\forall t \in [0, T(R)], \ \ \varrho(t)=0.
\end{align*}

$\bullet$ Step 2: let us now prove that $f_1=f_2$ and $u_1=u_2$ on $[0,T(R)]$. It is in fact a direct consequence of $\varrho=\varrho_1-\varrho_2=0$ on $[0,T(R)]$. Indeed, the previous step has shown that for all $t \in [0,T(R)]$

\begin{align*}
&\LRVert{u(t)}_{\Ld^2}^2 \lesssim \int_0^t \LRVert{u(s)}_{\Ld^2}^2 \, \mathrm{d}s+  \Lambda(t,R)\LRVert{\varrho}_{\Ld^2(0,t; \Ld^2)} \lesssim \int_0^t \LRVert{u(s)}_{\Ld^2}^2 \, \mathrm{d}s,
\end{align*}

therefore we directly have $u=0$ on $[0,T(R)]$. The equations on (<ref>) on $f$ now turn into

\begin{equation*}
\left\{
\begin{aligned}
\partial_t f +v \cdot \nabla_x f -v \cdot \nabla_v f  +\big( u_2-\nabla_x p(\varrho_2) \big)  \cdot \nabla_v f=df, \\[2mm]
f_{\mid t=0}=0.
\end{aligned}
\right.
\end{equation*}

The method of characteristics thus shows that $f=0$ on $[0,T(R)]$.

In conclusion, we have obtained $(f, \varrho, u)=(0,0,0)$ on $[0,T(R)]$. We eventually observe that we can repeat this procedure starting from $t=T(R)$ instead of $t=0$. As a matter of fact, $f_1(T(R),\cdot)$ still satisfies the Penrose stability condition. Since the time $T(R)$ only depends on $R$, we obtain $\varrho=0$ on $[0,2T(R)]$ and then $(f, u)=(0,0)$ on $[0,2T(R)]$. After a finite number of steps, this yields $(f, \varrho, u)=(0,0,0)$ on $[0,\bm{T}]$. This finally concludes the proof of the uniqueness part of the statement.

§ GENERALIZATION TO THE NON-BAROTROPIC CASE

In this section, we show how to handle the case of the full Navier-Stokes system for non-barotropic fluids, where we consider the additional internal energy $\mathfrak{e} \in \R^+$ for the fluid and where the the pressure depends on $\varrho$ and $\mathfrak{e}$ . As explained in the introduction, the system which is at stake is the following:

\begin{equation}
\left\{
\begin{aligned}
\partial_t f +v \cdot \nabla_x f +{\rm div}_v \Big[ f (u-v)-f \nabla_x p(\varrho, \mathfrak{e}) \Big]&=0,  \\
\partial_t (\alpha \varrho ) + \mathrm{div}_x (\alpha \varrho u)&=0, \\
\partial_t (\alpha \varrho u) + \mathrm{div}_x(\alpha \varrho u \otimes u)+\alpha \nabla_x p(\varrho, \mathfrak{e}) -\Delta_x u - \nabla_x \mathrm{div}_x u&=j_f-\rho_f u, \\
\partial_t (\alpha \varrho \mathfrak{e}) + \mathrm{div}_x(\alpha \varrho \mathfrak{e}u) +p(\varrho, \mathfrak{e})\left(\partial_t \alpha+ \mathrm{div}_x(\alpha u) \right)&=\int_{\R^d}\vert u-v \vert^2 f \, \mathrm{d}v, \\
\alpha&=1-\rho_f.
\end{aligned}
\right.
\end{equation}

Here, we will assume[It is likely that more general pressure $p(\varrho, \mathfrak{e})$ could be treated by our method.] that the pressure law is given as

\begin{align*}
p(\varrho, \mathfrak{e})=\pi(\varrho \mathfrak{e}),
\end{align*}

for some given function $\pi : \R^+ \rightarrow \R$ such that $\pi \in \mathscr{C}( \R^+ ) \cap \mathscr{C}^{\infty}(\R^+ {\setminus} \lbrace 0 \rbrace)$. For instance, the relation $p(\varrho,\mathfrak{e})=b \varrho \mathfrak{e}$ (with $b>0$) is a perfect gas pressure law.

Similarly to the technical hypothesis (<ref>), we shall assume that

\begin{align*}
y \mapsto \pi'(y) (y+\pi(y)) \ \ \text{is nondecreasing on} \ \  \R.
\end{align*}

\begin{align*}
\vartheta:=\varrho  \mathfrak{e},
\end{align*}

the system in $(f,\varrho, u, \vartheta)$ can be rewritten as

\begin{equation}\label{eq:TSe}
\tag{TS$_e$}
\left\{
\begin{aligned}
\partial_t f +v \cdot \nabla_x f   +\mathrm{div}_v \big[f(u-v-\nabla_x \pi(\vartheta)) \big] &=0,   \\[2mm]
\partial_t (\alpha \varrho ) + \mathrm{div}_x (\alpha \varrho u)&=0, \\[2mm]
\partial_t (\alpha \varrho u) + \mathrm{div}_x(\alpha \varrho u \otimes u) +\alpha\nabla_x \pi(\vartheta)-\Delta_x u - \nabla_x \mathrm{div}_x u &=j_f-\rho_f u,  \\[2mm]
\partial_t (\alpha \vartheta) + \mathrm{div}_x(\alpha \vartheta u) +\pi(\vartheta) \left( \partial_t \alpha+ \mathrm{div}_x(\alpha u) \right)&=\int_{\R^d} \vert v-u \vert^2 f \, \mathrm{d}v,  \\
\alpha&=1-\rho_f.
\end{aligned}
\right.
\end{equation}

As we shall see below, it is significant to define the following Penrose symbol of a function $(f(x,v), \vartheta(x))$ as

\begin{align}\label{def:PenroseSymbol-energyintern}
\mathbf{P}_{f, \vartheta}^{\mathrm{energy}}(x,\gamma, \tau, \eta):=\frac{\pi'(\vartheta(x))\left[\vartheta(x)+\pi(\vartheta(x))\right]}{1-\rho_f(x)}\int_0^{+\infty} e^{-(\gamma+i \tau) s} \frac{i k}{1+ \vert k \vert^2} \cdot \left( \mathcal{F}_v \nabla_v f \right)(x,s\eta)  \, \mathrm{d}s.
\end{align}

We can now introduce the following Penrose condition, adapted to (<ref>), which is that there exists $c>0$ such that

\begin{align}\label{cond:Penrose-energyintern}
\tag{\textbf{P$^{\mathrm{energy}}$}}
\forall x \in \T^d, \ \ \underset{(\gamma, \tau,\eta)\in (0,+\infty) \times \R \times \R^d {\setminus} \lbrace 0 \rbrace}{\inf} \, \vert 1- \mathbf{P}_{f, \vartheta}^{\mathrm{energy}}(x,\gamma, \tau, \eta) \vert>c,
\end{align}

Our main result for the system (<ref>) reads as follows.

There exist $m_0>0$ and $r_0>0$, depending only on the dimension, such that the following holds for all $m \geq m_0$ and $r \geq r_0$. Let

\begin{align*}
f^{\mathrm{in}} \in  \mathcal{H}^{m}_{r}, \ \  \varrho^{\mathrm{in}} \in \mathrm{H}^{m-1}, \ \ u^{\mathrm{in}} \in \mathrm{H}^{m}, \ \ \vartheta^{\mathrm{in}} \in \mathrm{H}^{m+1},
\end{align*}

such that $(f^{\mathrm{in}}, \vartheta^{\mathrm{in}})$ satisfies the $c$-Penrose stability condition $\eqref{cond:Penrose-energyintern}_c$ (with $c>0$) and

\begin{align*}
0 &\leq f^{\mathrm{in}}, \ \ \ \rho_{f^{\mathrm{in}}}<\Upsilon<1, \ \ \ 0<\mu \leq \varrho^{\mathrm{in}}, \vartheta^{\mathrm{in}} \ \ \ 0<\underline{\nu} \leq (1-\rho_{f^{\mathrm{in}}})\varrho^{\mathrm{in}}, (1-\rho_{f^{\mathrm{in}}})\vartheta^{\mathrm{in}} \leq \overline{\nu},
\end{align*}

for some fixed constants $\Upsilon, \mu, \underline{\nu}, \overline{\nu}$.
Then there exist $T>0$ and a solution $(f,  \varrho, u, \vartheta)$ to (<ref>)  with initial condition $(f^{\mathrm{in}}, \varrho^{\mathrm{in}}, u^{\mathrm{in}}, \vartheta^{\mathrm{in}})$ such that

\begin{align*}
f \in  \mathscr{C}\left([0,T];\mathcal{H}^{m-1}_{r} \right), \ \ \varrho \in \mathscr{C}([0,T];\mathrm{H}^{m-1}), \ \  \ u \in \mathscr{C}\left([0,T]; \mathrm{H}^{m} \right)\cap \Ld^2(0,T;\mathrm{H}^{m+1}), \ \ \vartheta \in \Ld^2(0,T;\mathrm{H}^{m}),
\end{align*}

and with $(f(t), \vartheta(t))$ satisfying the $c/2$-Penrose stability condition $\eqref{cond:Penrose-energyintern}_{c/2}$ for all $t \in [0,T]$. In addition, this solution is unique in this class.

Let us explain the main strategy for the proof of Theorem <ref>. First, we observe that the equation on $\vartheta$ can be rewritten as

\begin{align}\label{eq:theta}
\partial_t \vartheta+ u \cdot \nabla_x \vartheta + \frac{\vartheta + \pi(\vartheta)}{1-\rho_f} \mathrm{div}(j_f-\rho_f u)=-\frac{\vartheta + \pi(\vartheta)}{1-\rho_f} \mathrm{div}_x \, u+ \frac{1}{1-\rho_f}\int_{\R^d} \vert v-u \vert^2 f \, \mathrm{d}v.
\end{align}

Appart from the last term, this equation has exactly the same structure as the one for $\varrho$ in Lemma <ref>. Hence, the following estimate holds

\begin{align*}
\Vert \vartheta(t) \Vert_{\H^{m}} \leq   \Vert \varrho^{\mathrm{in}} \Vert_{\H^{m}} \Phi \Big( T, \cdots, \Vert u \Vert_{\Ld^{\infty}(0,T;\H^{m+1})},  \Vert  f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m+1}_{r})} \Big),
\end{align*}

and features the same loss of derivative as for $\varrho$ in (<ref>). Note also that the equation on $\varrho$ can be directly solved once $f$ and $u$ are given. As in Section <ref>, we consider the regularization $-\pi'(\vartheta ) \mathrm{J}_\eps \nabla_x \vartheta$ of the term $-\nabla_x \pi(\vartheta)$ in the kinetic equation of (<ref>) and introduce the quantity

\begin{multline*}
\textbf{N}_{m,r}(f_\eps,\varrho_\eps,u_\eps, \vartheta_\eps,T):= \Vert f_\eps \Vert_{\Ld^{\infty}\left(0,T;\mathcal{H}^{m-1}_{r}\right)}+\Vert \varrho_\eps \Vert_{\Ld^{\infty}(0,T;\H^{m-1})}\\
+\Vert u_\eps \Vert_{\Ld^{\infty}(0,T; \H^m) \cap \Ld^{2}(0,T;\H^{m+1})}+\Vert \vartheta_\eps \Vert_{\Ld^2(0,T;\H^{m})}.
\end{multline*}

Following the bootstrap procedure we have set for the case of (<ref>), we mainly want to control the quantity $\Vert \vartheta_\eps \Vert_{\Ld^2(0,T; \H^m)}$.

Using (<ref>) (and dropping the dependency in $\eps$) it is possible to obtain the following equation on $h=\partial_x^{\alpha} \vartheta$ with $\vert \alpha \vert \leq m$ (see Proposition <ref>): one has

\begin{align*}
\left( \mathrm{Id}-\frac{\vartheta+ \pi(\vartheta)}{1-\rho_f}\mathrm{K}_G^{\mathrm{free}}\circ\mathrm{J}_{\eps} \right)\Big[\partial_t h + u \cdot \nabla_x h\Big]=\mathcal{R}, \ \ t \in (0,T),
\end{align*}

with $G(t,x,v)=  \pi'( \vartheta(t,x)) \nabla_v f(t,x,v)$ and

\| \mathcal{R} \|_{\Ld^2(0,T; \Ld^2(\T^d))} \leq \Lambda (T,R, \|h(0)\|_{\H^1(\T^d)}).

Following the arguments of Section <ref>, we are thus led to the study of the following pseudodifferential equation

\begin{align*}
H-\frac{\pi'(\vartheta)(\vartheta+ \pi(\vartheta))}{1-\rho_f}\mathrm{Op}^{\gamma}(a_{f})(\mathrm{J}_{\eps} H)=\mathcal{R}, \ \ \text{on} \ \ (0,T) \times \T^d,
\end{align*}

where $a_f$ is defined in (<ref>), that is

\begin{align*}
\mathrm{Op}^{\gamma, \eps}(1-\mathbf{P}_{f, \vartheta}^{\mathrm{energy}})(H)=\mathcal{R}.
\end{align*}

In particular, it explains the introduction of the Penrose symbol (<ref>) above, which allows to invert the previous equation on $H$, up to a small remainder.

Some additional arguments have also to be given to treat the last term in (<ref>). Since

\begin{align*}
\int_{\R^d} \vert v-u \vert^2 f \, \mathrm{d}v = \int_{\R^d} \vert  v \vert^2 f \, \mathrm{d}v + \vert u \vert^2 \rho_f- 2 u \cdot j_f,
\end{align*}

we have to include the treatment of the second order moment in velocity $m_2f(t,x):=\int_{\R^d} \vert  v \vert^2 f(t,x,v) \, \mathrm{d}v$ in the analysis of Section <ref>. In addition to Proposition <ref>, we have

For all $|I|\leq m$, we have for any $t \in (0,T)$

\begin{align*}
\partial_x^I  m_2f(t,x)&=p'(\varrho(t,x))\int_0^t  \int_{\R^d}  \vert v \vert^2 \nabla_x  [\mathrm{J}_\eps \partial_x^I \varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s+R^I[m_2f](t,x),
\end{align*}

where the remainder $R^I[m_2f]$ satisfies

\begin{align*}
\left\Vert R^I[M_2f] \right\Vert_{\Ld^2(0,T, \H^1_x)} \leq \Lambda(T,R).
\end{align*}

In particular, we have

\begin{align*}
\LRVert{\partial_x^I \int_{\R^d} \vert v-u \vert^2 f \, \mathrm{d}v}_{\Ld^2(0,T; \Ld^2)} \leq \Lambda(T,R).
\end{align*}

The fairly straightforward adaptation of the analysis of Section <ref> is left to the reader.

§ GENERALIZATION TO THE INELASTIC BOLTZMANN CASE

In this section, we consider the case where one takes into account inelastic collisions between particles. This corresponds to the following Vlasov-Boltzmann equation in the coupling (the other equations on $(\varrho,u)$ being unchanged):

\begin{equation}\label{eq:TS-Boltz}
\tag{TS-Coll}
\left\{
\begin{aligned}
\partial_t f +v \cdot \nabla_x f +{\rm div}_v \big[ f (u-v)-f \nabla_x p(\varrho) \big]&=\mathcal{Q}_{\lambda}(f,f),  \\
\partial_t (\alpha \varrho ) + \mathrm{div}_x (\alpha \varrho u)&=0, \\
\partial_t (\alpha \varrho u) + \mathrm{div}_x(\alpha \varrho u \otimes u)+\alpha\nabla_x p -\Delta_x u - \nabla_x \mathrm{div}_x \, u&=j_f-\rho_f u,
\end{aligned}
\right.
\end{equation}

where $\mathcal{Q}_{\lambda}(f,f)$ stands for a quadratic collision operator of Boltzmann-type, in a inelastic hard-spheres regime. Here, the fixed parameter $\lambda \in (0,1)$ corresponds to the so-called restitution coefficient: if $'v$ and $'v_{\star}$ denote the velocities of two particles before collision, their respective velocities  $v$ and $v_{\star}$ after collision are given by

\begin{equation*}
\left\{
\begin{aligned}
v&= {'v}-\frac{1+\lambda}{2}({'u}\cdot n)n, \\
v_{\star}&= {'v_{\star}}+\frac{1+\lambda}{2}({'u}\cdot n)n,
\end{aligned}
\right.
\end{equation*}

where ${'u}:={'v}-{'v_{\star}}$ is the relative pre-collision velocity and $n \in \mathbb{S}^{d-1}$ is a unit vector that points from the particle center with velocity $v$ to the particle center with velocity $v_{\star}$ at the impact. Note that $\lambda=1$ corresponds to the standard elastic case.

In this representation, given two distribution functions $f=f(v)$ and $g=g(v)$, we can consider the following expression for Boltzmann collision operator, as a difference of a gain and loss term

\begin{align*}
\mathcal{Q}_{\lambda}(f,g)=\mathcal{Q}_{\lambda}^+(f,g)-\mathcal{Q}_{\lambda}^-(f,g),
\end{align*}

where, setting $u=v-v_{\star}$ and $\widehat{u}=u/ \vert u \vert$, we define

\begin{align}\label{eq:collisionOp}
\begin{split}
\mathcal{Q}_{\lambda}^+(f,g)(v)&=\frac{1}{\lambda^2}\int_{ \R^d \times  \mathbb{S}^{d-1} }  \vert u \cdot n \vert  b(\widehat{u}\cdot n) f({'v})g({'v_{\star}}) \, \mathrm{d}v_{\star} \mathrm{d}n, \\[1mm]
\mathcal{Q}_{\lambda}^-(f,g)(v)&=f(v)\int_{ \R^d \times  \mathbb{S}^{d-1} }   \vert u \cdot n \vert b(\widehat{u}\cdot n)g(v_{\star})  \, \mathrm{d}v_{\star}  \mathrm{d}n.
\end{split}
\end{align}

Here, the function $b \in \Ld^1([-1,1])$ is a given angular collision kernel. For the sake of simplicity, we consider $b\equiv 1$.  Note that in the (truly) inelastic case $\lambda \in (0,1)$, we have

\begin{align*}
\vert v \vert^2+ \vert v_{\star} \vert^2=\vert {'v} \vert^2+ \vert {'v_{\star}} \vert^2-\frac{1-\lambda^2}{2}({'u}\cdot n)^2,
\end{align*}

thus inducing a loss of kinetic energy at each collision, while mass and momentum are conserved. We refer to [130] (see also the introduction of [5]) for more details on this model coming from the theory of granular media and which describes a cloud of macroscopic particles of size larger than that usually described by the usual Boltzmann equation with elastic collisions (for so-called molecular gases). To include dissipative effects, inelastic collisions are thus considered. Note that the presence of such a collision operator (with a large parameter $\eps^{-1}$ in front of it) formally leads to a biphasic fluid model when starting from (<ref>)) (see [61]). We also refer to [122, 10, 55]
for its use in the study of sprays.

Our main result (which also includes the elastic case $\lambda=1$)  reads as follows.

Let $\lambda \in (0,1]$. There exist $m_0>0$, depending only on the dimension, such that the following holds for all $m \geq m_0$. Let

\begin{align*}
e^{\vert v \vert^2}  f^{\mathrm{in}} \in  \mathcal{H}_{0}^{m}, \ \ \varrho^{\mathrm{in}} \in \mathrm{H}^{m+1}, \ \ u^{\mathrm{in}} \in \mathrm{H}^{m},
\end{align*}

such that $(f^{\mathrm{in}}, \varrho^{\mathrm{in}})$ satisfies the $c$-Penrose stability condition $\eqref{cond:Penrose}_c$ (with $c>0$) and

\begin{align*}
0 &\leq f^{\mathrm{in}} \ \ \ \rho_{f^{\mathrm{in}}}<\Theta<1, \ \ \ 0<\mu \leq \varrho^{\mathrm{in}}, \ \ \ 0<\underline{\theta} \leq (1-\rho_{f^{\mathrm{in}}})\varrho^{\mathrm{in}} \leq \overline{\theta},
\end{align*}

for some fixed constants $\Theta, \mu, \underline{\theta}, \overline{\theta}$.
Then there exist $T>0$ and a solution $(f, \varrho,u)$ to (<ref>) with initial condition $(f^{\mathrm{in}}, \varrho^{\mathrm{in}}, u^{\mathrm{in}})$ such that

\begin{align*}
e^{\vert v \vert^2} f \in  \mathscr{C}\left([0,T];\mathcal{H}^{m-1}_{0} \right), \ \ \varrho \in \Ld^2(0,T;\mathrm{H}^m), \ \ u \in \mathscr{C}\left([0,T]; \mathrm{H}^{m} \right)\cap \Ld^2(0,T;\mathrm{H}^{m+1}),
\end{align*}

and with $(f(t), \varrho(t))$ satisfying the $c/2$-Penrose stability condition $\eqref{cond:Penrose}_{c/2}$ for all $t \in [0,T]$. In addition, this solution is unique in this class.

As seen below, the friction term in the kinetic equation comes in handy in order to treat some of the new terms due to the collision operator. This was remarked already in [109].

Let us present the main changes that must be considered in our strategy of proof and that are due to the collision operator. It mainly concerns:

$-$ the energy estimates for $f$ from Section <ref>;

$-$ the integro-differential system for the derivatives of $f$ from Section <ref>.

The rest of Section <ref> and of Section <ref> then remains unchanged, as well as Sections <ref>–<ref>.

§.§ New energy estimates for the kinetic part

Let us focus on the energy estimates for the new kinetic equation

\begin{align}\label{eq:VlasovB-f}
\partial_t f +v \cdot \nabla_x f + {\rm div}_v [f (u-v)-f \nabla_x p(\varrho)f]&=\mathcal{Q}_{\lambda}(f,f).
\end{align}

Following [109], we first define $g(t,x,v)=e^{\vert v \vert^2} f(t,x,v)$ with $f$ solving (<ref>).

$$E^{u, \varrho}=u-\nabla_x p(\varrho),$$

it implies that $g$ satisfies the following modified Vlasov-Boltzmann equation

\begin{align}\label{eq:modifiedVlasovB-g}
\partial_t g +v \cdot \nabla_x g + {\rm div}_v [(E^{u, \varrho}-v)g]-2v \cdot (E^{u, \varrho}-v)g =\Gamma_{\lambda}[g,g],
\end{align}

where for all functions $h_1(v), h_2(v)$, the operator  $\Gamma_{\lambda}[g,g]=\Gamma_{\lambda}^+[g,g]-\Gamma_{\lambda}^-[g,g]$ is defined via

\begin{align*}
\Gamma_{\lambda}^+[h_1,h_2](v)&:=\frac{1}{\lambda^2}\int_{ \R^d \times  \mathbb{S}^{d-1} } e^{-\vert v_{\star} \vert^2-\frac{1-\lambda^2}{2}(({'v}-{'v_{\star}})\cdot n)^2}  \vert u \cdot n \vert   h_1({'v})h_2({'v_{\star}}) \, \mathrm{d}v_{\star} \mathrm{d}n, \\
\Gamma_{\lambda}^-[h_1,h_2](v)&:=h_2(v)\int_{ \R^d \times  \mathbb{S}^{d-1} }  e^{-\vert v_{\star} \vert^2} \vert u \cdot n \vert h_1(v_{\star})  \, \mathrm{d}v_{\star}  \mathrm{d}n.
\end{align*}

Note that the additional term $2v \cdot (E^{u, \varrho}-v)g$ comes from the friction term in (<ref>). Using

\begin{align*}
{'u}={'v_{\star}}-{'v}=v_{\star}-v-(1+\lambda)({'u} \cdot n) n = v_{\star}-v+(1+\lambda)\lambda(u \cdot n) n,
\end{align*}

we observe that for all $\lambda \in (0,1)$, there exists a constant $c(\lambda)>0$ (and $c(1)=0)$ such that

\begin{align*}
\Gamma_{\lambda}^+[h_1,h_2]&:=\frac{1}{\lambda^2}\int_{ \R^d \times  \mathbb{S}^{d-1} } e^{-\vert v_{\star} \vert^2-c(\lambda)(({v}-{v_{\star}})\cdot n)^2}  \vert (v-v_{\star})\cdot n \vert   h_1({'v})h_2({'v_{\star}}) \, \mathrm{d}v_{\star} \mathrm{d}n.
\end{align*}

The exponential inside the integral is roughly behaving as $e^{-\vert v_{\star} \vert^2-c(\lambda)\vert v-v_{\star} \vert^2}$. We have the following bilinear estimates on the previous collision operators, where some loss of weights in velocity classically shows up.

There exists $s=s(d)>0$ large enough such that for all $\sigma \geq 0$, we have for any smooth nonnegative function $g=g(x,v)$

\begin{align*}
\sum_{\vert \alpha \vert+ \vert \beta \vert\leq s }\int_{\T^d} \int_{\R^d} \langle v \rangle^{2\sigma} \partial_x^{\alpha} \partial_v^{\beta}[\Gamma_{\lambda}(g,g)] \partial_x^{\alpha} \partial_v^{\beta} g \, \mathrm{d}x \, \mathrm{d}v \lesssim \LRVert{g}_{\mathcal{H}^{s}_{\sigma}}^2 \LRVert{g}_{\mathcal{H}^{s}_{\sigma+1}}.
\end{align*}

We refer to <cit.> combined to the (proof of) <cit.>.

The key estimate allowing to recover the previous loss of weight then comes from the following lemma bearing on the extra term $2v \cdot (E^{u, \varrho}-v)g$.

Let $s \geq 0$ and $\sigma >0$. For any smooth nonnegative function $g=g(x,v)$ and $\delta \in (0,1)$, we have

\begin{multline*}
\sum_{\vert \alpha \vert+ \vert \beta \vert\leq s }\int_{\T^d} \int_{\R^d} \langle v \rangle^{2\sigma} \partial_x^{\alpha} \partial_v^{\beta} \left[2v \cdot (E^{u, \varrho}-v)g \right] \partial_x^{\alpha} \partial_v^{\beta} g \, \mathrm{d}x \, \mathrm{d}v \\
\lesssim -(1-\delta) \LRVert{g}_{\mathcal{H}^{s}_{\sigma+1}}^2 + \left( 1+ \frac{1}{\delta} \LRVert{E^{u, \varrho}}_{\H^s}^2\right)\LRVert{g}_{\mathcal{H}^{s}_{\sigma}}^2.
\end{multline*}

We refer to <cit.>.

Let us show how one can now obtain an a priori energy estimate for a solution $g$ to (<ref>), that reads as

\begin{align*}
\mathcal{T}^{u,\varrho}(g)-2v\cdot (E^{u, \varrho}-v)g=\Gamma_{\lambda}(g,g),
\end{align*}

$$\mathcal{T}^{u,\varrho}=\partial_t +v \cdot\nabla_x -v \cdot \nabla_v + E^{u,\varrho}(t,x)\cdot \nabla_v-d \mathrm{Id}.

The result is the following.

For all $r \geq 0$, $m >3+d/2$, $c>0$ and all $T>0$, for all smooth functions $(f,\varrho,u)$  satisfying

\partial_tg +v \cdot\nabla_x g -v \cdot \nabla_v g + E^{u,\varrho}(t,x)\cdot \nabla_v-2v\cdot (E^{u, \varrho}-v)g-d g=\Gamma_{\lambda}(g,g) \quad \text{on} \quad [0,T],

and  $\varrho \geq c$ on $[0,T]$, there holds, for all $t \in [0,T]$

\begin{align}\label{ineqEnergyE(g)}
\begin{split}
\mathcal{E}_{m, \sigma}[g(t)] &\leq  \Vert g(0) \Vert_{\mathcal{H}^{m}_{\sigma}}^2 \exp \Big[C\Big( (1+ \Vert u \Vert_{\Ld^{\infty}(0,T;\H^{m})}+ \LRVert{\mathcal{E}_{m, \sigma}[g]}_{\Ld^{\infty}(0,T)})T \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad+ {\sqrt{T}}  \Lambda\left( \Vert \varrho \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \Big) \Vert \varrho \Vert_{\Ld^2(0,T;\H^{m+1})} \right) \Big],
\end{split}
\end{align}

for some universal constant $C>0$ and where

\begin{align*}
\mathcal{E}_{m, \sigma}[g(t)]:= \Vert g(t) \Vert_{\mathcal{H}^{m}_{\sigma}}^2+ \frac{1}{4}\int_0^t \Vert g(\tau) \Vert_{\mathcal{H}^{m}_{\sigma+1}}^2 \, \mathrm{d}\tau.
\end{align*}

By Lemma <ref>, we first have

\begin{multline*}
\mathcal{T}^{u,\varrho}(\partial_x^{\alpha} \partial_v^{\beta} g)=- \sum_{\substack{i=1\\\beta_i \neq 0}}^d  \left(  \partial_x^{\widehat{\alpha}^i} \partial_v^{\overline{\beta}^i} g-\partial_x^{\alpha} \partial_v^{\beta} g\right)  - \left[\partial_x^{\alpha} \partial_v^{\beta}, E^{u,\varrho}(t,x)\cdot \nabla_v\right]g \\
+\partial_x^{\alpha} \partial_v^{\beta} \left[2v \cdot (E^{u, \varrho}-v)g \right]+\partial_x^{\alpha} \partial_v^{\beta} \left[\Gamma_{\lambda}(f,f) \right].
\end{multline*}

The analogue of inequality (<ref>) from Proposition <ref> is, thanks to Lemmas <ref>–<ref>, for all $\delta \in (0,1)$

\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\Vert g(t) \Vert_{\mathcal{H}^{m}_{\sigma}}^2+ (1-\delta) \LRVert{g(t)}_{\mathcal{H}^{m}_{\sigma+1}}^2  & \lesssim \left(1+\Vert E^{u,\varrho}(t) \Vert_{\H^{m}}+\frac{1}{\delta} \LRVert{E^{u, \varrho}(t)}_{\H^m}^2 \right) \Vert g(t)\Vert_{\mathcal{H}^{m}_{\sigma}}^2+\LRVert{g(t)}_{\mathcal{H}^{m}_{\sigma}}^2 \LRVert{g(t)}_{\mathcal{H}^{m}_{\sigma+1}} \\
& \lesssim \left(1+\Vert E^{u,\varrho}(t) \Vert_{\H^{m}}+\frac{1}{\delta} \LRVert{E^{u, \varrho}(t)}_{\H^m}^2 \right) \Vert g(t)\Vert_{\mathcal{H}^{m}_{\sigma}}^2 \\
& \quad +\frac{2}{1-\delta}\LRVert{g(t)}_{\mathcal{H}^{m}_{\sigma}}^4 + \frac{1-\delta}{2}\LRVert{g(t)}_{\mathcal{H}^{m}_{\sigma+1}}^2,
\end{align*}

therefore after absorbing the last term with $\delta=1/2$, we get

\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \mathcal{E}_{m, \sigma}[g(t)]
%\lesssim \left(1+ \LRVert{E^{u, \varrho}(t)}_{\H^m}^2 \right)\mathcal{E}_{m, \sigma}(t)+\mathcal{E}_{m, \sigma}[g(t)]^2
\lesssim \left(1+ \LRVert{E^{u, \varrho}(t)}_{\H^m}^2 +\mathcal{E}_{m, \sigma}[g(t)] \right)\mathcal{E}_{m, \sigma}[g(t)].
\end{align*}

Concluding as in the proof of Proposition <ref>, we finally obtain the result.

The bootstrap argument from Section <ref> is then performed with the quantity

\begin{align*}
\mathscr{N}_{m,r}(g,\varrho,u,T):=  \LRVert{\mathcal{E}_{m-1, \sigma}[g]}_{\Ld^{\infty}(0,T)} +\Vert \varrho \Vert_{\Ld^2(0,T;\H^{m})}+\Vert u \Vert_{\Ld^{\infty}(0,T; \H^m) \cap \Ld^{2}(0,T;\H^{m+1})},
\end{align*}

instead of $\mathcal{N}_{m,r}(f,\varrho,u,T)$, by considering the same regularization $\mathrm{J}_\eps \nabla_x \varrho$ in the equation (<ref>). Taking into account (<ref>) and the quantity $\mathscr{N}_{m,r}(g,\varrho,u,T)$, the content of Section <ref> can be modified accordingly. Concerning the local in time existence for the regularized system from Section <ref>,  we just modify the kinetic part of the scheme of approximation:

\begin{equation*}
\left\{
\begin{aligned}
&\partial_t g^{n+1} +v \cdot \nabla_x g^{n+1}  +\mathrm{div}_v\left[(E^n(t,x)-v) g^{n+1} \right]-2v \cdot (E^{u, \varrho}_{\mathrm{reg}, \eps}-v)g^{n+1} \\
& \qquad \qquad \qquad \qquad  \qquad \qquad \qquad \qquad  \qquad \qquad  \qquad \qquad   =\Gamma_{\lambda}^+[g^n,g^n]-\Gamma_{\lambda}^-[g^n,g^{n+1}], \\
&{f^{n+1}}_{\mid t=0}=f^{in}.
\end{aligned}
\right.
\end{equation*}

where $\eps>0$ is given. We refer to [109] for more details.

§.§ Equation on the augmented variable $\mathcal{F}$

Let us highlight the main changes that arise in the end of Subsection <ref>, more precisely in Definition <ref>. Recall that the goal is to consider a new unknown $\mathcal{F}=\left(\partial_x^{K} \partial_v^{K}f \right)_{\vert K \vert+\vert L \vert \in \lbrace m-1,m \rbrace }$. Here, one has to consider a new coupling matrix $\mathbb{M}=\left( \mathbb{M}_{(I,J),(K,L)} \right)$ which takes into account the collision operator $\mathcal{Q}$, defined by

\begin{align*}
\mathbb{M}_{(I,J),(K,L)}:=\mathcal{M}_{(I,J),(K,L)}+ \mathcal{M}_{(I,J),(K,L)}^{\mathcal{Q}},
\end{align*}

with $\vert I \vert+\vert J \vert, \vert K \vert+\vert L \vert \in \lbrace m-1,m \rbrace$ and where

$-$ $\mathcal{M}_{(I,J),(K,L)} \in \R$ stands for the former terms of the coupling matrix already appearing in Definition <ref>;

$-$ $\mathcal{M}_{(I,J),(K,L)}^{\mathcal{Q}}$ is an new operator term coming from the collision operator and defined by the relation

\begin{align*}
\partial_x^I \partial_v^J \mathcal{Q}_{\lambda}(f,f)
&= \sum_{\substack{0 \leq \alpha \leq  I \\ 0 \leq \beta\leq  J}} \binom{I}{\alpha} \binom{J}{\beta}  \mathcal{Q}_{\lambda}(\partial_x^{\alpha} \partial_v^{\beta}f,\partial_x^{I-\alpha} \partial_v^{J-\beta} f)\\
&=\sum_{\vert K \vert+\vert L \vert \in \lbrace m-1,m \rbrace} \mathcal{M}_{(I,J),(K,L)}^{\mathcal{Q}}\left[\partial_x^K \partial_v^L f \right],
\end{align*}

that is

\begin{align*}
\mathcal{M}_{(I,J),(K,L)}^{\mathcal{Q}}(\bullet):=\sum_{\substack{0 \leq \alpha \leq I \\ 0 \leq \beta \leq J \\ \vert \alpha \vert+ \vert \beta \vert \leq 1}}\binom{I}{\alpha} \binom{J}{\beta} \mathbf{1}_{\substack{K=I-\alpha \\ L=J-\beta}} \mathcal{Q}_{\lambda} \Big(\partial_x^{\alpha} \partial_v^{\beta}f, \bullet \Big).
\end{align*}

Hence, $\mathcal{M}^{\mathcal{Q}}$ is a matrix with operator coefficients, acting on $\mathcal{F}=\left(\partial_x^{K} \partial_v^{K}f \right)_{\vert K \vert+\vert L \vert \in \lbrace m-1,m \rbrace }$.

Using the same notations as in Section <ref>, we obtain the following equation on $\mathcal{F}=\left(\partial_x^{K} \partial_v^{K}f \right)_{\vert K \vert+\vert L \vert \in \lbrace m-1,m \rbrace }$ (see (<ref>))

\begin{align*}
\mathcal{T}_{\reg,\eps}^{u,\varrho} \mathcal{F}+\mathbb{M} \mathcal{F}+\mathcal{L}=-\mathcal{R}_0-\mathcal{R}_1.
\end{align*}

After the composition by $ (t,x,v) \mapsto (t,\mathrm{X}^{t;0}(x,v),\mathrm{V}^{t;0}(x,v))$ where $\mathrm{Z}=(\X, \V)$ is the solution to (<ref>), there holds the equation

\begin{align*}
\partial_t \widetilde{\mathcal{F}}+\widetilde{\mathbb{M}} \widetilde{\mathcal{F}}+\widetilde{\mathcal{L}}=d\widetilde{\mathcal{F}} -\widetilde{\mathcal{R}}_0-\widetilde{\mathcal{R}}_1,
\end{align*}

with $\widetilde{g}(t,x,v)=g(t,\X^{t;0}(x,v),\V^{t;0}(x,v))$. We can still consider the resolvant associated to the previous operator $\mathbb{M}-d \mathrm{Id}$, that is the solution $s \mapsto \mathfrak{N}^{s,t}(x,v)$ of

\begin{equation*}
\left\{
\begin{aligned}
\partial_s \mathfrak{N}^{s;t}+\left[\mathbb{M} \circ \mathrm{Z}^{s;0}-d \mathrm{Id} \right] \mathfrak{N}^{s;t}&=0,\\
\mathfrak{N}^{t;t}&= \mathrm{Id},
\end{aligned}
\right.
\end{equation*}

whose existence and uniqueness is still provided by the Cauchy-Lipschitz theorem,
Hence, this shows that the contribution of the collision operator $\mathcal{Q}$ can be handled by the modified operator $\mathbb{M}$. The strategy of proof which is performed in the rest of Section <ref> applies mutatis mutandis.

§ GENERALIZATION TO THE DENSITY-DEPENDENT DRAG CASE

In this section, we show how one can deal with the case of density-dependent drag in the force acting on the particles, that is, with the notation of the introduction, when the force in the kinetic equation is

\begin{align*}
\Gamma(t,x,v)= \varrho(t,x)(u(t,x)-v)-\nabla_x [p(\varrho)](t,x).
\end{align*}

This additional factor also appears in the feedback of particles on the fluid, that is in the ource term in the Navier-Stokes equations. We are  thus led to consider the following system:

\begin{equation}\label{eq:DragCase}
\left\{
\begin{aligned}
\partial_t f +v \cdot \nabla_x f   -\nabla_x p \cdot \nabla_v f + {\rm div}_v [f \varrho(u-v)-f\nabla_x p(\varrho)]&=0, \\[2mm]
\partial_t (\alpha \varrho ) + \mathrm{div}_x (\alpha \varrho u)&=0,  \\[2mm]
\partial_t (\alpha \varrho u) + \mathrm{div}_x(\alpha \varrho u \otimes u) +\alpha\nabla_x p -\Delta_x u - \nabla_x \mathrm{div}_x \, u &=\varrho(j_f-\rho_f u), \\[2mm]
\alpha&=1-\rho_f.
\end{aligned}
\right.
\end{equation}

Our main result of local well-posedness is the following.

Consider the same assumptions of Theorem <ref>. Assume also that $f^{\mathrm{in}}$ is compactly supported in velocity. Then the conclusion of Theorem <ref> holds for the density-dependent drag case of System (<ref>).

To fix notations, let us assume that

\begin{align}\label{assump-fcpt-Min}
\mathrm{supp}\, f^{\mathrm{in}} \subset \T^d \times \B(0,M^{\mathrm{in}}), \ \ M^{\mathrm{in}}>0.
\end{align}

In what follows, we shall only present the main modifications which have to be added to the strategy used in this article.

§.§ Modification of the energy estimates and of the bootstrap argument
Our first goal is to adapt the energy estimates of Section <ref> and the bootstrap procedure to the density-dependent drag case. The main change comes from the estimate for $f$ from Proposition <ref>.

Indeed, let us set

$$\mathcal{T}^{u,\varrho}_{\mathrm{drag}}=\partial_t +v \cdot\nabla_x -\varrho v \cdot \nabla_v + E^{u,\varrho}(t,x)\cdot \nabla_v-d\varrho \mathrm{Id},$$

where $E^{u,\varrho}_{\mathrm{drag}}:=\varrho u - \nabla_x p(\varrho) $. Observe that because of the term $\varrho v \cdot \nabla_v$ (coming from friction), a growth in velocity can occur in the analysis, that is if $f$ is controlled in $\mathcal{H}^m_r$, then this term would a priori require a control in  $\mathcal{H}^m_{r+1}$. This explains the additional assumption of compact support in velocity in the next proposition. We mention though that the use of exponential-weighted norms in velocity could relax this assumption (see the work [8] on the Vlasov-Maxwell equations, and also [45] in the context of fluid-kinetic equations).

For all $r \geq 0$, $m >3+d/2$, $c>0$ and all $T>0$, for all smooth functions $(f,\varrho,u)$ with $f$ having a compact support in velocity:

\begin{align*}
\forall t \in [0,T], \ \ \mathrm{supp}\, f(t) \subset \T^d \times \B(0,M(t)),
\end{align*}

for some $M \in \Ld^{\infty}(0,T)$, satisfying

\mathcal{T}_{\mathrm{drag}}^{u,\varrho} (f) =0 \quad \text{on  } [0,T],

and $\varrho \geq c$ on $[0,T]$, the following holds. For all $t \in [0,T]$, we have

\begin{align*}
\Vert f(t) \Vert_{\mathcal{H}^{m}_{r}}^2& \leq \Vert f(0) \Vert_{\mathcal{H}^{m}_{r}}^2 \exp \Big[C(1+\LRVert{M}_{\Ld^{\infty}(0,T)})\Big( T+\sqrt{T}\Vert u \Vert_{\Ld^{\infty}(0,T;\H^{m})}\Vert \varrho \Vert_{\Ld^{2}(0,T;\H^{m})}\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad  + \sqrt{T} \Lambda\left( \Vert \varrho \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \right) \Vert \varrho \Vert_{\Ld^2(0,T;\H^{m+1})} \Big) \Big],
\end{align*}

Let us suppose that there exists $M \in \Ld^{\infty}(0,T)$ sucht that

\begin{align*}
\forall t \in [0,T], \ \  \mathrm{supp}\, f(t) \subset \T^d \times \B(0,M(t)).
\end{align*}

Since $\mathcal{T}_{\mathrm{drag}}^{u,\varrho} (f) =0$, we have by Lemma <ref>

\begin{align*}
\mathcal{T}_{\mathrm{drag}}^{u,\varrho}(\partial_x^{\alpha} \partial_v^{\beta} f)&=-\sum_{\substack{i=1\\\beta_i \neq 0}}^d  \partial_x^{\widehat{\alpha}^i} \partial_v^{\widehat{\beta}^i} f +\left[\partial_x^{\alpha} \partial_v^{\beta}, \varrho(t,x) v \cdot \nabla_v\right]-\left[\partial_x^{\alpha} \partial_v^{\beta},E_{\mathrm{drag}}^{u,\varrho}(t,x)\cdot \nabla_v\right]f +d\left[\partial_x^{\alpha} \partial_v^{\beta}, \varrho \mathrm{Id} \right] f,
\end{align*}

for all $\alpha, \beta \in \N^d$. We now take the scalar product of this equality with $(1+\vert v \vert^2)^{r}\partial_x^{\alpha} \partial_v^{\beta} f$, sum for all $\vert \alpha \vert + \vert \beta \vert \leq m$ and then integrate on $\T^d \times \R^d$. For the left-hand side, we have as in Proposition <ref>

\begin{multline*}
\sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  (1+\vert v \vert^2)^{r}\mathcal{T}_{\mathrm{drag}}^{u,\varrho}(\partial_x^{\alpha} \partial_v^{\beta} f)\partial_x^{\alpha} \partial_v^{\beta} f  \\
%&=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert f(t) \Vert_{\mathcal{H}^{m}_{r}}^2-d \sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  (1+\vert v \vert^2)^{r} \varrho \vert \partial_x^{\alpha} \partial_v^{\beta} f \vert^2\\
%& \qquad \qquad \qquad \qquad+ \sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  (1+\vert v \vert^2)^{r}[v \cdot\nabla_x - \varrho v \cdot \nabla_v + E_{\mathrm{drag}}^{u,\varrho}(t,x)\cdot \nabla_v](\partial_x^{\alpha} \partial_v^{\beta} f)\partial_x^{\alpha} \partial_v^{\beta} f  \\
= \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert f(t) \Vert_{\mathcal{H}^{m}_{r}}^2-\sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  \nabla_v(1+\vert v \vert^2)^{r} \cdot (E_{\mathrm{drag}}^{u,\varrho}-\varrho v) \frac{\vert \partial_x^{\alpha} \partial_v^{\beta} f \vert^2}{2},
\end{multline*}

the last term satisfying

\begin{align*}
\sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  \nabla_v(1+\vert v \vert^2)^{r} \cdot (E_{\mathrm{drag}}^{u,\varrho}-\varrho v) \frac{\vert \partial_x^{\alpha} \partial_v^{\beta} f \vert^2}{2} \leq (\LRVert{\varrho(t)}_{\Ld^{\infty}}+\Vert E_{\mathrm{drag}}^{u,\varrho}(t) \Vert_{\Ld^{\infty}})\Vert f(t)\Vert_{\mathcal{H}^{m}_{r}}^2.
\end{align*}

We now look at the four terms of the right-hand side. For the first, third and fourth ones, we proceed as in the proof of Proposition <ref> (with a variant of (<ref>)) and get

\begin{multline*}
\sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  (1+\vert v \vert^2)^{r} \Bigg(-\sum_{\substack{i=1\\\beta_i \neq 0}}^d   \partial_x^{\widehat{\alpha}^i} \partial_v^{\widehat{\beta}^i} f+\left[\partial_x^{\alpha} \partial_v^{\beta},E_{\mathrm{drag}}^{u,\varrho}(t,x)\cdot \nabla_v+ d\varrho \mathrm{Id} \right]f \Bigg) \partial_x^{\alpha} \partial_v^{\beta} f \\
\lesssim (1+\Vert E_{\mathrm{drag}}^{u,\varrho}(t) \Vert_{\H^{m}}+ \Vert \varrho(t) \Vert_{\H^{m}})\Vert f(t)\Vert_{\mathcal{H}^{m}_{r}}^2.
\end{multline*}

The treatment of the third term requires the use of the compact support in velocity of $f$. Invoking the inequality (<ref>), we have

\begin{align*}
\sum_{\vert \alpha \vert + \vert \beta \vert \leq m}\int_{\T^d \times \R^d}  (1+\vert v \vert^2)^{r}\left[\partial_x^{\alpha} \partial_v^{\beta}, \varrho(t,x) v \cdot \nabla_v\right]\partial_x^{\alpha} \partial_v^{\beta} f    \lesssim  (1+M(t)) \Vert \varrho(t) \Vert_{\H^{m}} \Vert f(t)\Vert_{\mathcal{H}^{m}_{r}}^2.
\end{align*}

All in all, we obtain

\begin{align}\label{ineq:d/dt:fweight-DRAGCASE}
\frac{\mathrm{d}}{\mathrm{d}t}\Vert f(t) \Vert_{\mathcal{H}^{m}_{r}}^2  \lesssim (1+M(t))(1+\Vert \varrho(t) \Vert_{\H^{m}}+\Vert E^{u,\varrho}(t) \Vert_{\H^{m}} ) \Vert f(t)\Vert_{\mathcal{H}^{m}_{r}}^2,
\end{align}

if $m > d/2$. As in the proof of Proposition <ref>, an by Sobolev embedding, we have

\begin{align*}
\Vert  E^{u,\varrho}(t) \Vert_{\H^{m}}  \lesssim \Vert \varrho (t) \Vert_{\H^{m}}\Vert u(t) \Vert_{\H^{m}}
+ \Lambda\left(\Vert \varrho(t) \Vert_{\H^{m-2}}  \right)   \Vert \varrho (t) \Vert_{\H^{m+1}}.
\end{align*}

By integrating in time the inequality (<ref>), we get

\begin{multline*}
\Vert f(t) \Vert_{\mathcal{H}^{m}_{r}}^2 \leq \Vert f(0) \Vert_{\mathcal{H}^{m}_{r}}^2 \\
+ C\int_0^t (1+M(s))\left(1+\Vert \varrho(s) \Vert_{\H^{m}}  \Vert u(s) \Vert_{\H^{m}} +  \Lambda\left(\Vert \varrho \Vert_{\Ld^{\infty}(0,T; \H^{m-2})} \right)\Vert \varrho(s) \Vert_{\H^{m+1}} \right) \Vert f(s)\Vert_{\mathcal{H}^{m}_{r}}^2 \, \mathrm{d}s,
\end{multline*}

for all $t \in [0,T)$ and for some constant $C>0$ independent of $\eps$. Using the Cauchy-Schwarz inequality and the Grönwall's inequality, this implies for all $t \in  [0,T)$

\begin{align*}
\Vert f(t) \Vert_{\mathcal{H}^{m}_{r}}^2& \leq \Vert f(0) \Vert_{\mathcal{H}^{m}_{r}}^2 \exp \Big[C(1+\LRVert{M}_{\Ld^{\infty}(0,T)})\Big( T+\sqrt{T}\Vert u \Vert_{\Ld^{\infty}(0,T;\H^{m})})\Vert \varrho \Vert_{\Ld^{2}(0,T;\H^{m})})\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad  + \sqrt{T} \Lambda\left( \Vert \varrho \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \right) \Vert \varrho \Vert_{\Ld^2(0,T;\H^{m+1})} \Big) \Big],
\end{align*}

and this concludes the proof.

The proof of the other estimates from Section <ref> is mainly unchanged and details are left to the reader. We need to adapt the bootstrap procedure, by taking into account the need of a compact support in velocity.
We thus define the following modified condition.

Let $T>0$. For any nonnegative functions $f(t,x,v)$ and $\varrho(t,x)$ on $[0,T]$, we define the property

\begin{align}\label{def:Bound-HAUTBAS-Compactsupp}
\tag{B$^{\mu, \theta}_{\Theta, M^{\mathrm{in}}}(T)$}
\forall t \in [0,T], \
\left\{
\begin{aligned}
\rho_{f}(t)\leq \frac{\Theta+1}{2}, \ \ \ \frac{\mu}{2} \leq \varrho(t), \ \ \ \frac{\underline{\theta}}{2} \leq (1-\rho_{f}(t))\varrho(t) \leq 2\overline{\theta}, \\[2mm]
\mathrm{supp}\, f(t) \subset \T^d \times \B(0,1+M^{\mathrm{in}}),
\end{aligned}
\right.
\end{align}

where $\Theta, \mu, \underline{\theta}, \overline{\theta}$ are given in the statement of Theorem <ref> and where $M^{\mathrm{in}}$  has been introduced in (<ref>).

If $T^{\ast}_\eps>0$ is the maximal time of existence to the system (<ref>) (with density-dependent drag term), we introduce the following time for all $\eps>0$:

\begin{align}
T_\eps=T_\eps(R):=\sup \big\lbrace T \in [0,T^{\ast}_\eps[, \ \  \mathcal{N}_{m,r}(f_\eps,\varrho_\eps,u_\eps,T) \leq R \ \ \text{and} \ \ \eqref{def:Bound-HAUTBAS-Compactsupp} \ \ \text{holds}  \big\rbrace,
\end{align}

where $R>0$ has to be chosen large enough and independent of $\eps$. Here, the quantity $\mathcal{N}_{m,r}(f_\eps,\varrho_\eps,u_\eps,T)$ is exactly the same as in Definition <ref>.

§.§ Modification in the straightening change of variable
As a matter of fact, the main difference in the analysis appears in the part related to characteristics, namely Section <ref>. Our purpose here is to explain how to modify the arguments of Section <ref> about the  straightening change of variable in velocity, that is Lemma <ref>. It turns out that in the density-dependent drag case, obtaining a suitable diffeomorphism $\psi^{s;t}_x$ is not as straightforward as in Lemma <ref>. Indeed, again because of the term $-\varrho v \cdot \nabla_v f$, there could be a growth in velocity in the dynamics which prevents our proof of Lemma <ref>, which is based on a perturbative approach, to directly hold. This is where the assumption compact support in velocity appears to be crucial; we do not know whether it is possible to replace it here by an exponential moment assumption.

With the notations of Section <ref>, we will actually directly straighten the total kinetic operator

$$\mathcal{T}_{\mathrm{drag},\mathrm{F}}=\partial_t +v \cdot\nabla_x -\varrho (t,x)v \cdot \nabla_v + \mathrm{F}(t,x)\cdot \nabla_v-d \mathrm{Id},$$

into the free-transport operator

$$\mathcal{T}^{\mathrm{free}}=\partial_t +v \cdot\nabla_x.$$

For $(x,v) \in \T^d \times \B(0,1+\M^{\mathrm{in}})$ and $t \in [0,T]$, we consider the solution $s \mapsto (\X_{\mathrm{drag}}^{s;t},\V_{\mathrm{drag}}^{s;t})(x,v)$ to the following system of differential equations

\begin{equation*}
\left\{
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}s} \X_{\mathrm{drag}}^{s;t} &= \V_{\mathrm{drag}}^{s;t}, \ \ \X_{\mathrm{drag}}^{t;t}(x,v)=x, \\
\frac{\mathrm{d}}{\mathrm{d}s} \V_{\mathrm{drag}}^{s;t}  &= -\varrho(s,\X_{\mathrm{drag}}^{s;t})  \V_{\mathrm{drag}}^{s;t} + \mathrm{F}(s,\X_{\mathrm{drag}}^{s;t}), \ \  \V_{\mathrm{drag}}^{t;t}(x,v)=v.\\
\end{aligned}
\right.
\end{equation*}

Let $T>0$ and $k \geq 1$. Let $\mathrm{F} \in  \Ld^2(0,T; \W^{k, \infty}(\T^d)) $ be a vector field such that

\begin{align*}
\Vert \mathrm{F} \Vert_{\Ld^2(0,T; \W^{k, \infty}(\T^d))} \leq \Lambda(T,\mathrm{R}),
\end{align*}

for some $\mathrm{R}>0$.
There exists $\overline{T}(\mathrm{R})>0$ such that for all $x \in \T^d$ and $s,t \in [0, \min(\overline{T}(\mathrm{R}),T)]$, there exists a diffeomorphism $\psi_{s,t}(x, \cdot) : \B(0,1+\M^{\mathrm{in}}) \rightarrow \B(0,1+\M^{\mathrm{in}})$ satisfying for all $v \in \B(0,1+\M^{\mathrm{in}})$

\begin{align*}
\mathrm{X}^{s;t}_{\mathrm{drag}}\left(x,\psi_{s,t}(x, v) \right)=x-(t-s)v,
\end{align*}

which furthermore verifies the estimates

\begin{align*}
\frac{1}{C} \leq \det \left(\mathrm{D}_v  \psi_{s,t}(x,v) \right) &\leq C, \\
\underset{s,t \in [0,T]}{\sup} \, \left\Vert \partial_{x,v}^{\alpha} \left(\psi_{s,t}(x,v)-v \right) \right\Vert_{\Ld^{\infty}(\T^d \times \R^d)} &\leq \varphi(T) \Lambda(T,\mathrm{R}), \ \ \vert \alpha \vert \leq k, \\
\underset{s,t \in [0,T]}{\sup} \, \left\Vert \partial_{x,v}^{\beta} \partial_s \psi_{s,t}(x,v) \right\Vert_{\Ld^{\infty}(\T^d \times \R^d)} &\leq \varphi(T) \Lambda(T,\mathrm{R}), \ \ \vert \beta \vert \leq k-1,
\end{align*}

for some $C>0$ and some nondecreasing continuous function $\varphi : \R^+ \rightarrow \R^+$ vanishing at $0$.

Let us sketch the proof. Dropping the $(x,v)$ dependency in the trajectories, we have

\begin{align*}
\V_{\mathrm{drag}}^{s;t} =  \exp \left(\int_s^t  \varrho(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \right) v -  \int_s^t \exp \left( \int_s^\tau  \varrho(\tau',\X_{\mathrm{drag}}^{\tau';t}) \, \mathrm{d} \tau'  \right) \mathrm{F} (\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau,
\end{align*}

from which we deduce

\begin{align*}
\X_{\mathrm{drag}}^{s,t} &=x - \left(\int_s^t \exp \left(\int_{s'}^t  \varrho(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \right)  \, \mathrm{d} s'\right) v  +  \int_s^t \int_{s'}^t \exp \left( \int_{s'}^\tau  \varrho(\tau',\X_{\mathrm{drag}}^{\tau';t}) \, \mathrm{d} \tau'  \right) \mathrm{F}(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \, \mathrm{d} s' \\
&=x - (t-s)\Bigg[ v  + \left(\frac{1}{t-s}\int_s^t \exp \left(\int_{s'}^t  \varrho(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \right)  \, \mathrm{d} s'- 1\right)  v  \\
&\qquad \qquad \qquad \qquad \qquad \qquad - \frac{1}{t-s}  \int_s^t \int_{s'}^t \exp \left( \int_{s'}^\tau  \varrho(\tau',\X_{\mathrm{drag}}^{\tau';t}) \, \mathrm{d} \tau'  \right)   \mathrm{F} (\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \, \mathrm{d} s'  \Bigg].
\end{align*}

Taking one derivative in velocity, we observe that because of the term

$$\left(\frac{1}{t-s}\int_s^t \exp \left(\int_{s'}^t  \varrho(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \right)  \, \mathrm{d} s'- 1\right)  v,$$

we obtain a term where the derivative is not falling on the $v$ factor. Hence, the latter is a priori unbounded, inducing a potential linear growth in velocity of the derivative. But since we restrict to  $v \in \B(0,1+\M^{\mathrm{in}})$, we can however deduce a rough bound on $\na_v  \X_{\mathrm{drag}}^{s;t} $ by Grönwall's inequality, which takes the form

\begin{equation*}
|\na_v  \X_{\mathrm{drag}}^{s;t} | \lesssim  \exp(T \Lambda(T,R) (|v|+1)) T \Lambda(T,R) \lesssim \exp(T \Lambda(T,R) (2+M^{\mathrm{in}})) T \Lambda(T,R).
\end{equation*}

We can now check that for $T $ small enough, for all $0\leq s,t \leq T$, the map

\begin{multline*}
v \mapsto v  + \left(\frac{1}{t-s}\int_s^t \exp \left(\int_{s'}^t  \varrho(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \right)  \, \mathrm{d} s'- 1\right)  v  \\
- \frac{1}{t-s}  \int_s^t \int_{s'}^t \exp \left( \int_{s'}^\tau  \varrho(\tau',\X_{\mathrm{drag}}^{\tau';t}) \, \mathrm{d} \tau'  \right)   E^{u,\varrho}_{\mathrm{drag}}(\tau,\X_{\mathrm{drag}}^{\tau;t}) \, \mathrm{d} \tau  \, \mathrm{d} s'
\end{multline*}

is a $\mathscr{C}^1$ diffeomorphism from $\B(0,1+\M^{\mathrm{in}})$ onto its image, since it is a small Lipschitz perturbation of the identity map. Details are left to the reader.

As a result, for small times, we can directly come down to the case of free-transport case.

§.§ Modifications in remainder terms and conclusion of the bootstrap

To conclude, let us point out the last main modifications which have to be made to conclude the bootstrap argument.

In Section <ref>, one shall be careful when handling the remainder terms $\mathscr{R}_{I}^{\mathrm{Diff}}, \mathscr{R}_{I,1}^{\mathrm{Duha}}, \mathscr{R}_{I,2}^{\mathrm{Duha}}, \mathscr{R}_I$ in $\Ld^2(0,T;\H^1)$ because they now involve some terms with at most $m+1$ derivatives on $\varrho u$ stemming from the force field  $E^{u,\varrho}_{\mathrm{drag}}$. It was somehow harmless in the linear-drag case because the corresponding terms involved only $u$, which has an additional regularity provided by the Navier-Stokes equation, namely a control in $\Ld^{\infty}(0,T; \H^m) \cap \Ld^2(0,T; \H^{m+1})$ (see the terms $\mathbf{S_2}$, $\mathbf{S_5}$, $\mathbf{S_{12}}$, $\mathbf{S_{16}}$). Since we only have a control of $\varrho$ in $\Ld^2(0,T; \H^{m})$, this requires a additional argument.

The main idea is to rely on the same type of decomposition as in Lemma <ref> combined with the use of the smoothing estimates from Section <ref>. The expression $\partial_x^K(\varrho u)$ ($K \in \N^d$) will involve terms or sum of terms

* of the form

$$\varrho\partial_x^K u \ \ \text{or} \  \ \partial_x^{\overline{K}^i-\beta} u \nabla_x (\partial_x^{\beta} \varrho ) \cdot e_i,$$

with $i=1, \cdots, d$ and $\vert \beta \vert =0, \cdots, \vert \lfloor \frac{\vert K \vert-1}{2} \rfloor$. They are treated using $\Ld^{\infty}$ bounds on $\varrho$ (with Sobolev embeddings) and $\Ld^2$ bounds on $u$;

* of the form

$$\nabla_x (\partial_x^{\beta} \varrho ) \cdot e_i \partial_x^{\overline{K}^i-\beta}u,$$

with $i=1, \cdots, d$ and $\vert \beta \vert = \lfloor \frac{\vert K \vert-1}{2} \rfloor+1, \cdots, \vert K \vert-1$
These terms are adressed thanks to Proposition <ref>. We refer to Section <ref> for the same kind of procedure for the estimates on remainders via the smoothing estimates of Section <ref>.

Let us eventually briefly conclude by showing how the bootstrap procedure ends, when one considers the condition (<ref>). Namely, we have to show how to control the compact support in velocity for $f$, for short times. We rely on the Lagrangian structure of the equation which implies a finite propagation in time of the support of $f$ in velocity, namely

\begin{align*}
\forall t>0, \ \ \mathrm{supp} \, f(t) \subset \T^d \times \V^{t;0}_{\mathrm{drag}}( \mathrm{supp} \, f^{\mathrm{in}}).
\end{align*}

\begin{align*}
\V_{\mathrm{drag}}^{t;0} =  \exp \left(-\int_0^t  \varrho(\tau,\X_{\mathrm{drag}}^{\tau;0}) \, \mathrm{d} \tau  \right) v +  \int_0^t \exp \left(- \int_{\tau}^t  \varrho(\tau',\X_{\mathrm{drag}}^{\tau';0}) \, \mathrm{d} \tau'  \right) E^{u,\varrho}_{\mathrm{drag}}(\tau,\X_{\mathrm{drag}}^{\tau;0}) \, \mathrm{d} \tau,
\end{align*}

the same kind of estimates on the trajectories as those performed in the proof of Lemma <ref> give for all $(x,v) \in  \T^d \times \B_v(0,M^{\mathrm{in}})$

\begin{align*}
\vert \V_{\mathrm{drag}}^{t;0} \vert \leq  \vert v \vert + T \Lambda(T,R)  \leq  M^{\mathrm{in}} + T \Lambda(T,R).
\end{align*}

As a consequence, choosing $ T$ small enough (once $R$ is given) is sufficient to control the size of the support of $f(t)$ for short times.

§ SOME CLASSICAL (PARA-)DIFFERENTIAL INEQUALITIES ON $\T^D$ AND $\T^D \TIMES \R^D$

We recall and state several classic inequalities of (para-)differential type. First, we have the following tame estimate for commutators (see <cit.>).

Let $s \geq 1$. There exists $C_s>0$ such that for any functions $g,E \in \H^s \cap \Ld^{\infty}$, we have

\begin{align*}
\forall \vert \alpha \vert \leq s, \ \ \left\Vert [\partial^{\alpha}_x, E]g \right\Vert_{\Ld^2} \leq C_s \left( \Vert \nabla_x E \Vert_{\Ld^{\infty}} \Vert  g \Vert_{\H^{s-1}} + \Vert E \Vert_{\H^s} \Vert g \Vert_{\Ld^{\infty}} \right).
\end{align*}

The following result is about tame estimates in Sobolev spaces (see e.g. <cit.>).

Let $s >0$. There exists $C_s>0$ such that for all $w_1,w_2 \in \H^s \cap \Ld^{\infty}$, we have

\begin{align*}
\LRVert{w_1 w_2}_{\H^s_x} \leq C_s \left(\LRVert{w_1}_{\Ld^{\infty}}\LRVert{w_2}_{\H^s}+\LRVert{w_1}_{\H^s}\LRVert{w_2}_{\Ld^{\infty}} \right).
\end{align*}

We also recall the following result of Bony about Sobolev continuity of the composition by a smooth function  (see e.g. <cit.> or <cit.> for a more precise version)

Let $I$ be an open interval of $\R$ containing $0$. Let $s>0$ and $\sigma$ be the smallest integer such that $\sigma>s$. Let $F \in \W^{\sigma+1, \infty}(I; \R)$ such that $F(0)=0$. If $w \in \H^s$ has value in $J \Subset I$, then there exists $C_s>0$ such that

\begin{align*}
\Vert F(w) \Vert_{\H^s} \leq C_s(1+ \Vert w \Vert_{\Ld^{\infty}} )^{\sigma} \Vert F' \Vert_{\W^{\sigma, \infty}(I)}  \Vert w \Vert_{\H^s} .
\end{align*}

Note that if $0 \notin I$, and $w \in \H^s$ with value in $J \Subset I$, we can extend $F$ outside $I$ by a smooth extension such that $F(0)=0$, and the previous proposition remains valid. Indeed, by Faà di Bruno's formula, we observe that $\Vert F(w) \Vert_{\H^s}$ only involves $F$ through its derivatives evaluated at $w$.

For all $k \in \N$ and $g: \T^d \rightarrow \R^+$ such that $0<c \leq g \leq C <1$, we have

\begin{align*}
\left\Vert\frac{1}{1-g} \right\Vert_{\H^{k}} \leq 1+\mathrm{C}_{k}\left( \LRVert{g}_{\Ld^{\infty}} \right) \LRVert{g}_{\H^k},
\end{align*}

for some non-decreasing continuous functions $\mathrm{C}_{k} :\R^+ \longrightarrow \R^+$.

We rely on Proposition <ref> by writing

\begin{align*}
\left\Vert\frac{1}{1-g} \right\Vert_{\H^{k}}=\left\Vert F(g)-F(0) \right\Vert_{\H^{k}}+1,
\end{align*}

and this directly concludes the proof.

Let us finally state several product and commutator laws using weighted-Sobolev norms.

Let $s \geq 0$. Consider a smooth nonnegative function $\chi=\chi(v)$ such that $\vert \partial^{\gamma} \chi \vert \lesssim_{\gamma} \chi$ for all $\gamma  \in \N^d$ such that $\vert \gamma \vert \leq s$.

* For any functions $f=f(x,v)$, $g=g(x,v)$ and $k \geq s/2$, we have

\begin{align}\label{Sob:estim1}
\Vert \chi f g \Vert_{\H^k_{x,v}} \lesssim \Vert f \Vert_{W^{k,\infty}_{x,v}}\Vert \chi g \Vert_{\H^s_{x,v}} + \Vert g \Vert_{W^{k,\infty}_{x,v}} \Vert \chi f \Vert_{\H^s_{x,v}}.
\end{align}

* For any functions $a=a(x)$, $F=F(x,v)$ and $s_0>d$, we have

\begin{align}\label{Sob:estim2}
\Vert \chi  aF \Vert_{\H^s_{x,v}} \lesssim \Vert a \Vert_{\H^{s_0}_{x}} \Vert \chi F \Vert_{\H^s_{x,v}} + \Vert a \Vert_{\H^s_{x}} \Vert \chi F \Vert_{\H^s_{x,v}}.
\end{align}

* For any vector field $E=E(x)$ and any function $f=f(x,v)$, there holds for all $s_0>1+d$ and all $\alpha, \beta \in \N^d$ satisfying $\vert \alpha \vert + \vert \beta \vert =s \geq 1$

\begin{align}\label{Sob:estim3}
\left\Vert \chi \left[ \partial_x^{\alpha} \partial_v^{\beta}, E(x) \cdot \nabla_v\right]f \right\Vert_{\Ld^2_{x,v}} \lesssim \Vert E \Vert_{\H^{s_0}_{x}} \Vert \chi f \Vert_{\H^s_{x,v}} + \Vert E \Vert_{\H^s_{x}} \Vert \chi f \Vert_{\H^s_{x,v}}.
\end{align}

* For any functions $a=a(x)$, $f=f(x,v)$ such that $f$ has a compact suport in velocity, there holds for all $s_0>1+d$ and all $\alpha, \beta \in \N^d$ satisfying $\vert \alpha \vert + \vert \beta \vert =s \geq 1$

\begin{align}\label{Sob:estim4}
\left\Vert \chi \left[ \partial_x^{\alpha} \partial_v^{\beta}, a(x)v \cdot \nabla_v\right]f \right\Vert_{\Ld^2_{x,v}} \lesssim (1+M_f) \Vert a \Vert_{\H^{s_0}_{x}} \Vert \chi f \Vert_{\H^s_{x,v}} + \Vert a \Vert_{\H^s_{x}} \Vert \chi f \Vert_{\H^s_{x,v}},
\end{align}

\begin{align*}
\mathrm{supp} \, f \, \subset \T^d \times \B(0,M_f), \ \ \M_f \in (0, + \infty).
\end{align*}

We refer to <cit.> for the proof the (<ref>)–(<ref>)–(<ref>). Let us briefly sketch the proof of (<ref>). By expanding the commutator, we have to estimate terms of the form

\begin{align*}
I_{\gamma, \mu}=\LRVert{\chi \partial_x^{\gamma} a \partial_v^{\mu}(v_i) \partial_{v_i} \partial_x^{\alpha-\gamma} \partial_v^{\beta-\mu} f}_{\Ld^2_{x,v}}
\end{align*}

for some $1 \leq i \leq d$ and with $(\gamma, \mu) \neq (0,0)$, $\gamma \leq \alpha$, $\mu \leq \beta$ and $\vert \mu \vert<2$.  If $\gamma \neq 0$ and $\mu \neq 0$, then $I_{\gamma, \mu}=0$ or $\partial_v^{\mu}(v_i)=1$ and we can conclude as for (<ref>). If $\gamma=0$ and $\mu>0$ then $\partial_v^{\mu}(v_i)=0$ or $1$ and we conclude by Sobolev embedding in $x$ since
$I_{\gamma, \mu} \leq \LRVert{a}_{\Ld^{\infty}_x} \Vert \chi f \Vert_{\H^s_{x,v}}$. Lastly, if $\gamma>0$ and $\mu=0$, we rely on the compact support in velocity for $f$ to get

\begin{align*}
I_{\gamma, \mu} \leq M_f\LRVert{\chi \partial_x^{\gamma} a\partial_{v_i} \partial_x^{\alpha-\gamma} \partial_v^{\beta} f}_{\Ld^2_{x,v}},
\end{align*}

and we end the proof as for (<ref>).

§ LOCAL WELL-POSEDNESS FOR (<REF>): PROOF OF PROPOSITION <REF>

Recall that we want to find a solution $(f_\eps, \varrho_\eps, u_\eps)$ (with $\eps>0$) defined on some interval $[0,T_\eps^*)$ to the system

\begin{equation*}\label{S_eps-primitif}
\tag{$\widetilde{S}_{\eps}$}
\left\{
\begin{aligned}
\partial_t f_\eps +v \cdot \nabla_x f_\eps   -p'(\varrho_\eps)\nabla_x \left[(I-\eps^2 \Delta_x)^{-1}\varrho_\eps \right] \cdot \nabla_v f_\eps + {\rm div}_v [f_\eps (u_\eps-v)]=0,&   \\[2mm]
\partial_t (\alpha_\eps\varrho_{\eps})+\mathrm{div}_x( \alpha_\eps \varrho_\eps u_\eps)=0,&\\[2mm]
\partial_t u_\eps +(u_\eps \cdot \nabla_x) u_\eps +\frac{1}{\varrho_\eps}\nabla_x p(\varrho_\eps) -\frac{1}{\varrho_\eps(1-\rho_{f_\eps})}\Big( \Delta_x  + \nabla_x \mathrm{div}_x \Big)u_\eps=\frac{1}{\varrho_\eps(1-\rho_{f_\eps})}(j_{f_\eps}-\rho_{f_\eps} u_\eps),& \\[2mm]
{f_\eps}_{\mid t=0}=f^{\mathrm{in}}, \ \ {\varrho_\eps}_{\mid t=0}=\varrho^{\mathrm{in}}, \ \ {u_\eps}_{\mid t=0}=u^{\mathrm{in}},
\end{aligned}
\right.
\end{equation*}

\begin{align*}
\rho_{f_\eps}(t,x):= \int_{\R^d} f_\eps(t,x,v) \, \mathrm{d}v, \ \ j_{f_\eps}(t,x):=\int_{\R^d} f_\eps(t,x,v) v \, \mathrm{d}v, \ \  \alpha_\eps(t,x)&=1-\rho_{f_\eps}(t,x).
\end{align*}

To do so, we rely on a standard iterative scheme. We will derive two types of estimates on the inductive solutions to that scheme:

$-$ first, a uniform bound in high regularity which allows to obtain, through a weak compactness argument, a weak converging (sub)sequence in spaces with high regularity. This will be possible if we consider a small enough time of existence.

$-$ next, contraction estimates in low regularity which aim at proving that this sequence of solutions is a Cauchy sequence in spaces with lower regularity, thus strongly converging.

This will be enough in order the pass to the limit in the iterative scheme, obtaining a solution with the aforementioned high order regularity. Note that the first type of estimates is actually required to prove the second one. Uniqueness will classically follow from the same computations leading to the contraction estimates.

We fix $\eps>0$ and we now drop the dependency in $\eps$ (we can take $\eps=1$ for instance): we set

\begin{align*}
f^0=f^{in}, \ \ \varrho^{0}=\varrho^{in}, \ \ u^{0}=u^{in},
\end{align*}

and for all $n \in \N$, a triplet $(f^n,u^n, \varrho^n)$ and a time $T_n>0$ being given with

\begin{align*}
(f^n, \varrho^n,u^n) \in \mathscr{C}(0,T_n; \mathcal{H}^m_r) \times \mathscr{C}(0,T_n; \H^m) \times \mathscr{C}(0,T_n; \H^{m}) \cap \Ld^2(0,T_n; \H^{m+1}) ,
\end{align*}

we consider the system

\begin{equation}\label{S_{n+1}}
\tag{$\widetilde{S}^{n+1}$}
\left\{
\begin{aligned}
\partial_t f^{n+1} +v \cdot \nabla_x f^{n+1} + \mathrm{div}_v\left[f^{n+1}(E^n(t,x)-v) \right]=0,   \\[2mm]
\partial_t \mathfrak{m}^{n+1}+\mathrm{div}_x( \mathfrak{m}^{n+1}u^n)=0,\\[2mm]
\partial_t u^{n+1} -\frac{1}{\mathfrak{m}^n}\Big( \Delta_x  + \nabla_x \mathrm{div}_x \Big)u^{n+1} =-(u^{n}\cdot \nabla_x) u^{n} -\nabla_x \pi(\varrho^n)+ \frac{1}{\mathfrak{m}^n}(j_{f^n}-\rho_{f^n} u^n), \\[2mm]
\varrho^{n+1}= \frac{1}{1-\rho_{f^n}}\mathfrak{m}^{n+1}, \\[2mm]
E^n:=u^n-p'(\varrho^n)\nabla_x \left[(I-\eps^2 \Delta_x)^{-1}\varrho^n \right], \\
{f^{n+1}}_{\mid t=0}=f^{\mathrm{in}}, \ \ {\mathfrak{m}^{n+1}}_{\mid t=0}=\alpha^{\mathrm{in}}\varrho^{\mathrm{in}}, \ \ {u^{n+1}}_{\mid t=0}=u^{\mathrm{in}},
\end{aligned}
\right.
\end{equation}

\begin{align*}
\pi'(x):=\frac{p'(x)}{x}, \ \ \pi(0)=0,
\end{align*}

and where

\begin{align*}
\rho_{f^n}(t,x):= \int_{\R^d} f^n(t,x,v) \, \mathrm{d}v, \ \ j_{f^n}(t,x):=\int_{\R^d} f^n(t,x,v) v \, \mathrm{d}v.
\end{align*}

We also set

\begin{align*}
R_0:=100 \left( \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^{m}_r}+\Vert \mathfrak{m}^{\mathrm{in}} \Vert_{\H^{m}} + \Vert u^{\mathrm{in}} \Vert_{\H^{m}} \right).
\end{align*}

Step 1: construction of $(f^{n+1}, \varrho^{n+1}, u^{n+1})$ and uniform estimates. In what follows, we construct  the next iteration of the scheme which is a solution to (<ref>), thus proving that our inductive scheme is well-defined.  We derive at the same time some uniform estimates in high regularity on the sequence of solutions.

Let $n \in \N$. By induction, we assume that there exists $T>0$ (depending on $\eps$) such that for all $k =0, \cdots ,n$

\begin{align*}
\Vert f^k \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m}_r)} +\Vert \mathfrak{m}^k \Vert_{\Ld^{\infty}(0,T;\H^{m})} + \Vert u^{k} \Vert_{\Ld^{\infty}(0,T; \H^{m}) \cap \Ld^2(0,T; \H^{m+1})} < R_0,
\end{align*}

the functions $f_k$ and $\varrho_k$ are nonnegative, and for $t \in [0,T]$

\begin{align}\label{cond:NonAnnuleLWPn}
\forall t \in [0,T], \ \ \rho_{f_k}(t)\leq \frac{\Theta+1}{2}, \ \ \ \frac{\mu}{2} \leq \varrho_k(t), \ \ \ \frac{\underline{\theta}}{2} \leq \mathfrak{m}_k(t) \leq 2\overline{\theta},
\end{align}

where $\Theta, \mu, \underline{\theta}, \overline{\theta}$ are the constants given in Proposition <ref>. In particular, we have

\begin{align*}
1-\rho_{f_k}(t) \geq \frac{1-\Theta}{2}>0.
\end{align*}

Note also that via Sobolev embedding, Lemma <ref> and Lemma <ref>, we have

\begin{align*}
\Vert \varrho^n \Vert_{\Ld^{\infty}(0,T;\H^{m})} &\leq \left\Vert \frac{1}{1-\rho_{f^{n-1}}} \right\Vert_{\Ld^{\infty}(0,T;\H^{m})} \Vert \mathfrak{m}^n \Vert_{\Ld^{\infty}(0,T;\H^{m})} \\
& \leq \left( 1+ \Lambda(\Vert \rho_{f^{n-1}} \Vert_{\Ld^{\infty}(0,T;\H^{m})} \right) \Vert \rho_{f^{n-1}} \Vert_{\Ld^{\infty}(0,T;\H^{m})} \Vert \mathfrak{m}^n \Vert_{\Ld^{\infty}(0,t;\H^{m})} \\
& \leq \left( 1+ \Lambda(\Vert f^{n-1} \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m}_r)} \right) \Vert f^{n-1} \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m}_r)} \Vert \mathfrak{m}^n \Vert_{\Ld^{\infty}(0,T;\H^{m})} \\
&\leq \Lambda(R_0).
\end{align*}

In what follows, we rely on the a priori estimates of Section <ref>.

$\bullet$ We can obtain a unique nonnegative solution $f^{n+1} \in \mathscr{C}(0,T; \mathcal{H}^m_r)$ to the Vlasov equation by the method of characteristics. Since there is the regularization due to $\mathrm{J}_\eps$, we can invoke Proposition <ref> so that we have for all $t \in [0,T]$

\begin{align*}
\Vert f^{n+1}(t) \Vert_{\mathcal{H}^{m}_{r}} &\leq \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^{m}_{r}} \exp \left[C\left( (1+ \Vert u^n \Vert_{\Ld^{\infty}(0,t;\H^{m})})t+ \frac{t}{\eps} \Lambda\left( \Vert \varrho^n \Vert_{\Ld^{\infty}(0,t;\H^{m-2})} \right) \Vert \varrho^n \Vert_{\Ld^{\infty}(0,t;\H^{m})} \right) \right] \\
& \leq \frac{R_0}{100}\exp \left[C\left( (1+ R_0)t+ \frac{t}{\eps} \Lambda(R_0) \right) \right].
\end{align*}

We then define $T_{[1]}=T_{[1]}(R_0)>0$ with $T_{[1]}(R_0)<T$
(depending on $\eps$) such that

\begin{align*}
\frac{R_0}{100}\exp \left[C\left( (1+ R_0)T_{[1]}+ \frac{\sqrt{T_{[1]}}}{\eps}  \Lambda(R_0) \right) \right]<\frac{R_0}{3}.
\end{align*}

$\bullet$ Since $u^n \in \mathscr{C}(0,T; \H^{m+1})$ is given, we can construct a nonnegative solution $\mathfrak{m}^{n+1}$ to the second equation in (<ref>), relying on standard arguments for continuity equations.

We obtain a unique solution $\mathfrak{m}^{n+1} \in \mathscr{C}^1(0,T; \H^m)$, with for all $t \in [0,T]$

\begin{align*}
\Vert \mathfrak{m}^{n+1}(t) \Vert_{\H^{m}} &\leq e^{\Vert \mathrm{div}_x \, u^n \Vert_{\Ld^{\infty}((0,t) \times \T^d)}t/2} \left(\Vert \mathfrak{m}^{in} \Vert_{\H^{m}} +\int_0^t  \Vert  u^n(\tau) \Vert_{\H^{m+1}} \Vert \mathfrak{m}^{n+1}(\tau) \Vert_{\H^{m}}  \, \mathrm{d}\tau  \right),
\end{align*}

thanks to Proposition <ref>. Assuming that $e^{R_0t/2} \leq 2$, we infer by Grönwall's lemma

\begin{align*}
\Vert \mathfrak{m}^{n+1}(t) \Vert_{\H^{m}} &\leq 2 \Vert \mathfrak{m}^{in} \Vert_{\H^{m}} \exp\left(2\int_0^t  \Vert  u^n(\tau) \Vert_{\H^{m+1}}   \, \mathrm{d}\tau \right) \\
& \leq  2  \Vert \mathfrak{m}^{in} \Vert_{\H^{m}} \exp\left(2 \sqrt{t} \Vert  u^n \Vert_{\Ld^2(0,t;\H^{m+1})} \right),
\end{align*}

therefore for such times $t$, we have

\begin{align*}
\forall t \in [0,T], \ \ \
\Vert \mathfrak{m}^{n+1}(t) \Vert_{\H^{m}} \leq 2 \frac{R_0}{100} e^{2\sqrt{t} R_0}.
\end{align*}

We then define $T_{[2]}=T_{[2]}(R_0)>0$ such that $e^{R_0T_{[2]}/2} \leq 2$ and such that

\begin{align*}
2\frac{R_0}{100} e^{2\sqrt{T_{[2]}} R_0} <\frac{R_0}{3}.
\end{align*}

$\bullet$  We define $u^{n+1}$ as the unique solution of the parabolic equation

\begin{align*}
\partial_t w -\frac{1}{\mathfrak{m}^n}\Big( \Delta_x  + \nabla_x \mathrm{div}_x \Big)w =-(u^{n}\cdot \nabla_x) u^{n} -\nabla_x \pi(\varrho^n)+ \frac{1}{\mathfrak{m}^n}(j_{f^n}-\rho_{f^n} u^n),
\end{align*}

starting from $u^{in}$, which satisfies

\begin{align*}
&\Vert u^{n+1} \Vert_{\Ld^{\infty}(0,t; \H^{m})}+ \Vert u^{n+1} \Vert_{\Ld^2(0,t; \H^{m+1})} \\
&\leq (1+\sqrt{t}\Lambda(t,R_0))\Bigg(\Vert u^{in} \Vert_{\H^m} + \sqrt{t}\Lambda(\Vert \varrho^n \Vert_{\Ld^{\infty}(0,t;\H^{m})}) \Vert \varrho^n \Vert_{\Ld^{\infty}(0,t;\H^{m})}+ \sqrt{t}\left\Vert u^n\right\Vert_{\Ld^{\infty}(0,t;\H^{m})}^2 \\
&  \quad \qquad \qquad \quad \qquad \qquad \quad \qquad \qquad \qquad \qquad\qquad \qquad    +  \sqrt{t} \left\Vert \frac{1}{\mathfrak{m}^n}(j_{f^n}-\rho_{f^n} u^n)\right\Vert_{\Ld^{\infty}(0,t;\H^{m-1})} \Bigg).
\end{align*}

Here, we have applied the same argument as for the proof of Proposition <ref>, making $\Vert \varrho^n \Vert_{\Ld^{\infty}(0,t;\H^{m})}$ appear to get a factor $\sqrt{t}$. The second term in the parenthesis is controlled by $\Lambda(R_0)$ while for the third term, we can use the same of kind of arguments (with Remark <ref>) to get

\begin{align*}
\Big\Vert \frac{1}{\mathfrak{m}^n}(j_{f^n}-\rho_{f^n} u^n & )\Big\Vert_{\Ld^{\infty}(0,t;\H^{m-1})} \\
&\leq \left\Vert \frac{1}{\mathfrak{m}^n}\right\Vert_{\Ld^{\infty}(0,t;\H^{m-1})} \left(\left\Vert j_{f^n}\right\Vert_{\Ld^{\infty}(0,t;\H^{m-1})} + \left\Vert \rho_{f^n}\right\Vert_{\Ld^{\infty}(0,t;\H^{m-1})}\left\Vert u^n\right\Vert_{\Ld^{\infty}(0,t;\H^{m-1})}\right) \\
& \leq \Lambda(R_0).
\end{align*}

We eventually obtain

\begin{align*}
\Vert u^{n+1} \Vert_{\Ld^{\infty}(0,t; \H^{m})}+\Vert u^{n+1} \Vert_{\Ld^2(0,t; \H^{m+1})} &\leq (1+\sqrt{t}\Lambda(t,R_0)) \frac{R_0}{100} + \sqrt{t}\Lambda(R_0),
\end{align*}

We then define $T_{[3]}=T_{[3]}(R_0)>0$ with $T_{[3]}(R_0)<T$
such that

\begin{align*}
\left(1+\sqrt{T_{[3]}}\Lambda(T_{[3]},R_0) \right) \frac{R_0}{100} + \sqrt{T_{[3]}} \Lambda(R_0)<\frac{R_0}{3}.
%, \ \ T_{[3]} <\frac{1}{2M \Lambda \left(R_0 \right)}.
\end{align*}

$\bullet$ Lastly, we can rely on Lemmas <ref>–<ref> to find a time $T_{[4]}=T_{[4]}(R_0)>0$ such that the condition (<ref>) is satisfied for $(f^{n+1}, \mathfrak{m}^{n+1}, u^{n+1})$ on the interval $[0, \min(T_{[4]}, T))$.

All in all, we define

\begin{align*}
T(R_0):=\min \left(T_{[1]}(R_0),T_{[2]}(R_0),T_{[3]}(R_0), T_{[4]}(R_0)\right)<T,
\end{align*}

which may depend on $\eps$ but which is independent of $n$.
An induction procedure based on the three previous estimate shows that one can obtain a time $T^{[\eps]}>0$ and a sequence $(f^n, \mathfrak{m}^n,u^n)_{n \in \N}$ satisfying for all $n \in \N$

\begin{align*}
(f^n, \mathfrak{m}^n,u^n) \in \mathscr{C}(0,T^{[\eps]}; \mathcal{H}^m_r) \times \mathscr{C}(0,T^{[\eps]}; \H^m) \times \mathscr{C}(0,T^{[\eps]}; \H^{m}) \cap \Ld^2(0,T; \H^{m+1}) ,
\end{align*}

with (<ref>) and the uniform estimate

\begin{align}\label{ineq:BoundUnif-n}
\Vert f^n \Vert_{\Ld^{\infty}(0,T^{[\eps]};\mathcal{H}^{m}_r)} +\Vert \mathfrak{m}^n \Vert_{\Ld^{\infty}(0,T^{[\eps]};\H^{m})} + \Vert u^{n} \Vert_{\Ld^{\infty}(0,T^{[\eps]}; \H^{m}) \cap \Ld^2(0,T^{[\eps]}; \H^{m+1})} < R_0.
\end{align}

Step 2: contraction estimates in $\Ld^{\infty}_T \mathcal{H}^0_r \times \Ld^{\infty}_T \Ld^2  \times \Ld^{\infty}_T \Ld^2 \cap \Ld^{2}_T \H^1 $. For $n \in  \N {\setminus} \lbrace 0 \rbrace$, we set

\begin{align*}
g^n:=f^{n+1}-f^n, \ \ \mathfrak{M}^n=\mathfrak{m}^{n+1}-\mathfrak{m}^n, \ \ w^n:=u^{n+1}-u^n,
\end{align*}

which satisfy the system of equations:

\begin{equation}\label{Cauchy-S_{n+1}}
\left\{
\begin{aligned}
\partial_t g^{n} +v \cdot \nabla_x g^{n} +\mathrm{div}_v\left[g^{n}(E^n(t,x)-v) \right]+\left(E^{n}-E^{n-1} \right) \cdot \nabla_v f^n=0,   \\[2mm]
\partial_t \mathfrak{M}^{n}+\mathrm{div}_x( \mathfrak{M}^{n}u^n)=-\mathrm{div}_x( \mathfrak{m}^{n}w^{n-1}),\\[2mm]
\partial_t w^{n} -\frac{1}{\mathfrak{m}^{n+1}}\Big( \Delta_x  + \nabla_x \mathrm{div}_x \Big)w^{n} =\mathfrak{S}^n, \\[2mm]
\varrho^{n+1}:= \frac{1}{1-\rho_{f^n}}\mathfrak{m}^{n+1}, \\[2mm]
E^n:=u^n-p'(\varrho^n)\nabla_x \left[(I-\eps^2 \Delta_x)^{-1}\varrho^n \right], \\
{g^{n}}_{\mid t=0}=0, \ \ {\mathfrak{M}^{n}}_{\mid t=0}=0, \ \ {w^{n}}_{\mid t=0}=0,
\end{aligned}
\right.
\end{equation}

and where

\begin{align*}
\mathfrak{S}^n&:=\left (\frac{1}{\mathfrak{m}^{n+1}}-\frac{1}{\mathfrak{m}^{n}} \right) \Big( \Delta_x  + \nabla_x \mathrm{div}_x \Big)u^n-(w^{n-1}\cdot \nabla_x) u^{n}-(u^{n-1}\cdot \nabla_x) w^{n-1} -\nabla_x \left[\pi(\varrho^{n-1})- \pi(\varrho^{n}) \right]\\
&\quad + \frac{1}{\mathfrak{m}^{n}}(j_{f^{n}}-\rho_{f^{n}} u^{n})-\frac{1}{\mathfrak{m}^{n-1}}(j_{f^{n-1}}-\rho_{f^{n-1}} u^{n-1}).
\end{align*}

Let us derive some $\Ld^2$ estimates on $(g^n, \mathfrak{M}^n, w^n)$. They will be satisfied on $[0,\widetilde{T}]$ for some $\widetilde{T}^{[\eps]}<T^{[\eps]}$.

$\bullet$ We perform $\Ld^2_{x,v}$-weighted estimates in the first equation on $g^n$ and as before (see (<ref>)), we get

\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\Vert g^n(t) \Vert_{\mathcal{H}^{0}_{r}}^2  \lesssim (1+\Vert E^n(t) \Vert_{\H^{m}} ) \Vert g^n(t)\Vert_{\mathcal{H}^{0}_{r}}^2+  \LRVert{E^{n}(t)-E^{n-1}(t)}_{\Ld^2} \LRVert{f^n(t)}_{\mathcal{H}^m_r} \LRVert{g^n(t)}_{\mathcal{H}^0_r}.
\end{align*}

For $t \in (0,T)$, we can infer

\begin{align*}
\Vert g^n(t) \Vert_{\mathcal{H}^{0}_{r}} &\lesssim \int_0^t (1+\Vert E^n(s) \Vert_{\H^{m}} ) \Vert g^n(s)\Vert_{\mathcal{H}^{0}_{r}} \, \mathrm{d}s + \int_0^t\LRVert{E^{n}(s)-E^{n-1}(s)}_{\Ld^2} \LRVert{f^n(s)}_{\mathcal{H}^m_r} \, \mathrm{d}s \\
& \leq \sqrt{t} \Bigg( \left(\sqrt{t}+\LRVert{E^n}_{\Ld^2(0,t; \mathrm{H}^m)} \right)\Vert g^n\Vert_{\Ld^{\infty}(0,t;\mathcal{H}^{0}_{r})}+R_0\LRVert{E^{n}-E^{n-1}}_{\Ld^2(0,t;\Ld^2)} \Bigg) \\
& \lesssim \sqrt{t} \Bigg( \left(\sqrt{t}+\frac{\sqrt{t}}{\eps} R_0\right)\Vert g^n\Vert_{\Ld^{\infty}(0,t;\mathcal{H}^{0}_{r})}+R_0\LRVert{E^{n}-E^{n-1}}_{\Ld^2(0,t;\Ld^2)} \Bigg).
\end{align*}

Choosing $\widetilde{T}^{[\eps]}<T^{[\eps]}$ small enough independent of $n$, we can absorb the first term in the parenthesis in the left-hand side so that for all $t \in (0,\widetilde{T}^{[\eps]})$

\begin{align*}
\Vert g^n(t) \Vert_{\mathcal{H}^{0}_{r}} &\lesssim \sqrt{t} R_0\LRVert{E^{n}-E^{n-1}}_{\Ld^2(0,t;\Ld^2)}.
\end{align*}

Then observe that by Sobolev embedding and Proposition <ref>, we have

\begin{align*}
\Vert E^{n}-E^{n-1} \Vert_{\Ld^2(0,t;\Ld^2)} &\leq \Vert u^n-u^{n-1} \Vert_{\Ld^2(0,t;\Ld^2)}+ \left\Vert p'(\varrho^n)\nabla_x \left[\mathrm{J}_{\eps}\varrho^n \right]-p'(\varrho^{n-1})\nabla_x \left[\mathrm{J}_{\eps}\varrho^{n-1} \right] \right\Vert_{\Ld^2(0,t;\Ld^2)} \\
& \leq \Vert u^n-u^{n-1} \Vert_{\Ld^2(0,t;\Ld^2)}+ \LRVert{p'(\varrho^n)}_{\Ld^{\infty}(0,t; \Ld^{\infty})} \left\Vert \nabla_x \left[\mathrm{J}_{\eps}(\varrho^n -\varrho^{n-1} \right] \right\Vert_{\Ld^2(0,t \Ld^2)} \\
& \quad + \Vert p'(\varrho^{n})-p'(\varrho^{n-1}) \Vert_{\Ld^{2}(0,t; \Ld^{2})} \left\Vert \nabla_x \left[\mathrm{J}_{\eps}\varrho^{n-1} \right] \right\Vert_{\Ld^{\infty}(0,t;\Ld^{\infty})} \\
& \lesssim \Vert u^n-u^{n-1} \Vert_{\Ld^2(0,t;\Ld^2)}+ \Lambda(R_0) \frac{1}{\eps} \left\Vert \varrho^n -\varrho^{n-1} \right\Vert_{\Ld^2(0,t \Ld^2)} \\
& \quad + \Lambda(t,R_0) \frac{1}{\eps} \Vert p'(\varrho^{n})-p'(\varrho^{n-1}) \Vert_{\Ld^{2}(0,t; \Ld^{2})} \\
& \lesssim \Vert u^n-u^{n-1} \Vert_{\Ld^2(0,t;\Ld^2)}+ \Lambda(t,R_0) \frac{1}{\eps} \Vert \mathfrak{m}^{n}-\mathfrak{m}^{n-1} \Vert_{\Ld^{2}(0,t; \Ld^{2})}.
\end{align*}

This yields for all $t \in (0,\widetilde{T}^{[\eps]})$

\begin{align*}
\Vert g^n \Vert_{\Ld^{\infty}(0,t;\mathcal{H}^{0}_{r})}  \lesssim \sqrt{t} \left( R_0\Vert w^{n-1} \Vert_{\Ld^2(0,t;\Ld^2)}+ \Lambda(t,R_0) \frac{1}{\eps} \Vert \mathfrak{M}^{n-1} \Vert_{\Ld^{2}(0,t; \Ld^{2})} \right).
\end{align*}

$\bullet$  An $\Ld^2$ energy estimate on the second equation on $\mathfrak{M}^{n}$ also leads to

\begin{align*}
\LRVert{\mathfrak{M}^{n}(t)}_{\Ld^{\infty}(0,t;\Ld^2)} &\leq e^{\LRVert{\mathrm{div}_x \, u_n}_{\Ld^{\infty}(0,t; \Ld^{\infty})}t}\int_0^t \LRVert{\mathrm{div}_x \, \mathfrak{m}^n w^{n-1}(\tau)}_{\Ld^2} \, \mathrm{d}\tau \\
& \leq e^{R_0t} \left( \int_0^t \LRVert{w^{n-1}\cdot \nabla_x \mathfrak{m}^n (\tau)}_{\Ld^2} \, \mathrm{d}\tau  + \int_0^t \LRVert{\mathfrak{m}^n\mathrm{div}_x \,  w^{n-1}(\tau)}_{\Ld^2} \, \mathrm{d}\tau  \right) \\
& \leq \sqrt{t} e^{R_0t}  \left( \LRVert{\nabla_x \mathfrak{m}^n}_{\Ld^{\infty}(0,t; \Ld^{\infty})} \LRVert{w^{n-1}}_{\Ld^2(0,t; \Ld^2)} + \LRVert{ \mathfrak{m}^n}_{\Ld^{\infty}(0,t; \Ld^{\infty})} \LRVert{\mathrm{div}_x w^{n-1}}_{\Ld^2(0,t; \Ld^2)} \right) \\
& \lesssim \sqrt{t}\Lambda(t,R_0) \LRVert{w^{n-1}}_{\Ld^2(0,t; \H^1)},
\end{align*}

for all $t \in (0,\widetilde{T}^{[\eps]})$.

$\bullet$ Performing an $\Ld^{\infty}_T \Ld^2 \cap \Ld^{2}_T \H^1 $ estimate by multiplying by $w_n$ in the parabolic equation satisfied by $w^n$ yields

\begin{multline*}
\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t}\Vert w^n \Vert_{\Ld^2}^2+\int_{\T^d } \frac{1}{\mathfrak{m}^{n+1}} \left( \vert \nabla_x w^{n} \vert^2 + \vert \mathrm{div}_x w^n \vert^2\right) \\
=\sum_{k=1}^5\int_{\T^d} \mathfrak{S}^n_k \cdot w^n - \sum_{i=1}^d \left\lbrace\int_{\T^d } \nabla_x w^n_i \cdot \nabla_x \left(\frac{1}{\mathfrak{m}^{n+1}} \right) w^n_i -\int_{\T^d } \mathrm{div}_x \, w^{n} \partial_i \left(\frac{1}{\mathfrak{m}^{n+1}} \right) w^n_i \right\rbrace ,
\end{multline*}

\begin{align*}
\mathfrak{S}^n_1:=\left (\frac{1}{\mathfrak{m}^{n+1}}-\frac{1}{\mathfrak{m}^{n}} \right)\Big( \Delta_x  + \nabla_x \mathrm{div}_x \Big)u^n, \ \ \mathfrak{S}^n_2:=-(w^{n-1}\cdot \nabla_x) u^{n}, \ \ \mathfrak{S}^n_3:=-(u^{n-1}\cdot \nabla_x) w^{n-1}, \\
\mathfrak{S}^n_4:=-\nabla_x \left[\pi(\varrho^{n-1})- \pi(\varrho^{n}) \right], \ \ \mathfrak{S}^n_5:=\frac{1}{\mathfrak{m}^{n}}(j_{f^{n}}-\rho_{f^{n}} u^{n})-\frac{1}{\mathfrak{m}^{n-1}}(j_{f^{n-1}}-\rho_{f^{n-1}} u^{n-1}).
\end{align*}

For all $\eta >0$, we obtain by Young's inequality

\begin{multline*}
\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t}\Vert w^n \Vert_{\Ld^2}^2+\int_{\T^d } \left(\frac{1}{\mathfrak{m}^{n+1}}-\eta\LRVert{\nabla_x\left(\frac{1}{\mathfrak{m}^{n+1}} \right) }_{\Ld^{\infty}}^2 \right) \left( \vert \nabla_x w^{n} \vert^2 + \vert \mathrm{div}_x w^n \vert^2\right) \\
\leq \sum_{k=1}^5\int_{\T^d} \mathfrak{S}^n_k \cdot w^n +\frac{1}{\eta} \int_{\T^d}  \vert  w^{n} \vert^2.
\end{multline*}

Let us focus on the source terms $\mathfrak{S}^n_k$. We have by Sobolev embedding

\begin{align*}
\int_{\T^d} \mathfrak{S}^n_1 \cdot w^n &\lesssim \LRVert{( \Delta_x  + \nabla_x \mathrm{div}_x )u^n}_{\Ld^{\infty}}^2 \LRVert{\frac{1}{\mathfrak{m}^{n+1}}-\frac{1}{\mathfrak{m}^{n}}}_{\Ld^{2}}^2 + \int_{\T^d}  \vert  w^{n} \vert^2 \\
&\lesssim R_0^2 \LRVert{\frac{1}{\mathfrak{m}^{n+1} \mathfrak{m}^n}}_{\Ld^{\infty}}^2  \LRVert{\mathfrak{m}^{n+1}-\mathfrak{m}^{n}}_{\Ld^{2}}^2 + \int_{\T^d}  \vert  w^{n} \vert^2 \\
&\lesssim \Lambda(R_0)  \LRVert{\mathfrak{M}^{n}}_{\Ld^{2}}^2 + \LRVert{w^{n}}_{\Ld^{2}}^2.
\end{align*}

We next have

\begin{align*}
\int_{\T^d} \mathfrak{S}^n_2 \cdot w^n  \lesssim  \int_{\T^d}  \vert  \mathfrak{S}^n_2 \vert^2  + \int_{\T^d}  \vert  w^{n} \vert^2 & \leq \LRVert{\nabla_x u^{n-1}}_{\Ld^{\infty}}^2 \LRVert{w^{n-1}}_{\Ld^{2}}^2 + \int_{\T^d}  \vert  w^{n} \vert^2 \\
& \leq R_0^2\LRVert{w^{n-1}}_{\Ld^{2}}^2 + \LRVert{w^{n}}_{\Ld^{2}}^2,
\end{align*}

again by Sobolev embedding while for all $\delta_1>0$, we have by integration by parts and Sobolev embedding

\begin{align*}
\int_{\T^d} \mathfrak{S}^n_3 \cdot w^n &\lesssim \LRVert{\nabla_x u^{n-1}}_{\Ld^{\infty}}^2 \LRVert{w^{n-1}}_{\Ld^{2}}^2 + \LRVert{w^{n}}_{\Ld^{2}}^2 +\frac{1}{\delta_1}\LRVert{ u^{n-1}}_{\Ld^{\infty}}^2 \LRVert{ w^{n-1}}_{\Ld^{2}}^2 + \delta_1 \int_{\T^d}  \vert  \nabla_x w^{n} \vert^2 \\
& \lesssim \Lambda_{\delta_1}(R_0)\LRVert{ w^{n-1}}_{\Ld^{2}}+\LRVert{w^{n}}_{\Ld^{2}}^2+ \delta_1 \int_{\T^d}  \vert  \nabla_x w^{n} \vert^2.
\end{align*}

We also have for all $\delta_2>0$

\begin{align*}
\int_{\T^d} \mathfrak{S}^n_4 \cdot w^n &\lesssim \frac{1}{\delta_2} \int_{\T^d}  \vert  \left[\pi(\varrho^{n-1})- \pi(\varrho^{n}) \right] \vert^2  + \delta_2 \int_{\T^d}  \vert  \mathrm{div}_x \, w^{n} \vert^2  \\
& \lesssim \Lambda_{\delta_2}(R_0) \LRVert{\varrho^{n-1}-\varrho^{n}}_{\Ld^2}^2  + \delta_2 \int_{\T^d}  \vert  \mathrm{div}_x \, w^{n} \vert^2.
\end{align*}

For $\mathfrak{S}^n_5$, we write

\begin{multline*}
\mathfrak{S}^n_5
%=\frac{1}{\mathfrak{m}^{n}}(j_{f^{n}}-\rho_{f^{n}} u^{n})-\frac{1}{\mathfrak{m}^{n-1}}(j_{f^{n-1}}-\rho_{f^{n-1}} u^{n-1})\\
=\left(\frac{1}{\mathfrak{m}^{n}}-\frac{1}{\mathfrak{m}^{n-1}} \right)(j_{f^{n}}-\rho_{f^{n}} u^{n})+\frac{1}{\mathfrak{m}^{n-1}}(j_{f^n}-j_{f^{n-1}})\\
+\frac{1}{\mathfrak{m}^{n}}\left(\rho_{f^{n-1}}-\rho_{f^n} \right)u^{n-1}+\frac{1}{\mathfrak{m}^{n-1}}\rho_{f^n}(u^{n-1}-u^n),
\end{multline*}

therefore by Sobolev embedding and Lemma <ref>

\begin{align*}
\int_{\T^d} \mathfrak{S}^n_5 \cdot w^n & \lesssim   \LRVert{j_{f^{n}}-\rho_{f^{n}} u^{n}}_{\Ld^{\infty}}^2\LRVert{\frac{1}{\mathfrak{m}^{n}}-\frac{1}{\mathfrak{m}^{n-1}}}_{\Ld^{2}}^2+\LRVert{\frac{1}{\mathfrak{m}^{n}}}_{\Ld^{\infty}}^2 \ \LRVert{j_{f^{n}}-j_{f^{n-1}}}_{\Ld^2}^2\\
&\quad +\LRVert{\frac{1}{\mathfrak{m}^{n}}}_{\Ld^{\infty}}^2\LRVert{u^{n-1}}_{\Ld^{\infty}}^2\LRVert{\rho_{f^{n}}-\rho_{f^{n-1}}}_{\Ld^2}^2+ \LRVert{\frac{1}{\mathfrak{m}^{n-1}}}_{\Ld^{\infty}}^2 \LRVert{\rho_{f^n}}_{\Ld^{\infty}}^2 \LRVert{u^{n-1}-u^n}_{\Ld^2}^2 \\
& \quad +\int_{\T^d}  \vert   w^{n} \vert^2 \\
& \lesssim \Lambda(R_0) \left(\LRVert{\mathfrak{m}^n-\mathfrak{m}^{n-1}}_{\Ld^2}^2+ \LRVert{f^n-f^{n-1}}_{\mathcal{H}^0_r}^2+ \LRVert{u^{n-1}-u^n}_{\Ld^2}^2 \right)+ \LRVert{w^{n}}_{\Ld^{2}}^2.
\end{align*}

We finally obtain

\begin{align}\label{ineq:dtw_n-uniquenessLWP}
\begin{split}
&\frac{\mathrm{d}}{\mathrm{d}t} \Vert w^n(t) \Vert_{\Ld^2}^2+\int_{\T^d } \left(\frac{1}{\mathfrak{m}^{n+1}}-\eta R_0^2-\delta_1 -\delta_2 \right) \left( \vert \nabla_x w^{n} \vert^2 + \vert \mathrm{div}_x w^n \vert^2\right) \\
&\leq \Lambda_{\delta_{1,2}, \eta}(R_0) \left( \LRVert{\mathfrak{M}^{n-1}(t)}_{\Ld^2}^2+ \LRVert{g^{n-1}(t)}_{\mathcal{H}^0_r}^2+ \LRVert{w^{n-1}(t)}_{\Ld^2}^2\right)+ \Lambda(R_0)\LRVert{\mathfrak{M}^{n}(t)}_{\Ld^{2}}^2 \\
& \quad + \Lambda_{\eta} \Vert w^n(t) \Vert_{\Ld^2}^2 .
\end{split}
\end{align}

for some constant $\Lambda_{\eta} >0$. A good choice of $\eta, \delta_1$ and $\delta_2$ with respect to $\inf \frac{1}{\mathfrak{m}^{n+1}}$ combined with Grönwall's lemma shows that

\begin{align*}
\Vert w^n(t) \Vert_{\Ld^2}^2 \lesssim  e^{\Lambda_{\eta} t}\int_0^t \left( \LRVert{\mathfrak{M}^{n-1}(\tau)}_{\Ld^2}^2+ \LRVert{g^{n-1}(\tau)}_{\mathcal{H}^0_r}^2+ \LRVert{w^{n-1}(\tau)}_{\Ld^2}^2+\LRVert{\mathfrak{M}^{n}(\tau)}_{\Ld^{2}}^2 \right) \, \mathrm{d}\tau.
\end{align*}

Reducing $\widetilde{T}^{[\eps]}$ such that $\widetilde{T}^{[\eps]} \leq 1/\Lambda_{\eta}$, we can integrate in time the previous differential inequality (<ref>) to obtain for all $t \in (0,\widetilde{T}^{[\eps]})$

\begin{multline*}
\Vert w^n(t) \Vert_{\Ld^{\infty}(0,t; \Ld^2)}^2+\Vert w^n(t) \Vert_{\Ld^{2}(0,t; \H^1)}^2 \\ \lesssim \Lambda(R_0)t\left( \LRVert{\mathfrak{M}^{n-1}}_{\Ld^{\infty}(0,t;\Ld^2)}^2+ \LRVert{g^{n-1}}_{\Ld^{\infty}(0,t;\mathcal{H}^0_r)}^2+ \LRVert{w^{n-1}}_{\Ld^{\infty}(0,t;\Ld^2)}^2\right)+ \Lambda(R_0)t\LRVert{\mathfrak{M}^{n}}_{\Ld^{\infty}(0,t;\Ld^{2})}^2.
\end{multline*}

Combining the three previous points, we reach the following inequality valid for all $t \in (0,\widetilde{T}^{[\eps]})$

\begin{align*}
&\Vert g^n(t) \Vert_{\Ld^{\infty}(0,t;\mathcal{H}^{0}_{r})}+\LRVert{\mathfrak{M}^{n}(t)}_{\Ld^{\infty}(0,t;\Ld^2)}+\Vert w^n(t) \Vert_{\Ld^{\infty}(0,t; \Ld^2)\cap \Ld^{2}(0,t; \H^1)} \\
& \lesssim \sqrt{t} \left( R_0 \sqrt{t}\Vert w^{n-1} \Vert_{\Ld^{\infty}(0,t;\Ld^2)}+ \Lambda(t,R_0) \frac{1}{\eps} \Vert \mathfrak{M}^{n-1} \Vert_{\Ld^{2}(0,t; \Ld^{2})} \right)+\sqrt{t}\Lambda(t,R_0) \LRVert{w^{n-1}}_{\Ld^2(0,t; \H^1)}\\
& \quad + \Lambda(R_0)t\left( \LRVert{\mathfrak{M}^{n-1}}_{\Ld^{\infty}(0,t;\Ld^2)}+ \LRVert{g^{n-1}}_{\Ld^{\infty}(0,t;\mathcal{H}^0_r)}+ \LRVert{w^{n-1}}_{\Ld^{\infty}(0,t;\Ld^2)}\right)+ \Lambda(R_0)t\LRVert{\mathfrak{M}^{n}}_{\Ld^{\infty}(0,t;\Ld^{2})}.
\end{align*}

Reducing again $\widetilde{T}^{[\eps]}$, it means that that for all $n \in  \N {\setminus} \lbrace 0 \rbrace$

\begin{multline}\label{ineq:Contract-n-1/2}
\Vert g^n \Vert_{\Ld^{\infty}(0,\widetilde{T}^{[\eps]};\mathcal{H}^{0}_{r})}+\LRVert{\mathfrak{M}^{n}}_{\Ld^{\infty}(0,\widetilde{T}^{[\eps]};\Ld^2)}+\Vert w^n \Vert_{\Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \Ld^2)\cap \Ld^{2}(0,\widetilde{T}^{[\eps]}; \H^1)} \\  \leq \frac{1}{2} \left( \Vert g^{n-1} \Vert_{\Ld^{\infty}(0,\widetilde{T}^{[\eps]};\mathcal{H}^{0}_{r})}+\LRVert{\mathfrak{M}^{n-1}}_{\Ld^{\infty}(0,\widetilde{T}^{[\eps]};\Ld^2)}+\Vert w^{n-1} \Vert_{\Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \Ld^2)\cap \Ld^{2}(0,\widetilde{T}^{[\eps]}; \H^1)}  \right).
\end{multline}

Note that $\widetilde{T}^{[\eps]}$ has been chosen independent of $n \in  \N {\setminus} \lbrace 0 \rbrace$.

Step 3: obtaining a unique solution to $\eqref{S_eps}$. Combining the uniform bound (<ref>) with the contraction estimate (<ref>), we deduce that the sequence $(f_n, \mathfrak{m}_n, u_n)$ is weakly$(-\star)$ compact in the space
$$\Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \mathcal{H}^m_r) \times \Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \H^m) \times \Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \H^{m}) \cap \Ld^2(0,\widetilde{T}^{[\eps]}; \H^{m+1}), $$
and is a Cauchy sequence in the space
$$\Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \mathcal{H}^0_r) \times \Ld^{\infty}(0,\widetilde{T}^{[\eps]}; \Ld^2)  \times \Ld^{\infty}(0,\widetilde{T}^{[\eps]};\Ld^2) \cap \Ld^{2}(0,\widetilde{T}^{[\eps]};\H^1) .$$

This shows that the whole sequence converges weakly$(-\star)$ in the first space. The limit $(f, m,u)$ belongs in particular to the first space with high regularity. Using weak-strong convergence principles then allows to prove that the limit is a solution to the system $\eqref{S_eps}$ on $(0,\widetilde{T}^{[\eps]})$ in the sense of distribution. We don't detail this part of the proof. Using the equation and the time derivative of the solution, one can show that the solution actually belongs to

$$\mathrm{X}_{\widetilde{T}^{[\eps]}}^m:=\mathscr{C}(0,\widetilde{T}^{[\eps]}; \mathcal{H}^m_r) \times \mathscr{C}(0,\widetilde{T}^{[\eps]}; \H^m) \times \mathscr{C}(0,\widetilde{T}^{[\eps]}; \H^{m}) \cap \Ld^2(0,\widetilde{T}^{[\eps]}; \H^{m+1}).$$

The uniqueness of the Cauchy problem for $\eqref{S_eps}$ in the former space is eventually obtained by mimicking the contraction estimates of the Step 2. In fact, if $(f_1, \mathfrak{m}_1, u_1)$ and $(f_2, \mathfrak{m}_2, u_2)$ are two solutions in $\mathrm{X}_{\widetilde{T}^{[\eps]}}^m$ starting at the same initial condition, performing the same computations prove that there exists $\overline{T}^{[\eps]}$ (depending on $\Vert f_{1,2}, \mathfrak{m}_{1,2}, u_{1,2} \Vert_{\mathrm{X}_{\widetilde{T}^{[\eps]}}^m}$) and with $\overline{T}^{[\eps]}<\widetilde{T}^{[\eps]}$ such that for all $t \in [0,\overline{T}^{[\eps]}]$, we have

\begin{multline*}
\Vert (f_1-f_2)(t) \Vert_{\mathcal{H}^{0}_{r}}^2+\LRVert{(\mathfrak{m}_1-\mathfrak{m}_2)(t)}_{\Ld^2}^2+\Vert (u_1-u_2)(t) \Vert_{\Ld^2}^2 \\
\leq \frac{1}{2}\left( \Vert (f_1-f_2)(t) \Vert_{\mathcal{H}^{0}_{r}}^2+\LRVert{(\mathfrak{m}_1-\mathfrak{m}_2)(t)}_{\Ld^2}^2+\Vert (u_1-u_2)(t) \Vert_{\Ld^2}^2 \right),
\end{multline*}

from which we directly infer $(f_1, \mathfrak{m}_1, u_1)=(f_2, \mathfrak{m}_2, u_2)$ on $[0, \overline{T}^{[\eps]}]$. Repeating this procedure starting from $\overline{T}^{[\eps]}<T_\eps'$, we obtain $(f_1, \mathfrak{m}_1, u_1)=(f_2, \mathfrak{m}_2, u_2)$ on $[0, 2\overline{T}^{[\eps]}]$. After a finite number of steps, we eventually obtain uniqueness on $[0, \widetilde{T}^{[\eps]}]$.

§ PSEUDODIFFERENTIAL CALCULUS WITH A LARGE PARAMETER ON $\R \TIMES \T^D$

In this section, we collect several results on pseudodifferential calculus that we shall need in this article. We refer to [2, 111] for a more general and complete approach. Here, our framework is adapted to the physical space $\R \times \T^d$.

For any symbol of the form $a(t,x,\gamma, \tau,k)$ defined on $\R \times  \T^d \times (0, + \infty) \times \R \times \R^d {\setminus} \lbrace 0 \rbrace$, we use the quantization

\begin{align*}
\mathrm{Op}^{\gamma}(a)(h)(t,x)=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}e^{i(\tau t+ k  \cdot x)}a(t,x,\gamma,\tau,k) \mathcal{F}_{t,x}h(\tau,k) \, \mathrm{d}\tau \, \mathrm{d}k.
\end{align*}

Here, we use the discrete measure on $\Z^d$.

Above, the variable $\gamma>0$ should be seen as a parameter. The Fourier transform of any function $h(t,x)$ defined on $\R \times \T^d$ is denoted by

\begin{align*}
\mathcal{F}_{t,x}h(\tau,k)&=\int_{\R \times \Z^d} e^{-i (\tau t+k  \cdot x)} h(t,x)   \, \mathrm{d}t \, \mathrm{d}x,
\end{align*}

while the Fourier transform of any symbol $a(t,x,\gamma, \tau,k)$

defined on $\R \times  \T^d \times (0, +\infty) \times \R \times \R^d {\setminus} \lbrace 0 \rbrace$

is written as

\begin{align*}
\mathcal{F}_{t,x} a(\theta, \ell, \gamma, \tau, k)&=\int_{\R \times \Z^d} e^{-i (\theta t+\ell  \cdot x)} a(t,x,\gamma, \tau,k)   \, \mathrm{d}t \, \mathrm{d}x.
%\mathcal{F}_{x} a(\ell, \gamma, \tau, k)&=\int_{\Z^d} e^{-i\ell  \cdot x} a(x,\gamma, \tau,k)  \, \mathrm{d}x.
\end{align*}

For the sake of readibility, we introduce the notation

\begin{align*}
\eta&:=(\gamma, \tau,k) \in (0, + \infty) \times \R \times \R^d {\setminus} {\lbrace 0 \rbrace}, \\
\vert \eta \vert&:=\left(\gamma^2+\tau^2 + \vert k \vert^2 \right)^{\frac{1}{2}},
\end{align*}
and we will use the notation $\Ld^{\infty}_{t,x, \eta}$ to denote $\Ld^{\infty}(\R \times \T^d \times  (0, + \infty) \times \R \times \R^d {\setminus} {\lbrace 0 \rbrace})$.

We state $\Ld^2$ continuity results of Calderón-Vaillancourt-type: namely, we shall ask for $\Ld^{\infty}$ bounds in all the variables for the symbols of the operators (see [49, 97]). We first introduce the following family of seminorms for our symbols.

For any $\mathrm{M} \geq 0$ and for any symbol $a(t,x,\eta)$ with $\eta=(\gamma, \tau, k)$, we set

\begin{align}
\label{seminorm0}
%\Vert a \Vert_{0,1,d}
\omega[a]&:= \underset{\substack{\alpha \in \N^d \\ \alpha_i \in \lbrace 0,1 \rbrace}}{\sup} \,  \left\Vert (1+t)\partial_{x}^{\alpha}   a \right\Vert_{\Ld^{\infty}_{t,x,\eta}}+\underset{\substack{\alpha \in \N^d \\ \alpha_i \in \lbrace 0,1 \rbrace}}{\sup} \,  \left\Vert (1+t)\partial_{t}\partial_x^{\alpha}   a \right\Vert_{\Ld^{\infty}_{t,x,\eta}}, \\
\label{seminormN}\Omega[a]&:=\underset{\substack{\alpha \in \N^d \\ \alpha_i \in \lbrace 0,1 \rbrace}}{\sup} \,   \left\lbrace \left\Vert  \vert \eta \vert \partial_{x}^{\alpha}   \nabla_{\tau,k} a \right\Vert_{\Ld^{\infty}_{t,x,\eta}}+\left\Vert  \vert \eta \vert \partial_{x}^{\alpha}   \nabla_{\tau,k} \partial_t a \right\Vert_{\Ld^{\infty}_{t,x,\eta}} \right \rbrace \\
& \notag \quad +\underset{\substack{\alpha \in \N^d \\ \alpha_i \in \lbrace 0,1 \rbrace}}{\sup} \,   \left\lbrace \left\Vert  \vert \eta \vert \partial_{x}^{\alpha}   \partial_{\tau}\nabla_{\tau,k} a \right\Vert_{\Ld^{\infty}_{t,x,\eta}}+\left\Vert  \vert \eta \vert \partial_{x}^{\alpha}   \partial_{\tau}\nabla_{\tau,k} \partial_t a \right\Vert_{\Ld^{\infty}_{t,x,\eta}} \right \rbrace, \\
\label{seminormA4}\Xi [a]_{\mathrm{M}}&:= \underset{\substack{1 \leq \vert \alpha \vert \leq 1+\mathrm{M}\\ \beta =0,1,2,3,4 } }{\sup} \, \left\Vert \partial_{x}^{\alpha} \partial_t^{\beta} a  \right\Vert_{\Ld^{\infty}_{t,x,\eta}}.
\end{align}

The following result states the $\Ld^2$ continuity of pseudodifferential operator with symbol having a finite seminorm $\omega[\cdot]$. We refer to <cit.> for a proof for symbols with compact support in $x$, whose adaptation to the physical space $\R \times \T^d$ is fairly straightforward[More precisely, one can introduce a weight $(1+t)^{-2}$ in <cit.> to get some integrability in time.

This turns out to be sufficient to consider the seminorm $\omega[\cdot]$ in our statement.].

There exists $C_d>0$ such that if $a$ is a symbol satisfying $\omega[a]<\infty$

then the following holds: for every $\gamma >0$, we have

\begin{align*}
\forall  h \in \Ld^2(\R \times \T^d), \ \ \Vert \mathrm{Op}^{\gamma}(a)(h) \Vert_{\Ld^2(\R \times \T^d)} \leq C_d \omega[a] \Vert h \Vert_{\Ld^2(\R \times \T^d)},
\end{align*}

Next, we have the following symbolic calculus result, with the use of the parameter $\gamma >0$. Note that taking $\gamma$ large can obviously be useful in view of an absorption argument. Here, we need to assume that one symbol has a compact support in time because the time variable $t \in \R$ in unbounded.

There exists $C_d>0$ and continuous nonnegative and nondecreasing function $\Lambda$ such that for any symbols $a,b$ satisfying

\begin{align*}
\Omega[a] , \Xi[b]_{\mathrm{M}} < \infty, \ \ \mathrm{M}>1+2d, \\
\nabla_x b  \text{ has compact support in time},
\end{align*}

the following holds: for all $\gamma>0$ and all $ h \in \Ld^2(\R \times \T^d)$, we have

\begin{align*}
\Vert \mathrm{Op}^{\gamma}(a)\mathrm{Op}^{\gamma}(b)(h)-\mathrm{Op}^{\gamma}(ab)(h) \Vert_{\Ld^2(\R \times \T^d)} \leq \frac{C_d}{\gamma}  \Lambda(\vert \mathrm{supp}_t \, \nabla_x b \vert)  \Omega[a]\Xi[b]_{\mathrm{M}} \Vert h \Vert_{\Ld^2(\R \times \T^d)}.
\end{align*}

A standard formula about the composition of pseudodifferential operators first shows that

\begin{align*}
\mathrm{Op}^{\gamma}(a)\mathrm{Op}^{\gamma}(b)=\mathrm{Op}^{\gamma}(c),
\end{align*}

\begin{align*}
c(t,x, \gamma, \tau, k)&=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \T^d}\int_{\R \times \Z^d}e^{i(\tau'-\tau) (t-t')}e^{i(k'-k)\cdot (x-x')}a(t,x,\gamma,\tau', k')b(t',x',\gamma, \tau, k) \, \mathrm{d} t'  \mathrm{d}x'  \mathrm{d}\tau'  \mathrm{d}k' \\
& =\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}e^{i(\tau't+k'\cdot x) }a(t,x,\gamma,\tau+\tau', k+k')\mathcal{F}_{t,x}b(\tau',k',\gamma, \tau, k) \, \mathrm{d}\tau'  \mathrm{d}k'.
\end{align*}

Therefore for $\eta=(\gamma, \tau, k) \in (0, + \infty) \times \R \times \R^d {\setminus} {\lbrace 0 \rbrace}$, we have

\begin{align*}
c(t,x, \eta)-a(t,x, \eta)b(t,x, \eta)
%&=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \T^d} \int_{\R \times \Z^d}e^{i(\tau'-\tau) (t-t')}e^{i(k'-k)\cdot (x-x')} b(t',x',\eta) \\
%&  \qquad \left\lbrace \int_0^1 \nabla_{\tau,k}a(t,x,\gamma,\tau+s(\tau'-\tau), k+s(k'-k)) \cdot (\tau'-\tau, k'-k) \, \mathrm{d}s \right\rbrace \, \mathrm{d} t'  \mathrm{d}x'  \mathrm{d}\tau'  \mathrm{d}k' \\
&=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}e^{i(\tau't+k'\cdot x) } \mathcal{F}_{t,x}b(\tau',k',\eta) \\
&  \qquad \qquad\qquad\left\lbrace \int_0^1 \nabla_{\tau,k}a(t,x,\gamma,\tau+s\tau', k+sk') \cdot (\tau', k') \, \mathrm{d}s \right\rbrace \, \mathrm{d}\tau'  \mathrm{d}k' \\
%&=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}e^{i(\tau't+k'\cdot x) } \mathcal{F}_{t,x}b(\tau',k',\eta) \mathcal{J}(a)(t,x, \eta, \tau', k') \, \mathrm{d}\tau'  \mathrm{d}k' \\
%&=:  \mathrm{m}(t,x, \eta).
&=: \frac{1}{\gamma }\mathrm{m}(t,x, \eta),
\end{align*}

\begin{align*}
\mathrm{m}(t,x, \eta):=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}e^{i(\tau't+k'\cdot x) } \mathcal{F}_{t,x}b(\tau',k',\eta) \mathcal{J}(a)(t,x, \eta, \tau', k') \, \mathrm{d}\tau'  \mathrm{d}k',
\end{align*}

\begin{align*}
%\mathcal{J}(a)(t,x, \eta, \tau', k'):=\int_0^1 \vert \eta \vert \nabla_{\tau,k}a(t,x,\gamma,\tau+s\tau', k+sk') \cdot (\tau', k') \, \mathrm{d}s
\mathcal{J}(a)(t,x, \eta, \tau', k'):=\int_0^1 \gamma \nabla_{\tau,k}a(t,x,\gamma,\tau+s\tau', k+sk') \cdot (\tau', k') \, \mathrm{d}s.
\end{align*}

Since $\mathrm{Op}^{\gamma}(a)\mathrm{Op}^{\gamma}(b)-\mathrm{Op}^{\gamma}(ab)=\frac{1}{\gamma}\mathrm{Op}^{\gamma}( \mathrm{m})$, and in view of the continuity property stated in Theorem <ref>, it remains to estimate the seminorm $\omega[\mathrm{m}]$. For all $\alpha \in \N^d$ such that $\alpha_i  \in \lbrace 0, 1 \rbrace$, we now have

\begin{align*}
(1+t)\mathrm{m}(t,x, \eta)&=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}(1-i\partial_{\tau'})\big(e^{i\tau't}\big) e^{ik'\cdot x } \mathcal{F}_{t,x}b(\tau',k',\eta) \mathcal{J}(a)(t,x, \eta, \tau', k') \mathrm{d}\tau'  \mathrm{d}k',
\end{align*}

\begin{align*}
(1+t)\partial_t\mathrm{m}(t,x, \eta)&=\frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}i \tau'(1-i\partial_{\tau'})\big(e^{i\tau't}\big) e^{ik'\cdot x } \mathcal{F}_{t,x}b(\tau',k',\eta) \mathcal{J}(a)(t,x, \eta, \tau', k') \mathrm{d}\tau'  \mathrm{d}k' \\
& +\quad \frac{1}{(2\pi)^{d+1}}\int_{\R \times \Z^d}(1-i\partial_{\tau'})\big(e^{i\tau't}\big) e^{ik'\cdot x } \mathcal{F}_{t,x}b(\tau',k',\eta) \partial_t\left(\mathcal{J}(a) \right)(t,x, \eta, \tau', k') \mathrm{d}\tau'  \mathrm{d}k'.
\end{align*}

Tedious but standard computations then show that for $\mathrm{M}>d$, we have

\begin{align*}
\omega[\mathrm{m}] &\lesssim \Omega[a]\underset{\substack{\alpha \in \N^d \\ \alpha_i \in \lbrace 0,1 \rbrace}}{\sup} \,  \left\Vert  (1+\vert \tau' \vert^2) \vert k'\vert^{1+\vert \alpha \vert }  \mathcal{F}_{t,x}\big[(1+t)b \big] \right\Vert_{\Ld^1(\R_{\tau'} \times \Z^d_{k'};\Ld^{\infty}_{\eta})} \\
& \lesssim \Omega[a] \left( \underset{1 \leq \vert \alpha \vert \leq 1+\mathrm{M}  }{\sup} \, \left\Vert  \mathcal{F}_{t,x}\big[(1+t)\partial_{x}^{\alpha} b \big]\right\Vert_{\Ld^1(\R \times \Z^d;\Ld^{\infty}_{\eta})} +\underset{\substack{1\leq \vert \alpha \vert \leq 1+\mathrm{M} \\ \beta =1,2  }}{\sup} \, \left\Vert  \mathcal{F}_{t,x}\big[(1+t)\partial_{x}^{\alpha} \partial_t^{\beta} b \big] \right\Vert_{\Ld^1(\R \times \Z^d;\Ld^{\infty}_{\eta})}  \right).
\end{align*}

As a consequence, we obtain for $\mathrm{M}>1+2d$

\begin{align*}
& \lesssim   \Omega[a] \left( \underset{\substack{1 \leq \vert \alpha \vert \leq 1+\mathrm{M} \\ \beta=0,1,2}}{\sup} \, \left\Vert  \mathcal{F}_{t,x}\big[(1+t)\partial_{x}^{\alpha} \partial_t^{\beta} b \big]\right\Vert_{\Ld^{\infty}(\R \times \Z^d;\Ld^{\infty}_{\eta})} +\underset{\substack{1 \leq \vert \alpha \vert \leq 1+\mathrm{M} \\ \beta =0,1,2,3,4  }}{\sup} \, \left\Vert  \mathcal{F}_{t,x}\big[(1+t)\partial_{x}^{\alpha} \partial_t^{\beta} b \big] \right\Vert_{\Ld^{\infty}(\R \times \Z^d;\Ld^{\infty}_{\eta})}  \right) \\
& \lesssim \Omega[a] \underset{\substack{1 \leq \vert \alpha \vert \leq 1+ \mathrm{M}\\ \beta =0,1,2,3,4 } }{\sup} \, \left\Vert  (1+t)\partial_{x}^{\alpha} \partial_t^{\beta} b  \right\Vert_{\Ld^{1}(\R \times \T^d;\Ld^{\infty}_{\eta})},
\end{align*}

and therefore

\begin{align*}
\omega[\mathrm{m}]  & \lesssim  \Lambda(\vert \mathrm{supp}_t \, \nabla_x b \vert)\Omega[a]\Xi[b]_{\mathrm{M}}.
\end{align*}

From Theorem <ref>, we have for all $ h \in \Ld^2(\R \times \T^d)$

\begin{align*}
\Vert \mathrm{Op}^{\gamma}(\mathrm{m})h \Vert_{\Ld^2(\R \times \T^d)} \leq C_d \omega[\mathrm{m}]  \Vert h \Vert_{\Ld^2(\R \times \T^d)} \lesssim C_d \Lambda(\vert \mathrm{supp}_t \, \nabla_x b \vert)  \Omega[a]  \Xi[b]_{\mathrm{M}} \Vert h \Vert_{\Ld^2(\R \times \T^d)}.
\end{align*}

We obtain the conclusion since $\mathrm{Op}^{\gamma}(a)\mathrm{Op}^{\gamma}(b)-\mathrm{Op}^{\gamma}(ab)=\frac{1}{\gamma}\mathrm{Op}^{\gamma}( \mathrm{m})$.

§ ACKNOWLEDGEMENTS

We are grateful to Laurent Boudin, Laurent Desvillettes, Julien Mathiaud and Ayman Moussa for inspiring discussions concerning spray equations during the preparation of this article. This work originated from our collaboration with Aymeric Baradat, whom we warmly thank.

Partial support by the grant ANR-19-CE40-0004 is acknowledged.

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