\in\mathbb{R}^{n}$ and $i\in\\{1,...,n\\}.$

We denote

$\displaystyle\alpha_{0}\colonequals\inf_{x\in\mathbb{R}^{n}}\alpha(x)$ (8)

and assume in the sequel that $\alpha_{0}>-\infty.$ Hence Assumption 2.2 can
only hold for some $c\geq-\alpha_{0}.$

We now state our first result, i.e. exponential decay of a functional that is
a linear combination of the weighted $L^{2}-$norm and a Fisher information-
type functional:

######  Theorem 2.3.

Let $V$ be a $C^{\infty}$ potential in $\mathbb{R}^{n}$ satisfying Assumptions
2.1 and 2.2. Let $C_{PI},$ $c,$ $\tau,$ and $\alpha_{0}$ be the constants in
(5), (6), and (8). Suppose the initial data $f_{0}$ satisfies
$\displaystyle\frac{f_{0}}{f_{\infty}}\in H^{1}(\mathbb{R}^{2n},f_{\infty})$
and
$\displaystyle\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f_{0}}{f_{\infty}}\right)\frac{\partial^{2}V}{\partial
x^{2}}\nabla_{v}\left(\frac{f_{0}}{f_{\infty}}\right)f_{\infty}dxdv<\infty.$
Then there are explicitly computable constants $C>0$ and $\lambda>0$
(independent of $f_{0}$) such that

$\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\\\
+\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv\\\
\leq Ce^{-2\lambda
t}\left[\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f_{0}}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\right.\\\
+\left.\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f_{0}}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f_{0}}{f_{\infty}}\right)f_{\infty}dxdv\right]$
(9)

holds for $t\geq 0.$ Moreover, we have:

  1. (a)

if $\displaystyle\alpha_{0}>\frac{\nu^{2}}{4},\,\,\,c\leq-\frac{\nu^{2}}{4},$
then $\displaystyle 2\lambda={\nu-\tau};$

  2. (b)

if $\displaystyle c=-\alpha_{0}=-\frac{\nu^{2}}{4},$ then $\displaystyle
2\lambda={\nu-\tau-\varepsilon}$ for any
$\displaystyle\varepsilon\in(0,\nu-\tau);$

  3. (c)

if $\displaystyle c>-\frac{\nu^{2}}{4},\,\,c+2\alpha_{0}>\frac{\nu^{2}}{4},$
then

$\displaystyle
2\lambda=\begin{cases}\displaystyle\nu-\tau-\frac{c+\frac{\nu^{2}}{4}}{\sqrt{c+\alpha_{0}}}&\text{
if }\nu-\tau\geq A^{-1}_{1}+\frac{c+\frac{\nu^{2}}{4}}{\sqrt{c+\alpha_{0}}}\\\
\displaystyle\frac{(\nu-\tau)\sqrt{c+\alpha_{0}}-(c+\frac{\nu^{2}}{4})(\sqrt{1+s_{1}^{2}}-s_{1})}{{\sqrt{c+\alpha_{0}}+A_{1}s_{1}(c+\frac{\nu^{2}}{4}})}&\text{
if
}\nu-\tau<A^{-1}_{1}+\frac{c+\frac{\nu^{2}}{4}}{\sqrt{c+\alpha_{0}}}\end{cases},$

where $\displaystyle
A_{1}:=\frac{1+\frac{\nu^{2}}{4}+c+\alpha_{0}+\sqrt{(\frac{\nu^{2}}{4}+c+\alpha_{0}-1)^{2}+\nu^{2}}}{2\sigma
C_{PI}},$
$s_{1}:=\begin{cases}\frac{A^{2}_{1}(c+\frac{\nu^{2}}{4})^{2}-c-\alpha_{0}}{2A_{1}(c+\frac{\nu^{2}}{4})\sqrt{c+\alpha_{0}}}&\text{
if }(\nu-\tau)A_{1}=2\\\
\frac{1}{\nu-\tau}\left[\left|\frac{(\nu-\tau)A_{1}-1}{(\nu-\tau)A_{1}-2}\right|\sqrt{\frac{(c+\frac{\nu^{2}}{4})^{2}}{c+\alpha_{0}}+2(\nu-\tau)A^{-1}_{1}-(\nu-\tau)^{2}}-\frac{c+\frac{\nu^{2}}{4}}{((\nu-\tau)A_{1}-2)\sqrt{c+\alpha_{0}}}\right]&\text{
if }(\nu-\tau)A_{1}\neq 2\end{cases};$

  4. (d)

if $\displaystyle
c>-\frac{\nu^{2}}{4},\,\,c+2\alpha_{0}\leq\frac{\nu^{2}}{4},$ then

$\displaystyle
2\lambda=\begin{cases}\displaystyle\nu-\tau-\sqrt{{\nu^{2}}-4\alpha_{0}}&\text{
if }\nu-\tau\geq A^{-1}_{2}+\sqrt{{\nu^{2}}-4\alpha_{0}}\\\
\displaystyle\frac{\nu-\tau-\sqrt{{\nu^{2}}-4\alpha_{0}}(\sqrt{1+s_{2}^{2}}-s_{2})}{1+A_{2}s_{2}\sqrt{{\nu^{2}}-4\alpha_{0}}}&\text{
if }\nu-\tau<A^{-1}_{2}+\sqrt{{\nu^{2}}-4\alpha_{0}}\end{cases},$

where $\displaystyle
A_{2}:=\frac{1+\frac{\nu^{2}}{2}-\alpha_{0}+\sqrt{(\frac{\nu^{2}}{2}-\alpha_{0}-1)^{2}+\nu^{2}}}{2\sigma
C_{PI}},$
$\displaystyle
s_{2}:=\begin{cases}\frac{A^{2}_{2}({\nu^{2}-4\alpha_{0}})-1}{2A_{2}\sqrt{{\nu^{2}}-4\alpha_{0}}}&\text{
if }(\nu-\tau)A_{2}=2\\\
\frac{1}{\nu-\tau}\left[\left|\frac{(\nu-\tau)A_{2}-1}{(\nu-\tau)A_{2}-2}\right|\sqrt{{{\nu^{2}}-4\alpha_{0}}+2(\nu-\tau)A^{-1}_{2}-(\nu-\tau)^{2}}-\frac{\sqrt{{\nu^{2}}-4\alpha_{0}}}{((\nu-\tau)A_{2}-2)}\right]&\text{
if }(\nu-\tau)A_{2}\neq 2\end{cases};$

  5. (e)

if $V(x)$ is a quadratic polynomial of $x$ and $\frac{\partial^{2}V}{\partial
x^{2}}$ is positive definite, then Assumptions 2.1 and 2.2 are satisfied with
$\tau=0,$ $c=-\alpha_{0}$ (this rules out the conditions in the case of (c)).
Moreover, the decay rates $\lambda$ in (a) and (d) are sharp and, in the case
of (d), $\nu\geq{A^{-1}_{2}}+\sqrt{{\nu^{2}}-4\alpha_{0}}$ holds and so
$2\lambda=\nu-\sqrt{{\nu^{2}}-4\alpha_{0}}.$ In the case of (b), the decay
rate $2\lambda=\nu-\varepsilon$ is sharp in the sense that (9) holds with the
rate $2\lambda=\nu-\varepsilon$ for any small fixed $\varepsilon\in(0,\nu),$
but it does not hold with the rate $2\lambda=\nu.$

######  Remark 2.4.

  1. 1.

It is possible to make weaker regularity hypothesis on the potential $V,$ but
we maintain the assumption that $V\in C^{\infty}$ to keep the presentation
simple.

  2. 2.

Since $\frac{\partial^{2}V}{\partial x^{2}}+(1-\alpha_{0})I\geq I,$ (9)
implies that the solution converges exponentially to the steady state in
$H^{1}(\mathbb{R}^{2n},f_{\infty}).$ If the eigenvalues of
$\frac{\partial^{2}V}{\partial x^{2}}$ are uniformly bounded, then (9) is
equivalent to the exponential decay of the solution to the steady state in
$H^{1}(\mathbb{R}^{2n},f_{\infty}).$ Due to the Poincaré inequality (5), the
$L^{2}-$term on the right hand side of (9) could be omitted.

  3. 3.

If $V$ satisfies Assumption 2.2 with some constants $c\in\mathbb{R}$ and
$\tau\in[0,\nu),$ then $V$ also satisfies Assumption 2.2 with any
$\tilde{c}\geq c$ and $\tilde{\tau}\in[\tau,\nu).$ Therefore, these constants
are not unique. But the exponential decay rate $\lambda$ obtained in Theorem
2.3 depends on the choice of $c$ and $\tau.$ To obtain a better rate, one has
to optimize $\lambda=\lambda(c,\tau)$ with respect to all $c$ and $\tau$
satisfying Assumption 2.2.

  4. 4.

In Theorem 2.3 $(b),$ the constant $C$ in (9) depends on $\varepsilon,$ and
$C=C(\varepsilon)\to\infty$ as $\varepsilon\to 0.$

  5. 5.

The highest exponential rate is $\frac{\nu}{2}$ which can be attained by the
quadratic potentials $V$ with $\frac{\partial^{2}V}{\partial
x^{2}}\geq\frac{\nu^{2}}{4}I.$

When $V$ is a quadratic polynomial as in Theorem 2.3 $(e),$ we prove the
following sharp estimates.

######  Proposition 2.5.

Let $V$ be a quadratic polynomial and $\frac{\partial^{2}V}{\partial x^{2}}$
be positive definite. Let $\alpha_{0}>0$ be the smallest eigenvalue of
$\frac{\partial^{2}V}{\partial x^{2}},$ then111For functions
$\varphi=\varphi(t)$ and $\phi=\phi(t),$ $\varphi\asymp\phi$ as $t\to\infty$
means $\varphi=\mathcal{O}(\phi)$ and $\phi=\mathcal{O}(\varphi)$ as
$t\to\infty.$

$\sup_{1\neq\frac{f_{0}}{f_{\infty}}\in
L^{2}(\mathbb{R}^{d},f_{\infty})}\frac{||f(t)/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}{||f_{0}/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}\asymp\begin{cases}e^{-\frac{\nu}{2}t},&\text{
if }\alpha_{0}>\frac{\nu^{2}}{4}\\\ (1+t)e^{-\frac{\nu}{2}t},&\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4}\\\
e^{-{\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2}t}},&\text{ if
}\alpha_{0}<\frac{\nu^{2}}{4}\end{cases}\,\,\,\,\,\text{ as }\,\,t\to\infty.$
(10)

We shall use this proposition to prove the sharpness of the decay rates in
Theorem 2.3 $(e).$ When $V$ is a quadratic polynomial and
$-\alpha_{0}=-\frac{\nu^{2}}{4}\equalscolon c,$ Theorem 2.3 $(e)$ shows that
the decay in (9) can be $e^{-(\nu-\varepsilon)t}$ for any small fixed
$\varepsilon\in(0,\nu),$ but it can not be $e^{-\nu t}.$ In this case, it is
natural to expect a decay between $e^{-\nu t}$ and $e^{-(\nu-\varepsilon)t}:$
Proposition 2.5 shows that this is indeed the case for the square of the
$L^{2}-$norm, with the decay $(1+t)^{2}e^{-{\nu}t}.$ But an analogous
extension of this result for the functional on the left hand side of (9)
(i.e., to replace the term $Ce^{-(\nu-\varepsilon)t}$ with $C(1+t)^{2}e^{-\nu
t}$ ) has not been obtained so far.

######  Remark 2.6.

Under assumptions of Proposition 2.5, we can construct special solutions
$f_{s}(t)$ (see [3, Section 6]) which satisfy

$\frac{||f_{s}(t)/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}{||f_{0}/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}\asymp\begin{cases}e^{-\frac{\nu}{2}t},&\text{
if }\alpha_{0}>\frac{\nu^{2}}{4}\\\ (1+t)e^{-\frac{\nu}{2}t},&\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4}\\\
e^{-{\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2}t}},&\text{ if
}\alpha_{0}<\frac{\nu^{2}}{4}\end{cases}\,\,\,\,\,\text{ as }\,\,t\to\infty.$

Our next result is about the estimates on the hypoelliptic regularization.

######  Theorem 2.7.

Assume $V$ is a $C^{\infty}$ potential on $\mathbb{R}^{n}$ and there are
constants $c\in\mathbb{R}$ and $\tau\geq 0$ such that the matrix (6) is
positive semi-definite for all $x\in\mathbb{R}^{n}.$ Suppose the initial data
$f_{0}$ satisfies
$\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv<\infty.$ Then, for any
$t_{0}>0,$ there are explicitly computable constants $C_{1}=C_{1}(t_{0})>0$
and $C_{2}=C_{2}(t_{0})>0$ (independent of $f_{0}$) such that the inequalities

$\displaystyle\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\leq\frac{C_{1}}{t^{3}}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv$ (11)

and

$\displaystyle\int_{\mathbb{R}^{2n}}\nabla_{v}^{T}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv\\\
\leq\frac{C_{2}}{t}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv$ (12)

hold for all $t\in(0,t_{0}].$

In Theorem 2.3 we assumed that the initial data $f_{0}/f_{\infty}$ is in
$H^{1}(\mathbb{R}^{d},f_{\infty}).$ If we use the estimates in Theorem 2.7,
this condition can be relaxed:

######  Corollary 2.8.

Let $V$ be a $C^{\infty}$ potential in $\mathbb{R}^{n}$ satisfying Assumptions
2.1 and 2.2. Suppose the initial data $f_{0}$ satisfies
$\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv<\infty.$ Then, for any
$t_{0}>0,$ there is an explicitly computable constant $C=C(t_{0})>0$
(independent of $f_{0}$) such that

$\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\\\
+\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv\\\
\leq Ce^{-2\lambda
t}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv$ (13)

holds for all $t\geq t_{0}$ with $\lambda$ defined in Theorem 2.3.

######  Remark 2.9.

  1. 1.

In contrast to Theorem 2.3, Theorem 2.7 holds even if the Poincaré inequality
(5) is not satisfied by $f_{\infty}.$ Also, $\tau$ can be larger than $\nu.$

  2. 2.

The exponents of $t$ in (11) and (12) are optimal when $V$ is a quadratic
polynomial (see [30, Appendix A]).

To illustrate our result, we present concrete examples of potentials $V$
satisfying our Assumption 2.1 and Assumption 2.2:

######  Example 2.10 (Polynomial confining potentials).

_ $a)$ As mentioned in Theorem 2.3, if $V(x)=\frac{{x}^{T}M^{-1}x}{2}+p\cdot
x+q,\,$ $x\in\mathbb{R}^{n}$ with a positive definite covariance matrix
$M^{-1}\in\mathbb{R}^{n\times n},$ a constant vector $p\in\mathbb{R}^{n}$ and
a constant $q\in\mathbb{R},$ the convergence rate is_

$\lambda=\begin{cases}\frac{\nu}{2},&\text{ if
}\alpha_{0}>\frac{\nu^{2}}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{
(case (a))}\\\ \frac{\nu-\varepsilon}{2},&\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4},\text{ for any
}\varepsilon\in(0,\nu)\,\,\,\,\text{ (case (b))}\\\
\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2},&\text{ if
}\alpha_{0}<\frac{\nu^{2}}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{
(case (d))}\end{cases},$

_and it is sharp for $\alpha_{0}\neq\frac{\nu^{2}}{4},$ where $\alpha_{0}$ is
the smallest eigenvalue of $M^{-1}$ (see Theorem 2.3 $(e)$)._

_$b)$ More generally, we consider potentials of the form_

$V(x)=r|x|^{2k}+V_{0}(x)$

_where $r>0,$ $k\in\mathbb{N}$ and $V_{0}\colon\mathbb{R}^{n}\to\mathbb{R}$ is
a polynomial of degree $j<2k.$ Since we have already considered quadratic
potentials, we assume $k\geq 2.$ $V$ satisfies the Poincaré inequality (5);
this can be proven, for example, by showing that $V$ satisfies one of the
sufficient conditions given in [6, Corollary 1.6]. Concerning Assumption 2.2’
we have_

$r\frac{\partial^{2}|x|^{2k}}{\partial
x^{2}}=2kr|x|^{2k-2}I+2k(2k-2)r|x|^{2k-4}\begin{pmatrix}x_{1}^{2}&x_{1}x_{2}&...&x_{1}x_{n}\\\
x_{1}x_{2}&x^{2}_{2}&...&x_{2}x_{n}\\\ ...&...&...&...\\\
x_{1}x_{n}&x_{2}x_{n}&...&x_{n}^{2}\\\ \end{pmatrix}\geq 2kr|x|^{2k-2}I.$

_Since $V_{0}$ has degree $j<2k,$ there is a constant $A>0$ such that_

$-A(1+|x|^{2k-3})I\leq\frac{\partial^{2}V_{0}(x)}{\partial x^{2}}\leq
A(1+|x|^{2k-3})I.$

_Therefore, we can estimate_

$\frac{\partial^{2}V(x)}{\partial
x^{2}}\geq\left(2kr|x|^{2k-2}-A|x|^{2k-3}-A\right)I.$ (14)

_We also observe that there exists a positive constant $B$ such that_

$-B(1+|x|^{2k-3})I\leq\frac{\partial^{2}(\partial_{x_{i}}V(x))}{\partial
x^{2}}\leq B(1+|x|^{2k-3})I$

_for all $i\in\\{1,...,n\\}.$ (14) shows that the smallest eigenvalue of
$\frac{\partial^{2}V(x)}{\partial x^{2}}$ satisfies $\alpha(x)\geq
2kr|x|^{2k-2}-A|x|^{2k-3}-A.$ Since $2kr|x|^{2k-2}-A|x|^{2k-3}-A$ grows faster
than $B(1+|x|^{2k-3})$ as $|x|\to\infty,$ there are constants $c$ and
$\tau\in[0,\nu)$ such that (7) is satisfied. Thus, Theorem 2.3 applies to this
type of potentials. In particular, it applies to double-well potentials of the
form $V(x)=r_{1}|x|^{4}-r_{2}|x|^{2},\,\,\,r_{1},r_{2}>0.$ _

######  Remark 2.11.

  1. 1.

Our decay and regularization results above extend those of [20], where a
stronger assumption, i.e.
$\partial^{2}_{x_{i}x_{j}}V\in\bigcap_{p=1}^{\infty}W^{p,\infty}(\mathbb{R}^{n})$
for all $i,j\in\\{1,...,n\\},$ was made. By contrast, we did not require the
boundedness of the second and higher derivatives of $V.$

  2. 2.

Most of the previous works on the exponential convergence $f(t)\to f_{\infty}$
as $t\to\infty$ (e.g. [29, 18, 9, 10, 30, 12, 13]) used the growth condition
(4) to get some weighted Poincaré type inequalities (see [29, Lemma A.24]),
which are crucial in these works – and additional to the Poincaré inequality
(5). Our technique is rather different, based on construction of appropriate
state dependent matrices and state dependent matrix inequalities so that the
(modified) dissipation functional (see (20) below) decays exponentially.

  3. 3.

Most of the previous methods for proving the exponential convergence do not
give an accurate decay rate, $\lambda$ is typically much too small there (see
[29, Section 7.2], [18, Section 1.4]). For example, in [29, Section 7.2], the
exponential decay rate $\lambda=\frac{1}{40}$ was obtained for
$V(x)=\frac{|x|^{2}}{2}$ and $\nu=\sigma=1.$ Since our decay rates are sharp
for quadratic potentials, in this setting, the true rate $\lambda=\frac{1}{2}$
is given by Theorem 2.3 $(a)$ and $(e).$

##  3 Modified entropy methods for degenerate Fokker-Planck equations

We first consider the following degenerate and non-symmetric Fokker-Planck
equation [2, 1]:

$\begin{cases}\partial_{t}f=\text{div}(D\nabla f+(D+R)\nabla
Ef),\,\,\xi\in\mathbb{R}^{d},\,t>0,\\\ f(t=0)=f_{0}\in
L_{+}^{1}(\mathbb{R}^{d}),\,\,\int_{\mathbb{R}^{d}}f_{0}\,d\xi=1\end{cases}$
(15)

where $D\in\mathbb{R}^{d\times d}$ is a constant, symmetric, positive semi-
definite $(\text{rank}(D)<d)$ matrix, $R\in\mathbb{R}^{d\times d}$ is a
constant skew-symmetric matrix. $E:\mathbb{R}^{d}\to\mathbb{R}$ is a function
which only depends on the state variable $\xi.$ We assume that $E$ is
confining (i.e. $E(\xi)\to\infty$ for $|\xi|\to\infty$) and smooth enough so
that (15) has a unique and smooth solution. If $E$ grows fast enough, (15) has
a normalized steady state $f_{\infty}=c_{E}e^{-E},\,c_{E}>0.$ The weak maximum
principle for degenerate parabolic equations [22] can be applied to (15) and
we can prove that $f(t,\xi)\geq 0$ for all $t>0,\,\xi\in\mathbb{R}^{d}.$ The
divergence structure implies that the initial mass is conserved and
$f(t,\cdot)$ describes the evolution of a probability density

$\int_{\mathbb{R}^{d}}f(t,\xi)d\xi=\int_{\mathbb{R}^{d}}f_{0}(\xi)d\xi=1,\,\,\,\forall
t\geq 0.$

We are interested in the large-time behavior of the solution, in particular,
when $\text{rank}(D)$ is less than the dimension $d.$ When $D$ is positive
definite (rank$(D)=d$), the large time behavior and exponential convergence
have been studied comprehensively (see [7], [4], [2]). One of the well-know
conditions which provides the exponential decay of the solution to the steady
state is called the Bakry-Emery condition (see (16) below) leading to:

######  Theorem 3.1.

Assume
$\displaystyle\int_{\mathbb{R}^{d}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi<\infty$
and

$\exists{\lambda}>0\,\text{ such that
}\,\frac{\partial^{2}E(\xi)}{\partial{\xi}^{2}}\geq\lambda
D^{-1}\,\,\,\forall\xi\in\mathbb{R}^{d}.$ (16)

Then

$\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi\leq
e^{-2\lambda
t}\int_{\mathbb{R}^{d}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi.$

To prove the theorem above, one considers the time derivative of the
$L^{2}-$norm and we see that it decreases

$\frac{d}{dt}\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi=-2\int_{\mathbb{R}^{d}}\nabla^{T}\left(\frac{f}{f_{\infty}}\right){D}\nabla\left(\frac{f}{f_{\infty}}\right)f_{\infty}d\xi=:-I(f(t)|f_{\infty})\leq
0.$ (17)

$I(f(t)|f_{\infty})$ is called the dissipation functional and since $D$ is
positive definite it vanishes if and only if $f=f_{\infty}.$ It can be proven
that, under the Bakry-Emery condition,

$\frac{d}{dt}I(f(t)|f_{\infty})\leq-2\lambda I(f(t)|f_{\infty}).$ (18)

Integrating this inequality from $(t,\infty)$ and using the convergences
$I(f(t)|f_{\infty})\to 0$ and
$\displaystyle\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi\to
0$ as $t\to\infty,$ it follows that

$\frac{d}{dt}\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi\leq-2\lambda\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi$
(19)

and, by Grönwall’s lemma, we get the desired result.

When $D$ is only positive semi-definite, i.e. rank$(D)<d,$ one observes that
$I(f(t)|f_{\infty})$ may vanish for certain probability densities $f\neq
f_{\infty}.$ Hence the inequalities (18) and (19) will not hold in general.
Since the above problems stem from the singularity of $D,$ one can modify the
dissipation function and define a modified dissipation functional (see also
[1, 3])

$S(f):=2\int_{\mathbb{R}^{d}}\nabla^{T}_{\xi}\left(\frac{f}{f_{\infty}}\right){P(\xi)}\nabla_{\xi}\left(\frac{f}{f_{\infty}}\right)f_{\infty}d\xi$
(20)

where $P:\mathbb{R}^{d}\to\mathbb{R}^{d\times d}$ is a symmetric positive
definite matrix which will be chosen later. Extending the approach of [1, 3],
we allow the matrix $P$ here to depend on $\xi\in\mathbb{R}^{d}.$ Our goal is
to derive a differential inequality similar to (18) (like the dissipation
functional satisfied for non-degenerate equations), i.e.

$\frac{d}{dt}S(f(t))\leq-2\lambda S(f(t)),\,\,\,\,$ (21)

for some $\lambda>0$ and a ”good” choice of the matrix $P.$ If this holds
true, we would obtain

$S(f(t))\leq S(f_{0})e^{-2\lambda t}.$

If we can choose such $P=P(\xi)\geq\eta I$ for some $\eta>0$ and all
$\xi\in\mathbb{R}^{d}$, under the validity of the Poincaré inequality (5) for
$f_{\infty}(\xi)=c_{E}e^{-E(\xi)}$ (where $(x,v)^{T}$ in (5) is replaced with
$\xi$) we have

$\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi\leq\frac{1}{C_{PI}}\int_{\mathbb{R}^{d}}\left|\nabla_{\xi}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}d\xi\leq\frac{1}{2C_{PI}\eta}S(f(t)),\,\,\,$

which implies the exponential decay of the $L^{2}-$norm

$\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi\leq\frac{1}{2C_{PI}\eta}S(f_{0})e^{-2\lambda
t}.$

More generally, since the quadratic entropy is also a decreasing function of
time $t$, instead of proving (21), we can consider the functional

$\Phi(f(t))\colonequals\gamma\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi+S(f(t))\\\
=\gamma\int_{\mathbb{R}^{d}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}d\xi+2\int_{\mathbb{R}^{d}}\nabla^{T}\left(\frac{f}{f_{\infty}}\right){P(\xi)}\nabla\left(\frac{f}{f_{\infty}}\right)f_{\infty}d\xi$
(22)

and choose a suitable parameter $\gamma\geq 0$ and a matrix $P$ such that

$\frac{d\Phi(f(t))}{dt}\leq-2\lambda\Phi(f(t))\leq 0$ (23)

for some $\lambda>0.$ This idea and method were successfully applied in [3] to
(15) when the potential $E$ is quadratic.

We shall apply this method to the kinetic Fokker-Planck equation with non-
quadratic $V(x).$ First, we denote $\xi\colonequals\begin{pmatrix}x\\\
v\end{pmatrix}\in\mathbb{R}^{2n},$
$E(\xi):=\frac{\nu}{\sigma}[V(x)+\frac{|v|^{2}}{2}],\,f_{\infty}=e^{-E}.$ Then
the kinetic Fokker-Planck equation (1) can be written in the form of (15),

$\partial_{t}f=\text{div}_{\xi}(D\nabla_{\xi}f+(D+R)\nabla_{\xi}Ef)$ (24)

with

$D=\begin{pmatrix}0&0\\\ 0&\sigma I\end{pmatrix}\in\mathbb{R}^{2n\times
2n}\,\,\,\,\,\,\text{and}\,\,\,\,\,\,R=\frac{\sigma}{\nu}\begin{pmatrix}0&-I\\\
I&0\end{pmatrix}\in\mathbb{R}^{2n\times 2n}.$ (25)

The rank of the diffusion matrix $D$ is $n<d=2n.$ Thus, (1) is both non-
symmetric and degenerate and the arguments above apply to the equation.

We will develop a modified entropy method. We will choose $\xi-$dependent
matrix $P$ in the modified dissipation functional (20) so that (23) holds and
$\lambda>0$ is as large as possible.

We also mention that when the potential $E$ is quadratic in (15), the question
about the long time behavior can be reduced to an ODE problem:

######  Theorem 3.2.

Let $0\neq D\in\mathbb{R}^{d\times d}$ be positive semi-definite,
$R\in\mathbb{R}^{d\times d}$ be skew-symmetric and $\mathbb{R}^{d}\ni\xi\to
E(\xi)=\frac{{\xi}^{T}K^{-1}\xi}{2}$ for some positive definite matrix $K.$
Assume $(D+R)K^{-1}$ is positive stable and there is no non-trivial subspace
of $\text{\emph{Ker}}D$ which is invariant under $K^{-1}(D-R).$ If $f$ is the
solution of (15) and $\xi(t)\in\mathbb{R}^{d}$ is the solution of the ODE
$\dot{\xi}(t)=-K^{-\frac{1}{2}}(D+R)K^{-\frac{1}{2}}\xi$ with initial datum
$\xi(0)=\xi_{0},$ then

$\sup_{1\neq\frac{f_{0}}{f_{\infty}}\in
L^{2}(\mathbb{R}^{d},f_{\infty})}\frac{||f(t)/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}{||f_{0}/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}=\sup_{0\neq\xi_{0}\in\mathbb{R}^{d}}\frac{||\xi(t)||_{2}}{||\xi_{0}||_{2}},\,\,\,t\geq
0.$ (26)

###### Proof.

See [5, Theorem 3.4]. ∎

One consequence of Theorem 3.2 is that the decay estimate of the ODE-solution
carries over to the corresponding Fokker-Planck equation.

##  4 The choice of the matrix P

The following lemmas will play a crucial role in our arguments.

######  Lemma 4.1.

Let $P:\mathbb{R}^{2n}\to\mathbb{R}^{2n\times 2n}$ be smooth and $f$ be the
solution of (1), then

$\frac{d}{dt}S(f(t))=-4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}\right\\}uf_{\infty}dxdv\\\
-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\right\\}uf_{\infty}dxdv,$ (27)

where $u\colonequals\nabla_{x,v}\left(\frac{f}{f_{\infty}}\right),$
$Q=Q(x)\colonequals\begin{pmatrix}0&I\\\ -\frac{\partial^{2}V(x)}{\partial
x^{2}}&\nu I\end{pmatrix},$ and
$[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]$ denotes a scalar differential operator
that is applied to each element of the matrix $P=P(x,v).$

###### Proof.

We denote $\displaystyle
u_{1}\colonequals\nabla_{x}\left(\frac{f}{f_{\infty}}\right),$ $\displaystyle
u_{2}\colonequals\nabla_{v}\left(\frac{f}{f_{\infty}}\right),$ then $u_{1}$
and $u_{2}$ satisfy

$\partial_{t}u_{1}=\sigma\Delta_{v}u_{1}-\nu\sum_{i=1}^{n}v_{i}\partial_{v_{i}}u_{1}+\sum_{i=1}^{n}\partial_{x_{i}}V\partial_{v_{i}}u_{1}-\sum_{i=1}^{n}v_{i}\partial_{x_{i}}u_{1}+\frac{\partial^{2}V}{\partial
x^{2}}u_{2},$
$\partial_{t}u_{2}=\sigma\Delta_{v}u_{2}-\nu\sum_{i=1}^{n}v_{i}\partial_{v_{i}}u_{2}+\sum_{i=1}^{n}\partial_{x_{i}}V\partial_{v_{i}}u_{2}-\sum_{i=1}^{n}v_{i}\partial_{x_{i}}u_{2}-u_{1}-\nu
u_{2}.$

These equations can be written with respect to $u=\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}:$

$\partial_{t}u=\sigma\Delta_{v}u-\nu\sum_{i=1}^{n}v_{i}\partial_{v_{i}}u+\sum_{i=1}^{n}\partial_{x_{i}}V\partial_{v_{i}}u-\sum_{i=1}^{n}v_{i}\partial_{x_{i}}u-Q^{T}u.$

It allows us to compute the time derivative of the modified dissipation
functional

$\frac{d}{dt}S(t)=4\int_{\mathbb{R}^{2n}}u^{T}P\partial_{t}uf_{\infty}dxdv\\\
=4\sigma\int_{\mathbb{R}^{2n}}u^{T}P\Delta_{v}uf_{\infty}dxdv-4\nu\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{v_{i}}uv_{i}f_{\infty}dxdv\\\
+4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{v_{i}}u\partial_{x_{i}}Vf_{\infty}dxdv-4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{x_{i}}uv_{i}f_{\infty}dxdv\\\
-2\int_{\mathbb{R}^{2n}}u^{T}\\{QP+PQ^{T}\\}uf_{\infty}dxdv.$ (28)

First, we consider the term in the second line of (28) and use
$\partial_{v_{i}}f_{\infty}=-\frac{\nu}{\sigma}v_{i}f_{\infty}:$

$4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial^{2}_{v_{i}v_{i}}uf_{\infty}dxdv-4\nu\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{v_{i}}uv_{i}f_{\infty}dxdv\\\
=-4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}\partial_{v_{i}}u^{T}P\partial_{v_{i}}uf_{\infty}dxdv-4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)\partial_{v_{i}}uf_{\infty}dxdv.$
(29)

By integrating by parts the last term of (29) we obtain

$-4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)\partial_{v_{i}}uf_{\infty}dxdv\\\
=4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)\partial_{v_{i}}uf_{\infty}dxdv+4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial^{2}_{v_{i}v_{i}}P)uf_{\infty}dxdv-4\nu\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)uv_{i}f_{\infty}dxdv$

and we find

$-4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)\partial_{v_{i}}uf_{\infty}dxdv=2\sigma\int_{\mathbb{R}^{2n}}u^{T}(\Delta_{v}P)uf_{\infty}dxdv-2\nu\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(v_{i}\partial_{v_{i}}P)uf_{\infty}dxdv.$

If we use this equality in (29), we get

$4\sigma\int_{\mathbb{R}^{2n}}u^{T}P\Delta_{v}uf_{\infty}dxdv-4\nu\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}Pv_{i}\partial_{v_{i}}uf_{\infty}dxdv\\\
=-4\sigma\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}uf_{\infty}dxdv-2\int_{\mathbb{R}^{2n}}u^{T}\\{[\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\\}uf_{\infty}dxdv.$ (30)

Next, we consider the terms in the third line of (28):

$4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{v_{i}}u\partial_{x_{i}}Vf_{\infty}dxdv\\\
=-4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{v_{i}}u\partial_{x_{i}}Vf_{\infty}dxdv-4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)u\partial_{x_{i}}Vf_{\infty}dxdv+\frac{4\nu}{\sigma}\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}Pu\partial_{x_{i}}Vv_{i}f_{\infty}dxdv,$
(31)
$-4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{x_{i}}uv_{i}f_{\infty}dxdv\\\
=4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}P\partial_{x_{i}}uv_{i}f_{\infty}dxdv+4\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{x_{i}}P)uv_{i}f_{\infty}dxdv-\frac{4\nu}{\sigma}\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}Pu\partial_{x_{i}}Vv_{i}f_{\infty}dxdv.$
(32)

(31) and (32) show that the third line of (28) equals

$-2\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{v_{i}}P)u\partial_{x_{i}}Vf_{\infty}dxdv+2\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}u^{T}(\partial_{x_{i}}P)uv_{i}f_{\infty}dxdv\\\
=-2\int_{\mathbb{R}^{2n}}u^{T}\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}]P\\}uf_{\infty}dxdv.$
(33)

Combining (28), (30), and (33) we obtain the statement (27). ∎

The key question for using the modified entropy dissipation functional $S(f)$
is how to choose the matrix $P.$ To determine $P$ we shall need the following
algebraic result:

######  Lemma 4.2.

For any fixed matrix $\mathcal{Q}\in\mathbb{R}^{d\times d},$ let
$\mu:=\min\\{\textsf{Re}(\beta):\beta\text{ is an eigenvalue of
}\mathcal{Q}\\}.$ Let $\\{\beta_{m}:1\leq m\leq m_{0}\\}$ be all the
eigenvalues of $\mathcal{Q}$ with $\mu=\textsf{Re}(\beta),$ only counting
their geometric multiplicity.

(a) If $\beta_{m}$ is non-defective for all $m\in\\{1,...,m_{0}\\},$ then
there exists a symmetric, positive definite matrix $P\in\mathbb{R}^{d\times
d}$ with

$\mathcal{Q}P+P\mathcal{Q}^{T}\geq 2\mu P.$

(b) If $\beta_{m}$ is defective for at least one $m\in\\{1,...,m_{0}\\},$ then
for any $\varepsilon>0$ there exists a symmetric, positive definite matrix
$P(\varepsilon)\in\mathbb{R}^{d\times d}$ with

$\mathcal{Q}P(\varepsilon)+P(\varepsilon)\mathcal{Q}^{T}\geq
2(\mu-\varepsilon)P(\varepsilon).$

###### Proof.

See [3, Lemma 4.3]. ∎

We consider the matrix function

$Q(x)\colonequals\begin{pmatrix}0&I\\\ -\frac{\partial^{2}V(x)}{\partial
x^{2}}&\nu I\end{pmatrix},\,\,\,\,x\in\mathbb{R}^{n},$ (34)

which appears in (27). We want to construct a symmetric positive definite
matrix $P(x)$ such that $Q(x)P(x)+P(x)Q^{T}(x)$ is positive definite and

$Q(x)P(x)+P(x)Q^{T}(x)\geq 2\mu P(x)$

for some $\mu>0$ and for all $x\in\mathbb{R}^{n}.$ We recall

$\displaystyle\alpha(x)\colonequals\min_{i\in\\{1,..,n\\}}\left\\{\alpha_{i}(x)\,:\,\alpha_{i}(x)\,\text{
is an eigenvalue of }\frac{\partial^{2}V(x)}{\partial x^{2}}\right\\},$
$\displaystyle\alpha_{0}\colonequals\inf_{x\in\mathbb{R}^{n}}\alpha(x),$
$\mu\colonequals\inf_{x\in\mathbb{R}^{n},\,i\in\\{1,..,n\\}}\\{\textsf{Re}(\beta_{i}(x)):\beta_{i}(x)\text{
is an eigenvalue of }Q(x)\\}.$

######  Lemma 4.3.

1) The matrix $Q(x)$ is positive stable at any fixed $x\in\mathbb{R}^{n},$ if
and only if $\frac{\partial^{2}V(x)}{\partial x^{2}}$ is positive definite.
2) Let $\frac{\partial^{2}V(x)}{\partial x^{2}}$ be positive definite for some
$x\in\mathbb{R}^{n}.$ Then:

  1. (a)

If $\alpha_{0}>\frac{\nu^{2}}{4},$ then $\mu=\frac{\nu}{2}$ and there exists a
symmetric positive definite matrix $P(x)$ such that

$Q(x)P(x)+P(x)Q^{T}(x)=2\mu P(x).$

  2. (b)

If $0<\alpha_{0}<\frac{\nu^{2}}{4},$ then
$\mu=\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2}$ and there exists a symmetric
positive definite matrix $P(x)$ such that

$Q(x)P(x)+P(x)Q^{T}(x)\geq 2\mu P(x).$

  3. (c)

If $\alpha_{0}=\frac{\nu^{2}}{4},$ then $\mu=\frac{\nu}{2}$ and, for any
$\varepsilon\in(0,\nu),$ there exists a symmetric positive definite matrix
$P(x,\varepsilon)$ such that

$Q(x)P(x,\varepsilon)+P(x,\varepsilon)Q^{T}(x)\geq(2\mu-\varepsilon)P(x,\varepsilon).$

###### Proof.

Part $1)$ Let $x$ be any point of $\mathbb{R}^{n},$ we compute the eigenvalues
$\beta(x)$ of $Q(x).$ If $\beta(x)\neq 0$ we have the condition

$\det(Q(x)-\beta(x)I)=\begin{vmatrix}-\beta(x)I&I\\\
-\frac{\partial^{2}V(x)}{\partial x^{2}}&(\nu-\beta(x))I\end{vmatrix}\\\
=\frac{1}{(\beta(x))^{n}}\begin{vmatrix}-\beta(x)I&0\\\
-\frac{\partial^{2}V(x)}{\partial x^{2}}&-\frac{\partial^{2}V(x)}{\partial
x^{2}}+\beta(x)(\nu-\beta(x))I\end{vmatrix}\\\
={(-1)^{n}}\det\left(-\frac{\partial^{2}V(x)}{\partial
x^{2}}+\beta(x)(\nu-\beta(x))I\right)=0.$

Let $\alpha_{i}(x)\in\mathbb{R},\,\,i\in\\{1,...,n\\}$ denote the eigenvalues
of $\frac{\partial^{2}V(x)}{\partial x^{2}},$ then the above eigenvalue
condition reads

$\prod_{i=1}^{n}(\beta^{2}(x)-\nu\beta(x)+\alpha_{i}(x))=0.$

Hence the non-zero eigenvalues of $Q(x)$ are

$\beta^{\pm}_{i}(x)=\frac{\nu\pm\sqrt{{\nu}^{2}-4\alpha_{i}(x)}}{2},\,\,i\in\\{1,...,n\\}.$
(35)

Moreover, $\beta(x)=0$ can be an eigenvalue of $Q(x)$ iff one of the
eigenvalues of $\frac{\partial^{2}V(x)}{\partial x^{2}}$ is zero. This shows
that $Q(x)$ is positive stable iff $\frac{\partial^{2}V(x)}{\partial
x^{2}}>0.$

For Part $2)$ we shall construct matrices $P(x),$ which relies on the proof of
Lemma 4.2 (Lemma 4.3 in [3]).

$(a)$ Let $\alpha_{0}>\frac{\nu^{2}}{4}.$ In this case, because of (35) the
matrix $Q(x)$ is positive stable and $\mu=\frac{\nu}{2}>0.$ We define the
matrix

$P(x)\colonequals\begin{pmatrix}2I&\nu I\\\ \nu
I&2\frac{\partial^{2}V(x)}{\partial x^{2}}\end{pmatrix},$

and for this choice, it is easy to check that

$Q(x)P(x)+P(x)Q^{T}(x)=\nu P(x)=2\mu P(x).$

To make sure that $P(x)$ is positive definite, we compute the eigenvalues
$\eta(x)$ of $P(x)$ at each $x\in\mathbb{R}^{n}:$ For $\eta(x)\neq 2$ we have
the condition

$\det(P(x)-\eta(x)I)=\begin{vmatrix}(2-\eta(x))I&\nu I\\\ \nu
I&2\frac{\partial^{2}V(x)}{\partial x^{2}}-\eta(x)I\end{vmatrix}\\\
=\frac{1}{(2-\eta(x))^{n}}\begin{vmatrix}(2-\eta(x))I&0\\\ \nu
I&(2-\eta(x))\left(2\frac{\partial^{2}V(x)}{\partial
x^{2}}-\eta(x)I\right)-\nu^{2}I\end{vmatrix}\\\
=\det\left((2-\eta(x))\left(2\frac{\partial^{2}V(x)}{\partial
x^{2}}-\eta(x)I\right)-\nu^{2}I\right)=0.$

$\eta(x)=2$ is not an eigenvalue of $P(x)$ and so the eigenvalues of $P(x)$
satisfy

$\prod_{i=1}^{n}\left(\eta^{2}(x)-(2+2\alpha_{i}(x))\eta(x)+4\alpha_{i}(x)-\nu^{2}\right)=0.$

We conclude that the eigenvalues are

$\eta^{\pm}_{i}(x)=1+\alpha_{i}(x)\pm\sqrt{(\alpha_{i}(x)+1)^{2}-(4\alpha_{i}(x)-\nu^{2})},\,\,i\in\\{1,...,n\\}.$

Since we assumed $\alpha_{i}(x)\geq\alpha(x)\geq\alpha_{0}>\frac{\nu^{2}}{4}$
for all $i\in\\{1,...,n\\},$ the eigenvalues are positive and satisfy

$\eta\colonequals\inf_{x\in\mathbb{R}^{n},\,i\in\\{1,...,n\\}}{\eta^{\pm}_{i}(x)}=1+\alpha_{0}-\sqrt{(\alpha_{0}+1)^{2}-(4\alpha_{0}-\nu^{2})}>0.$

Thus, $P(x)$ is positive definite and $P(x)\geq\eta I$ for all
$x\in\mathbb{R}^{n}.$

$(b)-(c)$ Let $0<\alpha_{0}\leq\frac{\nu^{2}}{4}.$ Then (35) shows
$\mu=\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2}.$ Let $\varepsilon>0$ be a fixed
small number. We define

$\omega\colonequals\begin{cases}\alpha_{0},\,\,\text{ if
}\alpha_{0}<\frac{\nu^{2}}{4}\\\
\alpha_{0}-\frac{\varepsilon^{2}}{4},\,\,\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4}\end{cases}$

and consider the matrix

$P(x)\colonequals\begin{pmatrix}2I&\nu I\\\ \nu
I&2\frac{\partial^{2}V(x)}{\partial x^{2}}+(\nu^{2}-4\omega)I\end{pmatrix}.$

We compute its eigenvalues $\eta(x)$ by a similar computation as above:

$\eta^{\pm}_{i}(x)=1+\zeta_{i}(x)\pm\sqrt{(\zeta_{i}(x)+1)^{2}-(4\zeta_{i}(x)-\nu^{2})},$
(36)

where
$\zeta_{i}(x)\colonequals\alpha_{i}(x)+\frac{\nu^{2}}{2}-2\omega>\frac{\nu^{2}}{4}.$
We also have

$\eta:=\inf_{x\in\mathbb{R}^{n},\,i\in\\{1,...,n\\}}{\eta^{\pm}_{i}(x)}={1+\alpha_{0}+\frac{\nu^{2}}{2}-2\omega-\sqrt{(\alpha_{0}+\frac{\nu^{2}}{2}-2\omega-1)^{2}+\nu^{2}}}>0.$

Thus, $P(x)$ is positive definite and $P(x)\geq\eta I$ for all
$x\in\mathbb{R}^{n}.$ Then we compute

$Q(x)P(x)+P(x)Q^{T}(x)\\\
=(\nu-\sqrt{\nu^{2}-4\omega})P(x)+\sqrt{\nu^{2}-4\omega}\begin{pmatrix}2I&(\nu+\sqrt{\nu^{2}-4\omega})I\\\
(\nu+\sqrt{\nu^{2}-4\omega})I&2\frac{\partial^{2}V}{\partial
x^{2}}+\sqrt{\nu^{2}-4\omega}(\nu+\sqrt{\nu^{2}-4\omega})I\end{pmatrix}.$ (37)

Since $\frac{\partial^{2}V}{\partial x^{2}}\geq\omega I,$ the second matrix in
the last line of (37) is bounded below by

$\begin{pmatrix}2I&(\nu+\sqrt{\nu^{2}-4\omega})I\\\
(\nu+\sqrt{\nu^{2}-4\omega})I&2\omega+\sqrt{\nu^{2}-4\omega}(\nu+\sqrt{\nu^{2}-4\omega})I\end{pmatrix}\\\
=\begin{pmatrix}2I&(\nu+\sqrt{\nu^{2}-4\omega})I\\\
(\nu+\sqrt{\nu^{2}-4\omega})I&\frac{1}{2}(\nu+\sqrt{\nu^{2}-4\omega})^{2}I\end{pmatrix}\geq
0.$

Consequently, we get

$Q(x)P(x)+P(x)Q^{T}(x)\geq(\nu-\sqrt{\nu^{2}-4\omega})P(x)\,\,\,\text{ for all
}\,\,\,x\in\mathbb{R}^{n}.$ (38)

∎

Lemma 4.3 shows that, if $\frac{\partial^{2}V(x)}{\partial x^{2}}$ is not
positive definite at some $x\in\mathbb{R}^{n}$ (and hence $\alpha_{0}\leq 0$),
then $Q(x)$ is not positive stable. In this case, it is not possible to find a
positive constant $\mu$ and a positive definite matrix $P(x)$ such that
$Q(x)P(x)+P(x)Q^{T}(x)\geq\mu P(x).$ If $\alpha_{0}$ is just finite and not
necessarily positive, we have the following modified inequality.

######  Lemma 4.4.

Let $\alpha_{0}>-\infty.$ Then there exist $\gamma\geq 0,$ $\delta\in[0,\nu),$
and a symmetric positive definite matrix function $P(x)$ such that

$Q(x)P(x)+P(x)Q^{T}(x)+\gamma D\geq(\nu-\delta)P(x),\,\,\,\,\forall
x\in\mathbb{R}^{d},$ (39)

where $D=\begin{pmatrix}0&0\\\ 0&\sigma I\end{pmatrix}\in\mathbb{R}^{2n\times
2n}$ is the matrix defined in (24).

###### Proof.

Let $a\geq 0$ be any constant such that $a+\alpha_{0}>\frac{\nu^{2}}{4}.$ We
consider the matrix

$P(x)\colonequals\begin{pmatrix}2I&\nu I\\\ \nu
I&2\frac{\partial^{2}V(x)}{\partial x^{2}}+2aI\end{pmatrix}.$ (40)

In analogy to (36) we find its eigenvalues as

$\eta^{\pm}_{i}(x)=1+\zeta_{i}(x)\pm\sqrt{(\zeta_{i}(x)+1)^{2}-(4\zeta_{i}(x)-\nu^{2})},$

where $\zeta_{i}(x):=\alpha_{i}(x)+a\geq a+\alpha_{0}>\frac{\nu^{2}}{4},$ and
$\alpha_{i}(x)\in\mathbb{R},\,\,i\in\\{1,...,n\\}$ denote the eigenvalues of
$\frac{\partial^{2}V(x)}{\partial x^{2}}.$ We also have

$\eta\colonequals\inf_{x\in\mathbb{R}^{n},i\in\\{1,...,n\\}}{\eta^{\pm}_{i}(x)}=\frac{4(a+\alpha_{0}-\frac{\nu^{2}}{4})}{1+a+\alpha_{0}+\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}}>0.$
(41)

Thus, $P(x)$ is uniformly positive definite and $P(x)\geq\eta I$ for all
$x\in\mathbb{R}^{n}.$

Next we compute

$QP+PQ^{T}+\gamma D=\nu P+\begin{pmatrix}0&2aI\\\ 2aI&(2\nu
a+\gamma\sigma)I\end{pmatrix}$ $=(\nu-\delta)P+\begin{pmatrix}2\delta
I&(\nu\delta+2a)I\\\ (\nu\delta+2a)I&\delta(2\frac{\partial^{2}V}{\partial
x^{2}}+2aI)+(2\nu a+\gamma\sigma)I\end{pmatrix},$ (42)

where $\delta\in[0,\nu)$ will be chosen later. We compute the (real)
eigenvalues $\theta$ of the symmetric matrix

$\begin{pmatrix}2\delta I&(\nu\delta+2a)I\\\
(\nu\delta+2a)I&\delta(2\frac{\partial^{2}V}{\partial x^{2}}+2aI)+(2\nu
a+\gamma\sigma)I\end{pmatrix}$ (43)

which appears in (42):
For $\theta(x)\neq 2\delta$ we have the condition

$\begin{vmatrix}(2\delta-\theta)I&(\nu\delta+2a)I\\\
(\nu\delta+2a)I&\delta(2\frac{\partial^{2}V}{\partial x^{2}}+2aI)+(2\nu
a+\gamma\sigma-\theta)I\end{vmatrix}\\\
=\frac{1}{(2\delta-\theta)^{n}}\begin{vmatrix}(2\delta-\theta)I&0\\\
(\nu\delta+2a)I&(2\delta-\theta)\left(\delta(2\frac{\partial^{2}V}{\partial
x^{2}}+2aI)+(2\nu
a+\gamma\sigma-\theta)I\right)-(\nu\delta+2a)^{2}I\end{vmatrix}\\\
=\prod_{i=1}^{n}\left(\theta^{2}-\theta\left[2\delta(\alpha_{i}(x)+a)+2\delta+2\nu
a+\gamma\sigma\right]+4\delta^{2}(\alpha_{i}(x)+a-{\nu^{2}}/{4})+2\delta\gamma\sigma-4a^{2}\right)=0.$

Let us consider the following equations with $i\in\\{1,...,n\\}:$

$\theta^{2}-\theta[2\delta(\alpha_{i}(x)+a)+2\delta+2\nu
a+\gamma\sigma]+[4\delta^{2}(\alpha_{i}(x)+a-{\nu^{2}}/{4})+2\delta\gamma\sigma-4a^{2}]=0,$
(44)

and we shall show that they have non-negative solutions for an appropriate
choice of $\delta$ and $\gamma.$ To this end we see first that

$2\delta(\alpha_{i}(x)+a)+2\delta+2\nu a+\gamma\sigma\geq
2\delta(\alpha_{0}+a)+2\delta\geq\frac{\delta\nu^{2}}{2}+2\delta\geq 0.$

Next, we choose

$\displaystyle\delta=\delta(a,\gamma)\colonequals\frac{1}{\sqrt{a+\alpha_{0}-\frac{\nu^{2}}{4}}}\left[\sqrt{\left(\frac{\gamma\sigma}{4\sqrt{a+\alpha_{0}-\frac{\nu^{2}}{4}}}\right)^{2}+a^{2}}-\frac{\gamma\sigma}{4\sqrt{a+\alpha_{0}-\frac{\nu^{2}}{4}}}\right]\geq
0,$ (45)

which satisfies

$4\delta^{2}(a+\alpha_{0}-\frac{\nu^{2}}{4})+2\delta\gamma\sigma-4a^{2}=0.$
(46)

Hence, the last term of (44) satisfies

$4\delta^{2}(\alpha_{i}(x)+a-\frac{\nu^{2}}{4})+2\delta\gamma\sigma-4a^{2}\geq
4\delta^{2}(a+\alpha_{0}-\frac{\nu^{2}}{4})+2\delta\gamma\sigma-4a^{2}=0$

for all $i\in\\{1,...,n\\}.$ Therefore, the quadratic equations (44) have non-
negative coefficients and so their solutions, i.e. the eigenvalues of (43),
are non-negative. Consequently, we get (39).

We note that $\delta$ from (45) satisfies, for any fixed
$a>\frac{\nu}{4}-\alpha_{0},$ $\delta(a,\gamma)\to 0$ as $\gamma\to\infty.$
Hence, choosing $\gamma$ large enough, we have $\delta\in[0,\nu).$ ∎

######  Remark 4.5.

If $\alpha_{0}>0,$ we can take $\gamma=0$ in Lemma 4.4. This follows by
choosing in the proof of Lemma 4.4

$a=\begin{cases}0,&\text{ if }\alpha_{0}>\frac{\nu^{2}}{4}\\\
\frac{\varepsilon^{2}}{2},&\text{ if }\alpha_{0}=\frac{\nu^{2}}{4}\\\
\frac{\nu^{2}-4\alpha_{0}}{2},&\text{ if
}0<\alpha_{0}<\frac{\nu^{2}}{4}\end{cases},\,\,\,\,\delta=\begin{cases}0,&\text{
if }\alpha_{0}>\frac{\nu^{2}}{4}\\\ \frac{\varepsilon}{\sqrt{2}},&\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4}\\\ \sqrt{\nu^{2}-4\alpha_{0}},&\text{ if
}0<\alpha_{0}<\frac{\nu^{2}}{4}\end{cases},$

with any $\varepsilon\in(0,\nu).$ Therefore, Lemma 4.4 includes the second
part of Lemma 4.3. However, if $\alpha_{0}\leq 0,$ we have to choose
$\gamma>0.$

##  5 Proofs

###  5.1 Proof of Theorem 2.3

###### Proof.

We denote $\displaystyle
u_{1}\colonequals\nabla_{x}\left(\frac{f}{f_{\infty}}\right),\,\,u_{2}\colonequals\nabla_{v}\left(\frac{f}{f_{\infty}}\right),$
and $u\colonequals\begin{pmatrix}u_{1}\\\ u_{2}\end{pmatrix}.$ We consider the
modified dissipation functional

$S(f(t))=2\int_{\mathbb{R}^{2n}}u^{T}(t)Pu(t)f_{\infty}dxdv$

for some symmetric positive definite matrix $P=P(x,v)\in\mathbb{R}^{2n\times
2n}.$ By Lemma 4.1 we have

$\displaystyle\frac{d}{dt}S(f(t))=-4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}\right\\}uf_{\infty}dxdv\\\
-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\right\\}uf_{\infty}dxdv,$ (47)

with $Q(x)=\begin{pmatrix}0&I\\\ -\frac{\partial^{2}V(x)}{\partial x^{2}}&\nu
I\end{pmatrix}.$ Let $c\in\mathbb{R}$ and $\tau\in[0,\nu)$ are the constants
such that Assumption 2.2 is satisfied. Since (6) is positive semi-definite,
$\frac{\partial^{2}V(x)}{\partial x^{2}}+cI$ is also positive semi-definite
and so $\frac{\partial^{2}V(x)}{\partial x^{2}}\geq-cI$ for all
$x\in\mathbb{R}^{n}.$ We define the matrix $P$ depending on the constant $c.$

#### Case $(a)$:

Assume $c\leq-\frac{\nu^{2}}{4},\,\,\alpha_{0}>\frac{\nu^{2}}{4}.$ By Lemma
4.3 $(2a)$ and by its proof, the matrix $P(x)\colonequals\begin{pmatrix}2I&\nu
I\\\ \nu I&2\frac{\partial^{2}V(x)}{\partial x^{2}}\end{pmatrix}$ satisfies

$Q(x)P(x)+P(x)Q^{T}(x)=\nu P(x)\,\,\text{ and }\,\,P(x)\geq\eta I$

for all $x\in\mathbb{R}^{n}$ and $\eta\colonequals
1+\alpha_{0}-\sqrt{(\alpha_{0}+1)^{2}-(4\alpha_{0}-\nu^{2})}>0.$ For this
choice of the matrix $P,$

$[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P(x)=\begin{pmatrix}0&0\\\
0&-2\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial x^{2}}\end{pmatrix}.$ (48)

Then (47) can be written as

$\frac{d}{dt}S(f(t))=-4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv\\\
-2\nu\int_{\mathbb{R}^{2n}}u^{T}Puf_{\infty}dxdv+4\int_{\mathbb{R}^{2n}}u^{T}\begin{pmatrix}0&0\\\
0&\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial
x^{2}}\end{pmatrix}uf_{\infty}dxdv\\\
=-4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv-\nu
S(f(t))+4\int_{\mathbb{R}^{2n}}u^{T}\begin{pmatrix}0&0\\\
0&\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial
x^{2}}\end{pmatrix}uf_{\infty}dxdv.$ (49)

We shall now consider each term of this equation. First we compute

$S(f(t))=2\int_{\mathbb{R}^{2n}}\left\\{2|u_{1}|^{2}+2\nu u_{1}\cdot
u_{2}+2u^{T}_{2}\frac{\partial^{2}V}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\
=4\int_{\mathbb{R}^{2n}}|u_{1}+\frac{\nu}{2}u_{2}|^{2}f_{\infty}dxdv+4\int_{\mathbb{R}^{2n}}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv\\\ \geq
4\int_{\mathbb{R}^{2n}}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv.$ (50)

Then

$\displaystyle
4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv\\\
=4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}\left(2|\partial_{v_{i}}u_{1}|^{2}+2\nu\partial_{v_{i}}u_{1}\cdot\partial_{v_{i}}u_{2}+2(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}V}{\partial
x^{2}}\partial_{v_{i}}u_{2}\right)\right\\}f_{\infty}dxdv\\\
=8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}|\partial_{v_{i}}u_{1}+\frac{\nu}{2}\partial_{v_{i}}u_{2}|^{2}\right\\}f_{\infty}dxdv+8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv\\\
\geq
8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv.$
(51)

Now we consider the last term in (49)

$\displaystyle 4\int_{\mathbb{R}^{2n}}u^{T}\begin{pmatrix}0&0\\\
0&\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial
x^{2}}\end{pmatrix}uf_{\infty}dxdv=4\int_{\mathbb{R}^{2n}}u_{2}^{T}\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial
x^{2}}u_{2}f_{\infty}dxdv\\\
=4\int_{\mathbb{R}^{2n}}\left\\{\sum_{i,j=1}^{n}u_{2,i}v\cdot\nabla_{x}V_{ij}u_{2,j}\right\\}f_{\infty}dxdv=4\int_{\mathbb{R}^{2n}}\left\\{\sum_{i,j,k=1}^{n}u_{2,i}v_{k}V_{ijk}u_{2,j}\right\\}f_{\infty}dxdv\\\
=-\frac{4\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i,j,k=1}^{n}u_{2,i}V_{ijk}u_{2,j}(\partial_{v_{k}}f_{\infty})\right\\}dxdv=\frac{4\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i,j,k=1}^{n}\partial_{v_{k}}(u_{2,i}u_{2,j})V_{ijk}\right\\}f_{\infty}dxdv\\\
=\frac{4\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i,j,k=1}^{n}(\partial_{v_{k}}u_{2,i})u_{2,j}V_{ijk}+u_{2,i}(\partial_{v_{k}}u_{2,j})V_{ijk}\right\\}f_{\infty}dxdv\\\
=\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i,j,k=1}^{n}(\partial_{v_{k}}u_{2,i})u_{2,j}V_{ijk}\right\\}f_{\infty}dxdv=\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{k=1}^{n}(\partial_{v_{k}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv,$ (52)

where we integrated by parts and used
$\partial_{v_{k}}f_{\infty}=-\frac{\nu}{\sigma}v_{k}f_{\infty}$ and the
notations
$u_{2,i}\colonequals\partial_{v_{i}}\left(\frac{f}{f_{\infty}}\right),$
$V_{ij}\colonequals\partial_{x_{i}x_{j}}^{2}V,$
$V_{ijk}\colonequals\partial^{3}_{x_{i}x_{j}x_{k}}V.$ By (49), (51), (52), and
(50) we obtain

$\displaystyle\frac{d}{dt}S(f(t))+(\nu-\tau)S(f(t))\leq-\tau S(f(t))\\\
-8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv+\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\
\leq-4\tau\int_{\mathbb{R}^{2n}}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv-8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv\\\
+\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv$
$=-\frac{8\sigma}{\nu}\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}\left\\{\nu(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}-(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\
-\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\frac{\tau\nu}{2\sigma}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv.$

The right hand side of this inequality is a quadratic polynomial with respect
to $\partial_{v_{i}}u_{2},\,\,i\in\\{1,...,n\\},$ and $u_{2}.$ The
corresponding matrix of this quadratic polynomial is

$\begin{pmatrix}\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)&0&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V)}{\partial
x^{2}}\\\ 0&\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V)}{\partial
x^{2}}\\\ ...&...&...&...&...\\\
0&0&...&\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V)}{\partial
x^{2}}\\\ -\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V)}{\partial
x^{2}}&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V)}{\partial
x^{2}}&...&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V)}{\partial
x^{2}}&\frac{\tau\nu}{2\sigma}\left(\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\right)\end{pmatrix}.$ (53)

The assumption $\frac{\partial^{2}V}{\partial
x^{2}}-\frac{\nu^{2}}{4}I\geq\frac{\partial^{2}V}{\partial x^{2}}+cI$ and the
Assumption 2.2 imply that (53) is positive semi-definite.

Thus we have obtained

$\displaystyle\frac{d}{dt}S(f(t))+(\nu-\tau)S(f(t))\leq 0$

and by Grönwall’s lemma

$S(f(t))\leq e^{-(\nu-\tau)t}S(f_{0}).$ (54)

The estimate $P(x)\geq\eta I$ and the Poincaré inequality (5) imply

$\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv\leq\frac{1}{2C_{PI}\eta}S(f(t))\leq\frac{1}{2C_{PI}\eta}e^{-(\nu-\tau)t}S(f_{0}).$
(55)

The matrix inequalities (see Lemma 6.1 in Appendix 6.2)

$\frac{1}{1+\alpha_{0}+\sqrt{(1-\alpha_{0})^{2}+\nu^{2}}}P\leq\begin{pmatrix}I&0\\\
0&\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\end{pmatrix}\leq\frac{1+\alpha_{0}+\sqrt{(1-\alpha_{0})^{2}+\nu^{2}}}{4\alpha_{0}-\nu^{2}}P$
(56)

show that $S(f(t))$ is equivalent to the functional

$\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv.$

This equivalence, and (55) let us obtain (9).

#### Case $(b)$:

Assume $c=-\alpha_{0}=-\frac{\nu^{2}}{4}.$ Then by Lemma 4.3 $(2c),$ for any
$\varepsilon\in(0,\nu-\tau),$ the matrix

$P(x)\colonequals\begin{pmatrix}2I&\nu I\\\ \nu
I&2\frac{\partial^{2}V(x)}{\partial x^{2}}+\varepsilon^{2}I\end{pmatrix}$

satisfies

$Q(x)P(x)+P(x)Q^{T}(x)\geq(\nu-{\varepsilon})P(x)\,\,\,\,\text{and}\,\,\,P(x)\geq\eta
I$ (57)

for all $x\in\mathbb{R}^{n}$ and
$\eta\colonequals{1+\frac{\nu^{2}+2\varepsilon^{2}}{4}-\sqrt{(\frac{\nu^{2}+2\varepsilon^{2}}{4}-1)^{2}+\nu^{2}}}>0.$
With this matrix we have

$S(f(t))=4\int_{\mathbb{R}^{2n}}|u_{1}+\frac{\nu}{2}u_{2}|^{2}f_{\infty}dxdv+4\int_{\mathbb{R}^{2n}}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv\\\ \geq
4\int_{\mathbb{R}^{2n}}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv,$ (58)
$4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv\\\
=8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}|\partial_{v_{i}}u_{1}+\frac{\nu}{2}\partial_{v_{i}}u_{2}|^{2}\right\\}f_{\infty}dxdv+8\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv,$
(59)

and by using (48),
$\partial_{v_{i}}f_{\infty}=-\frac{\nu}{\sigma}v_{i}f_{\infty}:$

$-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\right\\}uf_{\infty}dxdv\\\
=\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv.$ (60)

(47), (57), (58), (59), (60), and similar estimates as for Case $a)$ show that

$\displaystyle\frac{d}{dt}S(f(t))+(\nu-\tau-\varepsilon)S(f(t))\\\
\leq-\frac{8\sigma}{\nu}\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}\left\\{\nu(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}-(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\
-\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\frac{\tau\nu}{2\sigma}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv.$

The right hand side of this inequality is a quadratic polynomial with respect
to $\partial_{v_{i}}u_{2},\,\,i\in\\{1,...,n\\},$ and $u_{2}.$ The
corresponding matrix of this quadratic polynomial is

$\begin{pmatrix}\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)&0&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V)}{\partial
x^{2}}\\\ 0&\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V)}{\partial
x^{2}}\\\ ...&...&...&...&...\\\
0&0&...&\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V)}{\partial
x^{2}}\\\ -\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V)}{\partial
x^{2}}&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V)}{\partial
x^{2}}&...&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V)}{\partial
x^{2}}&\frac{\tau\nu}{2\sigma}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I\right)\end{pmatrix}.$ (61)

Because of $\frac{\partial^{2}V}{\partial
x^{2}}+\frac{2\varepsilon^{2}-\nu^{2}}{4}I>\frac{\partial^{2}V}{\partial
x^{2}}+cI$ and Assumption 2.2, (61) is positive definite and we get

$\displaystyle\frac{d}{dt}S(f(t))+(\nu-\tau-\varepsilon)S(f(t))\leq 0$

and by Grönwall’s lemma

$S(f(t))\leq e^{-(\nu-\tau-\varepsilon)t}S(f_{0}).$ (62)

Similar to (55), we have

$\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv\leq\frac{1}{2C_{PI}\eta}S(f(t))\leq\frac{1}{2C_{PI}\eta}e^{-(\nu-\tau-\varepsilon)t}S(f_{0}).$
(63)

The functional

$\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv$

and $S(f(t))$ are equivalent because of (see Lemma 6.1 in Appendix 6.2)

$\frac{1}{1+\frac{\nu^{2}+2\varepsilon^{2}}{4}+\sqrt{\left(1-\frac{\nu^{2}+2\varepsilon^{2}}{4}\right)^{2}+\nu^{2}}}P\leq\begin{pmatrix}I&0\\\
0&\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\end{pmatrix}\leq\frac{1+\frac{\nu^{2}+2\varepsilon^{2}}{4}+\sqrt{\left(1-\frac{\nu^{2}+2\varepsilon^{2}}{4}\right)^{2}+\nu^{2}}}{2\varepsilon^{2}}P.$
(64)

This equivalence, and (63) imply (9).

#### Case $(c)$ and $(d),$ exponential decay:

Assume $c>-\frac{\nu^{2}}{4}.$ For some $\gamma\geq 0$ to be chosen later, we
consider the functional

$\displaystyle\Phi(f(t))\colonequals$
$\displaystyle\gamma\int_{\mathbb{R}^{2n}}\left(\frac{f}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+S(f(t))$
$\displaystyle=$
$\displaystyle\gamma\int_{\mathbb{R}^{2n}}\left(\frac{f}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+2\int_{\mathbb{R}^{2n}}u^{T}Puf_{\infty}dxdv.$
(65)

Using (17) and (47) its time derivative reads

$\frac{d\Phi(f(t))}{dt}=-4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}+\gamma
D\right\\}uf_{\infty}dxdv$
$-2\int_{\mathbb{R}^{2n}}u^{T}\left\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\right\\}uf_{\infty}dxdv.$ (66)

Let $a,$ to be chosen later, be any number such that $a\geq
c+\frac{\nu^{2}}{4}>0$ and $a+\alpha_{0}>\frac{\nu^{2}}{4}.$ We consider the
matrix

$P(x)\colonequals\begin{pmatrix}2I&\nu I\\\ \nu
I&2\frac{\partial^{2}V(x)}{\partial x^{2}}+2aI\end{pmatrix}.$ (67)

Then, by Lemma 4.4 we have

$Q(x)P(x)+P(x)Q^{T}(x)+\gamma D\geq(\nu-\delta)P(x),\,\,\,\,\,\forall
x\in\mathbb{R}^{d},$ (68)

with a constant $\delta$ defined in (45). If $\gamma$ is large enough, (45)
shows that $\delta\in(0,\nu-\tau).$

The choice of the matrix $P$ in (67), (66), and (68) lets us estimate

$\frac{d\Phi(f(t))}{dt}\leq-4\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv\\\
-(\nu-\delta)S(f(t))+4\int_{\mathbb{R}^{2n}}u^{T}\begin{pmatrix}0&0\\\
0&\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial
x^{2}}\end{pmatrix}uf_{\infty}dxdv.$ (69)

Similar computations as for Case $(a)$ as well as (58) (but with
$\varepsilon^{2}=2a$) lead to

$\displaystyle\frac{d}{dt}\Phi(f(t))+(\nu-\delta-\tau)S(f(t))\\\
\leq-\frac{8\sigma}{\nu}\sum_{i=1}^{n}\int_{\mathbb{R}^{2n}}\left\\{\nu(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{4a-\nu^{2}}{4}I\right)\partial_{v_{i}}u_{2}-(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\
-\frac{8\sigma}{\nu}\int_{\mathbb{R}^{2n}}\frac{\tau\nu}{2\sigma}u^{T}_{2}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{4a-\nu^{2}}{4}I\right)u_{2}f_{\infty}dxdv.$

The two integrands of the right hand side are together a quadratic polynomial
of $\partial_{v_{i}}u_{2},$ $i\in\\{1,...,n\\},$ and $u_{2},$ and its
corresponding matrix is

$\begin{pmatrix}\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{4a-\nu^{2}}{4}I\right)&0&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V)}{\partial
x^{2}}\\\ 0&\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{4a-\nu^{2}}{4}I\right)&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V)}{\partial
x^{2}}\\\ ...&...&...&...&...\\\
0&0&...&\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{4a-\nu^{2}}{4}I\right)&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V)}{\partial
x^{2}}\\\ -\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V)}{\partial
x^{2}}&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V)}{\partial
x^{2}}&...&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V)}{\partial
x^{2}}&\frac{\tau\nu}{2\sigma}\left(\frac{\partial^{2}V}{\partial
x^{2}}+\frac{4a-\nu^{2}}{4}I\right)\end{pmatrix}.$ (70)

Because of $a-\frac{\nu^{2}}{4}\geq c$ and Assumption 2.2, the matrix (70) is
positive semi-definite, thus, we have

$\displaystyle\frac{d}{dt}\Phi(f(t))+(\nu-\tau-\delta)S(f(t))\leq 0.$ (71)

The estimate $P(x)\geq\eta I$ ($\eta>0$ defined in (41)) and the Poincaré
inequality (5) imply

$\int_{\mathbb{R}^{2n}}\left(\frac{f}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv\leq\frac{1}{2\eta
C_{PI}}S(f(t))$

and so

$\frac{1}{1+\frac{\gamma}{2\eta C_{PI}}}\Phi(f(t))\leq S(f(t)).$

This estimate and (71) let us conclude

$\frac{d}{dt}\Phi(f(t))+2\lambda\Phi(f(t))\leq 0$ (72)

for

$2\lambda=\frac{\nu-\tau-\delta}{1+\frac{\gamma}{2\eta C_{PI}}}>0.$ (73)

By Grönwall’s lemma we obtain

$\Phi(f(t))\leq e^{-2\lambda t}\Phi(f_{0}).$ (74)

One can check that (see Lemma 6.1 in Appendix 6.2)

$\frac{1}{a+\alpha_{0}+1+\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}}P\leq\begin{pmatrix}I&0\\\
0&\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\end{pmatrix}\leq\frac{a+\alpha_{0}+1+\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}}{4(a+\alpha_{0})-\nu^{2}}P.$
(75)

Hence, $S(f(t))$ is equivalent to the functional

$\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv.$

Subsequently, $\Phi(f(t))$ and the functional on the left hand side of (9) are
equivalent. This equivalence and (74) let us obtain (9).

#### Case $(c)$ and $(d),$ estimated decay rate:

Next, we shall estimate $\lambda$ from (73) explicitly, and we shall choose
the parameters $a$ and $\gamma$ such that $\lambda$ is (rather) large. By (41)
and (46), $\eta=\eta(a)$ and $\delta=\delta(a,\gamma)$ are functions of
$a\in[c+\frac{\nu^{2}}{4},\infty)\bigcap(\frac{\nu^{2}}{4}-\alpha_{0},\infty)$
and $\gamma\in[0,\infty).$ Since $\delta>0,$ and $\eta$ is monotonically
increasing up to 2, we have the following uniform estimate and choice of the
decay rate:

$2\lambda\colonequals\sup_{a\in[c+\frac{\nu^{2}}{4},\infty)\bigcap(\frac{\nu^{2}}{4}-\alpha_{0},\infty),\,\gamma\geq
0}\frac{\nu-\tau-\delta(a,\gamma)}{1+\frac{\gamma}{2\eta(a)C_{PI}}}\leq\sup_{\gamma\geq
0}\frac{\nu-\tau}{1+\frac{\gamma}{4C_{PI}}}\leq\nu-\tau.$

Next, we shall estimate this supremum (in fact it is a maximum). First we
introduce a new variable
$s:=\displaystyle\frac{\gamma\sigma}{4a\sqrt{a+\alpha_{0}-\frac{\nu^{2}}{4}}}\in[0,\infty),$
then

$\delta(a,\gamma)=\frac{a}{\sqrt{a+\alpha_{0}-\frac{\nu^{2}}{4}}}(\sqrt{1+s^{2}}-s).$

With the notations $\displaystyle
A(a):=\frac{1+a+\alpha_{0}+\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}}{2\sigma
C_{PI}}>0$ and $\displaystyle
B(a):=\frac{a}{\sqrt{a+\alpha_{0}-\frac{\nu^{2}}{4}}}>0,$ we have

$2\lambda=\max_{a\in[c+\frac{\nu^{2}}{4},\infty)\bigcap(\frac{\nu^{2}}{4}-\alpha_{0},\infty),\,s\geq
0}\frac{\nu-\tau-B(a)(\sqrt{1+s^{2}}-s)}{1+A(a)B(a)s}.$

Next, we shall fix the parameter $a.$ To estimate $\lambda$ as accurately as
possible, we choose $a$ as the argmin of $B(a)$ such that
$\nu-\tau-B(a)(\sqrt{1+s^{2}}-s)$ is maximal with respect to $a.$ We find the
minimal value of $B(a)$ is

$\displaystyle\min_{a\in[c+\frac{\nu^{2}}{4},\infty)\bigcap(\frac{\nu^{2}}{4}-\alpha_{0},\infty)}B(a)=\begin{cases}\displaystyle
B(a_{1})=\frac{c+\frac{\nu^{2}}{4}}{\sqrt{c+\alpha_{0}}}\,\,\,\,\,\,\,\,\text{
if }\,\,c+2\alpha_{0}>\frac{\nu^{2}}{4}\\\ \displaystyle
B(a_{2})=\sqrt{{\nu^{2}}-4\alpha_{0}}\,\,\text{ if
}\,\,c+2\alpha_{0}\leq\frac{\nu^{2}}{4}\end{cases},$

and this minimum is attained at $a_{1}\colonequals{c+\frac{\nu^{2}}{4}}$ if
$c+2\alpha_{0}>\frac{\nu^{2}}{4}$ (i.e. in Case $(c)$), and $a_{2}\colonequals
2(\frac{\nu^{2}}{4}-\alpha_{0})$ if $c+2\alpha_{0}\leq\frac{\nu^{2}}{4}$ (i.e.
in Case $(d)$).

If $c+2\alpha_{0}>\frac{\nu^{2}}{4},$ then $c>-\alpha_{0}$ and so $a$ varies
in

$[c+\frac{\nu^{2}}{4},\infty)\bigcap(\frac{\nu^{2}}{4}-\alpha_{0},\infty)=[c+\frac{\nu^{2}}{4},\infty)=[a_{1},\infty).$

Since $A(a)$ is increasing, both $A(a)$ and $B(a)$ attain their minimal values
at $a_{1}.$ Thus, $a_{1}$ is optimal, i.e.

$\max_{a}\frac{\nu-\tau-B(a)(\sqrt{1+s^{2}}-s)}{1+A(a)B(a)s}=\frac{\nu-\tau-B(a_{1})(\sqrt{1+s^{2}}-s)}{1+A(a_{1})B(a_{1})s}.$

If $c+2\alpha_{0}\leq\frac{\nu^{2}}{4},$
$a_{2}=2(\frac{\nu^{2}}{4}-\alpha_{0})$ may not be optimal as $A(a)$ does not
attain its minimum at this point, i.e.

$\max_{a}\frac{\nu-\tau-B(a)(\sqrt{1+s^{2}}-s)}{1+A(a)B(a)s}\geq\frac{\nu-\tau-B(a_{2})(\sqrt{1+s^{2}}-s)}{1+A(a_{2})B(a_{1})s}.$

But it is the optimal choice when $s=0$ and so it gives a good approximation
if $s$ is small. From now on we assume that $a$ is fixed as

$\displaystyle
a\colonequals\begin{cases}a_{1}={c+\frac{\nu^{2}}{4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\text{
if }\,\,c+2\alpha_{0}>\frac{\nu^{2}}{4}\\\
a_{2}=2(\frac{\nu^{2}}{4}-\alpha_{0})\,\,\text{ if
}\,\,c+2\alpha_{0}\leq\frac{\nu^{2}}{4}\end{cases}.$ (76)

Note that this choice is independent of $s.$
Let
$\displaystyle\Lambda(a,s)\colonequals\frac{\nu-\tau-B(a)(\sqrt{1+s^{2}}-s)}{1+A(a)B(a)s}$
and we seek its maximum with respect to $s\in[0,\infty).$ We compute

$\partial_{s}\Lambda(a,s)\\\
=\frac{B(a)}{(1+A(a)B(a)s)^{2}\sqrt{s^{2}+1}}\left([1-(\nu-\tau-B(a))A(a)]\sqrt{s^{2}+1}-A(a)B(a)(\sqrt{s^{2}+1}-1)-s\right).$
(77)

If $1-(\nu-\tau-B(a))A(a)\leq 0,$ then $\partial_{s}\Lambda(a,s)\leq 0$ which
implies that $\Lambda(a,s)$ is a decreasing function of $s$ and the maximum in
$[0,\infty)$ is attained at $s=0.$

If $1-(\nu-\tau-B(a))A(a)>0,$ then
$\partial_{s}\Lambda(a,0)=B(a)[1-(\nu-\tau-B(a))A(a)]>0$ and $\Lambda(a,s)$ is
increasing in a neighborhood of $s=0.$ We also see $\partial_{s}\Lambda(a,s)$
is negative if $s$ is large enough (since $\nu-\tau>0$). This means that
$\Lambda(a,s)$ starts to grow at $s=0$ and it decreases as $s\to\infty.$
Therefore, there is a point in $(0,\infty)$ at which $\Lambda(a,s)$ takes its
maximum. Setting $\partial_{s}\Lambda(a,s)=0$ we obtain

$[1-(\nu-\tau)A(a)]\sqrt{s^{2}+1}-s+A(a)B(a)=0.$

It has only one solution in $(0,\infty)$ given by

$s(a)=\begin{cases}\frac{A^{2}(a)B^{2}(a)-1}{2A(a)B(a)}&\text{ if
}(\nu-\tau)A(a)=2\\\
\frac{1}{\nu-\tau}\left[\left|\frac{(\nu-\tau)A(a)-1}{(\nu-\tau)A(a)-2}\right|\sqrt{B^{2}(a)+2(\nu-\tau)A^{-1}(a)-(\nu-\tau)^{2}}-\frac{B(a)}{(\nu-\tau)A(a)-2}\right]&\text{
if }(\nu-\tau)A(a)\neq 2\end{cases}$ (78)

and at this point $\Lambda(a,s)$ attains its maximum with respect to $s.$

Considering the computations above, we conclude that the decay rate can be
estimated by:

$\displaystyle 2\lambda=\begin{cases}\nu-\tau-B(a)&\text{ if }\nu-\tau\geq
A^{-1}(a)+B(a)\\\
\frac{\nu-\tau-B(a)(\sqrt{1+s^{2}(a)}-s(a))}{1+A(a)B(a)s(a)}&\text{ if
}\nu-\tau<A^{-1}(a)+B(a)\end{cases},$ (79)

where two cases correspond to the two cases discussed after (77). Moreover,
$a$ and $s(a)$ are defined in (76) and (78), respectively. If we denote
$A_{1}:=A(a_{1}),$ $A_{2}:=A(a_{2}),$ $s_{1}:=s(a_{1})$ and $s_{2}:=s(a_{2})$
and take into account that
$B(a_{1})=\frac{c+\frac{\nu^{2}}{4}}{\sqrt{c+\alpha_{0}}}$ and
$B(a_{2})=\sqrt{{\nu^{2}}-4\alpha_{0}},$ we obtain the explicit decay rates
stated in the theorem.

#### Case (e):

Let $V(x)$ be a quadratic function of $x$ and $\frac{\partial^{2}V}{\partial
x^{2}}$ be positive definite. Then,
$\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial x^{2}}$ are zero matrices for
all $i\in\\{1,...,n\\}.$ Thus, $V$ satisfies Assumption 2.2 with $\tau=0,$
$-c=\alpha_{0}>0.$

If $\alpha_{0}<\frac{\nu^{2}}{4},$ then
$c+2\alpha_{0}=\alpha_{0}<\frac{\nu^{2}}{4}$ which falls into Case $(d).$ The
constant in the Poincaré inequality (5) equals
$C_{PI}=\frac{\nu}{\sigma}\min\\{1,\alpha_{0}\\}$ (see [4]). It lets us
compute $A_{2}^{-1}$ explicitly:

$\displaystyle
A_{2}^{-1}=\frac{2\nu\min\\{1,\alpha_{0}\\}}{1+\frac{\nu^{2}}{2}-\alpha_{0}+\sqrt{(\frac{\nu^{2}}{2}-\alpha_{0}-1)^{2}+\nu^{2}}}.$

In Appendix 6.3 we prove the following inequality:

$\nu\geq A^{-1}_{2}+\sqrt{{\nu^{2}}-4\alpha_{0}}.$ (80)

Thus Case $(d)$ implies

$\lambda=\frac{\nu-\sqrt{{\nu^{2}}-4\alpha_{0}}}{2}.$ (81)

If $\alpha_{0}\geq\frac{\nu^{2}}{4},$ the decay rate is explicit by Case $(a)$
and Case $(b):$

$\lambda=\begin{cases}{\frac{\nu}{2}}&\text{ if
}\alpha_{0}>\frac{\nu^{2}}{4}\\\ \frac{\nu-\varepsilon}{2}&\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4},\text{ for any
}\varepsilon\in(0,\nu)\end{cases}.$ (82)

We now prove that the decay rates in (81) and (82) are sharp: From Corollary
2.8

$\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv\leq
Ce^{-2\lambda
t}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv,\,\,\,\,\,\,\forall t\geq
t_{0}$

holds with the same $\lambda$ given in (81) and (82). Since
$\left|\left|\frac{\partial^{2}V}{\partial x^{2}}\right|\right|+1$ is
constant, this estimate implies

$\sup_{1\neq\frac{f_{0}}{f_{\infty}}\in
L^{2}(\mathbb{R}^{d},f_{\infty})}\frac{||f(t)/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}{||f_{0}/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}\leq\tilde{C}e^{-\lambda
t},\,\,\,\,\,\,\forall t\geq t_{0}$ (83)

for some constant $\tilde{C}>0.$ One the one hand this means that the
estimated decay rate $\lambda$ can not be larger than the (true) decay rate of
the propagator norm given on the left hand side of (83). On the other hand,
Proposition 2.5 gives the sharp decay rates for this propagator norm. The
decay rates in (81) and (82) coincide with the ones in Proposition 2.5 except
in the case of $\alpha_{0}=\frac{\nu^{2}}{4}.$ Thus, the exponential decay
rates in Case $(a)$ and Case $(d)$ are sharp. When
$\alpha_{0}=\frac{\nu^{2}}{4},$ Proposition 2.5 provides the sharp decay
$(1+t)e^{-\frac{\nu}{2}t}$ for the propagator norm. Hence, (9) can hold with
rates $\lambda=\frac{\nu-\varepsilon}{2}$ for any small fixed
$\varepsilon\in(0,\nu),$ but it does not hold for $\varepsilon=0.$ ∎

###  5.2 Proof of Proposition 2.5

###### Proof of Proposition 2.5.

For the quadratic polynomial $V$ denote the positive definite matrix
$\frac{\partial^{2}V}{\partial x^{2}}\equalscolon M^{-1}\in\mathbb{R}^{n\times
n}.$ Then there are $x_{0}\in\mathbb{R}^{n}$ and $\mathcal{C}\in\mathbb{R}$
such that $V(x)=\frac{{(x-x_{0})}^{T}M^{-1}(x-x_{0})}{2}+\mathcal{C},$
$\forall x\in\mathbb{R}^{n}.$ Since the change $x\to x+x_{0}$ does not affect
the supremum in (10) and only the gradient of $V$ appeares in (1), without
loss of generality we assume that $x_{0}=0$ and $\mathcal{C}=0.$

#### Step 1, reformulation as an ODE-problem:

To this end we use Theorem 3.2. We check the conditions of this theorem for
the kinetic Fokker-Planck equation. With the notation $\xi=\begin{pmatrix}x\\\
v\end{pmatrix},$ we write

$E(\xi)=\frac{\nu}{\sigma}\left(V(x)+\frac{|v|^{2}}{2}\right)=\frac{\nu}{\sigma}\left(\frac{{x}^{T}M^{-1}x}{2}+\frac{|v|^{2}}{2}\right)=\frac{1}{2}\xi^{T}\begin{pmatrix}\frac{\nu}{\sigma}M^{-1}&0\\\
0&\frac{\nu}{\sigma}I\end{pmatrix}\xi=\frac{\xi^{T}K^{-1}\xi}{2}$ (84)

with $K^{-1}:=\frac{\nu}{\sigma}\begin{pmatrix}M^{-1}&0\\\ 0&I\end{pmatrix}.$
From (25) we see that
$\text{Ker}D=\\{(\psi,0)^{T}:\,\,\psi\in\mathbb{R}^{n}\\}.$ Let
$(\psi,0)^{T}\in\text{Ker}D,$ then its image under $K^{-1}(D-R)$ is

$K^{-1}(D-R)\begin{pmatrix}\psi\\\ 0\end{pmatrix}=\begin{pmatrix}0&M^{-1}\\\
-I&\nu I\end{pmatrix}\begin{pmatrix}\psi\\\ 0\end{pmatrix}=\begin{pmatrix}0\\\
-\psi\end{pmatrix}$

and it is in $\text{Ker}D$ iff $\psi=0.$ Therefore, there is no non-trivial
$K^{-1}(D-R)-$invariant subspace of $\text{Ker}D.$ Next we compute the
eigenvalues $\beta$ of $K^{-1/2}(D+R)K^{-1/2}=\begin{pmatrix}0&-M^{-1/2}\\\
M^{-1/2}&\nu I\end{pmatrix}:$

$\begin{vmatrix}-\beta I&-M^{-1/2}\\\
M^{-1/2}&(\nu-\beta)I\end{vmatrix}=\begin{vmatrix}-\beta I&0\\\
M^{-1/2}&(\nu-\beta)I-\beta^{-1}M^{-1}\end{vmatrix}\\\
=\text{det}(\beta(\beta-\nu)I+M^{-1})=\prod_{i=1}^{n}(\beta^{2}-\nu\beta+\alpha_{i})=0,$

where $\alpha_{i},$ $i\in\\{1,...,n\\}$ denote the eigenvalues of $M^{-1}.$ By
solving the latter equation, we find that the eigenvalues of
$K^{-1/2}(D+R)K^{-1/2}$ are
$\beta^{-}_{i}=\frac{\nu-\sqrt{\nu^{2}-4\alpha_{i}}}{2},$
$\beta^{+}_{i}=\frac{\nu+\sqrt{\nu^{2}-4\alpha_{i}}}{2},$ $i\in\\{1,...,n\\}.$
If $\alpha_{0}>0$ is the smallest eigenvalue of $M^{-1},$ then

$\mu\colonequals\min_{i}\\{\textsf{Re}(\beta_{i}):\beta_{i}\text{ is an
eigenvalue of }K^{-1/2}(D+R)K^{-1/2}\\}=\begin{cases}\frac{\nu}{2}&\text{ if
}\alpha_{0}\geq\frac{\nu^{2}}{4}\\\
\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2}&\text{ if
}\alpha_{0}<\frac{\nu^{2}}{4}\end{cases}.$

Hence $\mu$ is positive, so $K^{-1/2}(D+R)K^{-1/2}$ and $(D+R)K^{-1}$ are
positive stable. Therefore, Theorem 3.2 applies to the kinetic Fokker-Planck
equation.

#### Step 2, decay rates of the ODE-solution:

We consider the ODE

$\dot{\xi}(t)=-K^{-1/2}(D+R)K^{-1/2}\xi$

with the initial data $\xi(0)=\xi_{0}.$ Since $K^{-1/2}(D+R)K^{-1/2}$ is
positive stable, the solution ${\xi}(t)$ is stable. To quantify the decay
rate, we continue to analyze the eigenvalues of $K^{-1/2}(D+R)K^{-1/2}.$ Let
$m_{i}$ be the multiplicity of $\alpha_{i}>0$ as an eigenvalue of $M^{-1}$
(now the $\alpha_{i}$ with $i\in\\{1,...,\tilde{n}\\}$ are labeled without
multiplicity). Since $M^{-1}$ is symmetric, there are linearly independent
eigenvectors $\psi_{ij}\in\mathbb{R}^{n},\,\,j\in\\{1,...,m_{i}\\}$ of
$M^{-1}$ corresponding to $\alpha_{i}.$ Then we can check that the vectors

$\begin{pmatrix}-\frac{\alpha_{i}^{1/2}}{\beta^{-}_{i}}\psi_{ij}\\\
\psi_{ij}\end{pmatrix}\in\mathbb{R}^{2n},\,\,j\in\\{1,...,m_{i}\\}$ (85)

are linearly independent eigenvectors of $K^{-1/2}(D+R)K^{-1/2}$ corresponding
to $\beta^{-}_{i},$ $i\in\\{1,...,\tilde{n}\\}.$ Moreover, these vectors form
a basis of the space of eigenvectors corresponding to $\beta^{-}_{i}.$
Similarly, the vectors

$\begin{pmatrix}-\frac{\alpha_{i}^{1/2}}{\beta^{+}_{i}}\psi_{ij}\\\
\psi_{ij}\end{pmatrix}\in\mathbb{R}^{2n},\,\,j\in\\{1,...,m_{i}\\}.$ (86)

satisfy the same property for $\beta^{+}_{i}.$

If $\alpha_{i}\neq\frac{\nu^{2}}{4}$ for all $i\in\\{1,...,\tilde{n}\\}$
(i.e., $\beta^{-}_{i}\neq\beta^{+}_{i}),$ the algebraic multiplicities of
$\beta^{-}_{i}$ and $\beta^{+}_{i}$ are equal to $m_{i}.$ Then $\beta^{-}_{i}$
(resp. $\beta^{+}_{i}$) has $m_{i}$ eigenvectors given by (85) (resp. (86)).
Thus, the geometric multiplicities of $\beta^{-}_{i}$ and $\beta^{+}_{i}$ also
equal $m_{i}.$ In particular, $K^{-1/2}(D+R)K^{-1/2}$ is diagonalizable.

If $\alpha_{i_{0}}=\frac{\nu^{2}}{4}$ for some
$i_{0}\in\\{1,...,\tilde{n}\\},$ then the algebraic multiplicity of
$\beta^{-}_{i_{0}}=\beta^{+}_{i_{0}}=\frac{\nu}{2}$ equals $2m_{i_{0}}.$ Since
the vectors (85) and (86) coincide in this case, the geometric multiplicity of
$\frac{\nu}{2}$ equals $m_{i_{0}}.$ Thus, in this case, $\frac{\nu}{2}$ is a
defective222An eigenvalue is defective if its geometric multiplicity is
strictly less than its algebraic multiplicity. eigenvalue of
$K^{-1/2}(D+R)K^{-1/2}$ with the corresponding eigenvectors

$\begin{pmatrix}-\psi_{i_{0}j}\\\
\psi_{i_{0}j}\end{pmatrix}\in\mathbb{R}^{2n},\,\,j\in\\{1,...,m_{i_{0}}\\}.$
(87)

By solving the following linear system (with respect to $\xi$)

$K^{-1/2}(D+R)K^{-1/2}\xi-\frac{\nu}{2}\xi=\begin{pmatrix}-\frac{\nu}{2}I&-M^{-1/2}\\\
M^{-1/2}&\frac{\nu}{2}I\end{pmatrix}\xi=\begin{pmatrix}-\psi_{i_{0}j}\\\
\psi_{i_{0}j}\end{pmatrix},\,\,\,\,\xi\in\mathbb{R}^{2d},$

we find that the solution $\xi=\begin{pmatrix}0\\\
\frac{2}{\nu}\psi_{i_{0}j}\end{pmatrix}$ is a generalized eigenvector of
$\frac{\nu}{2}$ corresponding to the eigenvector
$\begin{pmatrix}-\psi_{i_{0}j}\\\ \psi_{i_{0}j}\end{pmatrix}.$ Since
$\psi_{i_{0}j},$ $j\in\\{1,...,m_{i_{0}}\\}$ are linearly independent, the
vectors

$\begin{pmatrix}0\\\
\frac{2}{\nu}\psi_{i_{0}j}\end{pmatrix},\,\,j\in\\{1,...,m_{i_{0}}\\}$ (88)

form a set of linearly independent generalized eigenvectors of
$\frac{\nu}{2}.$ Since the vectors in (87) and (88) are linearly independent
and their total number equals $2m_{i_{0}}$ (which is the algebraic
multiplicity of $\frac{\nu}{2}$), we conclude that each eigenvector of
$\frac{\nu}{2}$ has only one generalized eigenvector. Therefore, all Jordan
blocks associated to $\frac{\nu}{2}$ have the same size $2\times 2.$ In
particular, if $\alpha_{0}=\frac{\nu^{2}}{4},$ then the eigenvalue
$\mu=\frac{\nu}{2}$ is defective and the maximal size of the Jordan blocks
associated to $\frac{\nu}{2}$ is 2.

Then, the classical stability theory for ODEs shows that

$\sup_{1\neq\frac{f_{0}}{f_{\infty}}\in
L^{2}(\mathbb{R}^{d},f_{\infty})}\frac{||f(t)/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}{||f_{0}/f_{\infty}-1||_{L^{2}(\mathbb{R}^{d},f_{\infty})}}=\sup_{0\neq\xi_{0}\in\mathbb{R}^{d}}\frac{||\xi(t)||_{2}}{||\xi_{0}||_{2}}\asymp\begin{cases}e^{-\frac{\nu}{2}t},&\text{
if }\alpha_{0}>\frac{\nu^{2}}{4}\\\ (1+t)e^{-\frac{\nu}{2}t},&\text{ if
}\alpha_{0}=\frac{\nu^{2}}{4}\\\
e^{-{\frac{\nu-\sqrt{\nu^{2}-4\alpha_{0}}}{2}t}},&\text{ if
}\alpha_{0}<\frac{\nu^{2}}{4}\end{cases}$

as $t\to\infty.$

∎

######  Remark 5.1.

With the eigenvalues of $C\colonequals(D+R)K^{-1}$ $($see (24), (84)$)$
obtained at the end of Step 1 in the above proof, the sharpness of the decay
rate $\mu$ in the cases 1 and 3 of (10) would also follow from [3, Theorem
6.1].

###  5.3 Proof of Theorem 2.7 and Corollary 2.8

###### Proof of Theorem 2.7.

#### Step 1, an auxiliary inequality:

As we assume the matrix (6) is positive semi-definite, then the following
submatrices of (6) are positive semi-definite:

$Y_{k}\colonequals\begin{pmatrix}\nu\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial
x^{2}}\\\ -\frac{1}{2}\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial
x^{2}}&\frac{\tau\nu}{2\sigma}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\end{pmatrix}\in\mathbb{R}^{2n\times
2n},\,\,\,\,\,\,k\in\\{1,...,n\\}.$

Letting $\delta\in\mathbb{R},$ we consider

$X_{\delta}\colonequals\begin{pmatrix}I&\delta I\\\ \delta
I&\delta^{2}I\end{pmatrix}\otimes\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)=\begin{pmatrix}\frac{\partial^{2}V}{\partial
x^{2}}+cI&\delta\frac{\partial^{2}V}{\partial x^{2}}+\delta cI\\\
\delta\frac{\partial^{2}V}{\partial x^{2}}+\delta
cI&\delta^{2}\frac{\partial^{2}V}{\partial
x^{2}}+\delta^{2}cI\end{pmatrix}\in\mathbb{R}^{2n\times 2n}.$

$X_{\delta}$ is positive semi-definite as it is the Kronecker product [24,
Corollary 4.2.13] of two positive semi-definite matrices. Hence, we have for
all $k\in\\{1,...,n\\}:$

$\mathrm{Tr}(X^{1/2}_{\delta}Y_{k}X^{1/2}_{\delta})=\mathrm{Tr}(X_{\delta}Y_{k})=(\nu+\delta^{2}\frac{\tau\nu}{2\sigma})\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right]-\delta\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial
x^{2}}\right]\geq 0.$

This implies

$\frac{2\sigma\nu+\delta^{2}\tau\nu}{2\sigma\delta}\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right]\geq\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial x^{2}}\right]$
(89)

and by minimizing the constant on the left hand side of (89) with respect to
$\delta,$ we obtain

$\sqrt{\frac{2\tau\nu^{2}}{\sigma}}\mathrm{Tr}\left[\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)^{2}\right]\geq\mathrm{Tr}\left[\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)\frac{\partial^{2}(\partial_{x_{k}}V(x))}{\partial
x^{2}}\right]\,\,\,\,\,\,\,\text{for all}\,\,\,\,x\in\mathbb{R}^{n}.$ (90)

#### Step 2, growth estimate for the r.h.s. of (11), (12):

We denote $\displaystyle
u_{1}\colonequals\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right),$
$\displaystyle
u_{2}\colonequals\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right),$ and
$u\colonequals\begin{pmatrix}u_{1}\\\ u_{2}\end{pmatrix}.$ Since
$\displaystyle\frac{f(t)}{f_{\infty}}-1$ satisfies

$\partial_{t}\left(\frac{f(t)}{f_{\infty}}-1\right)=-v\cdot\nabla_{x}\left(\frac{f(t)}{f_{\infty}}-1\right)+\nabla_{x}V\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)+\sigma\Delta_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)-\nu
v\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)$

and by integrating by parts, we obtain

$\frac{d}{dt}\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv=-2\sigma\displaystyle\int_{\mathbb{R}^{2n}}|u_{2}|^{2}f_{\infty}dxdv.$
(91)

Next, we compute (with $||\cdot||$ denoting the Frobenius norm)

$\frac{d}{dt}\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
=2\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\partial_{t}\left(\frac{f(t)}{f_{\infty}}-1\right)\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
=2\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left[-v\cdot\nabla_{x}\left(\frac{f(t)}{f_{\infty}}-1\right)+\nabla_{x}V\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)\right]\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
+2\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left[\sigma\Delta_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)-\nu
v\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)\right]\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv.$ (92)

Integrating by parts with respect to $v,$ we obtain

$2\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left[\sigma\Delta_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)-\nu
v\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)\right]\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
=-2\sigma\displaystyle\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv.$ (93)

Next, we work on the term in the second line of (92):

$2\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left[-v\cdot\nabla_{x}\left(\frac{f(t)}{f_{\infty}}-1\right)+\nabla_{x}V\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)\right]\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
=\int_{\mathbb{R}^{2n}}\left(-v\cdot\nabla_{x}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}+\nabla_{x}V\cdot\nabla_{v}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\right)\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
=\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left[v\cdot\nabla_{x}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}\right)-\nabla_{x}V\cdot\nabla_{v}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}\right)\right]dxdv\\\
=\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}v\cdot\nabla_{x}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}\right)f_{\infty}dxdv\\\
=\frac{2\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)u_{2}\cdot\nabla_{x}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}\right)f_{\infty}dxdv\\\
=\frac{2\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\sum_{k=1}^{n}u_{2,k}\partial_{x_{k}}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}\right)f_{\infty}dxdv\\\
=\frac{4\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left\\{\sum_{k=1}^{n}u_{2,k}\sum_{i,j=1}^{n}(\partial^{2}_{x_{i}x_{j}}V+\delta_{ij}c)\partial^{2}_{x_{i}x_{j}}(\partial_{x_{k}}V)\right\\}f_{\infty}dxdv,$
(94)

where we integrated by parts twice, and used
$-\frac{\nu}{\sigma}vf_{\infty}=\nabla_{v}f_{\infty}$ and the notations

$u_{2,k}\colonequals\partial_{v_{k}}\left(\frac{f(t)}{f_{\infty}}\right)\,\,\,\,\text{
and }\,\,\,\,\delta_{ij}\colonequals\begin{cases}1\,\,\,\,\text{ if
}\,\,i=j\\\ 0\,\,\,\,\text{ if }\,\,i\neq j\end{cases}.$

Using the identity

$\displaystyle\sum_{i,j=1}^{n}(\partial^{2}_{x_{i}x_{j}}V+\delta_{ij}c)\partial^{2}_{x_{i}x_{j}}(\partial_{x_{k}}V)=\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial
x^{2}}\right],$

the estimate (90), and the discrete Hölder inequality, (94) can be estimated
as

$\frac{4\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left\\{\sum_{i,j,k=1}^{n}u_{2,k}(\partial^{2}_{x_{i}x_{j}}V+\delta_{ij}c)\partial^{2}_{x_{i}x_{j}}(\partial_{x_{k}}V)\right\\}f_{\infty}dxdv\\\
=\frac{4\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left\\{\sum_{k=1}^{n}u_{2,k}\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\frac{\partial^{2}(\partial_{x_{k}}V)}{\partial
x^{2}}\right]\right\\}f_{\infty}dxdv\\\ \leq
4\sqrt{2\sigma\tau}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)\left\\{\sum_{k=1}^{n}|u_{2,k}|\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right]\right\\}f_{\infty}dxdv\\\ \leq 4\sqrt{2\sigma\tau
n}\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)|u_{2}|\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right]f_{\infty}dxdv\\\
\leq\sigma\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right]f_{\infty}dxdv+8\tau
n\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right]f_{\infty}dxdv.$ (95)

Combining the equations from (92) to (95) and the identity

$\displaystyle\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}=\mathrm{Tr}\left[\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)^{2}\right],$

we get

$\frac{d}{dt}\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\\\
\leq-\sigma\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv+8\tau
n\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv.$ (96)

(96) can be reformulated as

$\frac{d}{dt}\left(e^{-8\tau
nt}\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv\right)\leq-\sigma e^{-8\tau
nt}\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv.$ (97)

#### Step 3, $t-$dependent functional $\Psi$:

In order to prove the short-time regularization of (11) and (12) we introduce
now an auxiliary functional that depends explicitly on time. Our strategy is
the generalization of the approach in [29, Theorem A.12], [20, Theorem 1.1],
[3, Theorem 4.8].

For $t\in(0,t_{0}],$ we consider the following functional

$\displaystyle\Psi(t,f(t))\colonequals\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left(\gamma_{1}e^{-8\tau
nt}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+\gamma_{2}\right)f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}u^{T}Puf_{\infty}dxdv,$
(98)

with the $t-$ and $x-$dependent matrix in $\mathbb{R}^{2n\times 2n},$

$P=P(t,x)\colonequals\begin{pmatrix}2\varepsilon^{3}t^{3}I&\varepsilon^{2}t^{2}I\\\
\varepsilon^{2}t^{2}I&2\varepsilon tI+t(\frac{\partial^{2}V}{\partial
x^{2}}+cI)\end{pmatrix}.$ (99)

$\varepsilon,$ $\gamma_{1},$ and $\gamma_{2}$ are positive constants which we
shall fix later. We note that, for all $t\in(0,t_{0}],$

$P(t,x)\geq\begin{pmatrix}\varepsilon^{3}t^{3}I&0\\\
0&t(\frac{\partial^{2}V}{\partial x^{2}}+cI)+\varepsilon
tI\end{pmatrix}>\begin{pmatrix}\varepsilon^{3}t^{3}I&0\\\
0&t(\frac{\partial^{2}V}{\partial x^{2}}+cI)\end{pmatrix}\geq 0$ (100)

as $\frac{\partial^{2}V}{\partial x^{2}}+cI$ is positive semi-definite. Thus,
$\Psi(t,f(t))$ is non-negative and satisfies

$\displaystyle\Psi(t,f(t))\geq\displaystyle\displaystyle\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}\left(\gamma_{1}e^{-8\tau
nt}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+\gamma_{2}\right)f_{\infty}dxdv+\varepsilon^{3}t^{3}\displaystyle\int_{\mathbb{R}^{2n}}|u_{1}|^{2}f_{\infty}dxdv\\\
+t\int_{\mathbb{R}^{2n}}u_{2}^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+(c+\varepsilon)I\right)u_{2}f_{\infty}dxdv.$ (101)

Our goal is to show that $\Psi(t,f(t))$ decreases. To this end we compute the
time derivative of the second term in (98). Since this term has the same
structure as $S(f)$ from (20), we can follow the proof of Lemma 4.1 but with
the matrix $P$ now defined in (99), and taking in account that $P$ also
depends on time $t.$ Then, (27) yields

$\frac{d}{dt}\int_{\mathbb{R}^{2n}}u^{T}Puf_{\infty}dxdv\\\
=-2\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv-\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}-\partial_{t}P\right\\}uf_{\infty}dxdv\\\
-\int_{\mathbb{R}^{2n}}u^{T}\left\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\right\\}uf_{\infty}dxdv,$ (102)

with $Q=\begin{pmatrix}0&I\\\ -\frac{\partial^{2}V(x)}{\partial x^{2}}&\nu
I\end{pmatrix}.$ We consider each terms of (102). Because of (100), the first
term can be estimated as

$-2\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u)^{T}P\partial_{v_{i}}u\right\\}f_{\infty}dxdv\leq-2t\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv.$ (103)

For the third term of (102) we have

$[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P=\begin{pmatrix}0&0\\\
0&-t\frac{\partial^{2}(v\cdot\nabla_{x}V)}{\partial x^{2}}\end{pmatrix}$

and using $vf_{\infty}=-\frac{\sigma}{\nu}\nabla_{v}f_{\infty}$ yields

$-\int_{\mathbb{R}^{2n}}u^{T}\left\\{[\nabla_{x}V\cdot\nabla_{v}-v\cdot\nabla_{x}+\nu
v\cdot\nabla_{v}-\sigma\Delta_{v}]P\right\\}uf_{\infty}dxdv\\\
=\frac{2t\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv.$ (104)

For the second term of (102) we compute

$-\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}-\frac{d}{dt}P\right\\}uf_{\infty}dxdv\\\
=-\int_{\mathbb{R}^{2n}}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}^{T}\small{\begin{pmatrix}0&(t-2\varepsilon^{3}t^{3})\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\\\
(t-2\varepsilon^{3}t^{3})\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)&(-1+2\nu
t-2\varepsilon^{2}t^{2})\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\end{pmatrix}}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}f_{\infty}dxdv\\\
-\int_{\mathbb{R}^{2n}}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}^{T}\small{\begin{pmatrix}2\varepsilon^{2}t^{2}(1-3\varepsilon)I&[2c\varepsilon^{3}t^{3}+\nu\varepsilon^{2}t^{2}+2(1-\varepsilon)\varepsilon
t]I\\\
[2c\varepsilon^{3}t^{3}+\nu\varepsilon^{2}t^{2}+2(1-\varepsilon)\varepsilon
t]I&[2c\varepsilon^{2}t^{2}+4\varepsilon\nu
t-2\varepsilon]I\end{pmatrix}}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}f_{\infty}dxdv.$ (105)

Using the estimates

$-(t-2\varepsilon^{3}t^{3})\int_{\mathbb{R}^{2n}}u_{1}^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)u_{2}f_{\infty}dxdv\\\
\leq\varepsilon^{3}t^{2}|1-2\varepsilon^{3}t^{2}|\int_{\mathbb{R}^{2n}}|u_{1}|^{2}f_{\infty}dxdv+\frac{|1-2\varepsilon^{3}t^{2}|}{4\varepsilon^{3}}\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}f_{\infty}dxdv$

and

$-(-1+2\nu
t-2\varepsilon^{2}t^{2})\int_{\mathbb{R}^{2n}}u_{2}^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)u_{2}f_{\infty}dxdv\\\ \leq|1-2\nu
t+2\varepsilon^{2}t^{2}|\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|f_{\infty}dxdv,$

we get

$-\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}-\frac{d}{dt}P\right\\}uf_{\infty}dxdv\\\
\leq\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left[\frac{|1-2\varepsilon^{3}t^{2}|}{2\varepsilon^{3}}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+|1-2\nu
t+2\varepsilon^{2}t^{2}|\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|\right]f_{\infty}dxdv\\\
-\int_{\mathbb{R}^{2n}}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}^{T}\small{\begin{pmatrix}2\varepsilon^{2}t^{2}(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)I&[2c\varepsilon^{3}t^{3}+\nu\varepsilon^{2}t^{2}+2(1-\varepsilon)\varepsilon
t]I\\\
[2c\varepsilon^{3}t^{3}+\nu\varepsilon^{2}t^{2}+2(1-\varepsilon)\varepsilon
t]I&[2c\varepsilon^{2}t^{2}+4\varepsilon\nu
t-2\varepsilon]I\end{pmatrix}}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}f_{\infty}dxdv.$ (106)

We fix $\varepsilon=\varepsilon(t_{0})>0$ so that the element in the upper
left corner of the matrix in (106) is positive for $t>0;$ more precisely we
require

$1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|>0\,\,\,\,\,\text{ for all
}\,\,\,t\in[0,t_{0}].$ (107)

Then, the matrix in the last line of (106) can be estimated as

$\begin{pmatrix}2\varepsilon^{2}t^{2}(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)I&[2c\varepsilon^{3}t^{3}+\nu\varepsilon^{2}t^{2}+2(1-\varepsilon)\varepsilon
t]I\\\
[2c\varepsilon^{3}t^{3}+\nu\varepsilon^{2}t^{2}+2(1-\varepsilon)\varepsilon
t]I&[2c\varepsilon^{2}t^{2}+4\varepsilon\nu t-2\varepsilon]I\end{pmatrix}\\\
\geq\begin{pmatrix}0&0\\\ 0&[2c\varepsilon^{2}t^{2}+4\varepsilon\nu
t-2\varepsilon]I-\frac{[2c\varepsilon^{2}t^{2}+\nu\varepsilon
t+2(1-\varepsilon)]^{2}}{2(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)}I\end{pmatrix}.$

Using this matrix inequality, we obtain from (106):

$-\int_{\mathbb{R}^{2n}}u^{T}\left\\{QP+PQ^{T}-\frac{d}{dt}P\right\\}uf_{\infty}dxdv\\\
\leq\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left[\frac{|1-2\varepsilon^{3}t^{2}|}{2\varepsilon^{3}}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+|1-2\nu
t+2\varepsilon^{2}t^{2}|\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|\right.\\\
\left.-2c\varepsilon^{2}t^{2}-4\varepsilon\nu
t+2\varepsilon+\frac{[2c\varepsilon^{2}t^{2}+\nu\varepsilon
t+2(1-\varepsilon)]^{2}}{2(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)}\right]f_{\infty}dxdv.$
(108)

(102), (103), (104), and (108) show that

$\frac{d}{dt}\int_{\mathbb{R}^{2n}}u^{T}Puf_{\infty}dxdv\leq-2t\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv\\\
+\frac{2t\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\
+\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left[\frac{|1-2\varepsilon^{3}t^{2}|}{2\varepsilon^{3}}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+|1-2\nu
t+2\varepsilon^{2}t^{2}|\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|\right.\\\
\left.-2c\varepsilon^{2}t^{2}-4\varepsilon\nu
t+2\varepsilon+\frac{[2c\varepsilon^{2}t^{2}+\nu\varepsilon
t+2(1-\varepsilon)]^{2}}{2(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)}\right]f_{\infty}dxdv.$

As the matrix (6) is positive semi-definite, we have

$-2t\sigma\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)\partial_{v_{i}}u_{2}\right\\}f_{\infty}dxdv+\frac{2t\sigma}{\nu}\int_{\mathbb{R}^{2n}}\left\\{\sum_{i=1}^{n}(\partial_{v_{i}}u_{2})^{T}\frac{\partial^{2}(\partial_{x_{i}}V)}{\partial
x^{2}}u_{2}\right\\}f_{\infty}dxdv\\\ \leq\tau
t\int_{\mathbb{R}^{2n}}u_{2}^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)u_{2}f_{\infty}dxdv\leq\tau
t\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|f_{\infty}dxdv.$

Subsequently,

$\frac{d}{dt}\int_{\mathbb{R}^{2n}}u^{T}Puf_{\infty}dxdv\\\
\leq\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left[\frac{|1-2\varepsilon^{3}t^{2}|}{2\varepsilon^{3}}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+(|1-2\nu t+2\varepsilon^{2}t^{2}|+\tau
t)\left|\left|\frac{\partial^{2}V}{\partial x^{2}}+cI\right|\right|\right.\\\
\left.-2c\varepsilon^{2}t^{2}-4\varepsilon\nu
t+2\varepsilon+\frac{[2c\varepsilon^{2}t^{2}+\nu\varepsilon
t+2(1-\varepsilon)]^{2}}{2(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)}\right]f_{\infty}dxdv.$
(109)

#### Step 4, decay of the functional $\Psi$:

We estimate the time derivative of (98): Combining (91), (97), and (109) yield

$\frac{d}{dt}\Psi(t,f(t))\\\
\leq-\int_{\mathbb{R}^{2n}}|u_{2}|^{2}\left[\left(\sigma e^{-8\tau
nt}\gamma_{1}-\frac{|1-2\varepsilon^{3}t^{2}|}{2\varepsilon^{3}}\right)\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}-(|1-2\nu t+2\varepsilon^{2}t^{2}|+\tau
t)\left|\left|\frac{\partial^{2}V}{\partial x^{2}}+cI\right|\right|\right.\\\
\left.+2\sigma\gamma_{2}+2c\varepsilon^{2}t^{2}+4\varepsilon\nu
t-2\varepsilon-\frac{[2c\varepsilon^{2}t^{2}+\nu\varepsilon
t+2(1-\varepsilon)]^{2}}{2(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)}\right]f_{\infty}dxdv.$
(110)

We fix $\gamma_{1}>0$ and $\gamma_{2}>0$ such that

$\left(\sigma e^{-8\tau
nt}\gamma_{1}-\frac{|1-2\varepsilon^{3}t^{2}|}{2\varepsilon^{3}}\right)\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}-(|-1+2\nu t-2\varepsilon^{2}t^{2}|+\tau
t)\left|\left|\frac{\partial^{2}V}{\partial x^{2}}+cI\right|\right|\\\
+2\sigma\gamma_{2}+2c\varepsilon^{2}t^{2}+4\varepsilon\nu
t-2\varepsilon-\frac{[2c\varepsilon^{2}t^{2}+\nu\varepsilon
t+2(1-\varepsilon)]^{2}}{2(1-3\varepsilon-\varepsilon|1-2\varepsilon^{2}t^{2}|)}\geq
0$ (111)

for all $x\in\mathbb{R}^{n}$ and $t\in[0,t_{0}].$ We recall that we have fixed
$\varepsilon=\varepsilon(t_{0})$ so that (107) holds, which makes the above
denominator positive. The existence of such $\gamma_{1}>0$ and $\gamma_{2}>0$
can be proven by the following arguments: We can consider the left hand side
of (111) as a quadratic polynomial of
$\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|\in[0,\infty).$ As time $t$ varies in a bounded
interval $[0,t_{0}],$ the terms containing $t$ are bounded. Therefore, we can
choose large values for $\gamma_{1}=\gamma_{1}(t_{0})$ and
$\gamma_{2}=\gamma_{2}(t_{0})$ so that this quadratic polynomial is non-
negative for all $t\in[0,t_{0}].$

Consequently, we obtain that

$\frac{d}{dt}\Psi(t,f(t))\leq 0.$

Hence $\Psi(t,f(t))$ is decreasing and

$\Psi(t,f(t))\leq\Psi(0,f_{0})\,\,\,\,\,\text{ for all
}\,\,\,\,t\in[0,t_{0}].$ (112)

(101) and (112) show that

$\displaystyle\int_{\mathbb{R}^{2n}}|u_{1}|^{2}f_{\infty}dxdv\leq\frac{1}{\varepsilon^{3}t^{3}}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\gamma_{1}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+\gamma_{2}\right)f_{\infty}dxdv,$ (113)
$\int_{\mathbb{R}^{2n}}|u_{2}|^{2}f_{\infty}dxdv\leq\frac{1}{\varepsilon
t}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\gamma_{1}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+\gamma_{2}\right)f_{\infty}dxdv,$ (114)

and

$\int_{\mathbb{R}^{2n}}u_{2}^{T}\left(\frac{\partial^{2}V}{\partial
x^{2}}+cI\right)u_{2}f_{\infty}dxdv\leq\frac{1}{t}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\gamma_{1}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+\gamma_{2}\right)f_{\infty}dxdv.$ (115)

It is clear that there is a positive constant $C$ such that

$\gamma_{1}\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}+cI\right|\right|^{2}+\gamma_{2}\leq
C\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right).$ (116)

(113), a proper linear combination of (114) and (115), and (116) imply the
claimed estimates (11), (12). ∎

###### Proof of Corollary 2.8.

Theorem 2.3 and Theorem 2.7 show that, for $t\geq t_{0}>0,$

$\int_{\mathbb{R}^{2n}}\left(\frac{f(t)}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t)}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\\\
+\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t)}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t)}{f_{\infty}}\right)f_{\infty}dxdv\\\
\leq
Ce^{-2\lambda(t-t_{0})}\left[\int_{\mathbb{R}^{2n}}\left(\frac{f(t_{0})}{f_{\infty}}-1\right)^{2}f_{\infty}dxdv+\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t_{0})}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\right.\\\
+\left.\int_{\mathbb{R}^{2n}}\nabla^{T}_{v}\left(\frac{f(t_{0})}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t_{0})}{f_{\infty}}\right)f_{\infty}dxdv\right]$
(117)

holds with the constant $C$ and the rate $\lambda$ given in Theorem 2.3. Using
(11) and (12) at $t=t_{0},$ we get

$\displaystyle\int_{\mathbb{R}^{2n}}\left|\nabla_{x}\left(\frac{f(t_{0})}{f_{\infty}}\right)\right|^{2}f_{\infty}dxdv\leq\frac{C_{1}}{t_{0}^{3}}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv$ (118)

and

$\displaystyle\int_{\mathbb{R}^{2n}}\nabla_{v}^{T}\left(\frac{f(t_{0})}{f_{\infty}}\right)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)\nabla_{v}\left(\frac{f(t_{0})}{f_{\infty}}\right)f_{\infty}dxdv\\\
\leq\frac{C_{2}}{t_{0}}\int_{\mathbb{R}^{2n}}\left(\frac{f_{0}}{f_{\infty}}-1\right)^{2}\left(\left|\left|\frac{\partial^{2}V}{\partial
x^{2}}\right|\right|^{2}+1\right)f_{\infty}dxdv.$ (119)

Combining (117), (118), and (119), we obtain (13). ∎

##  6 Appendix

###  6.1  Proof that Assumption 2.2’ implies Assumption 2.2

Assume Assumption 2.2’ is satisfied. Let $(u_{1},u_{2},...,u_{n+1})^{T}$ be
any vector in $\mathbb{R}^{n(n+1)},$ where $u_{i}$ is a vector in
$\mathbb{R}^{n}$ for all $i\in\\{1,...,n+1\\}.$ We compute the quadratic form
of the matrix (6)

$\begin{pmatrix}u_{1}\\\ u_{2}\\\ .\\\ .\\\ .\\\
u_{n+1}\end{pmatrix}^{T}\begin{pmatrix}\nu\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)&0&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V(x))}{\partial
x^{2}}\\\ 0&\nu\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)&...&0&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V(x))}{\partial
x^{2}}\\\ ...&...&...&...&...\\\
0&0&...&\nu\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V(x))}{\partial
x^{2}}\\\ -\frac{1}{2}\frac{\partial^{2}(\partial_{x_{1}}V(x))}{\partial
x^{2}}&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{2}}V(x))}{\partial
x^{2}}&...&-\frac{1}{2}\frac{\partial^{2}(\partial_{x_{n}}V(x))}{\partial
x^{2}}&\frac{\tau\nu}{2\sigma}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)\end{pmatrix}\begin{pmatrix}u_{1}\\\ u_{2}\\\ .\\\ .\\\ .\\\
u_{n+1}\end{pmatrix}$ $=\sum_{i=1}^{n}\left\\{\nu
u_{i}^{T}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)u_{i}-u^{T}_{i}\frac{\partial^{2}(\partial_{x_{i}}V(x))}{\partial
x^{2}}u_{n+1}\right\\}+\frac{\tau\nu}{2\sigma}u_{n+1}^{T}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)u_{n+1}.$

To show that (6) is positive semi-definite, it is enough to show the quadratic
form above is non-negative. Assumption 2.2’ implies

$\left|u^{T}_{i}\frac{\partial^{2}(\partial_{x_{i}}V(x))}{\partial
x^{2}}u_{n+1}\right|\leq|u_{i}||u_{n+1}|\sqrt{\frac{2\tau\nu^{2}}{n\sigma}}(\alpha(x)+c)\leq\nu(\alpha(x)+c)|u_{i}|^{2}+{\frac{\tau\nu}{2n\sigma}}(\alpha(x)+c)|u_{n+1}|^{2}.$

Therefore, we get the desired result

$\sum_{i=1}^{n}\left\\{\nu u_{i}^{T}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)u_{i}-u^{T}_{i}\frac{\partial^{2}(\partial_{x_{i}}V(x))}{\partial
x^{2}}u_{n+1}+\frac{\tau\nu}{2n\sigma}u_{n+1}^{T}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}+cI\right)u_{n+1}\right\\}$ $\geq\sum_{i=1}^{n}\left\\{\nu
u_{i}^{T}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}-\alpha(x)I\right)u_{i}+\frac{\tau\nu}{2n\sigma}u_{n+1}^{T}\left(\frac{\partial^{2}V(x)}{\partial
x^{2}}-\alpha(x)I\right)u_{n+1}\right\\}\geq 0.$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\square$

###  6.2  Matrix inequalities for Section 5.1

######  Lemma 6.1.

Let $\alpha_{0}>-\infty$ be the constant defined by (8), $a\in\mathbb{R}$ be
some constant such that $a+\alpha_{0}>\frac{\nu^{2}}{4},$ and
$P\colonequals\begin{pmatrix}2I&\nu I\\\ \nu I&2\frac{\partial^{2}V}{\partial
x^{2}}+2aI\end{pmatrix}.$ Then

$c_{1}P\leq\begin{pmatrix}I&0\\\ 0&\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\end{pmatrix}\leq c_{2}P$ (120)

holds with
$c_{1}\colonequals\frac{1}{a+\alpha_{0}+1+\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}}>0,\,\,\,\,\,c_{2}\colonequals\frac{a+\alpha_{0}+1+\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}}{4(a+\alpha_{0})-\nu^{2}}>0.$

###### Proof.

We consider, for some $k\in\mathbb{R}$ to be chosen later as
$\frac{1}{2c_{1,2}},$

$A\colonequals P-2k\begin{pmatrix}I&0\\\ 0&\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\end{pmatrix}=\begin{pmatrix}2(1-k)I&\nu I\\\ \nu
I&2(1-k)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)+2(a+\alpha_{0}-1)I\end{pmatrix}.$

We check the (real) eigenvalues $\eta$ of the symmetric matrix $A$ (depending
on $k$). It is easy to check that $\eta=2(1-k)$ is not an eigenvalue of $A.$
If $\eta\neq 2(1-k),$ then we have the condition

$\det(A-\eta I)=\begin{vmatrix}2(1-k)I-\eta I&\nu I\\\ \nu
I&2(1-k)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)+2(a+\alpha_{0}-1)I-\eta I\end{vmatrix}\\\
=\small\frac{1}{(2(1-k)-\eta)^{n}}\begin{vmatrix}2(1-k)I-\eta I&0\\\ \nu
I&(2(1-k)-\eta)\left[2(1-k)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)+2(a+\alpha_{0}-1)I-\eta
I\right]-\nu^{2}I\end{vmatrix}\\\
=\det\left((2(1-k)-\eta)\left[2(1-k)\left(\frac{\partial^{2}V}{\partial
x^{2}}+(1-\alpha_{0})I\right)+2(a+\alpha_{0}-1)I-\eta
I\right]-\nu^{2}I\right)=0.$

If $\alpha_{i},$ $i\in\\{1,...,n\\}$ are the eigenvalues of
$\frac{\partial^{2}V}{\partial x^{2}},$ then the eigenvalues $\eta$ of $A$
satisfy

$\prod_{i=1}^{n}\left(\eta^{2}-2\eta[(1-k)(\alpha_{i}-\alpha_{0}+2)+a+\alpha_{0}-1]+4(1-k)^{2}(\alpha_{i}-\alpha_{0}+1)+4(1-k)(a+\alpha_{0}-1)-\nu^{2}\right)=0.$
(121)

Right inequality of (120): From (121), we see that $A$ is positive semi-
definite (i.e., all $\eta\geq 0$) if the following three conditions hold:

$1-k\geq 0,\,\,\,\,\,\,\text{ (due to the first minor of $A$)}$ (122)
$(1-k)(\alpha_{i}-\alpha_{0}+2)+a+\alpha_{0}-1\geq 0,\,\,\,\forall
i\in\\{1,...,n\\},$ (123)
$4(1-k)^{2}(\alpha_{i}-\alpha_{0}+1)+4(1-k)(a+\alpha_{0}-1)-\nu^{2}\geq
0,\,\,\,\,\forall i\in\\{1,...,n\\}.$ (124)

We set

$k\colonequals\frac{1}{2c_{2}}>0.$

Then, (122) holds:

$1-k=\frac{\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}-(a+\alpha_{0}-1)}{2}>0.$ (125)

Using $\alpha_{i}\geq\alpha_{0}$ for all $i\in\\{1,...,n\\}$ we see that (123)
also holds:

$(1-k)(\alpha_{i}-\alpha_{0}+2)+a+\alpha_{0}-1\geq
2(1-k)+a+\alpha_{0}-1=\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}>0.$

To verify (124) we estimate using $\alpha_{i}\geq\alpha_{0}$ for all
$i\in\\{1,...,n\\}$ and (125)

$4(1-k)^{2}(\alpha_{i}-\alpha_{0}+1)+4(1-k)(a+\alpha_{0}-1)-\nu^{2}\\\ \geq
4(1-k)^{2}+4(1-k)(a+\alpha_{0}-1)-\nu^{2}=0.$

Therefore, for $k$ defined in (125), $A$ is positive semi-definite. Hence, the
inequality on the right hand side of (120) holds.

Left inequality of (120): Similarly, $A$ is negative semi-definite if the
following three conditions hold:

$1-k\leq 0,$ (126) $(1-k)(\alpha_{i}-\alpha_{0}+2)+a+\alpha_{0}-1\leq
0,\,\,\,\forall i\in\\{1,...,n\\},$ (127)
$4(1-k)^{2}(\alpha_{i}-\alpha_{0}+1)+4(1-k)(a+\alpha_{0}-1)-\nu^{2}\geq
0,\,\,\,\,\forall i\in\\{1,...,n\\}.$ (128)

Setting

$k\colonequals\frac{1}{2c_{1}}>0$

we find

$1-k=\frac{-\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}-(a+\alpha_{0}-1)}{2}<0$ (129)

and

$(1-k)(\alpha_{i}-\alpha_{0}+2)+a+\alpha_{0}-1\leq
2(1-k)+a+\alpha_{0}-1=-\sqrt{(a+\alpha_{0}-1)^{2}+\nu^{2}}<0.$

Finally, we check using $\alpha_{i}\geq\alpha_{0}$ for all $i\in\\{1,...,n\\}$
and (129)

$4(1-k)^{2}(\alpha_{i}-\alpha_{0}+1)+4(1-k)(a+\alpha_{0}-1)-\nu^{2}\geq
4(1-k)^{2}+4(1-k)(a+\alpha_{0}-1)-\nu^{2}=0.$

Therefore, for $k$ defined in (129), $A$ is negative semi-definite. Hence, the
inequality on the left hand side of (120) holds. ∎

######  Remark 6.2.

Lemma 6.1 proves the following matrix inequalities from Section 5.1:

  * (a)

If $a=0$ and $\alpha_{0}>\frac{\nu^{2}}{4},$ then (120) is the matrix
inequality (56).

  * (b)

If $a=\frac{\varepsilon^{2}}{2}$ and $\alpha_{0}=\frac{\nu^{2}}{4},$ then
(120) is the matrix inequality (64).

  * (c)

(120) coincides with the matrix inequality (75).

###  6.3  Proof of inequality (80)

We recall the assumption $\alpha_{0}<\frac{\nu^{2}}{4}.$ We first rewrite

$\displaystyle
A_{2}^{-1}=\frac{2\nu\min\\{1,\alpha_{0}\\}}{1+\frac{\nu^{2}}{2}-\alpha_{0}+\sqrt{(\frac{\nu^{2}}{2}-\alpha_{0}-1)^{2}+\nu^{2}}}=\frac{4\min\\{1,\alpha_{0}\\}}{{\nu}+2(1-\alpha_{0})\nu^{-1}+\sqrt{(\nu^{2}-4\alpha_{0})+4(\alpha_{0}+1)^{2}\nu^{-2}}},$
$\nu-\sqrt{\nu^{2}-4\alpha_{0}}=\frac{4\alpha_{0}}{\nu+\sqrt{\nu^{2}-4\alpha_{0}}}.$

Then (80) is equivalent to

$\frac{\alpha_{0}}{\nu+\sqrt{\nu^{2}-4\alpha_{0}}}\geq\frac{\min\\{1,\alpha_{0}\\}}{\nu+2(1-\alpha_{0})\nu^{-1}+\sqrt{(\nu^{2}-4\alpha_{0})+4(\alpha_{0}+1)^{2}\nu^{-2}}}.$
(130)

If $\min\\{1,\alpha_{0}\\}=\alpha_{0},$ then (130) is true because of

$\nu+2(1-\alpha_{0})\nu^{-1}+\sqrt{(\nu^{2}-4\alpha_{0})+4(\alpha_{0}+1)^{2}\nu^{-2}}>\nu+\sqrt{\nu^{2}-4\alpha_{0}}.$

If $\min\\{1,\alpha_{0}\\}=1,$ then (130) is equivalent to

$\alpha_{0}\nu-2\alpha_{0}(\alpha_{0}-1)\nu^{-1}+\alpha_{0}\sqrt{(\nu^{2}-4\alpha_{0})+4(\alpha_{0}+1)^{2}\nu^{-2}}\geq\nu+\sqrt{\nu^{2}-4\alpha_{0}},$

or equivalently

$(\alpha_{0}-1)(\nu^{2}-2\alpha_{0})\nu^{-1}+\alpha_{0}\sqrt{(\nu^{2}-4\alpha_{0})+4(\alpha_{0}+1)^{2}\nu^{-2}}\geq\sqrt{\nu^{2}-4\alpha_{0}}.$

The last inequality holds since

$(\alpha_{0}-1)(\nu^{2}-2\alpha_{0})\nu^{-1}\geq(\alpha_{0}-1)(\nu^{2}-4\alpha_{0})\nu^{-1}\geq
0$

and

$\alpha_{0}\sqrt{(\nu^{2}-4\alpha_{0})+4(\alpha_{0}+1)^{2}\nu^{-2}}>\sqrt{\nu^{2}-4\alpha_{0}}.$

These two cases show that inequality (80) holds.

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\square$

Acknowledgement. Both authors acknowledge support of the Austrian Science Fund
(FWF) through the SFB project F65 “Taming Complexity in Partial Differential
Systems”.

## References

  * [1] F. Achleitner, A. Arnold, D. Stürzer. Large-time behavior in non-symmetric Fokker-Planck equations. Rivista di Matematica della Universitá di Parma 6, no. 1 (2015): 1–68.
  * [2] A. Arnold, E. Carlen, Q. Ju. Large-time behavior of non-symmetric Fokker-Planck type equations. Comm. in Stochastic Analysis 2, no. 1 (2008): 153–175.
  * [3] A. Arnold, J. Erb. Sharp entropy decay for hypocoercive and non-symmetric Fokker-Planck equations with linear drift. arXiv preprint arXiv:1409.5425 (2014).
  * [4] A. Arnold, P. Markowich, G. Toscani, A. Unterreiter. On convex Sobolev inequalities and the rate of convergence to equilibrium for Fokker-Planck type equations. Comm. in Partial Differential Equations 26, no. 1-2 (2001): 43–100.
  * [5] A. Arnold, C. Schmeiser, B. Signorello. Propagator norm and sharp decay estimates for Fokker-Planck equations with linear drift. Comm. Math. Sci. 20, no. 4 (2022): 1047–1080.
  * [6] D. Bakry, F. Barthe, P. Cattiaux, A. Guillin. A simple proof of the Poincaré inequality for a large class of probability measures. Electron. Commun. Probab. 13 (2008): 60–66.
  * [7] D. Bakry, M. Emery. Diffusions hypercontractives. Séminaire de probabilités de Strasbourg 19 (1985): 177–206.
  * [8] D. Bakry, I. Gentil, M. Ledoux. Analysis and geometry of Markov diffusion operators.  Vol. 103. Cham: Springer, 2014.
  * [9] F. Baudoin. Bakry-Emery meet Villani. J. Funct. Anal. 273, no. 7 (2017): 2275–2291.
  * [10] F. Baudoin, M. Gordina, D. Herzog. Gamma calculus beyond Villani and explicit convergence estimates for Langevin dynamics with singular potentials. Arch. Rational. Mech. Anal. 241, no. 2 (2021): 765–804.
  * [11] J. A. Bittencourt. Fundamentals of plasma physics. Springer Science and Business Media, 2004.
  * [12] G. Brigati, G. Stoltz. How to construct decay rates for kinetic Fokker-Planck equations?. arXiv preprint arXiv:2302.14506 (2023).
  * [13] Y. Cao, J. Lu, L. Wang. On explicit $L^{2}-$convergence rate estimate for underdamped Langevin dynamics. Arch. Rational. Mech. Anal. 247, no. 5 (2023): 90.
  * [14] P. Cattiaux, A. Guillin, P. Monmarché, C. Zhang. Entropic multipliers methods for Langevin diﬀusion and weighted log Sobolev inequalities. J. Funct. Anal. 277, no. 11 (2019): 108288.
  * [15] S. Chandrasekhar. Stochastic problems in physics and astronomy. Reviews of Modern Physics. 15, no. 1, (1943): 1–89.
  * [16] S. Chandrasekhar. Brownian motion, dynamical friction, and stellar dynamics. Reviews of Modern Physics. 21, no. 3 (1949): 383–388.
  * [17] L. Desvillettes, C. Villani. On the trend to global equilibrium in spatially inhomogeneous entropy-dissipating systems: The linear Fokker-Planck equation. Comm. Pure Appl. Math. 54, no. 1 (2001): 1–42.
  * [18] J. Dolbeault, C. Mouhot, C. Schmeiser. Hypocoercivity for linear kinetic equations conserving mass. Transactions of the American Mathematical Society 367, no. 6 (2015): 3807–3828.
  * [19] B. Helffer, F. Nier. Hypoelliptic estimates and spectral theory for Fokker-Planck operators and Witten Laplacians. Springer Science and Business Media, 2005.
  * [20] F. Hérau. Short and long time behavior of the Fokker-Planck equation in a confining potential and applications. J. Funct. Anal. 244, no. 1 (2007): 95–118.
  * [21] F. Hérau, F. Nier. Isotropic Hypoellipticity and Trend to Equilibrium for the Fokker-Planck Equation with a High-Degree Potential. Arch. Rational Mech. Anal. 171, no. 1 (2004): 151–218.
  * [22] C.D. Hill. A sharp maximum principle for degenerate elliptic-parabolic equations. Indiana Univ. Math. J. 20, no. 3 (1970): 213–229.
  * [23] L. Hörmander. Hypoelliptic second order differential equations. Acta Mathematica 119, no. 1 (1967): 147–171.
  * [24] R. A. Horn, C. R. Johnson.  Topics in matrix analysis, Cambridge university press, 1994.
  * [25] A. Kolmogorov. Zufällige Bewgungen (zur Theorie der Brownschen Bewegung). Annals of Mathematics 35, no. 1 (1934): 116–117.
  * [26] R. L. Liboff. Kinetic theory: classical, quantum, and relativistic descriptions. Springer Science and Business Media, 2003.
  * [27] C. Mouhot, L. Neumann. Quantitative perturbative study of convergence to equilibrium for collisional kinetic models in the torus. Nonlinearity 19, no. 4, (2006): 969–998.
  * [28] H. Risken. The Fokker-Planck equation: Methods of solution and applications.  2nd. ed., Springer-Verlag, Berlin, Heidelberg, 1989.
  * [29] C.Villani. Hypocoercivity, Mem. Amer. Math. Soc. 202, 2009.
  * [30] C. Zhang.  Hypocoercivity and global hypoellipticity for the kinetic Fokker-Planck equation in $H^{k}$ spaces. Kinetic and Related Models, forthcoming.