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An \(L^2\) to \(L^\infty \) Framework for the Landau Equation

  • Jinoh Kim
  • Yan GuoEmail author
  • Hyung Ju HwangEmail author
Original Article
  • 41 Downloads

Abstract

Consider the Landau equation with Coulomb potential in a periodic box. We develop a new \(L^{2}\ \text{to}\ L^{\infty }\) framework to construct global unique solutions near Maxwellian with small \(L^{\infty }\) norm. The first step is to establish global \(L^{2}\) estimates with strong velocity weight and time decay, under the assumption of \(L^{\infty }\) bound, which is further controlled by such \(L^{2}\) estimates via De Giorgi’s method (Golse et al. in Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 19(1), 253–295 (2019), Imbert and Mouhot in arXiv:1505.04608 (2015)). The second step is to employ estimates in \(S_{p}\) spaces to control velocity derivatives to ensure uniqueness, which is based on Hölder estimates via De Giorgi’s method (Golse et al. in Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 19(1), 253–295 (2019), Golse and Vasseur in arXiv:1506.01908 (2015), Imbert and Mouhot in arXiv:1505.04608 (2015)).

Keywords

Landau equation Weak solution Existence and uniqueness L2 to \(L^\infty \) framework 

Mathematics Subject Classification

35Qxx 

1 Introduction

We consider the following Landau equation:
$$\begin{aligned} \partial _t F + v \cdot \nabla _{x} F& = Q(F,F) = \nabla _{v} \cdot \left\{ \int _{{\mathbb {R}}^3} \phi (v-v')[F(v')\nabla _{v} F(v) - F(v) \nabla _{v} F(v')] \mathrm{d}v'\right\} , \nonumber \\ F(0,x,v) & = F_0(x,v), \end{aligned}$$
(1.1)
where \(F(t,x,v)\ge 0\) is the spatially periodic distribution function for particles at time \(t\ge 0\), with spatial coordinates \(x=(x_1,x_2,x_3)\in [-\pi , \pi ]^3 = {\mathbb {T}}^3\) and velocity \(v=(v_1,v_2,v_3)\in {\mathbb {R}}^3\). The non-negative matrix \(\phi \) is
$$\begin{aligned} \phi ^{ij}(v)=\left\{ \delta _{i,j}-\frac{v_{i}v_{j}}{|v|^{2}}\right\} |v|^{-1}. \end{aligned}$$
(1.2)
As in the Boltzmann equation, it is well known that Maxwellians are steady states to (1.1) [13], etc. Let \(\mu \) be a normalized Maxwellian
$$\begin{aligned} \mu (v) = e^{-|v|^2}, \end{aligned}$$
(1.3)
and set
$$\begin{aligned} F(t,x,v) = \mu (v) + \mu ^{1/2}(v)f(t,x,v). \end{aligned}$$
(1.4)
Then the standard perturbation f(txv) to \(\mu \) satisfies
$$\begin{aligned} f_{t}+v\cdot \partial _{x}f+Lf=\Gamma (f,f), \end{aligned}$$
(1.5)
$$\begin{aligned} f(0,x,v)=f_{0}(x,v), \end{aligned}$$
(1.6)
where \(f_{0}\) is the initial data satisfying the conservation laws:
$$\begin{aligned} \int _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}f_{0}(x,v)\sqrt{\mu }=\int _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}v_{i\,}f_{0}(x,v)\sqrt{\mu }=\int _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}|v|^{2}f_{0}(x,v)\sqrt{\mu }=0. \end{aligned}$$
(1.7)
The linear operator L and the nonlinear part \(\Gamma \) are defined as
$$\begin{aligned} L = -A -K, \end{aligned}$$
(1.8)
$$\begin{aligned} Af & :=\mu ^{-1/2}\partial _{i} \{ \mu ^{1/2}\sigma ^{ij}[\partial _{j\,}f+v_{j\,}f]\} =\partial _{i}[\sigma ^{ij}\partial _{j\,}f]-\sigma ^{ij}v_{i}v_{j\,}f+\partial _{i}\sigma ^{i}f, \end{aligned}$$
(1.9)
$$\begin{aligned} Kf& :=-\mu ^{-1/2}\partial _{i} \{ \mu [ \phi ^{ij}*\{ \mu ^{1/2}[\partial _{j\,}f+v_{j\,}f] \} ] \}, \end{aligned}$$
(1.10)
$$\begin{aligned} \Gamma [g,f]& := \partial _{i} [ \{ \phi ^{ij}*[\mu ^{1/2}g] \} \partial _{j\,}f ] - \{ \phi ^{ij}*[v_{i}\mu ^{1/2}g] \} \partial _{j\,}f \nonumber \\&\quad -\partial _{i} [ \{ \phi ^{ij}*[\mu ^{1/2} \partial _{j}g]\} f] + \{ \phi ^{ij}*[v_{i}\mu ^{1/2}\partial _{j}g] \} f. \end{aligned}$$
(1.11)
Define the weighted norm and weighted energy associated with (1.5):
$$\begin{aligned} w & :=(1+|v|),\quad |f|_{p,\vartheta }^{p} :=\int _{{\mathbb {R}}^{3}}w^{p\vartheta }|f|^{p}\mathrm{d}v,\quad \Vert\; f\Vert _{p,\vartheta }^{p}:=\int _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}w^{p\vartheta }|f|^{p}\mathrm{d}x\mathrm{d}v. \end{aligned}$$
(1.12)
$$\begin{aligned} {\Vert } f\Vert _{\sigma ,\vartheta }^{2} & :=\iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}w^{2\vartheta } [ \sigma ^{ij}\partial _{i}f\partial _{j}f+\sigma ^{ij}v_{i}v_{j}f^{2}] \mathrm{d}v\mathrm{d}x, \end{aligned}$$
(1.13)
$$\begin{aligned} |f|_{\infty ,\vartheta } &=\sup _{{\mathbb {R}}^{3}}w^{\vartheta }(v)f(v),\quad \Vert\; f\Vert _{\infty ,\vartheta }=\sup _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}w^{\vartheta }(v)f(x,v). \end{aligned}$$
(1.14)
$$\begin{aligned} |f|_{2}& :=|f|_{2,0}, \quad \Vert\; f\Vert _{2}:=\Vert\; f\Vert _{2,0}, \nonumber \\ |f|_{\sigma }& :=|f|_{\sigma ,0}, \quad \Vert\; f\Vert _{\sigma }:=\Vert\; f\Vert _{\sigma ,0}, \nonumber \\ |f|_{\infty }& :=|f|_{\infty ,0}, \quad \Vert\; f\Vert _{\infty }:=\Vert\; f\Vert _{\infty ,0}, \nonumber \\ {\mathcal {E}}_{\vartheta }(f(t)) & := \frac{1}{2}\Vert\; f(t)\Vert _{2,\vartheta }^{2} + \int _{0}^{t} \Vert\; f(s)\Vert _{\sigma ,\vartheta }^{2} \mathrm{d}s. \end{aligned}$$
(1.15)
Here we introduce the main result.

Theorem 1.1

(Main result) There exist\(\vartheta ^{\prime }\)and\(0<\varepsilon _{0}\ll 1\)such that for some\(\vartheta \ge \vartheta ^{\prime }\)if\(f_{0}\)satisfies
$$\begin{aligned} \Vert\, f_{0}\Vert _{\infty ,\vartheta } \le \varepsilon _0, \quad \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert D_v f_0\Vert _{\infty , \vartheta } < \infty , \end{aligned}$$
(1.16)
where\(f_{0t} := -v\cdot \nabla _x f_{0} + {{\bar{A}}}_{f_{0}}f_{0}\), then there exists a unique weak solution (see Definition 4.1) fof (1.5), (1.6) on\((0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\)such that
  1. (1)

    if\(F(0) := \mu (v) + \sqrt{\mu }(v) f_0 \ge 0\), then\(F(t) := \mu (v) + \sqrt{\mu }(v) f(t) \ge 0\)for every\(t\ge 0.\)

     
  2. (2)
    Moreover for any\(t>0\), \(\vartheta _{0}\in {\mathbb {N}}\), and\(\vartheta \ge \vartheta '\), there existC, \(C_{\vartheta , \vartheta _{0}}\), \(l_0(\vartheta _0)\), and\(0<\alpha <1\)such thatfsatisfies
    $$\begin{aligned} \sup _{0 \le s\le \infty }{\mathcal {E}}_{\vartheta }(f(s)) \le C 2^{2\vartheta }{\mathcal {E}}_{\vartheta }(0), \end{aligned}$$
    (1.17)
    $$\begin{aligned} \Vert\; f(t)\Vert _{2,\vartheta } \le C_{\vartheta ,\vartheta _{0}} {\mathcal {E}}_{\vartheta +\vartheta _{0}/2}(0)^{1/2}\left( 1+ \frac{t}{\vartheta _{0}}\right) ^{-\vartheta _{0}/2}, \end{aligned}$$
    (1.18)
    $$\begin{aligned} \Vert\; f(t)\Vert _{\infty , \vartheta } \le C_{\vartheta ,\vartheta _{0}}(1+t)^{-\vartheta _{0}}\Vert\, f_{0}\Vert _{\infty ,\vartheta +l_{0}}, \end{aligned}$$
    (1.19)
    $$\begin{aligned} \Vert\; f\Vert _{C^{\alpha }\left( (0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\right) } \le C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) , \end{aligned}$$
    (1.20)
    and
    $$\begin{aligned} \Vert D_{v} f\Vert _{L^{\infty }\left( (0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\right) } \le C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert D_v f_0\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) . \end{aligned}$$
    (1.21)
     

Motivated by the study of global well-posedness for the Landau equation in a bounded domain with physical boundary conditions, our current study is the first step to develop an \(L^{2}\ \text{to}\ L^{\infty }\) framework with necessary analytic tools in a simpler periodic domain. There have been many results for Landau equations in either a periodic box or whole domain [2, 3, 6, 7, 8, 13, 15, 16, 17, 22, 23, 25, 26, 27], in which high-order Sobolev norms can be employed. On the other hand, in a bounded domain, even with the velocity diffusion, the solutions can not be smooth up to the grazing set [19]. New mathematical tools involving weaker norms are needed to be developed. In the case of Boltzmann equations, an \(L^{2}\ \text{to}\ L^{\infty }\) framework has been developed to construct unique global solutions in bounded domains [14].

Our work can be viewed as a similar \(L^{2}\ \text{to}\ L^{\infty }\) approach for the Landau equation. Our techniques are inspired by recent remarkable progresses of [10, 11, 20], in which a general machinery in the spirit of De Giorgi, has been developed for the Fokker–Planck equations, even for the Landau equation [20], to bootstrap \(L^{\infty }\) and Hölder space \(C^{0,\alpha }\) from a \(L^{2}\) weak solution. Unfortunately, to our knowledge, there is still no construction for \(L^{\infty }\) global weak solutions to the Landau equation.

Our paper settles the global existence and uniqueness for an \(L^{2}\) weak solution with a small weighted \(L^{\infty }\) perturbation of a Maxwellian initially. Our method is an intricate combination of different tools. Our starting point is a design of an iterating sequence
$$\begin{aligned} (\partial _{t}+v\cdot \nabla _{x}) f^{n+1} & = -Lf^{n+1}+\Gamma (f^{n},f^{n+1}) \\ & \equiv A_{f^{n}}(f^{n+1})+K_{f^{n}}(f^{n+1}), \end{aligned}$$
where all terms in \(A_{f^{n}}(f^{n+1})\) contain at least one momentum derivative of \(f^{n}\) or \(f^{n+1},\) so that \(f^{n}\) appears in the coefficients of the Landau operator for \(f^{n+1}. \) The crucial lemma states that if \( \Vert\; f^{n} \Vert_{\infty }\) is sufficiently small, the main part of \(A_{f^{n}}(f^{n+1})\) retains the same analytical properties of the linearized Landau operator A.

We first establish global energy estimates and time decay under the assumption \(f^{n}\) is small, in Sect. 4.

Let us consider the linearized Landau equation with a given g:
$$\begin{aligned} \partial _{t}f+v\cdot \nabla _{x}f+Lf=\Gamma (g,f). \end{aligned}$$
(1.22)

Theorem 1.2

Suppose that\(\Vert g\Vert _{\infty } < \varepsilon \). Let\(\vartheta \in 2^{-1}{\mathbb {N}} \cup \{0\}\)andfbe a classical solution of (1.6), (1.7), and (1.22). Then there existCand\(\varepsilon =\varepsilon (\vartheta )>0\)suchthat
$$\begin{aligned} \sup _{0 \le s< \infty }{\mathcal {E}}_{\vartheta }(f(s)) \le C 2^{2\vartheta } {\mathcal {E}}_{\vartheta }(0), \end{aligned}$$
(1.23)
and
$$\begin{aligned} \Vert\; f(t)\Vert _{2,\vartheta } \le C_{\vartheta ,k} \left( {\mathcal {E}}_{\vartheta +k/2}(0) \right) ^{1/2}\left( 1+ \frac{t}{k}\right) ^{-k/2} \end{aligned}$$
(1.24)
for any\(t>0\)and\(k\in {\mathbb {N}}\).

It is important to note that, thanks to the nonlinearity, the velocity weight can be arbitrarily strong. The proof of this step is a combination of energy estimates with positivity estimates for \(\mathbf{P }\,f\) [9, 13] and a timedecay estimate [25], but in the absence of high-order Sobolev regularity.

We next bootstrap such an \(L^{2}\) bound to an \(L^{\infty }\) bound.

Theorem 1.3

Letfbe a weak solution of (1.6), (1.7), and (1.22) in a periodic box and\(\vartheta \in {\mathbb {N}} \cup \{0\}\), \(\vartheta _{0} \in {\mathbb {N}}\). Then there exist\(\varepsilon \), \(l_{0}(\vartheta _{0})>0\)and\(C_{\vartheta , \vartheta _0}\)such that ifgsatisfies
$$\begin{aligned} \sup _{0\le s\le \infty }\Vert g(s)\Vert _{\infty }\le \varepsilon \end{aligned}$$
then
$$\begin{aligned} \Vert\; f(t)\Vert _{\infty , \vartheta +\vartheta _{0}}\le C_{\vartheta ,\vartheta _0}(1+t)^{-\vartheta _{0}}\Vert\, f_{0}\Vert _{\infty ,\vartheta +l_{0}}. \end{aligned}$$
(1.25)

It is important to note that even though there is a finite loss of velocity weight, we are still able to close the estimates thanks to the strong gain of velocity weight in (1.24). The proof of such an \(L^{\infty }\) estimate locally in x and v is an adaptation of recent work of [11, 20]. It is well known that the Landau operator is delicate to study and estimate for large velocities. Together with the maximum principle of the Landau operator as well as strong time decay for \(L^2\) norm in (1.24), we are able to control the ‘tails’ of solutions for large velocities, and obtain global (in x and v) \(L^{\infty }\) estimate.

Unfortunately, unlike in the Boltzmann case (see [14]), to establish the convergence of \(\{f^{n}\}\) and more importantly, uniqueness of our solution, such an \(L^{\infty }\) bound is not sufficient due to the presence of velocity derivative in the nonlinear Landau equation. We need to further control \( \Vert \nabla _{v}f^{n} \Vert_{\infty }\) as in Lemma 8.2, which follows from \(S_{p}\) estimates established in [5, 24]. One crucial requirement for such \(S_{p}\) estimates (as the classical \(W^{2,p}\) estimate in the elliptic theory), is the \( C^{0,\alpha }\) estimate (uniform in x and v) for the coefficients containing \(f^{n}.\) We establish

Theorem 1.4

Letfbe a solution of (1.6), (1.7), and (1.22). Then there exist\(\varepsilon \), \(\vartheta \), C, \(\alpha >0\)such that ifgsatisfies
$$\begin{aligned} \sup _{0\le s\le \infty }\Vert g(s)\Vert _{\infty }\le \varepsilon, \end{aligned}$$
then we have
$$\begin{aligned} \Vert\; f\Vert _{C^{\alpha }\left( (0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\right) } \le C(f_0), \end{aligned}$$
where
$$\begin{aligned} C(f_0) = C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) . \end{aligned}$$
(1.26)

Again, we follow the methods in [10, 11, 20] to establish such an estimate locally in (xv),  and use a delicate change of coordinates (6.16) locally to capture precisely the isotropic behavior of the Landau operator, thanks to Lemma 2.4 and our strong weighted \(L^{\infty }\) estimates to obtain uniform \( C^{0,\alpha }\) estimate. It is well known that the Landau equation is degenerate for \(|v|\rightarrow \infty \) and our strong energy estimate provides the control of velocity (tails) of the Landau solutions. An additional regularity condition \(\Vert\, f_{0t}\Vert_{\infty ,\theta }<+\infty \) is needed for such a Hölder estimate, but no smallness is required. A further bound \(\Vert\, f_{0v}\Vert_{\infty ,\theta }<+\infty \) is needed to apply the \(S_{p}\) theory in a non-divergent form.

Such an \(L^{2}\ \text{to}\ L^{\infty }\) framework is robust and is currently being applied to the study of several other problems in the kinetic theory.

2 Basic Estimates

For the reader’s convenience, we summarize and modify some basic estimates. We will adapt techniques in [13].

Proposition 2.1

There exists a uniform constantCsuch that for every functionfand a constant\(\vartheta \in {\mathbb {R}}\), we have
$$\begin{aligned} \Vert\; f\Vert _{2, \vartheta } \le C\Vert\; f\Vert _{\infty , \vartheta +2}. \end{aligned}$$

Proof

$$\begin{aligned} \begin{aligned} \Vert\; f\Vert _{2, \vartheta }^2&= \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} (w^{\vartheta }(v) f(x,v))^2 \mathrm{d}x\mathrm{d}v\\&= \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} w^{-4}(v) (w^{\vartheta +2 }(v) f(x,v))^2 \mathrm{d}x\mathrm{d}v\\&\le \Vert\; f\Vert _{\infty , \vartheta +2} \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} w^{-4}(v) \mathrm{d}x\mathrm{d}v\\&\le C\Vert\; f\Vert _{\infty , \vartheta +2}, \end{aligned} \end{aligned}$$
for some constant \(C>0\). \(\square \)

Lemma 2.2

(Lemma 2 in [13]) Let\(\vartheta >-3\), \(a(v) \in C^{\infty }({\mathbb {R}}^{3}{\setminus} \{0\})\)and\(b(v)\in C^{\infty }({\mathbb {R}}^{3})\). Assume for any positive multi-index\(\beta \), there is\(C_{\beta }>0\)such that
$$\begin{aligned} \begin{aligned} |\partial _{\beta }a(v)|&\le C_{\beta }|v|^{\vartheta -|\beta |},\\ |\partial _{\beta }b(v)|&\le C_{\beta }e^{-\tau _{\beta }|v|^{2}}, \end{aligned} \end{aligned}$$
with some\(\tau _{\beta }>0\). Then there is\(C^{*}_{\beta }>0\)such that
$$\begin{aligned} |\partial _{\beta }[a*b](v)| \le C^{*}_{\beta }[1+|v|]^{\vartheta - \beta }. \end{aligned}$$
For the reader’s convenience, we will use the following notation in this paper.
$$\begin{aligned} \sigma _{u}^{ij}(v) & := \phi ^{ij}*u = \int _{{\mathbb {R}}^{3}}\phi ^{ij}(v-v^{\prime })u(v^{\prime })\mathrm{d}v^{\prime }, \end{aligned}$$
(2.1)
$$\begin{aligned} \sigma ^{ij} &= \sigma ^{ij}_{\mu }, \quad \sigma ^{i} = \sigma ^{ij}v_{j}. \end{aligned}$$
(2.2)
For every \(v,\nu \in {\mathbb {R}}^{3}\), define
$$\begin{aligned} D_{u}(\nu ;v) & := \nu ^{T} \sigma _{u}(v) \nu \end{aligned}$$
(2.3)
and \(P_{v}\) is the projection onto the vector v as
$$\begin{aligned} P_{v} g & := \sum _{j=1}^{3}\langle g_{j}, v_{j} \rangle \frac{v_{i} }{|v|^{2}}, \quad 1\le i \le 3. \end{aligned}$$
(2.4)

Lemma 2.3

(Lemma 3 in [13]) If\(u = \mu \)or\(\sqrt{\mu }\), then
$$\begin{aligned} D_{u}(\nu ;v) & = \lambda _{1}(v)|P_{v} \nu |^{2} + \lambda _{2}(v) |(I-P_{v})\nu |^{2}. \end{aligned}$$
(2.5)
Moreover, there existsCsuch that
$$\begin{aligned} \frac{1}{C}(1+|v|)^{-3}\le \lambda _{1}(v) \le C(1+|v|)^{-3}, \end{aligned}$$
and
$$\begin{aligned} \frac{1}{C}(1+|v|)^{-1} \le \lambda _{2}(v) \le C(1+|v|)^{-1}. \end{aligned}$$

We can derive upper and lower bounds of eigenvalues for \(\sigma + \sigma _{\sqrt{\mu }g}\) by adapting ideas in the proof of Theorem 3 in [13].

Lemma 2.4

Letgbe a given function in\(L^\infty ((0,\infty )\times {\mathbb {T}}^3\times {\mathbb {R}}^3)\)and\(G = \mu + \sqrt{\mu }g\). Let\(\sigma _{G}\)be the matrix defined as in (2.1). Then there exists\(0<\varepsilon \ll 1\)such that ifgsatisfies
$$\begin{aligned} \sup _{0\le s\le \infty }\Vert g(s)\Vert _{\infty } \le \varepsilon, \end{aligned}$$
(2.6)
then
$$\begin{aligned} \begin{aligned} D_{G}(\nu ;v)&\ge \frac{1}{2C} ( (1+|v|)^{-3}|P_{v} \nu |^{2} + (1+|v|)^{-1}|(I-P_{v})\nu |^{2}) ,\\ D_{G}(\nu ;v)&\le 2C ( (1+|v|)^{-3}|P_{v} \nu |^{2} + (1+|v|)^{-1}|(I-P_{v})\nu |^{2}), \end{aligned} \end{aligned}$$
for every\(v \in {\mathbb {R}}^{3}\). Thus\(\sigma _{G}(v)\)has three non-negative eigenvalues. Moreover, \(\lambda (v)\), eigenvalue of\(\sigma _{G}(v)\), has the following estimate
$$\begin{aligned} \frac{1}{C}(1+|v|)^{-3}\le \lambda (v) \le C(1+|v|)^{-1}, \end{aligned}$$
for some constant\(C>0\).

Proof

Let \(u=\sqrt{\mu }g\). Then we claim that there exists \(C^{\prime }>0\) such that
$$\begin{aligned} |D_{u}(\nu ;v)|\le C^{\prime }\Vert g\Vert _{\infty } ( (1+|v|)^{-3}|P_{v} \nu |^{2} + (1+|v|)^{-1}|(I-P_{v})\nu |^{2}). \end{aligned}$$
(2.7)
Consider
$$\begin{aligned} \begin{aligned} D_{u}(\nu ;v)&= \sum _{i,j}\int _{2|v^{\prime }| > |v|} \nu _{i} \nu _{j} \phi ^{ij}(v-v^{\prime })\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime } \\ & \quad + \sum _{i,j}\int _{2|v^{\prime }| \le |v|} \nu _{i} \nu _{j} \phi ^{ij}(v-v^{\prime })\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime }\\&= (I) + (II). \end{aligned} \end{aligned}$$
Note that for \(2|v^{\prime }| > |v|\), \(\sqrt{\mu }(v^{\prime }) \le C^{\prime } \mu (v^{\prime }/4) \mu (v/4)\). Therefore,
$$\begin{aligned} |(I)| & \le C^{\prime }\mu \left( \frac{v}{4}\right) \Vert g\Vert _{\infty }|\nu |^{2}\int _{{\mathbb {R}}^{3}} \phi ^{ij}(v-v^{\prime }) \mu \left( \frac{v^{\prime }}{4}\right) \mathrm{d}v^{\prime } \nonumber \\ & \le C^{\prime -1}\mu \left( \frac{v}{4}\right) \Vert g\Vert _{\infty } |\nu |^{2}. \end{aligned}$$
(2.8)
To control (II), we expand \(\phi ^{ij}(v-v^{\prime })\) to get
$$\begin{aligned} \phi ^{ij}(v-v^{\prime }) &= \phi ^{ij}(v) - \sum _{k}\partial _{k} \phi ^{ij}(v)v_{k}^{\prime }+ \frac{1}{2}\sum _{k,l}\partial _{kl}\phi ^{ij}({{\bar{v}}})v_{k}^{\prime } v_{l}^{\prime }, \end{aligned}$$
for some \({{\bar{v}}}\) in a line segment of v and \(v-v^{\prime }\). Then we have
$$\begin{aligned} \begin{aligned} (II)&= \sum _{i,j} \nu _{i} \nu _{j} \phi ^{ij}(v)\int _{2|v^{\prime }| \le |v|}\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime }\\&\quad - \sum _{i,j} \nu _{i} \nu _{j} \sum _{k} \partial _{k}\phi ^{ij} (v)\int _{2|v^{\prime }| \le |v|}v_{k}^{\prime }\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime }\\&\quad + \frac{1}{2}\sum _{i,j}\int _{2|v^{\prime }| \le |v|} \nu _{i} \nu _{j} \sum _{k,l}\partial _{kl}\phi ^{ij}({{\bar{v}}})v_{k}^{\prime }v_{*}^{\prime }\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime }\\&=(II)_{1} + (II)_{2} + (II)_{3}. \end{aligned} \end{aligned}$$
Since
$$\begin{aligned} \sum _{i} \phi ^{ij}(v)v_{i} = \sum _{j} \phi ^{ij}(v)v_{j} = 0, \end{aligned}$$
we have
$$\begin{aligned} |(II)_{1}| & = \left( (I-P_{v}) \nu \right) ^{T} \phi (v) \left( (I-P_{v}) \nu \right) \int _{2|v^{\prime }|\le |v|}\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime } \nonumber \\ & \le C\Vert g\Vert _{\infty }(1+|v|)^{-1}|(I-P_{v}) \nu |^{2}. \end{aligned}$$
(2.9)
Note that
$$\begin{aligned} \sum _{i,j} \partial _{k} \phi ^{ij}(v)v_{i}v_{j} = 0. \end{aligned}$$
Therefore
$$\begin{aligned} |(II)_{2}| & \le \left| \sum _{k} \left( (I-P_{v})\nu \right) ^{T}\partial _{k} \phi (v) \left( (I-P_{v})\nu \right) \int _{2|v^{\prime }| \le |v|} v_{k}^{\prime } \sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime }\right| \nonumber \\& \quad + 2\left| \sum _{k} (P_{v}\nu )^{T}\partial _{k}\phi (v) \left( (I-P_{v} )\nu \right) \int _{2|v^{\prime }| \le |v|} v_{k}^{\prime }\sqrt{\mu }(v^{\prime }) g(v^{\prime }) \mathrm{d}v^{\prime }\right| \nonumber \\ & \le C\Vert g\Vert _{\infty }(1+|v|)^{-2}\left( |(I-P_{v})\nu |^{2} + |P_{v} \nu ||(I-P_{v})\nu |\right) \nonumber \\& \le C\Vert g\Vert _{\infty }\left( (1+|v|)^{-3}|P_{v} \nu |^{2} + (1 + |v|)^{-1}|(I-P_{v})\nu |^{2}\right) . \end{aligned}$$
(2.10)
Since \({{\bar{v}}}\) is in \((v,v-v^{\prime })\) and \(2|v^{\prime }| \le |v|\), we have \(|v|/2 \le |{{\bar{v}}}| \le 3|v|/2\). Therefore, \(\partial _{kl}\phi ^{ij}({{\bar{v}}}) \le C^{\prime } |v|^{-3}\). Thus we have
$$\begin{aligned} |(II)_{3}| \le C^{\prime }\Vert g\Vert _{\infty }(1+|v|)^{-3} |\nu |^{2}. \end{aligned}$$
(2.11)
Combining (2.8)–(2.11), we have (2.7).
Now we can compute \(D_{G}(\nu ;v)\). Since \(\varepsilon >0\) is a given small enough constant, from (2.5), (2.7), we have
$$\begin{aligned} D_{G}(\nu ;v)&\ge \frac{1}{2C} ( (1+|v|)^{-3}|P_{v} \nu |^{2} + (1+|v|)^{-1}|(I-P_{v})\nu |^{2}),\\ D_{G}(\nu ;v)&\le 2C ( (1+|v|)^{-3}|P_{v} \nu |^{2} + (1+|v|)^{-1}|(I-P_{v})\nu |^{2}). \end{aligned}$$
Therefore,
$$\begin{aligned} \frac{1}{2C} (1+|v|)^{-3}\le \lambda \le 2C(1+|v|)^{-1}. \end{aligned}$$
\(\square \)
Let
$$\begin{aligned} |f|_{\sigma ,\vartheta }^{2}& :=\int _{{\mathbb {R}}^{3}}w^{2\vartheta }\left[ \sigma ^{ij}\partial _{i}f\partial _{j}f+\sigma ^{ij}v_{i}v_{j}f^{2}\right] \mathrm{d}v. \end{aligned}$$
(2.12)

Lemma 2.5

(Corollary 1 in [13]) There exists\(c=c_{\vartheta }>0\), such that
$$\begin{aligned} |g|_{\sigma ,\vartheta }^2 & \ge c \{ | w^{\vartheta }[1+|v|]^{-3 /2} \{ P_{v} \partial _{i} g \} | _{2}^{2} + | w^{\vartheta }[1+|v|]^{-1/2} \{ [I-P_{v}] \partial _{i} g \} | _{2}^{2} \\& \quad + | w^{\vartheta }[1+|v|]^{-1/2}g | _{2}^{2} \}. \end{aligned}$$
Define
$$\begin{aligned} \langle f,g\rangle & :=\int _{{\mathbb {R}}^{3}}fg\mathrm{d}v,\quad (f,g):=\int _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}fg\mathrm{d}x\mathrm{d}v. \end{aligned}$$
(2.13)
For any real-valued function f(v), we define the projection onto the \(\text {span} \{\sqrt{\mu }, v \sqrt{\mu }, |v|^{2} \sqrt{\mu }\}\) in \(L^{2} ({\mathbb {R}}^{3})\) as
$$\begin{aligned} \mathbf{P } f & := \left( {\mathbf {a}}_{f}(t,x) + v \cdot {\mathbf {b}}_{f} (t,x) + \left( |v|^{2} -\frac{3}{2}\right) {\mathbf {c}}_{f}(t,x)\right) \sqrt{\mu }, \end{aligned}$$
(2.14)
where
$$\begin{aligned} \begin{aligned} {\mathbf {a}}_{f}&:= \frac{\langle f, \sqrt{\mu }\rangle }{|\langle \sqrt{\mu }, \sqrt{\mu }\rangle |^{2}},\\ {\mathbf {b}}_{f}^{i}&:= \frac{\langle f, v_{i} \sqrt{\mu }\rangle }{|\langle v_{i} \sqrt{\mu }, v_{i} \sqrt{\mu }\rangle |^{2}},\\ {\mathbf {c}}_{f}&:= \frac{\langle f, (|v|^{2} -3/2) \sqrt{\mu }\rangle }{|\langle (|v|^{2} -3/2)\sqrt{\mu }, (|v|^{2} -3/2)\sqrt{\mu }\rangle |^{2}}. \end{aligned} \end{aligned}$$

Lemma 2.6

(Lemma 5 in [13]) LetL, K, and\(\sigma ^{i}\)be defined as in (1.8), (2.2), and (1.10). Let\(\vartheta \in {\mathbb {R}}\). For any\(m>1\), there is\(0<C(m)<\infty \), such that
$$\begin{aligned} \begin{aligned}&|\langle w^{2 \vartheta }\partial _{i} \sigma ^{i} g_{1}, g_{2} \rangle | + |\langle w^{2 \vartheta } Kg_{1}, g_{2} \rangle |\\&\quad \le \frac{C}{m}|g_{1}|_{\sigma ,\vartheta }|g_{2}|_{\sigma ,\vartheta } + C(m) \left\{ \int _{|v|\le C(m)} |w^{\vartheta }g_{1}|^{2} \mathrm{d}v \right\} ^{1/2} \left\{ \int _{|v|\le C(m)} |w^{\vartheta }g_{2}|^{2} \mathrm{d}v\right\} ^{1/2}. \end{aligned} \end{aligned}$$
Moreover, there is\(\delta >0\), such that
$$\begin{aligned} \langle Lg,g \rangle \ge \delta |(I-{\mathbf {P}}) g|_{\sigma }^{2}. \end{aligned}$$

Lemma 2.7

(Lemma 6 in [13]) LetL, A, andKbe defined as in (1.8), (1.9), and (1.10). Let\(\vartheta \in {\mathbb {R}}\)and\(|\beta |\ge 0\). For small\(\delta >0\), there exists\(C_{\delta } = C_{\delta }(\vartheta )>0\)such that
$$\begin{aligned}&-\langle w^{2 \vartheta } Ag, g \rangle \ge |g|_{\sigma , \vartheta }^{2} - \delta |g|_{\sigma , \vartheta }^{2} - C_{\delta }|\mu g|_{2}^{2},\\& \quad |\langle w^{2 \vartheta } K g_{1}, g_{2} \rangle | \le \left\{ \delta |g_{1}|_{\sigma ,\vartheta } +C_{\delta }|\mu g_{1}|_{2} \right\} |g_{2} |_{\sigma , \vartheta }. \end{aligned}$$
Thus we have
$$\begin{aligned} \frac{1}{2}|g|_{\sigma ,\vartheta }^{2} - C_{\vartheta }|g|_{\sigma }^{2} \le \langle w^{2 \vartheta } Lg, g \rangle \le \frac{3}{2}|g|_{\sigma ,\vartheta }^{2} + C_{\vartheta }|g|_{\sigma }^{2}. \end{aligned}$$
For the nonlinear estimate in Theorem 3 in [13], they estimated
$$\begin{aligned} \left( w^{2 \vartheta }\Gamma [g_{1},g_{2}], g_{3} \right) \end{aligned}$$
in terms of \(\Vert g_{i}\Vert _{2, \vartheta }\) and \(\Vert g_{i}\Vert _{\sigma , \vartheta }\) for \(i=1,2,3\) and \(\vartheta \ge 0\). To get such an \(L^{2}\) estimate, they need a higher order regularity like \(\Vert D_{\beta }^{\alpha } g_{i}\Vert _{2, \vartheta }\) and \(\Vert D_{\beta }^{\alpha } g_{i}\Vert _{\sigma , \vartheta }\) for \(i=1,2,3\), \(\vartheta \ge 0\), \(|\alpha | + |\beta | \le N\) and \(N\ge 8\).

The following lemma is a refinement of Theorem 3 in [13]. First, the range of \(\vartheta \) is extended to \({\mathbb {R}}\). Second, we estimate the nonlinear term in terms of \(\Vert \cdot \Vert _{\infty }\), \(\Vert \cdot \Vert _{2, \vartheta }\), and \(\Vert \cdot \Vert _{\sigma , \vartheta }\) without a higher order regularity.

Theorem 2.8

Let\(\Gamma \)be defined as in (1.11).
  1. (1)
    For every\(\vartheta \in {\mathbb {R}}\), there exists\(C_{\vartheta }\)such that
    $$\begin{aligned} |\langle w^{2 \vartheta } \Gamma [g_{1},g_{2}], g_{3} \rangle |\le C_{\vartheta }|g_{1}|_{\infty }|g_{2}|_{\sigma , \vartheta } |g_{3}|_{\sigma , \vartheta }, \end{aligned}$$
    (2.15)
    and
    $$\begin{aligned} \left| \left( w^{2 \vartheta } \Gamma [g_{1},g_{2}], g_{3} \right) \right| \le C_{\vartheta }\Vert g_{1}\Vert _{\infty }\Vert g_{2}\Vert _{\sigma , \vartheta }\Vert g_{3}\Vert _{\sigma , \vartheta }. \end{aligned}$$
    (2.16)
     
  2. (2)
    There exists\({{\bar{\vartheta }}}<0\)such that for any\(\vartheta \le {{\bar{\vartheta }}}\),
    $$\begin{aligned} \left| \left( w^{2 \vartheta } \Gamma [g_{1},g_{2}], g_{3} \right) \right| \le C_{\vartheta }\min \{ \Vert g_{1}\Vert _{2, \vartheta }, \Vert g_{1}\Vert _{\sigma ,\vartheta } \}(\Vert g_{2}\Vert _{\infty }+ \Vert D_{v} g_{2}\Vert _{\infty })\Vert g_{3}\Vert _{\sigma ,\vartheta }. \end{aligned}$$
    (2.17)
     

Proof

  1. (1)
    By the integration by parts, we have
    $$\begin{aligned} |\langle w^{2 \vartheta } \Gamma [g_{1},g_{2}],g_{3} \rangle | & \le | \langle \partial _{i} w^{2 \vartheta }\{\phi ^{ij}*[\sqrt{\mu }g_{1}]\}\partial _{j} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta }\{\partial _{j} \phi ^{ij}*[\sqrt{\mu }g_{1}]\} g_{2}, \partial _{i} g_{3} \rangle | \nonumber \\& \quad + | \langle \partial _{i} w^{2 \vartheta }\{\partial _{j}\phi ^{ij} *[\sqrt{\mu }g_{1}]\} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle \partial _{i} w^{2 \vartheta }\{\phi ^{ij}*[v_{j} \sqrt{\mu }g_{1}]\} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta }\{\partial _{j} \phi ^{ij}*[v_{i} \sqrt{\mu }g_{1}]\} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta }\{\phi ^{ij}*[ \partial _{j} \{v_{i} \sqrt{\mu }\} g_{1}]\} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta }\{\phi ^{ij}* [v_{i}\sqrt{\mu }g_{1} ]\}\partial _{j} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta }\{\phi ^{ij}*[ v_{j} \sqrt{\mu }g_{1}]\} g_{2}, \partial _{i} g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta }\{\phi ^{ij}*[\sqrt{\mu }g_{1}]\}\partial _{j} g_{2}, \partial _{i} g_{3} \rangle | \nonumber \\ & = (I) + (II) + \dots + (IX), \end{aligned}$$
    (2.18)
    where \(\phi \) is the matrix defined as in (1.2). Clearly, \(|\partial _{i} w^{2 \vartheta }| \le C_{\vartheta } (1+|v|)^{-1}w^{2 \vartheta }\) and by Lemma 2.2, we have
    $$\begin{aligned}&|\phi ^{ij} * [\sqrt{\mu }g_{1}] | + |\phi ^{ij} * [v_{i} \sqrt{\mu }g_{1}] | + |\phi ^{ij} * [v_{j} \sqrt{\mu }g_{1}] | + |\phi ^{ij} * [\partial _{j} \{v_{i} \sqrt{\mu }\} g_{1}] | \le C(1+|v|)^{-1}\Vert g_{1}\Vert _{\infty },\\&|\partial _{j} \phi ^{ij} * [\sqrt{\mu }g_{1}] | + |\partial _{j} \phi ^{ij} * [v_{i} \sqrt{\mu }g_{1}]| \le C(1+|v|)^{-2}\Vert g_{1}\Vert _{\infty }. \end{aligned}$$
    Therefore, by Lemma 2.5 and the Hölder inequality,
    $$\begin{aligned} (I) & \le C_{\vartheta } |g_{1}|_{\infty }| \langle w^{\vartheta } (1+|v|)^{-3/2}\partial _{j} g_{2}, w^{\vartheta } (1+|v|)^{-1/2} g_{3} \rangle |\\ & \le C_{\vartheta } |g_{1}|_{\infty }|g_{2}|_{\sigma ,\vartheta } |g_{3} |_{\sigma , \vartheta },\\ (II) & \le C |g_{1}|_{\infty }| \langle w^{\vartheta } (1+|v|)^{-1/2} g_{2}, w^{\vartheta } (1+|v|)^{-3/2} \partial _{i} g_{3} \rangle |\\ & \le C |g_{1}|_{\infty }|g_{2}|_{\sigma ,\vartheta } |g_{3}|_{\sigma , \vartheta },\\ (III) + (IV) + (V) + (VI) & \le C_{\vartheta }|g_{1}|_{\infty }| \langle w^{\vartheta } (1+|v|)^{-1/2} g_{2}, w^{\vartheta } (1+|v|)^{-1/2} g_{3} \rangle |\\& \le C_{\vartheta } |g_{1}|_{\infty }|g_{2}|_{\sigma ,\vartheta } |g_{3} |_{\sigma , \vartheta }. \end{aligned}$$
    By (2.7) and the Hölder inequality,
    $$\begin{aligned} (VII) & \le C|g_{1}|_{\infty }\int w^{2 \vartheta } \big| (1+|v|)^{-3/2}|P_{v} \partial _{j} g_{2}| + (1+|v|)^{-1/2} |(I-P_{v}) \partial _{j} g_{2}| \big| (1+|v|)^{-1/2} |g_{3}| \mathrm{d}v\\ & \le C|g_{1}|_{\infty }|g_{2}|_{\sigma , \vartheta } |g_{3}|_{\sigma , \vartheta },\\ (VIII) &\le C|g_{1}|_{\infty }\int w^{2 \vartheta } (1+|v|)^{-1/2} |g_{2}| \big| (1+|v|)^{-3/2}|P_{v} \partial _{j} g_{3}| + (1+|v|)^{-1/2}|(I-P_{v}) \partial _{j} g_{3}| \big| \mathrm{d}v\\ &\le C|g_{1}|_{\infty }|g_{2}|_{\sigma , \vartheta } |g_{3}|_{\sigma , \vartheta }, \end{aligned}$$
    and
    $$\begin{aligned} (IX) &\le C|g_{1}|_{\infty }\int w^{2 \vartheta } \big| (1+|v|)^{-3/2}|P_{v} \partial _{j} g_{2}| + (1+|v|)^{-1/2} |(I-P_{v}) \partial _{j} g_{2}| \big| \\& \quad \times \big| (1+|v|)^{-3/2}|P_{v} \partial _{j} g_{3}| + (1+|v|)^{-1/2}|(I-P_{v})\partial _{j} g_{3}| \big| \mathrm{d}v\\& \le C|g_{1}|_{\infty }|g_{2}|_{\sigma , \vartheta } |g_{3}|_{\sigma ,\vartheta }. \end{aligned}$$
    Thus we obtain (2.15). By applying the Hölder inequality to (2.15),
    $$\begin{aligned} \left| \left( w^{2 \vartheta } \Gamma [g_{1},g_{2}], g_{3} \right) \right| & = \int |\langle w^{2 \vartheta } \Gamma [g_{1},g_{2}], g_{3} \rangle |\mathrm{d}x\\ & \le \int C_{\vartheta }|g_{1}|_{\infty }|g_{2}|_{\sigma , \vartheta }|g_{3}|_{\sigma , \vartheta } \mathrm{d}x\\ &\le C_{\vartheta }\Vert g_{1}\Vert _{\infty }\Vert g_{2}\Vert _{\sigma , \vartheta }\Vert g_{3}\Vert _{\sigma , \vartheta }. \end{aligned}$$
    Thus we have (2.16).
     
  2. (2)
    By the integration by parts again, we have
    $$\begin{aligned} |\langle w^{2 \vartheta } \Gamma [g_{1},g_{2}], g_{3} \rangle | & := | \langle w^{2 \vartheta } \{\phi ^{ij}*[\mu ^{1/2}g_{1}]\}\partial _{j} g_{2}, \partial _{i} g_{3} \rangle | \nonumber \\& \quad +|\langle w^{2 \vartheta } \{\phi ^{ij}*[v_{i} \mu ^{1/2} g_{1}] \}\partial _{j} g_{2} , g_{3} \rangle | \nonumber \\& \quad + | \langle w^{2 \vartheta } \{\phi ^{ij}*[\mu ^{1/2}\partial _{j} g_{1}] \} g_{2}, \partial _{i} g_{3} \rangle | \nonumber \\& \quad + |\langle w^{2 \vartheta } \{\phi ^{ij}*[v_{i} \mu ^{1/2} \partial _{j} g_{1}] \}g_{2}, g_{3} \rangle | \nonumber \\& \quad + |\langle \partial _{i} w^{2 \vartheta } \{\phi ^{ij}*[\mu ^{1/2} g_{1}] \} \partial _{j} g_{2}, g_{3} \rangle | \nonumber \\& \quad + | \langle \partial _{i} w^{2 \vartheta } \{\phi ^{ij}*[\mu ^{1/2} \partial _{j} g_{1}] \} g_{2}, g_{3} \rangle | \nonumber \\ & = (i) + (ii) + \cdots + (vi). \end{aligned}$$
    (2.19)
    By the Hölder inequality and the integration by parts, we have
    $$\begin{aligned}&|\langle \{\phi ^{ij}*[\mu ^{1/2}g_{1}]\} \rangle | + |\langle \{\phi ^{ij}*[v_{i} \mu ^{1/2} g_{1}] \} \rangle | + |\langle \{\phi ^{ij}*[\mu ^{1/2}\partial _{j} g_{1}] \} \rangle | + |\langle \{\phi ^{ij}*[v_{i} \mu ^{1/2} \partial _{j} g_{1}] \} \rangle |\\&\quad \le C_{\vartheta } (1+|v|)^{-1}\min \{ |g_{1}|_{2, \vartheta },|g_{1}|_{\sigma , \vartheta }\}. \end{aligned}$$
    Let \({{\bar{\vartheta }}} := -2\), then by applying the Hölder inequality to (i) and Lemma 2.5, we have
    $$\begin{aligned} {(i)} & \le C_{\vartheta } \min \{ |g_{1}|_{2, \vartheta },|g_{1}|_{\sigma , \vartheta }\} \left| \int _{{\mathbb {R}}^{3}} w^{2\vartheta } (1+|v|)^{-1} \partial _{j} g_{2}(v) \partial _{i} g_{3}(v) \mathrm{d}v\right| \\ & \le C_{\vartheta } \min \{ |g_{1}|_{2, \vartheta },|g_{1}|_{\sigma , \vartheta }\} |\partial _{j} g_{2}|_{\infty } \left( \int _{{\mathbb {R}}^{3}} (1+|v|)^{2 \vartheta + 1} \mathrm{d}v\right) ^{1/2}\\ & \quad \times \left( \int _{{\mathbb {R}}^{3}} w^{2\vartheta } (1+|v|)^{-3} |\partial _{i} g_{3}|^{2} \mathrm{d}v \right) ^{1/2}\\ & \le C_{\vartheta } \min \{ |g_{1}|_{2, \vartheta },|g_{1}|_{\sigma , \vartheta }\} |D_{v} g_{2}|_{\infty }|g_{3}|_{\sigma , \vartheta }. \end{aligned}$$
    Similarly,
    $$\begin{aligned} (ii) + (iii) + \cdots + (vi) \le C_{\vartheta } \min \{ |g_{1}|_{2, \vartheta },|g_{1}|_{\sigma , \vartheta }\}(|g_{2}|_{\infty }+ |D_{v} g_{2}|_{\infty })|g_{3}|_{\sigma , \vartheta }. \end{aligned}$$
    Therefore, by the Hölder inequality again, we have
    $$\begin{aligned} \left| \left( w^{2 \vartheta } \Gamma [g_{1}, g_{2}], g_{3} \right) \right| & = \int | \langle w^{2 \vartheta } \Gamma [g_{1}, g_{2}], g_{3} \rangle | \mathrm{d}x\\ & \le \int C_{\vartheta } \min \{ |g_{1}|_{2, \vartheta },|g_{1}|_{\sigma ,\vartheta }\}(|g_{2}|_{\infty } + |D_{v} g_{2}|_{\infty }) |g_{3}|_{\sigma ,\vartheta } \mathrm{d}x\\ &\le C_{\vartheta } \min \{ \Vert g_{1}\Vert _{2, \vartheta },\Vert g_{1}\Vert _{\sigma ,\vartheta }\}(\Vert g_{2}\Vert _{\infty } + \Vert D_{v} g_{2}\Vert _{\infty }) \Vert g_{3}\Vert _{\sigma ,\vartheta }. \end{aligned}$$
     
\(\square \)

Lemma 2.9

LetKbe defined as in (1.10). Then there exists\(C=C_{\vartheta }>0\)such that for every\(N,M>0\),
$$\begin{aligned} \Vert w^{\theta }Kf\Vert _{L^{\infty } ({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}\le C\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}, \end{aligned}$$
(2.20)
$$\begin{aligned} \Vert w^{\theta }K1_{|v|>M} f\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}\le C(1+M)^{-1}\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}, \end{aligned}$$
(2.21)
and
$$\begin{aligned} \Vert w^{\theta }Kf\Vert _{L^{2}({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})} \le C N^{2}\Vert\; f^{\vartheta } \Vert _{L^{2}({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})} + \frac{C}{N}\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}. \end{aligned}$$
(2.22)

Proof

After the integration by parts, we have
$$\begin{aligned} w^{\vartheta }Kf & = - w^{\vartheta }\mu ^{-1/2}\partial _{i} \{ \mu [ \phi ^{ij}* \{ \mu ^{1/2}[\partial _{j} f + v_{j} f]\}] \} \\ & = 2w^{\vartheta }v_{i} \mu ^{1/2}[ \phi ^{ij}* \{ \mu ^{1/2}[\partial _{j} f + v_{j} f] \} ] - w^{\vartheta }\mu ^{1/2}[ \partial _{i} \phi ^{ij}* \{ \mu ^{1/2}[\partial _{j} f + v_{j} f] \} ] \\ & = 2w^{\vartheta }v_{i} \mu ^{1/2}[ \phi ^{ij}*(v_{j} \mu ^{1/2} f)] - 2w^{\vartheta }v_{i} \mu ^{1/2}[ \phi ^{ij}*(\partial _{j}\mu ^{1/2} f)] + 2w^{\vartheta }v_{i} \mu ^{1/2}[ \partial _{j} \phi ^{ij}*(\mu ^{1/2} f)] \\& \quad -w^{\vartheta }\mu ^{1/2}[ \partial _{i} \phi ^{ij}* (v_{j}\mu ^{1/2}f)] + w^{\vartheta }\mu ^{1/2}[ \partial _{i} \phi ^{ij}* (\partial _{j}\mu ^{1/2}f)] -w^{\vartheta }\mu ^{1/2}[ \partial _{ij} \phi ^{ij}* (\mu ^{1/2}f)] \\ & = 4w^{\vartheta }v_{i} \mu ^{1/2}[ \phi ^{ij}*(v_{j} \mu ^{1/2} f)] + 2w^{\vartheta }v_{i} \mu ^{1/2}[ \partial _{j} \phi ^{ij}*(\mu ^{1/2} f)] \\& \quad -w^{\vartheta }2\mu ^{1/2}[ \partial _{i} \phi ^{ij}* (v_{j} \mu ^{1/2}f)] - w^{\vartheta }\mu ^{1/2}[ \partial _{ij} \phi ^{ij}* (\mu ^{1/2}f)] \\ & = 4w^{\vartheta }v_{i} \mu ^{1/2}[ \phi ^{ij}*(v_{j} w^{-\vartheta }\mu ^{1/2} f^{\vartheta })] + 2w^{\vartheta }v_{i} \mu ^{1/2}[ \partial _{j} \phi ^{ij}*(w^{-\vartheta }\mu ^{1/2} f^{\vartheta })] \\& \quad -w^{\vartheta }2\mu ^{1/2}[ \partial _{i} \phi ^{ij}* (v_{j} w^{-\vartheta }\mu ^{1/2}f^{\vartheta })] - w^{\vartheta }\mu ^{1/2}[ \partial _{ij} \phi ^{ij}* (w^{-\vartheta }\mu ^{1/2}f^{\vartheta })]\\ & = {(I)} + {(II)} + { (III)} + {(VI)}. \end{aligned}$$
Applying Lemma 2.2 to \((I) + (II) + (III)\), we have \((I) + (II) + (III) \le C\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^3 \times {\mathbb {R}}^3)}\). Note that \(\partial _{ij}\phi ^{ij}*(\mu ^{1/2}f) = -8 \pi \mu ^{1/2}f\). Thus we also have \({(VI)} \le C\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^3 \times {\mathbb {R}}^3)}\). Therefore, we have (2.20).

Since every convolution term of Kf contains \(\mu ^{1/2}\), we have (2.21).

For (2.22), clearly we have
$$\begin{aligned} \Vert 1_{|v|\ge N}w^{\vartheta }K f\Vert _{2} \le \frac{C}{N} \Vert w^{\vartheta }f\Vert _{\infty }. \end{aligned}$$
(2.23)
Now we will estimate \(\Vert 1_{|v|<N}w^{\vartheta }K f\Vert _{2}\). First, consider \( \Vert 1_{|v|<N} v_{i} w^{\vartheta }\mu ^{1/2} [ \partial _{j} \phi ^{ij}*(w^{-\vartheta }\mu ^{1/2} f^{\vartheta })] \Vert _{L^{2}({\mathbb {T}}^3 \times {\mathbb {R}}^3)}\).
$$\begin{aligned}&\left\| 1_{|v|<N} v_{i} w^{\vartheta }\mu ^{1/2}\left[ \partial _{j} \phi ^{ij}*(w^{-\vartheta }\mu ^{1/2} f^{\vartheta })\right] \right\| _{2}^{2}\\&\quad = \int _{{\mathbb {T}}^{3}}\int _{|v|<N} \left( \int \partial _{j} \phi ^{ij}(v-v^{\prime })v_{i}w^{\vartheta }(v)\mu ^{1/2}(v) \mu ^{1/2}(v^{\prime })w^{- \vartheta }(v^{\prime })f^{\vartheta }(v^{\prime }) \mathrm{d}v^{\prime }\right) ^{2} \mathrm{d}v \mathrm{d}x\\&\quad \le C \left( \int _{{\mathbb {T}}^{3}} \int _{|v|<N} \left( \int _{1/N<|v-v^{\prime }|<2N}\mathrm{d}v^{\prime }\right) ^{2} \mathrm{d}v\mathrm{d}x\right. + \int _{{\mathbb {T}}^{3}}\int _{|v|<N} \left( \int _{|v-v^{\prime }|>2N}\mathrm{d}v^{\prime }\right) ^{2} \mathrm{d}v \mathrm{d}x\\&\qquad \left. + \int _{{\mathbb {T}}^{3}}\int _{|v|<N} \left( \int _{|v-v^{\prime }|<1/N}\mathrm{d}v^{\prime }\right) ^{2} \mathrm{d}v \mathrm{d}x\right) \\&\quad ={(i)} + { (ii)} + {(iii)}. \end{aligned}$$
Since \(|\partial _{j} \phi ^{ij}(v-v^{\prime })|\le C|v-v^{\prime }|^{-2}\), by the Minkowski and Hölder inequalities,
$$\begin{aligned} {(i)}&\le \int _{{\mathbb {T}}^{3}}\left( \int _{|v^{\prime }|<3N}\left( \int _{1/N<|v-v^{\prime }|<2N}|v-v^{\prime }|^{-4} w^{2 \vartheta }(v)v_{i}^2\mu (v)w^{-2 \vartheta }(v^{\prime })\mu (v^{\prime })(f^{\vartheta })^{2}(v^{\prime })\mathrm{d}v\right) ^{1/2}\mathrm{d}v^{\prime }\right) ^{2} \mathrm{d}x\\&= \int _{{\mathbb {T}}^{3}}\left( \int _{|v^{\prime }|<3N}w^{- \vartheta }(v^{\prime })\mu ^{1/2}(v^{\prime })f^{\vartheta }(v^{\prime })\left( \int _{1/N<|v-v^{\prime }|<2N}|v-v^{\prime }|^{-4} w^{2 \vartheta }(v)v_{i}^2\mu (v) \mathrm{d}v\right) ^{1/2}\mathrm{d}v^{\prime }\right) ^{2}\mathrm{d}x\\&\le C\int _{{\mathbb {T}}^{3}}\left( \int _{|v^{\prime }|<3N} \mu ^{1/2}(v^{\prime })(f^{\vartheta })^{2}(v^{\prime })\mathrm{d}v^{\prime }\right) \left( \int _{|v^{\prime }|<3N}\int _{1/N<|v-v^{\prime }|<2N}|v-v^{\prime }|^{-4}w^{2\vartheta }(v) v_{i}^{2}\mu (v) \mathrm{d}v\mathrm{d}v^{\prime }\right) \mathrm{d}x\\&\le C N^{4}\Vert\; f^{\vartheta }\Vert _{2}^{2}. \end{aligned}$$
Note that if \(|v|<N\) and \(|v-v^{\prime }|>2N\), then \(|v^{\prime }|>N\). Since the integrand of (ii) contains a Maxwellian and \(|v^{\prime }|>N\), for every \(\beta >0\) we have
$$\begin{aligned} { (ii)} \le \frac{C_{\beta }}{N^{2\beta }}\Vert\; f^{\vartheta }\Vert _{\infty }^{2}. \end{aligned}$$
Finally,
$$\begin{aligned} {(iii)} \le C \Vert\; f^{\vartheta }\Vert _{\infty }^{2} \iint _{{\mathbb {T}}^{3} \times [0,\infty )}\int _{{\mathbb {R}}^{3}} \mu (v)\left( \int _{|v-v^{\prime } |<1/N}|v-v^{\prime }|^{-2}\mathrm{d}v^{\prime }\right) ^{2}\mathrm{d}v \mathrm{d}x\mathrm{d}t \le C\frac{1}{N^{2}}\Vert\; f^{\vartheta }\Vert _{\infty }^{2}. \end{aligned}$$
So we have
$$\begin{aligned} \Vert 1_{|v|<N} w^{\vartheta }\mu ^{1/2} [ \partial _{ij} \phi ^{ij}* (w^{-\vartheta }\mu ^{1/2} f^{\vartheta }) ] \Vert _{2} \le CN^{2}\Vert\; f^{\vartheta }\Vert _{2} + \frac{C}{N}\Vert\; f^{\vartheta }\Vert _{\infty }. \end{aligned}$$
(2.24)
In a similar manner,
$$\begin{aligned} \Vert 1_{|v|<N} w^{\vartheta }\mu ^{1/2} [ \partial _{i} \phi ^{ij}* (v_{j} w^{-\vartheta }\mu ^{1/2}f^{\vartheta })] \Vert _{2} \le CN^{2}\Vert\; f^{\vartheta }\Vert _{2} + \frac{C}{N}\Vert\; f^{\vartheta }\Vert _{\infty } \end{aligned}$$
(2.25)
and
$$\begin{aligned} \Vert 1_{|v|<N} v_{i} w^{\vartheta }\mu ^{1/2} [ \phi ^{ij}*(v_{j} w^{-\vartheta }\mu ^{1/2} f^{\vartheta })] \Vert _{2}\le CN\Vert\; f^{\vartheta }\Vert _{2} + \frac{C}{N^{2}}\Vert\; f^{\vartheta }\Vert _{\infty }. \end{aligned}$$
(2.26)
Note that
$$\begin{aligned} w^{\vartheta }\mu ^{1/2} [\partial _{ij}\phi ^{ij}*(w^{-\vartheta }\mu ^{1/2}f^{\vartheta })]&= w^{\vartheta }\mu ^{1/2} [-8 \pi w^{-\vartheta }\mu ^{1/2}f^{\vartheta } ]\\&=-8 \pi \mu f^{\vartheta }. \end{aligned}$$
Thus,
$$\begin{aligned} \Vert 1_{|v|<N} w^{\vartheta }\mu ^{1/2} [\partial _{ij}\phi ^{ij}*(w^{-\vartheta }\mu ^{1/2}f^{\vartheta })]\Vert _{2} \le C \Vert\; f^{\vartheta }\Vert _{2}. \end{aligned}$$
(2.27)
From (2.23)–(2.27), we have (2.22).

So the proof is complete. \(\square \)

3 Maximum Principle

To get \(L^{\infty }\) estimates, we rearrange (1.5) as follows:
$$\begin{aligned} f_{t}+v\cdot \nabla _{x}f={\bar{A}}_{f}f+{\bar{K}}_{f}f, \end{aligned}$$
(3.1)
where
$$\begin{aligned} {\bar{A}}_{g}f &:= \partial _{i} [ \{ \phi ^{ij}*[\mu +\mu ^{1/2}g] \} \partial _{j}f ] \nonumber \\&\quad\; - \{ \phi ^{ij}*[v_{i}\mu ^{1/2}g]\} \partial _{j}f- \{ \phi ^{ij}*[\mu ^{1/2}\partial _{j}g]\} \partial _{i}f \nonumber \\ &=: \nabla _{v}\cdot (\sigma _{G}\nabla _{v}f)+a_{g}\cdot \nabla _{v}f, \end{aligned}$$
(3.2)
$$\begin{aligned} {\bar{K}}_{g}f& := Kf+\partial _{i}\sigma ^{i}f -\sigma ^{ij}v_{i}v_{j}f \nonumber \\& \quad - \partial _{i} \{ \phi ^{ij}*[\mu ^{1/2}\partial _{j}g] \} f+ \{ \phi ^{ij}*[v_{i}\mu ^{1/2}\partial _{j}g]\} f. \end{aligned}$$
(3.3)
If f satisfies (1.22), then for every \(\vartheta \in {\mathbb {R}}\), \(f^{\vartheta }:=w^{\vartheta }f\) satisfies
$$\begin{aligned}&\partial _{t}f^{\vartheta }+v\cdot \nabla _{x}f^{\vartheta }={\bar{A}}_{g} ^{\vartheta }f^{\vartheta }+{\bar{K}}_{g}^{\vartheta }f,\\&f^{\vartheta }(0)=w^{\vartheta }f_{0}=:f_{0}^{\vartheta }, \end{aligned}$$
where
$$\begin{aligned} {\bar{K}}_{g}^{\vartheta }f=w^{\vartheta }{\bar{K}}_{g}f+\left( 2\frac{\partial _{i}w^{\vartheta }\partial _{j}w^{\vartheta }}{w^{2\vartheta }}\sigma _{G}^{ij}-\frac{\partial _{ij}w^{\vartheta }}{w^{\vartheta }}\sigma _{G}^{ij}-\frac{\partial _{j}w^{\vartheta }}{w^{\vartheta }}\partial _{i}\sigma _{G}^{ij}-\frac{\partial _{i}w^{\vartheta }}{w^{\vartheta }}a_{g}^{i}\right) f^{\vartheta }, \end{aligned}$$
(3.4)
$$\begin{aligned} {\bar{A}}_{g}^{\vartheta }:={\bar{A}}_{g}-2\frac{\partial _{i}w^{\vartheta } }{w^{\vartheta }}\sigma _{G}^{ij}\partial _{j}. \end{aligned}$$
(3.5)
Note that \({\bar{A}}_{g}^{0}={\bar{A}}_{g}\) and \(f^{0}=f\). In this section, we first define a weak solution for
$$\begin{aligned} h_{t}+v\cdot \nabla _{x}h={\bar{A}}_{g}^{\vartheta }h \end{aligned}$$
(3.6)
and obtain the well-posedness and the maximum principle of the weak solution for (3.6). Due to the lack of regularity, we cannot use a direct contradiction argument for the weak solution as in the case of strong solutions. Therefore, we first construct a smooth approximated solution and then pass to the limit to obtain the maximum principle for the weak solution.

Remark 3.1

If \(g\in C^{1}_{v}\), \(f\in C^{2}_{v}\), and \(\varphi \in C^{2}_{v}\) with a compact support in \({\mathbb {R}} ^{3}\), then
$$\begin{aligned} \int _{{\mathbb {R}}^3} \left( {{\bar{A}}}_{g}(f) + {{\bar{K}}}_{g}(f) \right) \varphi & = - \langle f, \varphi \rangle _{\sigma } + \int _{{\mathbb {R}}^3} \left[ - \sigma _{\mu ^{1/2}g} \partial _{i}f \partial _{j} \varphi - (a_{g} \cdot \nabla _{v} \varphi ) f\right. \\&\left. \quad + \left( K\varphi + \partial _{i} \sigma ^{i} \varphi - \partial _{i} \left\{ \phi ^{ij} * [\mu ^{1/2} \partial _{j} g] \right\} \varphi + \left\{ \phi ^{ij}* [v_{i} \mu ^{1/2}\partial _{j}g] \right\} \varphi \right) f \right] \mathrm{d}v. \end{aligned}$$

Definition 3.2

Let \(h(t,x,v) \in L^{\infty }((0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3},w^{\vartheta }(v)\mathrm{d}t\mathrm{d}x\mathrm{d}v)\) be a periodic function in \(x\in {\mathbb {T}}^{3} = [-\pi ,\pi ]^{3}\) satisfying
$$\begin{aligned} \int _{0}^{t}\iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} \left( \sigma ^{ij}\partial _{i}h \partial _{j}h\right) (s,x,v) \mathrm{d}x\mathrm{d}v\mathrm{d}s< \infty , \end{aligned}$$
where \(\sigma \) is defined as in (2.2). We say that h is a weak solution of (3.6), with \(h(0) = h_0\) on \((0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\) if for all \(t\in (0,\infty )\) and all \(\varphi \in C^{1,1,1}_{t,x,v}\left( (0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\right) \) such that \(\varphi (t,x,v)\) is a periodic function in \(x\in {\mathbb {T}}^{3} = [-\pi ,\pi ]^{3}\) and \(\varphi (t,x,\cdot )\) is compactly supported in \({\mathbb {R}}^{3}\), it satisfies
$$\begin{aligned}&\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} h(t,x,v) \varphi (t,x,v) \mathrm{d}x\mathrm{d}v - \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} h_{0}(x,v) \varphi (0,x,v) \mathrm{d}x\mathrm{d}v \nonumber \\&\quad = \iiint _{(0,t)\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \bigg[ h(s,x,v)\left( \partial _{s} \varphi + v \cdot \nabla _{x}\varphi - \left( a_g + 2 \frac{\nabla w^{\vartheta }}{w^{\vartheta }} \sigma _{G}\right) \cdot \nabla _v \varphi \right) (s,x,v) \nonumber \\&\qquad - \nabla _{v}h(s,x,v) \cdot (\sigma _{G} \nabla _{v}\varphi )(s,x,v) \bigg]\mathrm{d}s\mathrm{d}x\mathrm{d}v, \end{aligned}$$
(3.7)
where \(\sigma _G\) is defined as in (2.1) with \(G = \mu + \mu ^{1/2} g\).

Lemma 3.3

Assume (2.6). Let\(\sigma _G\)be the matrix defined as in (2.1) with\(G=\mu + \mu ^{1/2}g\). Let\(\vartheta \in {\mathbb {N}}\cup \{0\}\), \(\delta \ge 0\), andhbe a classical solution of (3.6). Then there exist\(C = C(\vartheta )\), \(0<\varepsilon \ll 1\)such that if\(\Vert g\Vert _{\infty } < \varepsilon \), then
$$\begin{aligned} \sup _{0\le s\le t}\Vert h(s)\Vert _{L^{2}}^{2}+\int _{0}^{t}\iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} \left( \sigma ^{ij}\partial _{i}h \partial _{j}h\right) (s,x,v) \mathrm{d}x\mathrm{d}v\mathrm{d}s\le C(\vartheta )\Vert h(0)\Vert _{L^{2}}^{2}. \end{aligned}$$
(3.8)

Proof

Multiplying (3.6) by h and integrating both sides of the resulting equation, we have
$$\begin{aligned} \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} \frac{1}{2} \left( h^2(t,x,v) - h^2(0,x,v) \right) \mathrm{d}x\mathrm{d}v = \int _0^t \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} ({{\bar{A}}}_{g}^{\vartheta } h(s,x,v) )h(s,x,v) \mathrm{d}x\mathrm{d}v\mathrm{d}s. \end{aligned}$$
(3.9)
By Lemma 2.4, we have
$$\begin{aligned} C^{-1} \sigma ^{ij}\partial _{i}h \partial _{j}h \le \sigma _{G}^{ij}\partial _{i}h \partial _{j}h \le C\sigma ^{ij}\partial _{i}h \partial _{j}h, \end{aligned}$$
(3.10)
for some C. By Lemma 2.4 and the Young inequality, we have
$$\begin{aligned} \left| \frac{\partial _{i}w^{\vartheta }}{w^{\vartheta }}\sigma _{G}^{ij} (\partial _{j}h) h\right| & \le C (1+|v|)^{-1} (\sigma _{G}^{ij}\partial _{i}h\partial _{j}h)^{1/2}(\sigma _{G}^{ij}h^2)^{1/2} \nonumber \\ &\le \varepsilon \sigma _{G}^{ij}\partial _{i}h\partial _{j}h + C_{\varepsilon } (1+|v|)^{-1}\sigma _{G}^{ij}h^2 \nonumber \\ &\le \varepsilon \sigma ^{ij}\partial _{i}h\partial _{j}h + C_{\varepsilon } h^2. \end{aligned}$$
(3.11)
In a similar manner, by (2.7) and the Young inequality, we have
$$\begin{aligned} | \{\phi * [v_i \mu ^{1/2}g] \} (\partial _{i}h) h | &\le C\Vert g\Vert _{\infty } D_{\mu }(\nabla _{v} h ; v)^{1/2}(\sigma ^{ij}h^2)^{1/2} \nonumber \\ &\le \varepsilon \sigma ^{ij}\partial _{i}h\partial _{j}h + \varepsilon h^2 \end{aligned}$$
(3.12)
and
$$\begin{aligned} | \{ \phi ^{ij}*[\mu ^{1/2}\partial _{j}g] \} \partial _{i}h h | \le \varepsilon \sigma ^{ij}\partial _{i}h\partial _{j}h + \varepsilon h^2. \end{aligned}$$
(3.13)
Thus from (3.9)–(3.13), we have
$$\begin{aligned}&\iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} h^2(t,x,v)\mathrm{d}x\mathrm{d}v + \int _{0}^{t}\iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} \left( \sigma ^{ij}\partial _{i}h \partial _{j}h\right) (s,x,v) \mathrm{d}x\mathrm{d}v\mathrm{d}s \\&\quad \le \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} h^2(0,x,v)\mathrm{d}x\mathrm{d}v + \varepsilon \int _{0}^{t}\iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} \left( \sigma ^{ij}\partial _{i}h \partial _{j}h\right) (s,x,v) \mathrm{d}x\mathrm{d}v\mathrm{d}s \\& \qquad + C_{\varepsilon } \int _{0}^{t}\iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} h^2(s,x,v) \mathrm{d}x\mathrm{d}v\mathrm{d}s. \end{aligned}$$
Absorbing the second term of the RHS to the LHS and applying the Gronwall inequality to the resulting equation, we have (3.8). \(\square \)

Lemma 3.4

Assume (2.6). Then there exists a unique weak solution to (3.6) which satisfies (3.8).

Proof

We approximate g by \(g^{\delta } \in C^{\infty }\) and \(h_{0}\) by \(h^{\delta }_{0}\in C^\infty \) such that \(\Vert g^{\delta }\Vert _{\infty }\le \Vert g\Vert _{\infty }\), \(\Vert h^{\delta }_{0}\Vert _{\infty }\le \Vert h_{0}\Vert _{\infty }\) and
$$\begin{aligned} \lim \Vert g^{\delta }-g\Vert _{\infty }=0,\quad \lim \Vert h^{\delta }_{0}-h_{0}\Vert _{1}=0. \end{aligned}$$
Consider
$$\begin{aligned}&\partial _{t} h^{\delta }+v\cdot \nabla _{x}h^{\delta }={\bar{A}}_{g^{\delta } }^{\vartheta }h^{\delta } \nonumber, \\&h^{\delta }(0,x,v) = h^{\delta }_{0}(x,v). \end{aligned}$$
(3.14)
By Lemma 2.4, \(\sigma _G \ge 0\). Since \(\sigma _G \ge 0\), it is rather standard (for instance, by adding regularization \(\varepsilon (\nabla _{x,v})^{2m}\), for some large integer m, then letting \(\varepsilon \rightarrow 0\), if necessary) that there exists a solution \(h^{\delta }\) to the linear equation (3.14). Since \(g^{\delta }\) and \(h^{\delta }_{0}\) are smooth, we can derive a similar energy estimate for the derivatives of \(h^{\delta }\) by taking derivatives of the above equation and multiplying by the derivatives of \(h^{\delta }\) and integrating both sides of the resulting equation as in [13]. For more details, see [13]. Therefore, \(h^{\delta }\) is smooth.
By (3.8), \(\Vert h^{\delta }(s)\Vert _{L^2}^2\) is uniformly bounded on \(0\le s\le t\). Therefore, there exists h such that \(h^{\delta }\rightarrow h\) weakly in \(L^2\). Multiplying (3.14) by a test function \(\varphi \), integrating both sides of the resulting equation, and taking the integration by parts, we have
$$\begin{aligned}&\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} h^{\delta }(t,x,v) \varphi (t,x,v) \mathrm{d}x\mathrm{d}v - \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} h^{\delta }_{0}(x,v) \varphi (0,x,v) \mathrm{d}x\mathrm{d}v \nonumber \\&\quad = \iiint _{(0,t)\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} h^{\delta }(s,x,v)\left( \partial _{s} \varphi (s,x,v) + v \cdot \nabla _{x}\varphi (s,x,v)- \left( a_{g^{\delta }} + 2 \frac{\nabla w^{\vartheta }}{w^{\vartheta }} \sigma _{G^{\delta }}\right) \cdot \nabla _v \varphi \right. \nonumber \\&\qquad \left.\vphantom{ \iiint _{(0,t)\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}} + \nabla _v \cdot (\sigma _{G^{\delta }} \nabla _v \varphi ) \right)\mathrm{d}s\mathrm{d}x\mathrm{d}v, \end{aligned}$$
(3.15)
where \(G^{\delta } = \mu + \sqrt{\mu }g^{\delta }\). Since \(h^{\delta } \rightarrow h\) weakly in \(L^2\), taking \(\delta \rightarrow 0\) in (3.15) we have (3.7). Therefore h is a weak solution of (3.6). The second assertion is an analogue of Lemma 3.3. Let h and \({{\tilde{h}}}\) be weak solutions to (3.6). Then \(h-{{\tilde{h}}}\) is also a weak solution to (3.6) with zero initial data. Therefore, we have \(\sup _{0\le s \le t} \Vert (h-{{\tilde{h}}})(s)\Vert _{L^2}^2 = 0\). Thus we obtain the uniqueness. \(\square \)
Before we derive the maximum principle for weak solutions, we establish the maximum principle for strong solutions. We first derive the maximum principle for strong solutions in bounded domains. The following technique is similar to that in [18] and [19]. Let
$$\begin{aligned} {\mathcal {M}}_{g}^{\vartheta }h \,{:=} \,(\partial _{t} + v\cdot \nabla _{x} - {{\bar{A}}}_{g}^{\vartheta })h. \end{aligned}$$

Lemma 3.5

Assume (2.6). Let\(h\in C^{1,1,2}_{t,x,v}\left( [0,T] \times {\mathbb {T}}^{3} \times B(0;M) \right) \)be a periodic function satisfying\({\mathcal {M}}_{g}^{\vartheta }h \le 0\). Thenhattains its maximum only at\(t=0\)or\(|v|=M\).

Proof

Let us assume that \(\max _{(t,x,v)\in [0,T] \times {\mathbb {T}}^{3} \times B(0;M)} h(t,x,v) > 0\) and \({\mathcal {M}}_{g}^{\vartheta }h <0\). Suppose that h attains its maximum at an interior point \((t,x,v) \in [0,T] \times {\mathbb {T}}^{3} \times B(0;M)\) or at (Txv) with (xv) lying in the interior. Since \(\sigma _G \ge 0\) by Lemma 2.4, we have \(\partial _{t} h \ge 0\), \(\nabla _{x} h =0\), and \(\nabla _{v} h =0\) while \(\sigma _{G}^{ij}\partial _{ij} h \le 0\) and \(h(t,x,v)>0\). Thus \({\mathcal {M}}_{g}^{\vartheta }h(t,x,v) \ge 0\) and this gives a contradiction.

In the case of \({\mathcal {M}}_{g}^{\vartheta }h \le 0\), define \(h^{k} := h-kt\) for \(k>0\), then \({\mathcal {M}}_{g}^{\vartheta }h^{k} <0\). Thus we have
$$\begin{aligned} \sup _{(t,x,v)\in [0,T] \times {\mathbb {T}}^{3} \times B(0;M)} h^{k}(t,x,v) = \sup _{t=0 \text { or }|v|=M} h^{k}(t,x,v). \end{aligned}$$
Taking \(k \rightarrow 0\), we complete the proof. \(\square \)

Lemma 3.6

Assume (2.6). There exists\(\varphi \in C^{1,1,2}\left( [0,T]\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\right) \)with\(\varphi \ge 0\), which satisfies\({\mathcal {M}}_{g}^{\vartheta }\varphi \ge 0\)and\(\varphi \rightarrow \infty \) as \(|v| \rightarrow \infty \)uniformly in\(t \in [0,T]\).

Proof

Define
$$\begin{aligned} \varphi (t,x,v):= \varphi (t,v) = \alpha _{1}(t) + \alpha _{2}(t) |v|^{2}. \end{aligned}$$
(3.16)
Then
$$\begin{aligned} {\mathcal {M}}_{g}^{\vartheta }\varphi & = \alpha _{1}^{\prime }(t) + \alpha _{2}^{\prime }(t)|v|^{2}- 2 \alpha _{2}(t) \nabla _{v}\cdot (\sigma _{G} v)\\& \quad - 2 \alpha _{2}(t) a_{g}\cdot v -2 \alpha _{2}(t) \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta } }\sigma ^{ij}v_{j}\\ & = \alpha _{1}^{\prime }(t) + \alpha _{2}^{\prime }(t)|v|^2- 2 \alpha _{2}(t) \partial _{i} \sigma _{G}^{ij}v_{j} - 2 \alpha _{2}(t) \sigma _{G}^{ii}\\& \quad - 2 \alpha _{2}(t)a_{g}\cdot v-2 \alpha _{2}(t) \frac{\partial _{i} w^{\vartheta } }{w^{\vartheta }}\sigma ^{ij}v_{j}. \end{aligned}$$
Note that
$$\begin{aligned}&|\partial _{i} \sigma _{G}^{ij}| \le C \Vert g\Vert _{\infty }(1+|v|)^{-2}, \quad |\sigma _{G}^{ii}|\le C \Vert g\Vert _{\infty }(1+|v|)^{-1},\\&|a_{g}| \le C \Vert g\Vert _{\infty }(1+|v|)^{-1},\text { and } \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}\sigma ^{ij}v_{j} \le C \Vert g\Vert _{\infty }(1+|v|)^{-3}. \end{aligned}$$
Choose \(\alpha _{1}(t) = \alpha _{2}(t) \,{:=} \, \exp (k t)\). Then \(\varphi \ge 0\). Moreover,
$$\begin{aligned} {\mathcal {M}}_{g}^{\vartheta }\varphi \ge e^{kt} \big( k(1+|v|)^{2} - C(1+|v|)^{-1} - C(1+|v|)^{-1}(1+|v|) \big) \ge 0 \end{aligned}$$
for a sufficiently large k. \(\square \)

Lemma 3.7

Assume (2.6). Let\(h\in C^{1,1,2}_{t,x,v}\left( [0,T] \times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3} \right) \)be aperiodic function satisfying\({\mathcal {M}}_{g}^{\vartheta }h \le 0\). Thenhattains its maximum only at\(t=0\).

Proof

Fix \(\lambda >0\). Let \(\varphi \) be a barrier function obtained in Lemma 3.6. Define \(\eta ^{\lambda }(t,x,v) := h(t,x,v) - \lambda \varphi (t,x,v)\), then \({\mathcal {M}}_{g}^{\vartheta }\eta ^{\lambda } \le 0\). Thus we can apply Lemma 3.5 on the domain \([0,T]\times {\mathbb {T}}^{3} \times B(0;M)\). Then we have
$$\begin{aligned} \eta ^{\lambda }(t,x,v) \le \sup _{t=0 \text { or } |v|=M} \eta ^{\lambda }(t,x,v), \quad \text {for }(t,x,v)\in [0,T]\times {\mathbb {T}}^{3} \times B(0;M). \end{aligned}$$
Note that
$$\begin{aligned} \eta ^{\lambda }(0,x,v) = h(0,x,v) - \lambda \varphi (0,x,v) \le h(0,x,v) \le \sup _{x,v}h(0,x,v). \end{aligned}$$
For a sufficiently large M, we have
$$\begin{aligned} \eta ^{\lambda }(t,x,v) = h(t,x,v) - \lambda \varphi (t,x,v) = h(t,x,v) - \lambda (\alpha _{1}(t) + \alpha _{2}(t)M^{2}) \le \sup _{x,v} h(0,x,v) \end{aligned}$$
for \(|v|=M\). Thus
$$\begin{aligned} \eta ^{\lambda }(t,x,v) \le \sup _{x,v}h(0,x,v), \quad \text {for }(t,x,v)\in [0,T]\times {\mathbb {T}}^{3} \times B(0;M). \end{aligned}$$
Since M is an arbitrary large enough constant, we can take \(M \rightarrow \infty \). Then we have
$$\begin{aligned} \eta ^{\lambda }(t,x,v) \le \sup _{x,v}h(0,x,v), \quad \text {for }(t,x,v)\in [0,T]\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}. \end{aligned}$$
Taking \(\lambda \rightarrow 0\), we have
$$\begin{aligned} h(t,x,v) \le \sup _{x,v}h(0,x,v). \end{aligned}$$
Thus we complete the proof. \(\square \)

Now we will derive the maximum principle for weak solutions.

Lemma 3.8

Assume (2.6) and\(g\in C^{0}\)and\(\Vert h_{0}\Vert_{\infty }<\infty .\)Then the weak solution to (3.6) satisfies
$$\begin{aligned} \sup _{t}\Vert h(t)\Vert _{\infty }\le \Vert h_{0}\Vert _{\infty }. \end{aligned}$$
(3.17)

Proof

Approximating g by \(g^{\delta } \in C^{\infty }\) and \(h_{0}\) by \(h^{\delta }_{0}\in C^\infty \) as in Lemma 3.4, we can obtain a smooth solution \(h^{\delta }\) to (3.6). Thus by Lemma 3.7, we have
$$\begin{aligned} \sup _{t} \Vert h^{\delta }(t)\Vert _{\infty } \le \Vert h^{\delta }_{0}\Vert _{\infty } \le \Vert h_{0}\Vert _{\infty }. \end{aligned}$$
In a similar manner to Lemma 3.3 we can derive an energy estimate for \(h^{\delta } - h^{\delta '}\) and we can show that \(h^{\delta }\) is a Cauchy sequence in \(L^2\). Therefore there exists h such that \(\Vert h^{\delta } - h\Vert _{L^2} \rightarrow 0\). In a similar manner to Lemma 3.4, we can show that h is a weak solution. Since \(\sup _{t} \Vert h^{\delta }(t)\Vert _{\infty } \le \Vert h_{0}\Vert _{\infty }\) and \(\Vert h^{\delta } - h\Vert _{L^2} \rightarrow 0\), we can obtain (3.17). \(\square \)

4 \(L^{2}\) Decay

In this section, we will establish a weighted \(L^{2}\) estimate for (1.22). We will adapt techniques in [9, 13, 25].

As a starting point, we prove that (1.22) has a unique weak solution globally in time. Define the inner product associated with weighted norm (2.12) and (1.13):
$$\begin{aligned} \langle f,g \rangle _{\sigma } := \int _{{\mathbb {R}}^{3}} [\sigma ^{ij}\partial _{i}f\partial _{j}g+\sigma ^{ij}v_{i}v_{j}fg] \mathrm{d}v, \end{aligned}$$
(4.1)
$$\begin{aligned} ( f, g)_{\sigma } := \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} [ \sigma ^{ij}\partial _{i}f\partial _{j}g+\sigma ^{ij}v_{i}v_{j}fg ] \mathrm{d}v\mathrm{d}x. \end{aligned}$$
(4.2)

Definition 4.1

Let \(f(t,x,v) \in L^{\infty }((0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3},w^{\vartheta }(v)\mathrm{d}t\mathrm{d}x\mathrm{d}v)\) be a periodic function in \(x\in {\mathbb {T}}^{3} = [-\pi ,\pi ]^{3}\) satisfying
$$\begin{aligned} \int _{0}^{t}\Vert\; f(s)\Vert _{\sigma , \vartheta }^2 \mathrm{d}s < \infty . \end{aligned}$$
(4.3)
We say that f is a weak solution of the Landau Eqs. (1.6), (1.22) on \((0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\) if for all \(t\in (0,\infty )\) and all \(\varphi \in C^{1,1,1}_{t,x,v}\left( (0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\right) \) such that \(\varphi (t,x,v)\) is a periodic function in \(x\in {\mathbb {T}}^{3} = [-\pi ,\pi ]^{3}\) and \(\varphi (t,x,\cdot )\) is compactly supported in \({\mathbb {R}}^{3}\), it satisfies
$$\begin{aligned}&\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} f(t,x,v) \varphi (t,x,v) \mathrm{d}x\mathrm{d}v - \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} f_{0}(x,v) \varphi (0,x,v) \mathrm{d}x\mathrm{d}v \nonumber \\&\quad = - (f,\varphi )_{\sigma } + \iiint _{(0,t)\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} [f(s,x,v) (\partial _{s} \varphi (s,x,v) + v \cdot \nabla _{x}\varphi (s,x,v)-a_{g}(s,x,v)\cdot \nabla _{v} \varphi (s,x,v) \nonumber \\&\qquad + K \varphi (s,x,v) + \partial _{i}\sigma ^{i}(s,x,v) \varphi (s,x,v) - \partial _{i} \{\phi ^{ij}*[\mu ^{1/2}\partial _{j}g] \}(s,x,v)\varphi (s,x,v) \nonumber \\&\qquad + \{\phi ^{ij}*[v_{i}\mu ^{1/2}\partial _{j}g] \}(s,x,v)\varphi (s,x,v) ) - \sigma _{\mu ^{1/2}g}(s,x,v) \partial _{i}f(s,x,v) \partial _{j} \varphi (s,x,v)] \mathrm{d}s\mathrm{d}x\mathrm{d}v. \end{aligned}$$
(4.4)
Moreover if \(g=f\), then we say that f is a weak solution of the Landau equation (1.5), (1.6) on \((0,\infty )\times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\).
Let \(f(t) = U(t,s)f_0\) be a solution of the following equation
$$\begin{aligned}&f_{t}+v\cdot \partial _{x}f={{\bar{A}}}_{g} f, \nonumber \\&f(s) = U(s,s)f_0 = f_0, \end{aligned}$$
(4.5)
where \({{\bar{A}}}_{g}\) is defined as in (3.2). Then by the Duhamel principle, the solution of (1.22) is
$$\begin{aligned} f(t) = U(t,0)f_{0} + \int _{0}^{t} U(t, \tau ) {{\bar{K}}}_{g} f(\tau ) \mathrm{d} \tau . \end{aligned}$$
(4.6)

Lemma 4.2

Assume (2.6). Then there exists a unique weak solution f to (1.22) in the sense of Definition 4.1with\(f(0) = f_{0}\), which satisfies
$$\begin{aligned} \sup _{0\le s\le t}\Vert\; f(s)\Vert _\infty \le C(t) \Vert\, f_0\Vert _\infty . \end{aligned}$$

Sketch of proof

It is clear from (4.6) and by the Gronwall inequality. \(\square \)

Let \({\mathbf {P}}\) be the projection onto the \(\text {span} \{\sqrt{\mu }, v \sqrt{\mu }, |v|^{2} \sqrt{\mu }\}\) in \(L^{2} ({\mathbb {R}}^{3})\) and \({\mathbf {a}}_{f}\), \({\mathbf {b}}_{f}\), and \({\mathbf {c}}_{f}\) are coefficient functions as we defined in (2.14). We will prove the positivity of L. By Lemma 2.6, L is only semi-positive;
$$\begin{aligned} \left( Lf, f \right) \ge C \Vert (I-{\mathbf {P}}) f\Vert _{\sigma }^{2}. \end{aligned}$$
Now we will estimate \({\mathbf {P}} f\) in terms of \((I-{\mathbf {P}})f\). The following lemma is an adaptation of Lemma 6.1 in [9].

Lemma 4.3

(Lemma 6.1 in [9]) Assume (2.6). Letfbe a weak solution of (1.6), (1.7), (1.22). Then there existCand a function\(\eta (t) \le C\Vert\; f(t)\Vert _{2}^{2}\), such that
$$\begin{aligned} \int _{s}^{t} \Vert {\mathbf {P}} f(\tau )\Vert _{\sigma }^{2} \le \eta (t)-\eta (s) + C \int _{s}^{t} \Vert (I-{\mathbf {P}})f(\tau )\Vert _{\sigma }^{2}. \end{aligned}$$

Proof

For every periodic test function \(\psi \), f satisfies
$$\begin{aligned}&-\int _{s}^{t}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v\cdot \nabla _{x}\psi f - \int _{s}^{t}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \partial _{t}\psi f \nonumber \\&\quad = -\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(t) +\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\psi f(s) + \int _{s}^{t}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \big[ -\psi L(I-{\mathbf {P}})f + \psi \Gamma (g,f) \big]. \end{aligned}$$
(4.7)
By convention, we denote \(a(t,x) = {\mathbf {a}}_{f}(t,x)\), \(b(t,x)={\mathbf {b}}_{f}(t,x)\), and \(c(t,x)= {\mathbf {c}}_{f}(t,x)\), where \({\mathbf {a}}_{f}\), \({\mathbf {b}}_{f}\), and \({\mathbf {c}}_{f}\) are defined as in (2.14). We note that, with such choices \(\eta (t) = -\int \int _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} \psi f(t) \mathrm{d}x\mathrm{d}v\), and \(\psi = p(v)\phi (t,x)\) for some \(|p(v)|\le \exp (-|v|^{2}/4)\) and \(\Vert \phi (t)\Vert _{2} \le C (\Vert a(t)\Vert _{2} + \Vert b(t)\Vert _{2} + \Vert c(t)\Vert _{2})\). Thus,
$$\begin{aligned} |\eta (t)| &\le \Vert\; f(t)\Vert _{2} \left( \int _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} p(v) |\phi (t,x)|^{2} \mathrm{d}x\mathrm{d}v\right) ^{1/2}\\ & = C\Vert\; f(t)\Vert _{2}\left( \int _{{\mathbb {T}}^{3}} |\phi (t,x)|^{2} \mathrm{d}x\right) ^{1/2}\\ &\le C\Vert\; f(t)\Vert _{2} \Vert \phi (t)\Vert _{2}\\ & \le C\Vert\; f(t)\Vert _{2}^{2}. \end{aligned}$$
Without loss of generality, we can take \(s=0\).
Step 1. Estimate of\(\nabla _{x} \Delta ^{-1} \partial _{t} a = \nabla _{x} \partial _{t} \phi _{a}\). Choosing a test function \(\psi = \phi \sqrt{\mu }\) with \(\phi \) dependent only on x, we have (note that \({ \int \sqrt{\mu }Lf = \int \sqrt{\mu }\Gamma (g,f) = 0}\))
$$\begin{aligned} \sqrt{2 \pi }\int _{{\mathbb {T}}^{3}} [a(t+ \varepsilon ) - a(t)]\phi (x) = 2 \pi \sqrt{2 \pi } \int _{t}^{t+ \varepsilon }\int _{{\mathbb {T}}^{3}}(b\cdot \nabla _{x})\phi (x). \end{aligned}$$
Therefore,
$$\begin{aligned} \int _{{\mathbb {T}}^{3}} \phi \partial _{t} a = \sqrt{2 \pi } \int _{{\mathbb {T}}^{3} }(b \cdot \nabla _{x})\phi . \end{aligned}$$
First, take \(\phi =1\). Then, we have \(\int _{{\mathbb {T}}^{3}} \partial _{t} a(t) \mathrm{d}x =0\) for all \(t>0\). On the other hand, for all \(\phi (x)\in H^{1} ({\mathbb {T}}^{3})\), we have
$$\begin{aligned} \left| \int _{{\mathbb {T}}^{3}} \phi (x) \partial _{t} a \mathrm{d}x \right| \lesssim \Vert b\Vert _{2} \Vert \phi \Vert _{H^{1}}. \end{aligned}$$
Therefore, for all \(t>0\), \(\Vert \partial _{t} a(t)\Vert _{(H^{1})^{*}} \lesssim \Vert b(t)\Vert _{2}\). Since \(\int _{{\mathbb {T}}^{3}} \partial _{t} a \mathrm{d}x =0\) for all \(t>0\), we can find a solution of the Poisson equation with the periodic boundary condition \(- \Delta \Phi _{a} = \partial _{t} a(t)\). Let \(\phi _{a}\) be a solution of the Poisson equation with the periodic boundary condition \(- \Delta \phi _{a} = a(t)\). Then \(\Phi _{a} = \partial _{t} \phi _{a}\). Moreover, we have
$$\begin{aligned} \Vert \nabla _{x} \partial _{t} \phi _{a} \Vert _{2} = \Vert \Phi _{a}\Vert _{H^{1}} \lesssim \Vert \partial _{t} a(t)\Vert _{(H^{1})^{*}} \lesssim \Vert b(t)\Vert _{2}. \end{aligned}$$
(4.8)
Step 2. Estimate of\(\nabla _{x} \Delta ^{-1} \partial _{t} b^{j} = \nabla _{x} \partial _{t} \phi _{b}^{j}\). Choosing a test function \(\psi = \phi (x)v_{i} \sqrt{\mu }\), we have
$$\begin{aligned}&\frac{3}{2} \pi \sqrt{\pi } \int _{{\mathbb {T}}^{3}} [b_{i}(t+ \varepsilon )-b_{i}(t)]\phi \\&\quad = \frac{3}{2} \pi \sqrt{ \pi }\int _{t}^{t+\varepsilon }\int _{{\mathbb {T}}^{3}} \partial _{i} \phi [a+c] + \int _{t}^{t+\varepsilon }\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\sum _{j=1}^{d} v_{j}v_{i} \sqrt{\mu }\partial _{j} \phi (I-{\mathbf {P}}) f\\&\qquad + \int _{t}^{t+ \varepsilon }\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\phi v_{i} \Gamma (g,f) \sqrt{\mu }. \end{aligned}$$
Therefore,
$$\begin{aligned} \int _{{\mathbb {T}}^{3}}\partial _{t} b_{i}(t) \phi & = \int _{{\mathbb {T}}^{3}}\partial _{i} \phi [a(t)+c(t)] + \frac{2}{3 \pi \sqrt{ \pi }} \bigg \{\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\sum _{j=1}^{d} v_{i}v_{j} \sqrt{\mu }\partial _{j} \phi (I-{\mathbf {P}})f(t) \\&+ \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\phi v_{i} \Gamma (f,f)(t)\sqrt{\mu } \bigg \}. \end{aligned}$$
By the Hölder inequality and Theorem 2.8,
$$\begin{aligned}&\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\sum _{j=1}^{d} v_{i}v_{j} \sqrt{\mu }\partial _{j} \phi (I-{\mathbf {P}})f(t)\\&\quad \le C\Vert (1+|v|)^{-1/2}(I-{\mathbf {P}})f(t)\Vert _{2}^{2}\left\| (1+|v|)^{-\frac{-1}{2}}\sum _{j=1}^{d} v_{i}v_{j} \sqrt{\mu }\partial _{j} \phi \right\| _{2}\\&\quad \le C\Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma }\Vert \nabla _x \phi \Vert _{2} \end{aligned}$$
and
$$\begin{aligned} \left\| \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\phi v_{i} \Gamma (g,f)(t) \sqrt{\mu }\right\| \le C \Vert g\Vert _{\infty }\Vert\; f\Vert _{\sigma }\Vert \phi v_{i} \mu \Vert _{\sigma } \le \Vert g\Vert _{\infty }\Vert\; f\Vert _{\sigma }\Vert \phi \Vert _{2}. \end{aligned}$$
For fixed \(t>0\), we choose \(\phi = \Phi _{b}^{i}\), where \(\Phi _{b}^{i}\) is a solution of the Poisson equation with the periodic boundary condition \(- \Delta \Phi _{b}^{i} = \partial _{t} b_{i}(t)\). Let \(\phi _{b}^{i}\) be a solution of the Poisson equation with the periodic boundary condition \(- \Delta \phi _{b}^{i} = b_{i}(t)\). Then \(\Phi _{b}^{i} = \partial _{t} \phi _{b}^{i}\). By the Poincaré inequality,
$$\begin{aligned} \int _{{\mathbb {T}}^{3}} |\nabla _{x} \partial _{t} \phi _{b}^{i}(t)|^{2} \mathrm{d}x & = \int _{{\mathbb {T}}^{3}} |\nabla _{x} \Phi _{b}^{i}|^{2} \mathrm{d}x = -\int _{T^{3}} \Delta _{x} \Phi _{b}^{i} \Phi _{b}^{i} \mathrm{d}x\\ & \le \varepsilon \{\Vert \nabla _{x} \Phi _{b}^{i}\Vert _{2}^{2} + \Vert \Phi _{b}^{i} \Vert _{2}^{2}\}\\& \quad + C_{\varepsilon } \{ \Vert a(t)\Vert _{2}^{2} + \Vert c(t)\Vert _{2}^{2} + \Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma }^{2} + \Vert g(t)\Vert _{\infty }^{2}\Vert\; f(t)\Vert _{\sigma }^{2} \}\\ & \le C\varepsilon \Vert \nabla _{x} \Phi _{b}^{i}\Vert _{2}^{2} + C_{\varepsilon } \{ \Vert a(t)\Vert _{2}^{2} + \Vert c(t)\Vert _{2}^{2} \\& \quad + \Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma }^{2} + \Vert g(t)\Vert _{\infty }^{2}\Vert\; f(t)\Vert _{\sigma }^{2} \}, \end{aligned}$$
for every \(\varepsilon >0\). Now, we choose small \(\varepsilon \), such that \(C \varepsilon \le 1/4\). Then we can absorb the first term in RHS to the LHS. Then we have for all \(t>0\),
$$\begin{aligned} \Vert \nabla _{x} \partial _{t} \phi _{b}^{i}(t)\Vert _{2} \le C_{\varepsilon }\{ \Vert a(t)\Vert _{2} + \Vert c(t)\Vert _{2} + \Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma } + \Vert g(t)\Vert _{\infty }\Vert\; f(t)\Vert _{\sigma } \}. \end{aligned}$$
(4.9)
Step 3. Estimate of\(\nabla _{x} \Delta ^{-1}\partial _{t} \phi _c = \nabla _{x} \partial _{t} \phi _c\). Choosing a test function \(\psi = \phi (x) \big( |v|^{2}-\frac{3}{2}\big) \sqrt{\mu }\), we have
$$\begin{aligned}&\frac{3}{2} \pi \sqrt{ \pi } \int _{{\mathbb {T}}^{3}} \phi (x)[c(t+ \varepsilon )-c(t)]\\&\quad = \frac{3}{2} \pi \sqrt{ \pi } \int _{t}^{t+ \varepsilon }\int _{{\mathbb {T}}^{3}}b\cdot \nabla _{x} \phi - \int _{t}^{t + \varepsilon }\iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}}(I-{\mathbf {P}})f \left( |v|^{2}-\frac{3}{2}\right) \sqrt{\mu }(v\cdot \nabla _{x})\phi \\&\qquad + \int _{t}^{t+\varepsilon }\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \phi \Gamma (g,f) \left( |v|^{2}-\frac{3}{2}\right) \sqrt{\mu }. \end{aligned}$$
Therefore,
$$\begin{aligned}&\int _{{\mathbb {T}}^{3}} \phi (x) \partial _{t} c(t)\\&\quad = \int _{{\mathbb {T}}^{3}} b(t)\cdot \nabla _{x} \phi + \frac{2}{3 \pi \sqrt{ \pi }}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} (I-{\mathbf {P}})f(t)\left( |v|^{2}-\frac{3}{2}\right) \sqrt{\mu }(v \cdot \nabla _{x})\phi \\&\qquad + \frac{2}{3 \pi \sqrt{ \pi }}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\phi \Gamma (g,f)(t) \left( |v|^{2}-\frac{3}{2}\right) \sqrt{\mu }. \end{aligned}$$
Similar to Step 2,
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} (I-{\mathbf {P}})f(t)\left( |v|^{2}-\frac{3}{2}\right) \sqrt{\mu }(v \cdot \nabla _{x})\phi \le C\Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma }\Vert \nabla _{x} \phi \Vert _{2} \end{aligned}$$
and
$$\begin{aligned} \left\| \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\phi \Gamma (g,f)(t) \left( |v|^{2}-\frac{3}{2}\right) \sqrt{\mu }\right\| \le \Vert g\Vert _{\infty }\Vert\; f\Vert _{\sigma }\Vert \phi \Vert _{2}. \end{aligned}$$
For fixed \(t>0\), we choose \(\phi = \Phi _{c}\), where \(\Phi _{c}\) is a solution of the Poisson equation with the periodic boundary condition \(- \Delta \Phi _{c} = \partial _{t} c(t)\). Let \(\phi _{c}\) be a solution of the Poisson equation with the periodic boundary condition \(- \Delta \phi _{c} = c(t)\). Then \(\Phi _{c} = \partial _{t} \phi _{c}\). By the Poincaré inequality,
$$\begin{aligned} \int _{{\mathbb {T}}^{3}} |\nabla _{x} \partial _{t} \phi _{c}(t)|^{2} \mathrm{d}x & = \int _{{\mathbb {T}}^{3}} |\nabla _{x} \Phi _{c}|^{2} \mathrm{d}x = -\int _{{\mathbb {T}}^{3}} \Delta _{x} \Phi _{c} \Phi _{c} \mathrm{d}x\\ &\le \varepsilon \{ \Vert \nabla _{x} \Phi _{c}\Vert _{2}^{2} + \Vert \Phi _{c}\Vert _{2} ^{2}\}\\& \quad + C_{\varepsilon }\{ \Vert b(t)\Vert _{2}^{2} + \Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma }^{2} + \Vert g(t)\Vert _{\infty }^{2}\Vert\; f(t)\Vert _{\sigma }^{2} \}\\ & \le C\varepsilon \Vert \nabla _{x} \Phi _{c}\Vert _{2}^{2} + C_{\varepsilon }\{ \Vert b(t)\Vert _{2}^{2} + \Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma }^{2} + \Vert g(t)\Vert _{\infty } ^{2}\Vert\; f(t)\Vert _{\sigma }^{2} \}. \end{aligned}$$
Therefore, for all \(t>0\),
$$\begin{aligned} \Vert \nabla _{x} \partial _{t} \phi _{c}(t)\Vert _{2} \le C_{\varepsilon }\{ \Vert b(t)\Vert _{2} + \Vert (I-{\mathbf {P}})f(t)\Vert _{\sigma } + \Vert g(t)\Vert _{\infty }\Vert\; f(t)\Vert _{\sigma }\}. \end{aligned}$$
(4.10)
Step 4. Estimate ofc. Choosing a test function \(\psi = \big( |v|^{2}-\frac{5}{2}\big) \sqrt{\mu } v\cdot \nabla _{x} \phi _{c}\), we have
$$\begin{aligned} -\frac{45}{2}\pi \sqrt{ \pi }\int _{0}^{t} \int _{{\mathbb {T}}^{3}} \Delta _{x} \phi _{c} c & = -\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(t) + \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(0)\\& \quad + \sum _{i=1}^{d}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \left( |v|^{2}-\frac{5}{2}\right) v_{i} \sqrt{\mu } \partial _{t} \partial _{i} \phi _{c} f\\& \quad + \int _{0}^{t} \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} v\cdot \nabla _{x}\psi (I- {\mathbf {P}})f\\& \quad - \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f + \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f).\\ & = -\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(t) + \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(0)\\& \quad + \sum _{i,j=1}^{d}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \left( |v|^{2}-\frac{5}{2}\right) v_{i} v_{j} \mu \partial _{t} \partial _{i} \phi _{c} b_{j}\\& \quad + \sum _{i=1}^{d}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \left( |v|^{2}-\frac{5}{2}\right) v_{i} \sqrt{\mu } \partial _{t} \partial _{i} \phi _{c} (I-{\mathbf {P}}) f\\& \quad + \int _{0}^{t} \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} v\cdot \nabla _{x}\psi (I- {\mathbf {P}})f\\& \quad - \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f + \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f). \end{aligned}$$
Note that \({ \int \big( |v|^{2}-\frac{5}{2}\big) |v|^2 \mu = 0}\). Therefore, the third term of RHS is zero. Moreover,
$$\begin{aligned}&\sum _{i=1}^{d} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \left( |v|^{2}-\frac{5}{2}\right) v_{i} \sqrt{\mu } \partial _{t} \partial _{i} \phi _{c} (I-{\mathbf {P}}) f\\&\quad \le C\Vert \nabla _{x} \partial _{t} \phi _{c}\Vert _{2} \Vert (I-{\mathbf {P}})f\Vert _{\sigma }\\&\quad \le C(C_{\varepsilon }\{ \Vert b\Vert _{2} + \Vert (I-{\mathbf {P}})f\Vert _{\sigma } + \Vert g\Vert _{\infty }\Vert\; f\Vert _{\sigma }\})\Vert (I-{\mathbf {P}})f\Vert _{\sigma }\\&\quad \le \varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\{\Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2} \}, \end{aligned}$$
$$\begin{aligned} &\int _{0}^{t} \iint _{{\mathbb {T}}^3 \times {\mathbb {R}}^3} v\cdot \nabla _{x}\psi (I- {\mathbf {P}})f \le C\Vert c\Vert _2\Vert (I- {\mathbf {P}})f\Vert _{\sigma } \le \varepsilon \Vert c\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}})\Vert _{\sigma }^{2},\end{aligned}$$
$$\begin{aligned} & \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f = \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} L \psi (I-{\mathbf {P}})f\\&\quad \le C\Vert \nabla _{x} \phi _{c}\Vert _{2}\Vert (I-{\mathbf {P}})f\Vert _{\sigma }\\&\quad \le C\Vert c\Vert _{2} \Vert (I-{\mathbf {P}})f\Vert _{\sigma }\\&\quad \le \varepsilon \Vert c\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}})\Vert _{\sigma }^{2} \end{aligned}$$
and
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f) \le C\Vert g\Vert _{\infty }\Vert\; f\Vert _{\sigma }\Vert c\Vert _{2}\le \varepsilon \Vert c\Vert _{2}^{2} + C_{ \varepsilon }\Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2}. \end{aligned}$$
For a small \(\varepsilon >0\), we can absorb \(\Vert c\Vert _{2}^{2}\) on the RHS to the LHS. By (4.10), we have
$$\begin{aligned} \int _{0}^{t} \Vert c(s)\Vert ^{2} \mathrm{d}s\le C( \eta (t)-\eta (0)) + \int _{0}^{t} C_{\varepsilon }\left\{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2}\right\} +\varepsilon \Vert b\Vert _{2}^{2} \mathrm{d}s. \end{aligned}$$
(4.11)
Step 5. Estimate ofb. We will estimate \((\partial _{ij} \phi ^{j} _{b})b_{i}\) for all \(i,j = 1,\ldots ,d,\) and \((\partial _{jj}\phi ^{i}_{b})b_{i}\) for \(i \ne j\).
We first estimate \((\partial _{ij} \phi ^{j}_{b})b_{i}\). Choosing a test function \(\psi = \big[ (v_{i})^{2} - \frac{1}{2}\big] \sqrt{\mu }\partial _{j} \phi _{b}^{j}\), we have
$$\begin{aligned}&-\sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{lj} \phi ^{j}_{b} f - \int _{0}^{t}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{t}\partial _{j} \phi ^{j}_{b} f \nonumber \\&\quad = -\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(t) + \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(0) \nonumber \\&\qquad - \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f + \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\psi \Gamma (g,f). \end{aligned}$$
(4.12)
Note that for \(i\ne k\)
$$\begin{aligned} \int \left[ (v_{i})^{2} - \frac{1}{2}\right] \mu = \int \left[ (v_{i})^{2} - \frac{1}{2}\right] (v_{k})^{2} \mu = 0, \end{aligned}$$
and
$$\begin{aligned} \int \left[ (v_{i})^{2} - \frac{1}{2}\right] (v_{i})^{2} \mu = \frac{1}{2}\pi \sqrt{\pi }. \end{aligned}$$
Therefore,
$$\begin{aligned}&\sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{lj} \phi ^{j}_{b} f \nonumber \\&\quad = \sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}(v_{l})^{2} \left[ (v_{i})^{2} - \frac{1}{2}\right] \mu \partial _{lj} \phi ^{j}_{b} b_{l} \nonumber \\&\qquad + \sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}v_{l} \left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{lj} \phi ^{j}_{b} (I-{\mathbf {P}})f \nonumber \\&\quad = \frac{1}{2} \pi \sqrt{\pi } \int _{{\mathbb {T}}^{3}} \partial _{ij} \phi ^{j}_{b} b_{i} +\sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{lj} \phi ^{j}_{b} (I-{\mathbf {P}})f, \end{aligned}$$
(4.13)
and
$$\begin{aligned} \left| \sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l} \left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{lj} \phi ^{j}_{b} (I-{\mathbf {P}})f\right| &\le C \int _{0}^{t} \Vert b\Vert _{2}\Vert (I-{\mathbf {P}})f\Vert _{\sigma } \nonumber \\ &\le \varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2}. \end{aligned}$$
(4.14)
Moreover
$$\begin{aligned} \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{t} \partial _{j} \phi ^{j}_{b} f & = \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\left[ (v_{i})^{2} - \frac{1}{2}\right] \mu \partial _{t} \partial _{j} \phi ^{j}_{b} \left( |v|^{2} - \frac{3}{2}\right) c\\& \quad + \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{t} \partial _{j} \phi ^{j}_{b} (I-{\mathbf {P}})f. \end{aligned}$$
By (4.9),
$$\begin{aligned} \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\left[ (v_{i})^{2} - \frac{1}{2}\right] \sqrt{\mu }\partial _{t} \partial _{j} \phi ^{j}_{b} f\right| & \le \int _{0}^{t} C_{\varepsilon } \left\{ \Vert a\Vert _{2} + \Vert c\Vert _{2} + \Vert (I-{\mathbf {P}} )f\Vert _{\sigma } + \Vert\; f\Vert _{\sigma } \Vert\; f\Vert _{\sigma } \right\} \nonumber \\& \quad \times \left\{ \Vert c\Vert _{2} + C_{\vartheta }\Vert (I-{\mathbf {P}} )f\Vert _{\sigma }\right\} \nonumber \\& \le \int _{0}^{t} [C_{\varepsilon } \left\{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2} + \Vert c\Vert _{2}^{2} \right\} + \varepsilon \Vert a\Vert _{2}^{2}]. \end{aligned}$$
(4.15)
In a similar way to Step 4,
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f \le \varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}} )\Vert _{\sigma }^{2} \end{aligned}$$
(4.16)
and
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f) \le \varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\Vert g\Vert _{\infty } ^{2}\Vert\; f\Vert _{\sigma }^{2}. \end{aligned}$$
(4.17)
Combining (4.12)–(4.17),
$$\begin{aligned}&\int \partial _{ij} \phi ^{j}_{b} b_{i} \nonumber \\&\quad \le C(\eta (t)-\eta (0)) + \int _{0}^{t} [C_{\varepsilon }\left\{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2} + \Vert c\Vert _{2}^{2}\right\} + \varepsilon \{ \Vert a\Vert _{2}^{2} + \Vert b\Vert _{2}^{2}\}]. \end{aligned}$$
(4.18)
Now we estimate \((\partial _{jj}\phi ^{i}_{b})b_{i}\). Choose test function \(\psi = |v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{j} \phi ^{i}_{b}\) for \(i\ne j\). Then
$$\begin{aligned}&-\sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l}|v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{lj}\phi ^{i}_{b} f - \int _{0}^{t}\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}|v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{t} \partial _{j} \phi ^{i}_{b} f \nonumber \\&\quad = -\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(t) + \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(0)- \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f + \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}\psi \Gamma (g,f). \end{aligned}$$
(4.19)
Note that
$$\begin{aligned}&\sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l}|v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{lj}\phi ^{i}_{b} f \nonumber \\&\quad = \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} |v|^{2}(v_{i})^{2} (v_{j})^{2} \sqrt{\mu }[\partial _{ij}\phi ^{i}_{b} b_{j} +\partial _{jj} \phi ^{i}_{b} b_{i}] \nonumber \\&\qquad + \sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}v_{l} |v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{lj}\phi ^{i}_{b} (I-{\mathbf {P}})f. \end{aligned}$$
(4.20)
From (4.18),
$$\begin{aligned}&\left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} |v|^{2} (v_{i})^{2} (v_{j})^{2} \sqrt{\mu }\partial _{ij}\phi ^{i}_{b} b_{j}\right| \nonumber \\&\quad \le C(\eta (t)-\eta (0)) + \int _{0}^{t} [C_{\varepsilon } \{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma , \vartheta }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma , \vartheta }^{2} + \Vert c\Vert _{2}^{2} \} + \varepsilon \{ \Vert a\Vert _{2}^{2} + \Vert b\Vert _{2}^{2}\}], \end{aligned}$$
(4.21)
and
$$\begin{aligned} \left| \sum _{l}\int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} v_{l} |v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{lj}\phi ^{i}_{b} (I-{\mathbf {P}})f\right| & \le \int _{0}^{t} C\Vert b\Vert _{2} \Vert (I-{\mathbf {P}})f\Vert _{\sigma } \nonumber \\ & \le \int _{0}^{t} [\varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}} )f\Vert _{\sigma }^{2}]. \end{aligned}$$
(4.22)
Moreover, by (4.9)
$$\begin{aligned} \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}|v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{t} \partial _{j} \phi ^{i}_{b} f\right| & = \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}}|v|^{2} v_{i} v_{j} \sqrt{\mu }\partial _{t} \partial _{j} \phi ^{i}_{b} (I-{\mathbf {P}})f\right| \nonumber \\ & \le \int _{0}^{t} C_{\varepsilon }\left\{ \Vert a\Vert _{2} + \Vert c\Vert _{2} + \Vert (I-{\mathbf {P}})f\Vert _{\sigma } + \Vert g\Vert _{\infty }\Vert\; f\Vert _{\sigma }\right\} \Vert (I-{\mathbf {P}} )f\Vert _{\sigma } \nonumber \\ & \le \int _{0}^{t} [C_{\varepsilon }\left\{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2}\right\} + \varepsilon \left\{ \Vert a\Vert _{2}^{2} + \Vert c\Vert _{2}^{2} \right\} ]. \end{aligned}$$
(4.23)
Similar to (4.16) and (4.17),
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f \le \varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}} )\Vert _{\sigma }^{2} \end{aligned}$$
(4.24)
and
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f) \le \varepsilon \Vert b\Vert _{2}^{2} + C_{\varepsilon }\Vert g\Vert _{\infty } ^{2}\Vert\; f\Vert _{\sigma }^{2}. \end{aligned}$$
(4.25)
Combining (4.19)–(4.25) yields
$$\begin{aligned}&\int \partial _{jj} \phi ^{i}_{b} b_{i} \nonumber \\&\quad \le C(\eta (t)-\eta (0)) + \int _{0}^{t} [C_{\varepsilon } \left\{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2} + \Vert c\Vert _{2}^{2} \right\} + \varepsilon \{ \Vert a\Vert _{2}^{2} + \Vert b\Vert _{2}^{2}\}]. \end{aligned}$$
(4.26)
From (4.18) to (4.26) for small \(\varepsilon \), we can absorb \(\Vert b\Vert _{2}^{2}\) term on RHS to the LHS. Then we can conclude that
$$\begin{aligned} \int _{0}^{t} \Vert b(s)\Vert ^{2} \mathrm{d}s \le C(\eta (t)-\eta (0)) + \int _{0}^{t} [C_{\varepsilon }\left\{ \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2} + \Vert g\Vert _{\infty }^{2}\Vert\; f\Vert _{\sigma }^{2} + \Vert c\Vert _{2}^{2} \right\} + \varepsilon \Vert a\Vert _{2}^{2} ] \mathrm{d}s. \end{aligned}$$
(4.27)
Step 6. Estimate ofa. Choosing a test function
$$\begin{aligned} \psi = (|v|^{2} - 5) v\cdot \nabla _{x} \phi _{a} \sqrt{\mu }, \end{aligned}$$
we have
$$\begin{aligned}&-\int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2} -5)v_{i}v_{j} \partial _{ij} \phi _{a} \sqrt{\mu }f -\int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2}-5)v_{i} \partial _{t} \partial _{i} \phi _{a} \sqrt{\mu }f \nonumber \\&\quad = -\iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(t) + \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi f(0)- \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f + \int _{0}^{t} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f). \end{aligned}$$
(4.28)
Note that
$$\begin{aligned} \int (|v|^{2} -5)\left( |v|^{2} - \frac{3}{2}\right) (v_{i})^{2} \mu = 0. \end{aligned}$$
Therefore,
$$\begin{aligned}&\int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2} -5)v_{i}v_{j} \partial _{ij} \phi _{a} \sqrt{\mu }f \nonumber \\&\quad =\int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2} -5)(v_{i})^{2} \mu \partial _{ii} \phi _{a} a+ \int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2}-5)v_{i}v_{j} \partial _{ij} \phi _{a} \sqrt{\mu }(I-{\mathbf {P}})f \end{aligned}$$
(4.29)
and
$$\begin{aligned} \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2} -5)v_{i}v_{j} \partial _{ij} \phi _{a} \sqrt{\mu }(I-{\mathbf {P}})f\right| & \le \int _{0}^{t} C\Vert a\Vert _{2} \Vert (I-{\mathbf {P}})f\Vert _{\sigma } \nonumber \\ & \le \int _{0}^{t} [\varepsilon \Vert a\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}} )f\Vert _{\sigma }^{2}]. \end{aligned}$$
(4.30)
Moreover, by (4.8)
$$\begin{aligned} \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2} -5)v_{i} \partial _{t}\partial _{i} \phi _{a} \sqrt{\mu }f\right| & \le \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2} -5)(v_{i})^{2} \mu \partial _{t} \partial _{i} \phi _{a} b_{i}\right| \nonumber \\& \quad + \left| \int _{0}^{t} \iint _{{\mathbb {T}}^{3}\times {\mathbb {R}}^{3}} (|v|^{2}-5)v_{i} \partial _{t}\partial _{i} \phi _{a} \sqrt{\mu }(I-{\mathbf {P}}) f\right| \nonumber \\ & \le \int _{0}^{t} C\Vert b\Vert _{2}\left\{ \Vert b\Vert _{2} + C\Vert (I-{\mathbf {P}})f\Vert _{\sigma }\right\} \nonumber \\&\le \int _{0}^{t} C\{ \Vert b\Vert _{2}^{2} + \Vert (I-{\mathbf {P}})f\Vert _{\sigma }^{2}\}. \end{aligned}$$
(4.31)
Similar to Steps 4 and 5, we have
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi L(I-{\mathbf {P}})f \le \varepsilon \Vert a\Vert _{2}^{2} + C_{\varepsilon }\Vert (I-{\mathbf {P}} )\Vert _{\sigma }^{2} \end{aligned}$$
(4.32)
and
$$\begin{aligned} \iint _{{\mathbb {T}}^{3} \times {\mathbb {R}}^{3}} \psi \Gamma (g,f) \le \varepsilon \Vert a\Vert _{2}^{2} + C_{\varepsilon }\Vert g\Vert _{\infty } ^{2}\Vert\; f\Vert _{\sigma }^{2}. \end{aligned}$$
(4.33)
Similarly, from (4.28) to (4.33) for a small \(\varepsilon \), we can absorb \(\Vert a\Vert _{2}^{2}\) on the RHS to the LHS. Then we have
$$\begin{aligned} \int _{0}^{t} \Vert a(s)\Vert ^{2} \mathrm{d}s \le C(\eta (t)-\eta (0)) + \int _{0}^{t} C_{\varepsilon } \left\{ \Vert (I-{\mathbf {P}})f(s)\Vert _{\sigma }^{2} + \Vert g(s)\Vert _{\infty }^{2}\Vert\; f(s)\Vert _{\sigma }^{2} + \Vert b(s)\Vert _{2}^{2}\right\} \mathrm{d}s. \end{aligned}$$
(4.34)
Combining (4.11), (4.27), and (4.34), we have
$$\begin{aligned} \int _{0}^{t} \Vert {\mathbf {P}} f\Vert _{\sigma }^{2} & \le C(\eta (t)-\eta (0) )+ \int _{0}^{t} C_{\varepsilon }\left\{ \Vert (I-{\mathbf {P}})f(s)\Vert _{\sigma }^{2} + \Vert g(s)\Vert _{\infty }^{2} \Vert\; f(s)\Vert _{\sigma }^{2} \right\} \mathrm{d}s\\& \quad + \int _{0}^{t} \varepsilon \Vert {\mathbf {P}} f(s)\Vert _{\sigma }^{2} \mathrm{d}s\\ & \le C(\eta (t)-\eta (0)) + \int _{0}^{t} C_{\varepsilon }\left\{ \Vert (I-{\mathbf {P}})f(s)\Vert _{\sigma }^{2} + \Vert g(s)\Vert _{\infty }^{2} \Vert (I-{\mathbf {P}} )f(s)\Vert _{\sigma }^{2}\right\} \mathrm{d}s\\& \quad + \int _{0}^{t} (C_{\varepsilon }\Vert g(s)\Vert _{\infty }^{2} + \varepsilon )\Vert {\mathbf {P}} f(s)\Vert _{\sigma }^{2} \mathrm{d}s. \end{aligned}$$
Note that \(C_{\varepsilon } = C \varepsilon ^{-1}\). Choosing \(\varepsilon _{0} = \varepsilon \), we have \(C_{\varepsilon } \Vert g(s)\Vert _{\infty }^{2} \le C_{\varepsilon } \varepsilon ^{2} = C \varepsilon \le 1/4\) so that \(\Vert {\mathbf {P}}f\Vert _{\sigma }^{2}\) term on the RHS can be absorbed to the LHS. Thus we complete the proof. \(\square \)

Corollary 4.4

Assume (2.6). Letf(txv) be a weak solution of (1.6), (1.7), and (1.22) in the sense of Definition 4.1. Then there exist a constant\(0< \delta ^{\prime }\le 1/4\)and a function\(0\le \eta (t) \le C\Vert\; f(t)\Vert _{2}^{2}\), such that
$$\begin{aligned} \int _{s}^{t} (L[f(\tau )],f(\tau ))\mathrm{d} \tau \ge \delta ^{\prime }\left( \int _{s}^{t} \Vert\; f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau - \{\eta (t)-\eta (s)\} \right) . \end{aligned}$$
(4.35)

Proof

By Lemma 2.6 and Lemma 4.3,
$$\begin{aligned} \int _{s}^{t} (L[f(\tau )],f(\tau ))\mathrm{d} \tau & \ge \delta \int _{s}^{t} \Vert (I-{\mathbf {P}})f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau \\ & \ge \delta \frac{C}{1+C} \int _{s}^{t} \Vert (I-{\mathbf {P}})f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau + \delta \frac{1}{1+C} \int _{s}^{t} \Vert (I-{\mathbf {P}})f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau \\ & \ge \delta \frac{C}{1+C} \int _{s}^{t} \Vert (I-{\mathbf {P}})f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau + \delta \frac{1}{1+C} C \left( \int _{s}^{t} \Vert {\mathbf {P}} f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau - \{\eta (t) - \eta (s)\}\right) \\ & = \frac{C \delta }{1+C} \left( \int _{0}^{t} \Vert\; f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau - \{\eta (t) - \eta (s)\} \right) . \end{aligned}$$
\(\square \)

Remark 4.5

Note that in Lemma 2.6, we can take \(\delta >0\) sufficiently small. Therefore we can also take \(\delta ^{\prime }\) small enough.

Now we will prove Theorem 1.2. The proof is a modification of Theorem 5.1 in [25].

Proof of Theorem 1.2

We will prove
$$\begin{aligned}&\sum _{0\le {{\bar{\vartheta }}}\le 2\vartheta }\left( \frac{C_{{{\bar{\vartheta }}}}^{*}}{2}\{ \Vert\; f(t)\Vert _{2, {{\bar{\vartheta }}}/2}^{2} - \Vert\; f(s)\Vert _{2, {{\bar{\vartheta }}}/2}^{2} \} + \delta _{{{\bar{\vartheta }}},2\vartheta }\int _{s}^{t} \Vert\; f(\tau )\Vert _{\sigma , {{\bar{\vartheta }}}/2}^{2} \mathrm{d} \tau \right) - \delta ^{\prime }\{\eta (t) - \eta (s)\} \nonumber \\&\quad \le C_{\vartheta } \int _{s}^{t} \Vert g(\tau )\Vert _{\infty } \Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} \mathrm{d} \tau \end{aligned}$$
(4.36)
by the induction on \(\vartheta \).
Basis Step (\(\vartheta =0\)). Multiplying (1.22) by f, integrating both sides of the resulting equation, by Theorem 2.8 and Corollary 4.4, we have
$$\begin{aligned} \frac{1}{2}\{ \Vert\; f(t)\Vert _{2}^{2} - \Vert\; f(s)\Vert _{2}^{2}\} + \delta ^{\prime }\left( \int _{s}^{t} \Vert\; f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau - \{\eta (t) - \eta (s)\}\right) \le C\int _{s}^{t} \Vert g(\tau )\Vert _{\infty } \Vert\; f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau . \end{aligned}$$
Inductive Step. Suppose that (4.36) holds for \(\vartheta -1/2\). Multiplying (1.22) by \(w^{2\vartheta }f\), integrating both sides of the resulting equation, by Lemma 2.7 and Theorem 2.8, we have
$$\begin{aligned} \frac{1}{2}\{ \Vert\; f(t)\Vert _{2,\vartheta }^{2} - \Vert\; f(s)\Vert _{2,\vartheta }^{2} \} + \int _{s}^{t}\left( \frac{1}{2} \Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} - C_{\vartheta } \Vert\; f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau \right) \le C_{\vartheta }\int _{s}^{t} \Vert g(\tau )\Vert _{\infty }\Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} \mathrm{d} \tau . \end{aligned}$$
(4.37)
Multiply (4.37) by \(\frac{\delta _{0,2\vartheta -1} }{2C_{\vartheta }}\) and add it to (4.36). Then we have
$$\begin{aligned}&\sum _{0\le {{\bar{\vartheta }}}\le 2\vartheta -1}\left( \frac{C_{{{\bar{\vartheta }}}}^{*} }{2}\{ \Vert\; f(t)\Vert _{2, {{\bar{\vartheta }}}/2}^{2} - \Vert\; f(s)\Vert _{2, {{\bar{\vartheta }}}/2}^{2} \} + \delta _{{{\bar{\vartheta }}},2\vartheta -1}\int _{s}^{t} \Vert\; f(\tau )\Vert _{\sigma , {{\bar{\vartheta }}}/2}^{2} \mathrm{d} \tau \right) - \delta ^{\prime }\{\eta (t) - \eta (s)\}\\&\qquad + \frac{\delta _{0,2\vartheta -1}}{2C_{\vartheta }} \left[ \frac{1}{2}\{\Vert\; f(t)\Vert _{2,\vartheta }^{2} - \Vert\; f(s)\Vert _{2,\vartheta }^{2}\} + \int _{s} ^{t}\left( \frac{1}{2} \Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} - C_{\vartheta } \Vert\; f(\tau )\Vert _{\sigma }^{2} \mathrm{d} \tau \right) \right] \\&\quad \le C_{\vartheta -1/2} \int _{s}^{t} \Vert g(\tau )\Vert _{\infty } \Vert\; f(\tau )\Vert _{\sigma ,\vartheta -1/2}^{2} \mathrm{d} \tau + \frac{\delta _{0,2\vartheta -1}}{2} \int _{s}^{t} \Vert g(\tau )\Vert _{\infty }\Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} \mathrm{d} \tau . \end{aligned}$$
Note that \(\Vert \cdot \Vert _{2,\vartheta -1/2} \le \Vert \cdot \Vert _{2,\vartheta }\), \(\Vert \cdot \Vert _{\sigma ,\vartheta -1/2} \le \Vert \cdot \Vert _{\sigma ,\vartheta }\). Choosing sequences of \(C_{2\vartheta }^{*}\), \(\delta _{{{\bar{\vartheta }}}, 2\vartheta }\), and \(C_{\vartheta }\) such that
$$\begin{aligned} C_{0}^{*} & = 1,\quad \delta _{0,0} = \delta ^{\prime }, \quad C_{0} = C, \nonumber \\ C_{\vartheta }^{*} & = \frac{\delta _{0,2\vartheta -1}}{2 C_{\vartheta }}, \end{aligned}$$
(4.38)
$$\begin{aligned} \delta _{{{\bar{\vartheta }}}, 2\vartheta } = {\left\{ \begin{array}{ll} \frac{\delta _{0, 2\vartheta -1}}{2}, & \text { if }{{\bar{\vartheta }}}=0\\ \delta _{{{\bar{\vartheta }}}, 2\vartheta -1}, & \text { if }{\bar{2}}\vartheta = 1,\ldots , \vartheta -1,\\ \frac{\delta _{0, 2\vartheta -1}}{4C_{\vartheta }}, & \text { if }{{\bar{\vartheta }}}= 2 \vartheta , \end{array}\right. } \end{aligned}$$
(4.39)
and
$$\begin{aligned} C_{\vartheta } = C_{\vartheta -1/2} + \frac{\delta _{0,2\vartheta -1}}{2}, \end{aligned}$$
(4.40)
we have (4.36) for all \(\vartheta \).
Note that from (4.38)–(4.40), we have
$$\begin{aligned}&\delta _{0,k} = \frac{\delta ^{\prime }}{2^{k}}, \text { for } k = 1,2, \ldots , 2 \vartheta .\\&C<C_{\vartheta } = C+\sum _{0 \le {{\bar{\vartheta }}}\le 2\vartheta -1} \frac{\delta _{0,{{\bar{\vartheta }}}}}{2}< C+ \delta ^{\prime }< C+1,\\&\frac{\delta ^{\prime }}{2^{2\vartheta }(C+1)}< C_{\vartheta }^{*} < \frac{\delta ^{\prime }}{2^{2\vartheta } C}, \end{aligned}$$
and
$$\begin{aligned} \delta _{2\vartheta , 2\vartheta } = \frac{\delta _{0,2\vartheta -1}}{4C_{\vartheta }} = \frac{C_{\vartheta }^{*}}{2}. \end{aligned}$$
Let \(\varepsilon = \frac{\delta _{2\vartheta ,2\vartheta }}{2 C_{\vartheta }}\). By Remark 4.5, we can choose \(\delta ^{\prime }\) small enough such that
$$\begin{aligned} \delta ^{\prime }\eta (t) \le \frac{C_{0}^{*}}{4}\Vert\; f(t)\Vert _{2}^{2} = \frac{1}{4}\Vert\; f(t)\Vert _{2}^{2}. \end{aligned}$$
(4.41)
From (4.36), we have
$$\begin{aligned} \frac{C_{\vartheta }^{*}}{2}\Vert\; f(t)\Vert _{2,\vartheta }^{2} + \frac{\delta _{2\vartheta ,2\vartheta }}{2}\int _{s}^{t} \Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} \mathrm{d} \tau & \le \frac{1}{4}\Vert\; f(s)\Vert _{2}^{2} + \sum _{1\le {{\bar{\vartheta }}}\le 2\vartheta } \frac{C_{{{\bar{\vartheta }}}}^{*}}{2}\Vert\; f(s)\Vert _{2,{{\bar{\vartheta }}}/2}^{2} \nonumber \\ & \le \left( \frac{1}{4} + \frac{\delta ^{\prime }}{C} \right) \Vert\; f(s)\Vert _{2,\vartheta }^{2} \nonumber \\ & \le \frac{1}{2}\Vert\; f(s)\Vert _{2,\vartheta }^{2}. \end{aligned}$$
(4.42)
Taking \(s=0\) and dividing by \(\frac{\delta _{2\vartheta ,2\vartheta }}{2}\) both sides of (4.42), we have
$$\begin{aligned} 2\Vert\; f(t)\Vert _{2,\vartheta }^{2} + \int _{0}^{t} \Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} \mathrm{d} \tau \le \frac{2}{C_{\vartheta }^{*}}\Vert\; f(0)\Vert _{2,\vartheta }^{2} \le C 2^{2\vartheta }\Vert\; f(0)\Vert _{2,\vartheta }^{2}. \end{aligned}$$
Therefore, we have (1.23).
Fix \(\vartheta ,k \ge 0\), by the Hölder inequality and (1.23),
$$\begin{aligned} \Vert\; f\Vert _{2,\vartheta }^{2} & = \int w^{2\vartheta } f^{2} \nonumber \\ & = \int \left( w^{2 \left( \vartheta - \frac{1}{2}\right) }f^{2}\right) ^{\frac{k}{k+1}}\left( w^{2 \left( \vartheta + \frac{k}{2}\right) }f^{2}\right) ^{\frac{1}{k+1}} \nonumber \\&\le \left( \int w^{2 \left( \vartheta - \frac{1}{2}\right) }f^{2}\right) ^{\frac{k}{k+1}}\left( \int w^{2 \left( \vartheta + \frac{k}{2}\right) }f^{2}\right) ^{\frac{1}{k+1}} \nonumber \\&\le \Vert\; f\Vert _{2,\vartheta -1/2}^{2\frac{k}{k+1}} \left( C 2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0) \right) ^{\frac{1}{k+1}}. \end{aligned}$$
(4.43)
By Lemma 2.5,
$$\begin{aligned} \Vert\; f\Vert _{\sigma ,\vartheta } \ge \Vert (1+|v|)^{-1/2} f\Vert _{2,\vartheta } = \Vert\; f\Vert _{2,\vartheta -1/2}. \end{aligned}$$
(4.44)
Combining (4.36), (4.43), and (4.44), we have
$$\begin{aligned}&\sum _{1 \le {{\bar{\vartheta }}}\le 2\vartheta } \frac{C_{{{\bar{\vartheta }}}}^{*}}{2} \left( \Vert\; f(t)\Vert _{2,{{\bar{\vartheta }}}/2}^{2} - \Vert\; f(s)\Vert _{2,{{\bar{\vartheta }}}/2} ^{2}\right) + \left\{ \frac{1}{2}\Vert\; f(t)\Vert _{2}^{2} - \delta ^{\prime } \eta (t)\right\} - \left\{ \frac{1}{2}\Vert\; f(s)\Vert _{2}^{2} - \delta ^{\prime } \eta (s)\right\} \nonumber \\&\quad \le -\frac{\delta _{2\vartheta ,2\vartheta }}{2} \int _{s}^{t} \Vert\; f(\tau )\Vert _{\sigma ,\vartheta }^{2} \mathrm{d} \tau \nonumber \\&\quad \le -\frac{\delta _{2\vartheta ,2\vartheta }}{2} \int _{s}^{t} \Vert\; f(\tau )\Vert _{2,\vartheta -1/2}^{2} \mathrm{d} \tau \nonumber \\&\quad \le -\frac{\delta _{2\vartheta ,2\vartheta }}{2} \int _{s}^{t} \left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}} \Vert\; f(\tau )\Vert _{2,\vartheta }^{2\frac{k+1}{k}} \mathrm{d} \tau . \end{aligned}$$
(4.45)
Let
$$\begin{aligned} y(t) \,{:=} \, \left\{ \frac{1}{2}\Vert\; f(t)\Vert _{2}^{2} - \delta ^{\prime }\eta (t)\right\} + \sum _{1 \le {{\bar{\vartheta }}}\le 2\vartheta } \frac{C_{{{\bar{\vartheta }}}}^{*}}{2} \Vert\; f(t)\Vert _{2,{{\bar{\vartheta }}}/2}^{2}. \end{aligned}$$
Then
$$\begin{aligned} \frac{C_{\vartheta }^{*}}{2}\Vert\; f(t)\Vert _{2,\vartheta }^{2} \le y(t) \le \left( \frac{1}{2} + \sum _{{{\bar{\vartheta }}}=1}^{2\vartheta } \frac{C_{{{\bar{\vartheta }}}}^{*}}{2}\right) \Vert\; f(t)\Vert _{2,\vartheta }^{2} \le \Vert\; f(t)\Vert _{2,\vartheta }^{2}. \end{aligned}$$
(4.46)
Combining (4.45), (4.46), we have
$$\begin{aligned} y(t) - y(s) \le - \frac{\delta _{2\vartheta ,2\vartheta }}{2} \int _{s}^{t} \left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}}\left( y(\tau )\right) ^{\frac{k+1}{k}} \mathrm{d} \tau . \end{aligned}$$
Therefore, we have
$$\begin{aligned} y^{\prime }(t) \le -\frac{1}{2} \delta _{2\vartheta ,2\vartheta }\left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}} y(t)^{\frac{k+1}{k}}\le - \frac{1}{2^{2\vartheta }C} \left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}} y(t)^{\frac{k+1}{k}}. \end{aligned}$$
(4.47)
Multiplying (4.47) by \(-\frac{1}{k}y^{-\frac{k+1}{k}}\), we have
$$\begin{aligned} \partial _{t} \left( y(t)^{-\frac{1}{k}}\right) \ge \frac{1}{2^{2\vartheta }Ck} \left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}}. \end{aligned}$$
Integrating above over [0, t] yields
$$\begin{aligned} y(t)^{-\frac{1}{k}} & \ge \frac{t}{2^{2\vartheta }Ck} \left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}} + y(0)^{-\frac{1}{k} }\\ &\ge \frac{t}{2^{2\vartheta }Ck} \left( C2^{2\vartheta +k}{\mathcal {E}} _{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}} + \left( \Vert\; f(0)\Vert _{2,\vartheta } ^{2}\right) ^{-\frac{1}{k}}\\ &\ge \frac{\left( C2^{2\vartheta +k}{\mathcal {E}}_{\vartheta +k/2}(0)\right) ^{-\frac{1}{k}}}{2^{2\vartheta }C} \left( \frac{t}{k} + 1\right) . \end{aligned}$$
Therefore,
$$\begin{aligned} \Vert\; f(t)\Vert _{2,\vartheta }^{2} \le \frac{2}{C_{\vartheta }^{*}} y(t) \le C_{\vartheta ,k} {\mathcal {E}}_{\vartheta +k/2}(0)\left( 1+ \frac{t}{k}\right) ^{-k}, \end{aligned}$$
where we use (4.46) in the first inequality. Thus we complete the proof. \(\square \)

Theorem 4.6

Assume (2.6). Let\(\vartheta \in 2^{-1}{\mathbb {N}} \cup \{0\}\)andfbe a classical solution of (1.7), (4.5). Then there exist\(C, \varepsilon (\vartheta )>0\)such that if \(\Vert g\Vert _{\infty }< \varepsilon \), then
$$\begin{aligned} \sup _{0 \le s< \infty }{\mathcal {E}} _{\vartheta }(f(s)) \le C 2^{2\vartheta } {\mathcal {E}}_{\vartheta }(0), \end{aligned}$$
(4.48)
and for any\(t>0\), \(k\in {\mathbb {N}}\),
$$\begin{aligned} \Vert\; f(t)\Vert _{2,\vartheta } \le C_{\vartheta ,k} {\mathcal {E}}_{\vartheta +k/2}(0)\left( 1+ \frac{t}{k}\right) ^{-k/2}. \end{aligned}$$
(4.49)

Sketch of proof

The proof can be done by choosing \(\Gamma = 0\) in Theorem 1.2. \(\square \)

5 \(L^{2}-L^{\infty }\) Estimate

5.1 Local \(L^2-L^\infty \) Estimate

In this section we will derive a local \(L^{\infty }\) estimate for h.
$$\begin{aligned} {\mathcal {M}}_{g}^{\vartheta }h:= (\partial _{t} + v\cdot \nabla _{x} - {{\bar{A}}}_{g}^{\vartheta })h, \end{aligned}$$
(5.1)
where \({{\bar{A}}}_{g}^{\vartheta }\) is defined as in (3.5).

Here we will refine the results about the \(L^{2}-L^{\infty }\) estimate in [20]. Comparing with [20], we have an additional term \(a_{g}\cdot \nabla _{v} f-2 \frac{\partial _{i} w^{\vartheta } }{w^{\vartheta }}\sigma _{G}^{ij} \partial _{j} f\) and a diffusion matrix of \({\mathcal {M}}_{g}^{\vartheta }\) which is not uniformly elliptic. Moreover, to get an \(L^{2}-L^{\infty }\) estimate for \({\mathbb {T}}^{3} \times {\mathbb {R}}^{3}\), we need to know the local \(L^{2}-L^{\infty }\) estimate more explicitly.

Define \(Q_{n} := [-t_{n},0]\times {\mathbb {T}}^{3} \times B(0;R_{n})\), for \(t_{n} \ge t_{n+1}\) and \(R_{n} \ge R_{n+1}\). The following estimates are refinements of Lemmas 4–6 and Theorem 2 and Theorem 7 in [20].

Lemma 5.1

(Lemma 4 in [20]) Assume (2.6). Lethbe a non-negative periodic function inxsatisfying\({\mathcal {M}}_{g}^{\vartheta }h \le 0\). Thenhsatisfies
$$\begin{aligned} \int _{Q_{1}}|\nabla _{v} h|^{2}&\le C \int _{Q_{0}}h^{2}, \end{aligned}$$
(5.2)
$$\begin{aligned} \Vert h\Vert _{L^{2}_{t}L^{2}_{x}L^{q}_{v}(Q_{1})}^{2}&\le C \int _{Q_{0}} h^{2}, \end{aligned}$$
(5.3)
$$\begin{aligned} \Vert h\Vert _{L^{\infty }_{t}L^{2}_{x}L^{2}_{v}(Q_{1})}^{2}&\le C \int _{Q_{0}}h^{2} \end{aligned}$$
(5.4)
for some \(q>2\) and \(C = {{\bar{C}}}(R_{0})\left( 1+ \frac{1}{t_{0} - t_{1}} + \frac{1}{R_{0} - R_{1}} + \frac{1}{(R_{0} - R_{1})^{2}} \right) \), \({{\bar{C}}}(R_{0}) = C'(1+R_{0})^3\).

Proof

Consider a test function \(\Phi \in C^{\infty }({\mathbb {R}} \times {\mathbb {T}}^{3} \times {\mathbb {R}}^{3})\), periodic with respect to x and \(\Phi (t,x,v)=0\) for \(|v|> R_{0}\). Multiplying (5.1) by \(2h \Phi ^{2}\) and integrating the resulting equation over \({\mathcal {R}}:= [-t_{0},s] \times {\mathbb {T}}^{3} \times B(0;R_{0})\) for some \(s\in [-t_{1},0]\), we have
$$\begin{aligned}&\int _{{\mathcal {R}}} \partial _{t}(h^{2})\Phi ^{2} + \int _{{\mathcal {R}}}v\cdot \nabla _{x}(h^{2})\Phi ^{2}\\&\quad \le 2\int _{{\mathcal {R}}} \nabla _{v}\cdot (\sigma _{G} \nabla _{v} h) h\Phi ^{2} + \int _{{\mathcal {R}}} a_{g} \cdot \nabla _{v}(h^{2})\Phi ^{2} -2\int _{{\mathcal {R}}} \frac{\nabla _{v} (w^{\vartheta })}{w^{\vartheta }}\cdot \sigma _{G} \nabla _{v}(h^{2})\Phi ^{2}, \end{aligned}$$
where \(\sigma _G\) is defined as in (2.1) with \(G = \mu + \mu ^{1/2}g\). Using the integration by parts and the positivity of \(\sigma _G\), we have
$$\begin{aligned}&\int _{{\mathcal {R}}} \partial _{t}(h^{2} \Phi ^{2}) + 2\int _{{\mathcal {R}} }(\nabla _{v} h \cdot \sigma _{G} \nabla _{v} h) \Phi ^{2}\\&\quad \le \int _{{\mathcal {R}}} h^{2}\left( \partial _{t}(\Phi ^{2}) + v\cdot \nabla _{x}(\Phi ^{2}) - \nabla _{v}\cdot (\Phi ^{2} a_{g}) + 2 \nabla _{v} \cdot \left( \Phi ^{2} \sigma _{G} \frac{\nabla _{v} (w^{\vartheta } )}{w^{\vartheta }}\right) \right) \\ &\qquad - 4 \int _{{\mathcal {R}}} h \Phi \nabla _{v} \Phi \cdot \sigma _{G} \nabla _{v} h\\&\quad \le \int _{{\mathcal {R}}} h^{2}\left( \partial _{t}(\Phi ^{2}) + v\cdot \nabla _{x}(\Phi ^{2}) - \nabla _{v}\cdot (\Phi ^{2} a_{g}) + 2 \nabla _{v} \cdot \left( \Phi ^{2} \sigma _{G} \frac{\nabla _{v} (w^{\vartheta } )}{w^{\vartheta }}\right) \right) \\&\qquad + \int _{{\mathcal {R}}}(\nabla _{v} h \cdot \sigma _{G} \nabla _{v} h) \Phi ^{2} + C\int _{{\mathcal {R}}}(\nabla _{v} \Phi \cdot \sigma _{G} \nabla _{v} \Phi ) h^{2}. \end{aligned}$$
Thus we have
$$\begin{aligned}&\int _{{\mathcal {R}}} \partial _{t}(h^{2} \Phi ^{2}) + \min (1,(1+R_{0})^{-3 })\int _{{\mathcal {R}}}|\nabla _{v} h|^{2} \Phi ^{2}\\&\quad \le C' \max (1, (1+R_{0})^{-1})\left( \Vert \partial _{t} \Phi \Vert _{\infty }\Vert \Phi \Vert _{\infty }+ R_{0}\Vert \nabla _{x} \Phi \Vert _{\infty } \Vert \Phi \Vert _{\infty }\right. \\&\qquad + \Vert \Phi \Vert _{\infty }\Vert a_{g}\Vert _{\infty }\Vert \nabla _{v} \Phi \Vert _{\infty }+ \Vert \Phi \Vert _{\infty }^{2}\Vert \nabla _{v}\cdot a_{g}\Vert _{\infty }\\&\qquad \left. + \Vert \nabla _{v}\Phi \Vert _{\infty }^{2} + \Vert \Phi \Vert _{\infty }\Vert \nabla _{v} \Phi \Vert _{\infty }\left\| \sigma \frac{\nabla _{v} (w^{\vartheta })}{w^{\vartheta } }\right\| _{\infty }+ \Vert \Phi \Vert _{\infty }^{2}\left\| \nabla _{v} \cdot \left( \sigma \frac{\nabla _{v} (w^{\vartheta })}{w^{\vartheta }}\right) \right\| _{\infty }\right) \int _{{\mathcal {R}}} h^{2}. \end{aligned}$$
Choosing \(\Phi \) such that \(\Phi (-t_{0},x,v)=0\) and \(\Phi = 1\) in \(Q_{1}\), we have
$$\begin{aligned} \int _{{\mathbb {T}}^{3} \times B(0;R_{1})}h^{2}(s)\mathrm{d}x\mathrm{d}v +\int _{{\mathcal {R}}} |\nabla _{v} h|^{2}\le {{\bar{C}}}(R_{0})\left( 1+ \frac{1}{t_{0} - t_{1}} + \frac{1}{R_{0}- R_{1}} + \frac{1}{(R_{0} - R_{1})^{2}} \right) \int _{{\mathcal {R}}}h^{2}. \end{aligned}$$
(5.5)
Especially,
$$\begin{aligned} \sup _{s\in [-t_{1},0]}\int _{{\mathbb {T}}^{3} \times B(0;R_{1})}h^{2}(s)\mathrm{d}x\mathrm{d}v \le {{\bar{C}}}(R_{0})\left( 1+ \frac{1}{t_{0} - t_{1}} + \frac{1}{R_{0} - R_{1}} + \frac{1}{(R_{0} - R_{1})^{2}} \right) \int _{Q_{0}}h^{2}. \end{aligned}$$
Therefore, we prove (5.4). Choosing \(s=0\) in (5.5), we have
$$\begin{aligned} \int _{Q_{1}} |\nabla _{v} h|^{2} \le {{\bar{C}}}(R_{0})\left( 1+ \frac{1}{t_{0} - t_{1}} + \frac{1}{R_{0} - R_{1}} + \frac{1}{(R_{0} - R_{1})^{2}} \right) \int _{Q_{0}}h^{2}, \end{aligned}$$
so we obtain (5.2). Moreover, the Sobolev inequality implies (5.3). \(\square \)

Lemma 5.2

(Lemma 5 in [20]) Assume (2.6). Ifhis a weak solution of (3.6), then
$$\begin{aligned} \begin{aligned} \Vert D_{x}^{1/3}h\Vert _{L^{2}(Q_{1})}^{2}&\le C\Vert h\Vert _{L^{2}(Q_{0})}^{2},\\ \Vert D_{t}^{1/3}h\Vert _{L^{2}(Q_{1})}^{2}&\le C\Vert h\Vert _{L^{2}(Q_{0})}^{2} \end{aligned} \end{aligned}$$
(5.6)
for some\(C = {{\bar{C}}}(R_{0})\left( 1+ \frac{1}{t_{0} - t_{1}} + \frac{1}{R_{0} - R_{1}} + \frac{1}{(R_{0} - R_{1})^{2}} \right) \).

Proof

Let \(R_{\frac{1}{2}} = \frac{R_{1} + R_{0}}{2}\) and \(Q_{\frac{1}{2}} = Q_{R_{\frac{1}{2}}}\). Define truncation functions \(\chi _{1}\) and \(\chi _{1/2}\) such that
$$\begin{aligned} \begin{aligned} \chi _{1}&= {\left\{ \begin{array}{ll} 1, & \text { if }(t,x,v)\in Q_{1},\\ 0, & \text { if }(t,x,v)\in Q_{\frac{1}{2}}^{c}, \end{array}\right. } \\ \chi _{\frac{1}{2}}&= {\left\{ \begin{array}{ll} 1, & \text { if }(t,x,v)\in Q_{\frac{1}{2}},\\ 0, & \text { if }(t,x,v)\in Q_{0}^{c}. \end{array}\right. } \end{aligned} \end{aligned}$$
Let \(h_{i} = h \chi _{i}\), for \(i = 1, \frac{1}{2}\). Then we have
$$\begin{aligned} \begin{aligned} (\partial _{t} + v \cdot \nabla _{x})h_{1}&= \nabla _{v} \cdot H_{1} + H_{0} \quad \text { in }(-\infty ,0]\times {\mathbb {R}}^{6},\\ H_{1}&= \chi _{1} \sigma _{G}\nabla _{v} h_{\frac{1}{2}},\\ H_{0}&= - \nabla _{v} \chi _{1} \cdot \sigma _{G}\nabla _{v} h_{\frac{1}{2}} + \alpha _{1} h_{\frac{1}{2}} + \chi _{1} a_{g} \cdot \nabla _{v} h_{1/2} -2 \chi _{1} \frac{\nabla _{v}(w^{\vartheta } )}{w^{\vartheta }}\cdot \sigma _{G}\nabla _{v} h_{1/2},\\ \alpha _{1}&= (\partial _{t} + v\cdot \nabla _{x})\chi _{1}, \end{aligned} \end{aligned}$$
where \(\sigma _G\) is defined as in (2.1) with \(G=\mu + \mu ^{1/2}g\). By Lemma 5.1,
$$\begin{aligned} \Vert H_{0}\Vert _{L^{2}({\mathbb {R}}^{7})} + \Vert H_{1}\Vert _{L^{2}({\mathbb {R}}^{7})} \le C\Vert h\Vert _{L^{2}(Q_{0})} \end{aligned}$$
with C as in the statement. Applying Theorem 1.3 in [4] with \(p=2\), \(r=0\), \(\beta =1\), \(m=1\), \(\kappa =1\) and \(\Omega = 1\) yields (5.6). \(\square \)

Lemma 5.3

(Lemma 6 in [20]) Under the assumptions of Lemma 5.1, there exists\(p>2\)such that
$$\begin{aligned} \Vert h\Vert _{L^{2}_{t}L^{p}_{x}L^{2} _{v}(Q_{1})}^{2} \le C \Vert h\Vert _{L^{2}(Q_{0})}^{2} \end{aligned}$$
(5.7)
with the same C as in Lemma 5.1.

Proof

The proof is exactly the same as in the proof of Lemma 6 in [20]. We omit the proof. \(\square \)

The following lemma is a consequence of Lemmas 5.1 and 5.2. We omit the proof.

Lemma 5.4

Under the assumptions of Lemma 5.2, we have
$$\begin{aligned} \Vert h\Vert _{H^{s}_{x,v,t}(Q_{1})} \le C\Vert h\Vert _{L^{2}(Q_{0})} \end{aligned}$$
with the same C as in Lemma 5.1and\(s=1/3\).

Lemma 5.5

(Theorem 2 in [20]) Under the assumptions of Lemma 5.1, there exists\(q>2\)such that
$$\begin{aligned} \Vert h\Vert _{L^{q}(Q_{1})}^{2} \le C \Vert h\Vert _{L^{2}(Q_{0})}^{2} \end{aligned}$$
(5.8)
with the same C as in Lemma 5.1.

Proof

The proof is exactly the same as in the proof of Theorem 2 in [20]. We omit the proof. \(\square \)

Lemma 5.6

(Theorem 7 in [20]) Assume (2.6). Lethbe a non-negative subsolution of (3.6). Then, there exists\(m>1\)such that
$$\begin{aligned} \Vert h\Vert _{L^{\infty }(Q_{\infty })} \le {{\bar{C}}} (R_{0})^{m} \left( 1+ \frac{1}{\min (t_{0}-t_{\infty }, (R_{0}-R_{\infty })^{2})}\right) ^{m} \Vert h\Vert _{L^{2}(Q_{0})}, \end{aligned}$$
where\(Q_{0} = [-t_{0},0]\times {\mathbb {T}}^{3} \times [-R_{0},R_{0}]\)and\(Q_{\infty }= [-t_{\infty },0] \times {\mathbb {T}}^{3} \times [-R_{\infty }, R_{\infty }]\).

Proof

Let \(\kappa := q/2>1\). Since \(|h|^{q_{n}}\), \(q_{n}>1\), is also a subsolution of (3.6), by Lemma 5.5
$$\begin{aligned} \Vert |h|^{q_{n}}\Vert _{L^{q}(Q_{n+1})}^{2} \le C_{n} \Vert |h|^{q_{n}}\Vert _{L^{2}(Q_{n} )}^{2}, \end{aligned}$$
where \(C_{n} = {{\bar{C}}}(R_{0})\left( 1+ \frac{1}{t_{n} - t_{n+1}} + \frac{1}{R_{n} - R_{n+1}} + \frac{1}{(R_{n} - R_{n+1})^{2}} \right) \). Changing \(\Vert \cdot \Vert _{q}\) to \(\Vert \cdot \Vert _{2}\) yields
$$\begin{aligned} \Vert |h|^{\kappa q_{n}}\Vert _{L^{2}(Q_{n+1})}^{2} \le C_{n}^{\kappa }\Vert |h|^{q_{n} }\Vert _{L^{2}(Q_{n})}^{2 \kappa }. \end{aligned}$$
Let \(q_{n} := \kappa ^{n}\), then after iteration we have
$$\begin{aligned} \Vert |h|^{q_{n}}\Vert _{L^{2}(Q_{n})}^{2} \le \prod _{j=1}^{n} C_{n-j}^{\kappa ^{j}} \Vert h\Vert _{L^{2}(Q_{0})}^{2\kappa ^{n}}. \end{aligned}$$
Changing \(\Vert \cdot \Vert _{2}\) to \(\Vert \cdot \Vert _{2q_{n}}\), we have
$$\begin{aligned} \Vert h\Vert _{L^{2 q_{n}}(Q_{n})}^{2}\le \prod _{j=1}^{n} C_{n-j}^{\kappa ^{j-n}} \Vert h\Vert _{L^{2}(Q_{0})}^{2} = \prod _{j=0}^{n-1} C_{j}^{\kappa ^{-j}} \Vert h\Vert _{L^{2}(Q_{0})}^{2}. \end{aligned}$$
Choosing \(t_{n}-t_{n+1} = \alpha (t_{0}-t_{\infty })n^{-4}\) and \(R_{n}-R_{n+1} = \beta (R_{0}-R_{\infty })n^{-2}\), we have
$$\begin{aligned} C_{j}^{\kappa ^{-j}} \le {{\bar{C}}}^{\kappa ^{-j}} \left( C^{\prime }\left( 1+ \frac{1}{\min (t_{0}-t_{\infty }, (R_{0}-R_{\infty })^{2})}\right) \right) ^{j^{4} \kappa ^{-j}}. \end{aligned}$$
Thus,
$$\begin{aligned} \prod _{j=0}^{\infty } C_{j}^{\kappa ^{-j}} \le C^{\prime m }{{\bar{C}}}(R_{0})^{m}\left( 1+ \frac{1}{\min (t_{0}-t_{\infty }, (R_{0}-R_{\infty })^{2})}\right) ^{m} \end{aligned}$$
for some \(m>1\). So the proof is complete. \(\square \)

Lemma 5.7

Assume (2.6). If\(h_{+}=\max \left\{ h ,0\right\} \), wherehis a subsolution of (3.6), then\(h_{+}\)is a subsolution.

Proof

Approximate a convex function \(Q\left( h\right) \rightarrow h_{+}\) and then use the convexity of \(Q\left( h\right) \) such that \(Q^{\prime }\left( h\right) >0\) and \(Q^{\prime \prime }\left( h\right) >0\). Applying \(Q\left( h\right) \) to Eq. (3.6), we complete the proof. \(\square \)

Let h be a weak solution. Then since \(\left| h\right| =h_{+}-h_{-}\) and \(h_{+}=\max \left\{ h,0\right\} \) are subsolutions (maximum of two subsolutions is a subsolution) and \(h_{-}=\max \left\{ -h,0\right\} \) is a supersolution (minimum of two supersolution is a supersolution), we can apply Lemma 5.7 to both \(h_{+}\) and \(-h_{-}\). Thus we obtain:

Lemma 5.8

(Theorem 7 in [20]) Assume (2.6). Lethbe a subsolution of (3.6). Then, there exists\(m>1\)such that
$$\begin{aligned} \Vert h\Vert _{L^{\infty }(Q_{\infty })} \le {{\bar{C}}} (R_{0})^{m} \left( 1+ \frac{1}{\min (t_{0}-t_{\infty }, (R_{0}-R_{\infty })^{2})}\right) ^{m} \Vert h\Vert _{L^{2}(Q_{0})}, \end{aligned}$$
where\(Q_{0} = [-t_{0},0]\times {\mathbb {T}}^{3} \times [-R_{0},R_{0}]\)and\(Q_{\infty }= [-t_{\infty },0] \times {\mathbb {T}}^{3} \times [-R_{\infty }, R_{\infty }]\).

5.2 \(L^2 - L^\infty \) Estimate for (1.22)

We now consider (1.22) and let f be a solution of (1.22). Then we split f into two parts:
$$\begin{aligned} f=f {\mathbf {1}} _{\left\{ |v| \le M\right\} }+f {\mathbf {1}} _{\left\{ |v| \ge M\right\} }=:f_{1}+f_{2}. \end{aligned}$$
Let \(U\left( t,s\right) h\) be a solution of (3.6) corresponding to the initial times s with the initial data h. Then
$$\begin{aligned} f_{1}(t,x,v) & = {\mathbf {1}}_{\left\{ |v| \le M\right\} }U(t,0) f_{0} + {\mathbf {1}}_{\left\{ |v| \le M\right\} }\int _{0}^{t}U(t,\tau ) \bar{K}_{g}^{\vartheta }f(\tau ) \mathrm{d}\tau \\ & = {\mathbf {1}}_{\left\{ |v| \le M\right\} }U(t,0) f_{0} + \int _{0} ^{t}{\mathbf {1}}_{\left\{ |v| \le M\right\} }U(t,\tau ) {{\bar{K}}}_{g}^{\vartheta }f(\tau ) \mathrm{d}\tau . \end{aligned}$$
To obtain the \(L^{\infty }\) estimates for \(f_{1}\), we will give some basic estimates for \({{\bar{K}}}_g^\vartheta \).

Lemma 5.9

Let\({{\bar{K}}}_g^\vartheta \)be defined as in (3.4). Suppose thatgsatisfies the assumption in Lemma 2.4. Then there exists\(C=C_{\vartheta }>0\)such that for every\(N,M>0\),
$$\begin{aligned} \Vert {{\bar{K}}}_{g}^{\vartheta }f\Vert _{L^{\infty } ({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}\le C\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}, \end{aligned}$$
(5.9)
$$\begin{aligned} \Vert {{\bar{K}}}_{g}^{\vartheta }1_{|v|>M} f\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}\le C(1+M)^{-1}\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}, \end{aligned}$$
(5.10)
and
$$\begin{aligned} \Vert {{\bar{K}}}_{g}^{\vartheta }f\Vert _{L^{2}({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})} \le C N^{2}\Vert\; f^{\vartheta } \Vert _{L^{2}({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})} + \frac{C}{N}\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}. \end{aligned}$$
(5.11)

Proof

Note that
$$\begin{aligned} {{\bar{K}}}^{\vartheta }_{g} f & = w^{\vartheta }{{\bar{K}}}_{g} f + \left( 2 \frac{\partial _{i} w^{\vartheta }\partial _{j} w^{\vartheta }}{w^{2 \vartheta } }\sigma _{G}^{ij} -\frac{\partial _{ij}w^{\vartheta }}{w^{\vartheta }}\sigma _{G}^{ij} - \frac{\partial _{j}w^{\vartheta }}{w^{\vartheta }}\partial _{i} \sigma _{G}^{ij} - \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}a_{g}^{i} \right) f^{\vartheta }\\ & = w^{\vartheta }Kf + \partial _{i} \sigma ^{i} f^{\vartheta }-v\cdot \sigma v f^{\vartheta }-\partial _{i}\left\{ \phi ^{ij}*[\mu ^{1/2}\partial _{j} g] \right\} f^{\vartheta }+ \left\{ \phi ^{ij}*[v_{i} \mu ^{1/2} \partial _{j} g] \right\} f^{\vartheta }\\& \quad +\left( 2 \frac{\partial _{i} w^{\vartheta }\partial _{j} w^{\vartheta } }{w^{2 \vartheta }}\sigma _{G}^{ij} -\frac{\partial _{ij}w^{\vartheta } }{w^{\vartheta }}\sigma _{G}^{ij} - \frac{\partial _{j}w^{\vartheta } }{w^{\vartheta }}\partial _{i}\sigma _{G}^{ij} - \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}a_{g}^{i} \right) f^{\vartheta }\\ & = w^{\vartheta }Kf + (I)\times f^{\vartheta }, \end{aligned}$$
where \(\phi \), K, \(\sigma _G\), \(\sigma ^i\) are defined as in (1.2), (1.10), (2.1), and (2.2) with \(G = \mu + \mu ^{1/2}g\). Since \(w^{\vartheta }Kf\) satisfies (2.20)–(2.22), it is enough to estimate (I).
By Lemmas 2.2 and 2.3, we have
$$\begin{aligned} |(I)| & \le |\partial _{i} \sigma ^{i}(v)| + |v\cdot \sigma v| + \left| \partial _{i}\left\{ \phi ^{ij}*[\mu ^{1/2}\partial _{j} g](v) \right\} \right| + \left| \left\{ \phi ^{ij}*[v_{i} \mu ^{1/2} \partial _{j} g](v) \right\} \right| \\& \quad + \left| 2 \frac{\partial _{i} w^{\vartheta }\partial _{j} w^{\vartheta }}{w^{2\vartheta }}\sigma _{G}^{ij}\right| +\left| \frac{\partial _{ij}w^{\vartheta }}{w^{\vartheta }}\sigma _{G}^{ij}\right| + \left| \frac{\partial _{j}w^{\vartheta }}{w^{\vartheta }}\partial _{i}\sigma _{G}^{ij}\right| + \left| \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}a_{g}^{i} \right| \\ & \le C(1+|v|)^{-1}. \end{aligned}$$
So, we complete the proof. \(\square \)

Now we can obtain the \(L^{\infty }\) estimates for \(f_{1}\):

Lemma 5.10

Assume (2.6). Letfbe a weak solution of (1.6), (1.7), and (1.22) in a periodic box in the sense of Definition 4.1, then there existC, \(\beta >0\)satisfying the following property: for any\(Z, s, k>1\), and\(\vartheta , l\in {\mathbb {N}} \cup \{0\}\), there exists\(C_{\vartheta ,l}\)such that
$$\begin{aligned}&\left\| 1_{|v|<Zs^{k}} f^{\vartheta }( s) \right\| _{\infty } \nonumber \\&\quad \le C_{\vartheta ,l}\left( Zs^{k}\right) ^{\beta }(1+s)^{-l} \Vert\, f_{0}\Vert _{2,\vartheta + l} + \frac{C}{1+Zs^{k}}\sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
(5.12)

Proof

By the Duhamel principle,
$$\begin{aligned} \Vert 1_{|v|<Zs^{k}}f^{\vartheta }(s)\Vert _{L^{\infty }} & \le \Vert 1_{|v|<Zs^{k} }U(s,s-1)f^{\vartheta }(s-1)\Vert _{L^{\infty }}\\& \qquad + \int _{0}^{1-\varepsilon } \Vert 1_{|v|\le Zs^{k}} U(s,s-1+ \tau ) \bar{K}^{\vartheta }_{g} f(s -1 +\tau )\Vert _{\infty }\mathrm{d} \tau \\& \qquad + \int _{1-\varepsilon }^{1} \Vert 1_{|v|\le Zs^{k}} U(s,s-1+ \tau ) \bar{K}^{\vartheta }_{g} f(s -1 +\tau )\Vert _{\infty }\mathrm{d} \tau \\ & = (i)+(ii)+(iii), \end{aligned}$$
where \({{\bar{K}}}_g^{\vartheta }\) is defined as in (3.4) and \(\varepsilon \) is a constant which will be chosen later. By Lemma 5.8, there exists \(m>0\) such that
$$\begin{aligned} (i) \le C\left( Zs^{k}\right) ^{m} \left( \int _{0}^{1} \Vert U(s^{\prime },s-1)f^{ \vartheta }(s-1)\Vert _{2}^{2} \mathrm{d}s^{\prime }\right) ^{1/2}. \end{aligned}$$
By Theorem 1.2 and Lemma 3.4, for every integer l, there exists \(C_{l}\) such that
$$\begin{aligned} \Vert U(s^{\prime },s-1)f^{\vartheta }(s-1)\Vert _{2} &\le C\Vert\; f^{\vartheta }(s-1)\Vert _{2} = C\Vert\; f(s-1)\Vert _{2,\vartheta }\\ &\le C_{\vartheta ,l}\left( 1+\frac{s-1}{l}\right) ^{-l}\Vert\, f_{0} \Vert _{2,\vartheta +l}\\ & \le C_{\vartheta ,l}(1+s)^{-l}\Vert\, f_{0}\Vert _{2, \vartheta +l}. \end{aligned}$$
Thus
$$\begin{aligned} (i) \le C_{\vartheta ,l} \left( Zs^{k}\right) ^{m}(1+s)^{-l}\Vert\, f_{0} \Vert _{2,\vartheta + l}. \end{aligned}$$
By the maximum principle and (5.9),
$$\begin{aligned} (iii) \le C \varepsilon \sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
By Lemma 5.8,
$$\begin{aligned}&\Vert 1_{|v|\le Zs^{k}} U(s,s-1+\tau ) {{\bar{K}}}^{\vartheta }_{g} f(s-1+\tau )\Vert _{\infty }\\&\quad \le C \left( Zs^{k}\right) ^{m}\left( 1+ \frac{1}{1- \tau }\right) ^{m} \left( \int _{s-1+\tau }^{s}\Vert 1_{|v|<2Zs^{k}} U(s^{\prime },s-1+\tau ){{\bar{K}}}^{\vartheta }_{g} f(s-1+ \tau )\Vert _{2}^{2} \mathrm{d}s^{\prime }\right) ^{1/2}. \end{aligned}$$
By (5.11) and Theorem 1.2, for any \(N>0\),
$$\begin{aligned} (ii) & \le C\int _{0}^{1- \varepsilon } \left( Zs^{k}\right) ^{m}\left( 1+\frac{1}{1- \tau }\right) ^{m} \left( \int _{s-1+\tau }^{s}\Vert {{\bar{K}}}^{\vartheta }_{f} f(s-1+ \tau )\Vert _{2}^{2} \mathrm{d}s^{\prime }\right) ^{1/2} \mathrm{d} \tau \\ & \le C\left( Zs^{k}\right) ^{m} \left( 1+ \frac{1}{\varepsilon }\right) ^{m} \int _{0}^{1} \left( N^{2} \Vert\; f^{\vartheta }(s-1+\tau )\Vert _{2} +\frac{1}{N}\Vert\; f^{\vartheta }(s-1+ \tau )\Vert _{\infty }\right) \mathrm{d}\tau \\ & \le \left( Zs^{k}\right) ^{m} \left( 1+ \frac{1}{\varepsilon }\right) ^{m} \left( C_{\vartheta ,l} N^{2}(1+s)^{-l} \Vert\, f_{0}\Vert _{2,\vartheta +l} +\frac{C}{N}\sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }\right) . \end{aligned}$$
Choose \(\varepsilon ^{-1} = 1+Zs^{k}\) and \(N = (1+Zs^{k} )^{2m+1}\). Then
$$\begin{aligned} (i)+(ii)+(iii) \le C_{\vartheta ,l}\left( Zs^{k}\right) ^{\beta } (1+s)^{-l}\Vert\, f_{0}\Vert _{2,\vartheta +l} + \frac{C}{1+Zs^{k}}\sup _{s^{\prime } \in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }, \end{aligned}$$
where \(\beta = 6m+2>0\). \(\square \)

Based on the above results, we will prove Theorem 1.3.

Proof of Theorem 1.3

Choose \(\varepsilon \) as in Lemma 2.4. By Lemma 5.10, there exists l such that
$$\begin{aligned} \left\| 1_{|v|<Zs^{k}} f^{\vartheta }( s) \right\| _{L^{\infty }}\le C_{\vartheta , l, Z} (1+s)^{-2}\Vert\, f_{0}\Vert _{2,\vartheta +l} + \frac{C}{1+Zs^{k} }\sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
Therefore by the Duhamel principle, the maximum principle, and (5.10),
$$\begin{aligned} \Vert\; f^{\vartheta }(n+1)\Vert _{L^{\infty }} & \le \Vert U(n+1,n)f^{\vartheta }(n)\Vert _{L^{\infty }}\\& \qquad + \int _{0}^{1} \left\| U(n+1, n + s){{\bar{K}}}^{\vartheta }_{g} \left( 1_{|v|< Z(n+s)^{k}}f(n+s) + 1_{|v|>Z(n+s)^{k}} f(n+s) \right) \right\| _{L^{\infty }} \mathrm{d}s\\& \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z} \int _{0}^{1} \bigg ( (1+n+s)^{-2}\Vert\, f_{0}\Vert _{2,\vartheta +l} \\& \qquad + \frac{C}{1+Z(n+s)^{k}} \sup _{s^{\prime }\in (n+s-1,n+s)} \Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty } \bigg ) \mathrm{d}s\\& \qquad + \int _{0}^{1} (1+Z(n+s)^{k})^{-1}\Vert\; f^{\vartheta }(n+s)\Vert _{\infty }\mathrm{d}s\\ & \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z} (1+n)^{-2} \Vert\, f_{0}\Vert _{2,\vartheta +l} \\ & \qquad + C (Zn^{k})^{-1}\sup _{s^{\prime }\in [n-1,n+1]} \Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
After iteration,
$$\begin{aligned} \Vert\; f^{\vartheta }(n+1)\Vert _{L^{\infty }} & \le \Vert\; f^{\vartheta }(1)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z} \sum _{{\bar{n}}=1}^{n} (1+{\bar{n}})^{-2} \Vert\, f_{0}\Vert _{2,\vartheta +l}\\& \qquad + C Z^{-1} \sum _{{\bar{n}}=1}^{n} {\bar{n}}^{-k}\sup _{s\in [0,n+1]} \Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
Choose large k and Z such that \(-k < -1\) and \(CZ^{-1} \sum _{{\bar{n}}=1}^{\infty }{\bar{n}}^{k(-1)} \le \varepsilon _{0}\), where \(\varepsilon _{0}\) will be determined later. Then
$$\begin{aligned} \Vert\; f^{\vartheta }(n+1)\Vert _{L^{\infty }} \le \Vert\; f^{\vartheta }(1)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z}\Vert\, f_{0}\Vert _{2,\vartheta +l} + \varepsilon _{0}\sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
Since n is an arbitrary integer,
$$\begin{aligned} \sup _{s=1,2, \ldots n+1}\Vert\; f^{\vartheta }(s)\Vert _{L^{\infty }} \le \Vert\; f^{\vartheta }(1)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z}\Vert\, f_{0}\Vert _{2,\vartheta ,+l} + \varepsilon _{0}\sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
By the Duhamel principle, the maximum principle, and (5.10),
$$\begin{aligned} \Vert\; f^{\vartheta }(n+t)\Vert _{L^{\infty }} &\le \Vert U(n+t,n)f^{\vartheta }(n)\Vert _{L^{\infty }} + \int _{0}^{t} \left\| U(n+t,n+s){{\bar{K}}}^{\vartheta }_{g} f(n+s)\right\| _{L^{\infty }} \mathrm{d}s\\ &\le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + \int _{0}^{t} \left\| \bar{K}^{\vartheta }_{g} f(n+s)\right\| _{L^{\infty }} \mathrm{d}s\\ & \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + C \int _{0}^{t} \left\| f^{\vartheta }(n+s)\right\| _{L^{\infty }} \mathrm{d}s, \end{aligned}$$
for \(t\in [0,1]\). By the Gronwall inequality,
$$\begin{aligned} \Vert\; f^{\vartheta }(n+t)\Vert _{L^{\infty }} \le C\Vert\; f^{\vartheta }(n)\Vert _{\infty }, \text { for all }t\in [0,1]. \end{aligned}$$
Therefore,
$$\begin{aligned} \sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{L^{\infty }} \le C\Vert\; f^{\vartheta } _{0}\Vert _{L^{\infty }} + C_{\vartheta ,l}\Vert\, f_{0}\Vert _{2,\vartheta +l} + C\varepsilon _{0}\sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
Now, we choose a small \(\varepsilon _{0}\) satisfying \(C\varepsilon _{0}< 1/2\), and then absorb the last term on the RHS to the LHS. Then, we have (1.25) in case of \(\vartheta _0 = 0\) by taking \(l_0(0) = l\).
By (5.12), there exist C, \(l_{1}(\vartheta _0)\) such that
$$\begin{aligned} \Vert 1_{|v| < (1+t)^{\vartheta _{0}}}f^{\vartheta }(t)\Vert _{\infty } \le C(1+t)^{-\vartheta _{0}}(\Vert\, f_{0}\Vert _{2,\vartheta +l_{1}(\vartheta _{0})} + \Vert\; f\Vert _{\infty , \vartheta }). \end{aligned}$$
Thus, by Proposition 2.1, we have
$$\begin{aligned} \Vert\; f(t)\Vert _{\infty , \vartheta } & \le \Vert 1_{|v|< (1+t)}f(t)\Vert _{\infty , \vartheta } + \Vert 1_{|v| \ge (1+t)} f(t)\Vert _{\infty , \vartheta }\\ & \le C(1+t)^{-\vartheta _{0}}\left( \Vert\, f_{0}\Vert _{2,\vartheta +l_{1}(\vartheta _{0})} + \sup _{0 \le s \le t}\Vert\; f(s)\Vert _{\infty ,\vartheta }\right) + C(1+t)^{-\vartheta _{0}}\Vert 1_{|v| \ge (1+t)}f(t)\Vert _{\infty , \vartheta +\vartheta _{0}}\\ & \le C(1+t)^{-\vartheta _{0}}\left( \Vert\, f_{0}\Vert _{2,\vartheta +l_{1}(\vartheta _{0})} + \sup _{0 \le s \le t}\Vert\; f(s)\Vert _{\infty ,\vartheta + \vartheta _{0}}\right) \\ & \le C(1+t)^{-\vartheta _{0}}(\Vert\, f_{0}\Vert _{2,\vartheta +l_{1}(\vartheta _{0})} + \Vert\, f_{0}\Vert _{2,\vartheta +\vartheta _{0}+l_{0}(0)} + \Vert\, f_{0}\Vert _{\infty , \vartheta +\vartheta _{0}})\\ & \le C(1+t)^{-\vartheta _{0}}\Vert\, f_{0}\Vert _{\infty ,\vartheta +l_{0}(\vartheta _0)}, \end{aligned}$$
where \(l_{0}(\vartheta _0) = \max \{l_{1}({\vartheta _{0}}), \vartheta _{0} + l_{0}(0)\} + 2\). \(\square \)

Lemma 5.11

Assume (2.6). Letfbe a strong solution of (1.6), (1.7), and (1.22) in a periodic box. Let\(\beta >0\)and\(p>2\)be given constants. Then there exist\(l\in {\mathbb {N}}\)and\(C_{\beta ,l}\)such that
$$\begin{aligned} \left( \int _{0}^{t} \Vert\; f(s)\Vert _{p,\beta }^{p} \mathrm{d}s \right) ^{1/p} & \le C_{\beta ,l_{0}} \Vert\, f_{0}\Vert _{2,\beta +l}^{2/p}\left( \Vert\, f_{0}\Vert _{\infty ,\beta }+\Vert\, f_{0}\Vert _{2,\beta +l}\right) ^{(p-2)/p} \nonumber \\ & \le C_{\beta ,l_{0}} \left( \Vert\, f_{0}\Vert _{\infty ,\beta } + \Vert\, f_{0}\Vert _{2,\beta +l}\right) . \end{aligned}$$
(5.13)

Proof

By Theorems 1.2 and 1.3, there exist \(l\in {\mathbb {N}}\) and \(C_{\beta ,l}\) such that
$$\begin{aligned} \Vert\; f(s)\Vert _{2,\beta } & \le C_{\beta ,l_{0}} (1+s)^{-l} \Vert\, f_{0}\Vert _{2,\beta +l},\\ \Vert\; f(s)\Vert _{\infty , \beta } & \le C_{\beta ,l_{0}} \left( \Vert\, f_{0}\Vert _{\infty , \beta } + \Vert\, f_{0}|_{2,\beta +l}\right) . \end{aligned}$$
By the interpolation, we have
$$\begin{aligned} \Vert\; f(s)\Vert _{p,\beta }^{p} \le (C_{\beta ,l_{0}})^{p} (1+s)^{-2l} \Vert\, f_{0}\Vert _{2,\beta +l} ^{2}\left( \Vert\, f_{0}\Vert _{\infty ,\beta } + \Vert\, f_{0}\Vert _{2,\beta +l}\right) ^{p-2}. \end{aligned}$$
Taking the integral over \(s\in (0,\infty )\), we have the first inequality of (5.13). The second inequality of (5.13) comes from the Young inequality, then we complete the proof. \(\square \)

5.3 \(L^2-L^\infty \) Estimate for (4.5)

We will derive another type of \(L^2-L^\infty \) estimate to obtain a uniform Hölder estimate for a weak solution of (1.22) in the sense of Definition 4.1. The proof is similar to the case of Subsect. 5.2.

Let us multiply (4.5) by \(w^{\vartheta }\), then \(h:= w^{\vartheta }f = f^{\vartheta }\) satisfies
$$\begin{aligned} (\partial _{t} + v\cdot \nabla _{x} - \bar{A}_{g}^{\vartheta })h = {{\tilde{K}}}_{g}^{\vartheta }h, \end{aligned}$$
(5.14)
where
$$\begin{aligned} {{\tilde{K}}}^{\vartheta }_{g} h = \left( 2 \frac{\partial _{i} w^{\vartheta }\partial _{j} w^{\vartheta }}{w^{2 \vartheta }} \sigma ^{ij}_{G} -\frac{\partial _{ij}w^{\vartheta }}{w^{\vartheta }}\sigma ^{ij}_{G} - \frac{\partial _{j}w^{\vartheta }}{w^{\vartheta }}\partial _{i} \sigma ^{ij}_{G} - \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}a_{g}^{i} \right) h. \end{aligned}$$
(5.15)
Similar to Definition 4.1, we can define a weak solution of (5.14).
Then we split \(f^{\vartheta }\) into two parts:
$$\begin{aligned} f^{\vartheta }=f^{\vartheta }{\mathbf {1}} _{\left\{ |v| \le M\right\} }+f^{\vartheta }{\mathbf {1}} _{\left\{ |v| \ge M\right\} }=:f_{1}+f_{2}. \end{aligned}$$
Let \(U^{\vartheta }(t,s)f_{0}\) be a weak solution of (5.1) in the sense of Definition 3.2 corresponding to the initial data \(f_{0}\) with the initial time \(t=s\), then we have
$$\begin{aligned} f_{1}(t)&= {\mathbf {1}}_{|v| \le M} U^{\vartheta }(t,0)f_{0}^{\vartheta } + {\mathbf {1}}_{|v| \le M} \int _{0}^{t}U^{\vartheta }(t, \tau ){{\tilde{K}}}_{g}^{\vartheta } f^{\vartheta }(\tau ) \mathrm{d} \tau \\&={\mathbf {1}}_{|v| \le M} U^{\vartheta }(t,0)f_{0}^{\vartheta } + \int _{0}^{t}{\mathbf {1}}_{|v| \le M} U^{\vartheta }(t, \tau ){{\tilde{K}}}_{g}^{\vartheta } f^{\vartheta }(\tau ) \mathrm{d} \tau . \end{aligned}$$

Lemma 5.12

Assume (2.6). There exists\(C=C_{\vartheta }>0\)such that
$$\begin{aligned} \Vert {{\tilde{K}}}_{g}^{\vartheta }f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}\le C\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}, \end{aligned}$$
(5.16)
$$\begin{aligned} \Vert {{\tilde{K}}}_{g}^{\vartheta } 1_{|v|>M}f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}\le C(1+M)^{-1}\Vert\; f^{\vartheta }\Vert _{L^{\infty }({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}, \end{aligned}$$
(5.17)
and
$$\begin{aligned} \Vert {{\tilde{K}}}_{g}^{\vartheta }f^{\vartheta }\Vert _{L^{2}({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})} \le C\Vert\; f^{\vartheta }\Vert _{L^{2}({\mathbb {T}}^{3} \times {\mathbb {R}}^{3})}. \end{aligned}$$
(5.18)

Proof

Since
$$\begin{aligned} {{\tilde{K}}}^{\vartheta }_{g} f^{\vartheta } = \left( 2 \frac{\partial _{i} w^{\vartheta }\partial _{j} w^{\vartheta }}{w^{2 \vartheta }}\sigma ^{ij}_{G} -\frac{\partial _{ij}w^{\vartheta }}{w^{\vartheta }}\sigma ^{ij}_{G} - \frac{\partial _{j}w^{\vartheta }}{w^{\vartheta }}\partial _{i}\sigma ^{ij}_{G} - \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}a_{g}^{i} \right) f^{\vartheta } \end{aligned}$$
and by Lemma 2.2,
$$\begin{aligned} \left| 2 \frac{\partial _{i} w^{\vartheta }\partial _{j} w^{\vartheta }}{w^{2 \vartheta }}\sigma _{G}^{ij}\right| +\left| \frac{\partial _{ij}w^{\vartheta } }{w^{\vartheta }}\sigma _{G}^{ij}\right| + \left| \frac{\partial _{j}w^{\vartheta }}{w^{\vartheta }}\partial _{i}\sigma _{G}^{ij}\right| + \left| \frac{\partial _{i} w^{\vartheta }}{w^{\vartheta }}a_{g}^{i} \right| \le C(1+|v|)^{-1}. \end{aligned}$$
So the proof is complete. \(\square \)

Lemma 5.13

Assume (2.6). Letfbe a weak solution of (4.5) in a periodic box in the sense of Definition 3.2, then there existC, \(\beta >0\)satisfying the following property: for anyZ, \(s>1\), \(\vartheta \), and\(k>0\), and\(l\in {\mathbb {N}}\), there exists\(C_{\vartheta ,l}\)such that
$$\begin{aligned}&\left\| 1_{|v|<Zs^{k}} f^{\vartheta }( s) \right\| _{\infty } \nonumber \\&\quad \le C_{\vartheta ,l}\left( Zs^{k}\right) ^{\beta }(1+s)^{-l} \Vert\, f_{0}\Vert _{2,\vartheta + l} + \frac{C}{1+Zs^{k}}\sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }, \end{aligned}$$
(5.19)
for any\(s\ge 1\).

Proof

By the Duhamel principle,
$$\begin{aligned} \Vert 1_{|v|<Zs^{k}}f^{\vartheta }(s)\Vert _{L^{\infty }} & \le \Vert 1_{|v|<Zs^{k}} U^{\vartheta }(s, s-1)f^{\vartheta }(s-1)\Vert _{L^{\infty }}\\&+ \int _{0}^{1-\varepsilon } \Vert 1_{|v|\le Zs^{k}}U^{\vartheta }(s, s-1+ \tau ) {{\tilde{K}}}^{\vartheta }_{g} f^{\vartheta }(s -1 +\tau )\Vert _{\infty } \mathrm{d} \tau \\&+ \int _{1-\varepsilon }^{1} \Vert 1_{|v|\le Zs^{k}}U^{\vartheta }(s,s-1+ \tau ) {{\tilde{K}}}^{\vartheta }_{g} f^{\vartheta }(s -1 +\tau )\Vert _{\infty } \mathrm{d} \tau \\ & = (i)+(ii)+(iii). \end{aligned}$$
By Lemma 5.8, there exists \(m>0\) such that
$$\begin{aligned} (i) \le C\left( Zs^{k}\right) ^{m} \left( \int _{s-1}^{s} \Vert U^{\vartheta }(s^{\prime },s-1)f^{\vartheta }(s-1)\Vert _{2}^{2} \mathrm{d}s^{\prime }\right) ^{1/2}. \end{aligned}$$
By Theorem 4.6 and Lemma 3.4, for every integer l, there exists \(C_{l}\) such that
$$\begin{aligned} \Vert U^{\vartheta }(s^{\prime },s-1)f^{\vartheta }(s-1)\Vert _{2} & \le C\Vert\; f^{\vartheta }(s-1)\Vert _{2} = C\Vert\; f(s-1)\Vert _{2,\vartheta }\\ & \le C_{\vartheta ,l}\left( 1+\frac{s-1}{l}\right) ^{-l}\Vert\, f_{0} \Vert _{2,\vartheta +l}\\ & \le C_{\vartheta ,l}(1+s)^{-l}\Vert\, f_{0}\Vert _{2, \vartheta +l}. \end{aligned}$$
Thus
$$\begin{aligned} (i) \le C_{\vartheta ,l} \left( Zs^{k}\right) ^{m}(1+s)^{-l}\Vert\, f_{0} \Vert _{2,\vartheta + l}. \end{aligned}$$
By the maximum principle and (5.16),
$$\begin{aligned} (iii) \le C \varepsilon \sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
By Lemma 5.8,
$$\begin{aligned}&\Vert 1_{|v|\le Zs^{k}} U^{\vartheta }(s,s-1+ \tau ) {{\tilde{K}}}^{\vartheta }_{g} f(s-1+\tau )\Vert _{\infty }\\&\quad \le C \left( Zs^{k}\right) ^{m}\left( 1+ \frac{1}{1- \tau }\right) ^{m} \left( \int _{s}^{s-1+ \tau }\Vert 1_{|v|<2Zs^{k}} U^{\vartheta }(s^{\prime }, s-1+\tau ){{\tilde{K}}}^{\vartheta }_{g} f(s-1+ \tau )\Vert _{2}^{2} \mathrm{d}s^{\prime }\right) ^{1/2}. \end{aligned}$$
By (5.18) and Theorem 4.6, for any \(N>0\),
$$\begin{aligned} (ii) & \le C\int _{0}^{1- \varepsilon } \left( Zs^{k}\right) ^{m}\left( 1+ \frac{1}{1- \tau }\right) ^{m} \left( \int _{0}^{1- \tau }\Vert \tilde{K}^{\vartheta }_{g} f^{\vartheta }(s-1+ \tau )\Vert _{2}^{2} \mathrm{d}s^{\prime }\right) ^{1/2} \mathrm{d} \tau \\ & \le C\left( Zs^{k}\right) ^{m} \left( 1+ \frac{1}{\varepsilon }\right) ^{m} \int _{0}^{1} C \Vert\; f^{\vartheta }(s-1+\tau )\Vert _{2} \mathrm{d} \tau \\ & \le \left( Zs^{k}\right) ^{m} \left( 1+ \frac{1}{\varepsilon }\right) ^{m} C_{\vartheta ,l}(1+s)^{-l} \Vert\, f_{0}\Vert _{2,\vartheta -l}. \end{aligned}$$
Choose \(\varepsilon ^{-1} = 1+Zs^{k}\). Then
$$\begin{aligned} (i)+(ii)+(iii) \le C_{\vartheta ,l}\left( Zs^{k}\right) ^{\beta } (1+s)^{-l}\Vert\, f_{0}\Vert _{2,\vartheta -l} + \frac{C}{1+Zs^{k}}\sup _{s^{\prime } \in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }, \end{aligned}$$
where \(\beta = 2m+1\). \(\square \)

Theorem 5.14

Assume (2.6). Letfbe a weak solution of (1.7), (4.5) in a periodic box in thesense of Definition 3.2. Then there existslsuch that for every\(\vartheta >0\),
$$\begin{aligned} \Vert\; f^{\vartheta }( t) \Vert _{L^{\infty }}\le C \Vert\, f_{0}^{\vartheta }\Vert _{L^{\infty }} + C_{\vartheta ,l}\Vert\, f_{0}\Vert _{2,\vartheta +l}\le C\Vert\, f_0\Vert _{\infty , \vartheta + l_{0}},\, \text { for any }t>0, \end{aligned}$$
(5.20)
where \(l_0 = l+2\).

Proof

By Lemma 5.13, there exists \(l_{0}\) such that for \(l> l_{0}\),
$$\begin{aligned} \left\| 1_{|v|<Zs^{k}} f^{\vartheta }( s) \right\| _{L^{\infty }}\le C_{\vartheta , l, Z} (1+s)^{-2}\Vert\, f_{0}\Vert _{2,\vartheta +l} + \frac{C}{1+Zs^{k} }\sup _{s^{\prime }\in (s-1,s)}\Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
Therefore by the Duhamel principle, the maximum principle, and (5.17),
$$\begin{aligned} \Vert\; f^{\vartheta }(n+1)\Vert _{L^{\infty }} & \le \Vert U^{\vartheta }(n+1,n)f^{\vartheta }(n)\Vert _{L^{\infty }}\\&+ \int _{1}^{0} \left\| U^{\vartheta }(n+1,n+s){{\tilde{K}}}^{\vartheta }_{g} \left( 1_{|v|< Z(n+s)^{k}}f^{\vartheta }(n+s) + 1_{|v|>Z(n+s)^{k}} f^{\vartheta }(n+s) \right) \right\| _{L^{\infty }} \mathrm{d}s\\ & \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z} \int _{0}^{1} (1+n+s)^{-2}\Vert\, f_{0}\Vert _{2,\vartheta +l}\\&+ \frac{C}{1+Z(n+s)^{k}} \sup _{s^{\prime }\in (n+s-1,n+s)} \Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }\mathrm{d}s\\&+ \int _{0}^{1} (1+Z(n+s)^{k})^{-1}\Vert\; f^{\vartheta }(n+s)\Vert _{\infty }\mathrm{d}s\\ & \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z} (1+n)^{-2} \Vert\, f_{0}\Vert _{2,\vartheta +l} + C (Zn^{k})^{-1}\sup _{s^{\prime }\in [n-1,n+1]} \Vert\; f^{\vartheta }(s^{\prime })\Vert _{\infty }. \end{aligned}$$
After iteration,
$$\begin{aligned} \Vert\; f^{\vartheta }(n+1)\Vert _{L^{\infty }} & \le \Vert\; f^{\vartheta }(1)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z} \sum _{{\bar{n}}=1}^{n} (1+{\bar{n}})^{-2} \Vert\, f_{0}\Vert _{2,\vartheta +l}\\&+ C Z^{-1} \sum _{{\bar{n}}=1}^{n} {\bar{n}}^{k(-1)}\sup _{s\in [0,n+1]} \Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
Choose large k and Z such that \(k(-1) < -1\) and \(CZ^{-1} \sum _{{\bar{n}}=1}^{\infty }{\bar{n}}^{k(-1)} \le \varepsilon \), where \(\varepsilon \) will be determined later. Then
$$\begin{aligned} \Vert\; f^{\vartheta }(n+1)\Vert _{L^{\infty }} \le \Vert\; f^{\vartheta }(1)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z}\Vert\, f_{0}\Vert _{2,\vartheta +l} + \varepsilon \sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
Since n is an arbitrary integer,
$$\begin{aligned} \sup _{s=1,2,\ldots n+1}\Vert\; f^{\vartheta }(s)\Vert _{L^{\infty }} \le \Vert\; f^{\vartheta }(1)\Vert _{L^{\infty }} + C_{\vartheta ,l,Z}\Vert\, f_{0}\Vert _{2,\vartheta ,+l} + \varepsilon \sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
By the Duhamel principle, the maximum principle, and (5.17),
$$\begin{aligned} \Vert\; f^{\vartheta }(n+t)\Vert _{L^{\infty }} & \le \Vert U^{\vartheta } (n+t,n)f^{\vartheta }(n)\Vert _{L^{\infty }} + \int _{0}^{t} \left\| U^{\vartheta }(n+t,n+s){{\tilde{K}}}^{\vartheta }_{g} f^{\vartheta }(n+s)\right\| _{L^{\infty }} \mathrm{d}s\\ & \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + \int _{0}^{t} \left\| \tilde{K}^{\vartheta }_{g} f^{\vartheta }(n+s)\right\| _{L^{\infty }} \mathrm{d}s\\ & \le \Vert\; f^{\vartheta }(n)\Vert _{L^{\infty }} + C \int _{0}^{t} \left\| f^{\vartheta }(n+s)\right\| _{L^{\infty }} \mathrm{d}s. \end{aligned}$$
By the Gronwall inequality,
$$\begin{aligned} \Vert\; f^{\vartheta }(n+t)\Vert _{L^{\infty }} \le C\Vert\; f^{\vartheta }(n)\Vert _{\infty }, \text { for all }t\in [0,1]. \end{aligned}$$
Therefore,
$$\begin{aligned} \sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{L^{\infty }} \le C\Vert\; f^{\vartheta } _{0}\Vert _{L^{\infty }} + C_{\vartheta ,l}\Vert\, f_{0}\Vert _{2,\vartheta +l} + C\varepsilon \sup _{s\in [0,n+1]}\Vert\; f^{\vartheta }(s)\Vert _{\infty }. \end{aligned}$$
Now, we choose small \(\varepsilon \) satisfying \(C\varepsilon < 1/2\), and then absorb the last term on the RHS to the LHS. Thus, we obtain the first inequality of (5.20). The second inequality of (5.20) is a consequence of Proposition 2.1. \(\square \)

6 \(L^\infty \) to Hölder Estimate

6.1 Local Hölder Estimate

In this section, we will derive a local Hölder estimate for (4.5). We redefine \(Q_{R}(z_{0}) := (t_{0}-R^{2},t_{0}]\times B(x_{0}; \, R^{3}) \times B(v_{0}; \, R)\), \(z_{0} = (t_{0}, x_{0}, v_{0})\), and \(Q_{R} := Q_{R}((0,0,0))\).

Since we consider the local properties of the solution on the interior part, we can use the technique in [10] for our modified operator \(\tilde{A_{g}}\). In this subsection, we assume that g satisfies the conditions in Lemma 2.4.

First, we introduce a De Giorgi-type lemma.

Lemma 6.1

(Lemma 13 in [10]) Assume (2.6). Let\({\hat{Q}}:= Q_{1/4}(0,0,-1)\). For any (universal) constants\(\delta _1 \in (0,1)\)and\(\delta _2 \in (0,1)\)there exist\(\nu > 0\)and\(\vartheta \in (0,1)\)(both universal) such that for any solutionfof (4.5) in\(Q_{2}\)with\(|f|\le 1\)and
$$\begin{aligned} \begin{aligned} |\{f\ge 1- \vartheta \} \cap Q_{1/4} |&\ge \delta _{1}|Q_{1/4}|,\\ |\{f\le 0\} \cap {\hat{Q}} |&\ge \delta _{2}|{\hat{Q}}|, \end{aligned} \end{aligned}$$
we have
$$\begin{aligned} |\{0<f<1- \vartheta \}\cap B_{1} \times B_{1} \times (-2,0]| \ge \nu . \end{aligned}$$

Proof

The proof is exactly the same as [10]. We omit the proof. \(\square \)

Lemma 6.2

(Lemma 17 in [10]) Assume (2.6). Let\({\hat{Q}}:= Q_{1/4}(0,0,-1)\)andfbe a weak solution of (4.5) in\(Q_{2}\)in the sense of Definition 3.2with\(|f| \le 1\). If
$$\begin{aligned} |\{f\le 0\}\cap {\hat{Q}}| \ge \delta _{2}|{\hat{Q}}|, \end{aligned}$$
then
$$\begin{aligned} \sup _{Q_{1/8}} f \le 1- \lambda \end{aligned}$$
for some\(\lambda \in (0,1)\), depending only on dimension and the eigenvalue of\(\sigma \).

Proof

The proof is exactly the same as [10]. We omit the proof. \(\square \)

The following lemma can be derived by the previous lemma.

Lemma 6.3

Assume (2.6). Letfbe a weak solution of (4.5) in\(Q_{2}\)in the sense of Definition 3.2with\(|f| \le 1\). Then
$$\begin{aligned} \sup _{Q_{1/8}} f - \inf _{Q_{1/8}} f \le 2- \lambda \end{aligned}$$
for some\(\lambda \in (0,2)\), depending only on dimension and the eigenvalue of\(\sigma \).

By the scaling argument, \(Q_{2}\) and \(Q_{1/8}\) can be replaced by \(Q_{2r}\) and \(Q_{r/8}\).

Lemma 6.4

Assume (2.6). Letfbe a weak solution of (4.5) in\(Q_{2r}\)in the sense of Definition 3.2with\(|f| \le 1\). For any subset\(Q\subset {\mathbb {R}}^{7}\), define
$$\begin{aligned} \underset{Q}{\text {Osc}} f := \sup _{(t^{\prime },x^{\prime },v^{\prime })\in Q} f(t^{\prime },x^{\prime },v^{\prime }) - \inf _{(t^{\prime },x^{\prime },v^{\prime })\in Q} f(t^{\prime },x^{\prime },v^{\prime }). \end{aligned}$$
Then for every\(r\le 1,\)
$$\begin{aligned} \underset{Q_{r/8}}{\text {Osc}} f \le \left( 1- \frac{\lambda }{2}\right) \underset{Q_{2r}}{\text {Osc}} f \end{aligned}$$
for some\(\lambda \in (0,2)\), depending only on dimension and the eigenvalue of\(\sigma \).

Proof

Define
$$\begin{aligned} {{\bar{F}}}(t,x,v):=\frac{2}{\underset{Q_{2r}}{\text {Osc}} f} \left( f(r^{2} t, r^{3} x, r v) - \frac{\sup _{Q_{2r}} f + \inf _{Q_{2r}} f}{2}\right) . \end{aligned}$$
Then \({{\bar{F}}}\) satisfies
$$\begin{aligned}&{{\bar{F}}}_{t} + v\cdot \partial _{x} {{\bar{F}}} = \tilde{A_{g}}^{r} {{\bar{F}}},\\&\tilde{A_{g}}^{r} {{\bar{F}}}(t,x,v) := \nabla _{v}(\sigma _{G}(r^{2} t, r^{3} x, rv) \nabla _{v} {{\bar{F}}}(t,x,v)) + r a_{g}(r^{2} t, r^{3} x, rv) \cdot \nabla _{v} {{\bar{F}}}(t,x,v) ,\end{aligned}$$
and then apply Lemma 6.3. \(\square \)

Now we establish the Hölder continuity at \(v=0\).

Lemma 6.5

(Hölder continuity near \(v=0\)) Assume (2.6). Letfbe a weak solution of (4.5) in\(Q_{R}(t_{0},x_{0},0)\)in the sense of Definition 3.2. Then there exist a uniform constant\(C>0\)and a constant\(\alpha \in (0,1)\)depending only on dimension and the eigenvalue of\(\sigma _{G}\)such that
$$\begin{aligned} \Vert\; f\Vert _{C^{\alpha }(Q_{R/128}(t_{0},x_{0},0))} \le \frac{C}{R^{3\alpha } }\Vert\; f\Vert _{L^{\infty }(Q_{R}(t_{0},x_{0},0))}, \end{aligned}$$
for every\(R < 1\).

Proof

We first prove
$$\begin{aligned} \sup _{(s,y,w)\in Q_{R/16}} \frac{|f(s,y,w)-f(0,0,0)|}{(|s|+|y|+|w|)^{\alpha }} \le \frac{C}{R^{3\alpha }}\Vert\; f\Vert _{L^{\infty }(Q_{R})}. \end{aligned}$$
Define \({\text {Osc}_{Q}} f\) as in Lemma 6.4 and
$$\begin{aligned} \varphi (r):= r^{-\alpha _{0}} \underset{Q_{r}}{\text {Osc}} f, \end{aligned}$$
where \(\alpha _{0} >0\) can be chosen later. By Lemma 6.4,
$$\begin{aligned} \underset{Q_{r/16}}{\text {Osc}} f \le \left( 1- \frac{\lambda }{2}\right) \underset{Q_{r}}{\text {Osc}} f. \end{aligned}$$
(6.1)
Choose \(\alpha _{0}\) such that \(16^{\alpha _{0}}\big( 1- \frac{\lambda }{2}\big) <1\). Then by (6.1),
$$\begin{aligned} \begin{aligned} \varphi \left( \frac{r}{16}\right)&= r^{-\alpha _{0}}16^{\alpha _{0} }\underset{Q_{r/4}(t,x,v)}{\text {Osc}} f\\&\le 16^{\alpha _{0}}\left( 1- \frac{\lambda }{2}\right) r^{-\alpha _{0} }\underset{Q_{r}}{\text {Osc}} f\\&<\varphi (r). \end{aligned} \end{aligned}$$
Therefore, we have
$$\begin{aligned} \begin{aligned} \sup _{0<r\le R/16} \varphi (r)&\le \sup _{\frac{R}{16}<r\le R}\varphi (r)\\&\le 2 \frac{16^{\alpha _{0}}}{R^{\alpha _{0}}}\sup _{(t,x,v) \in Q_{R} }|f(t,x,v)|. \end{aligned} \end{aligned}$$
(6.2)
If \((t,x,v)\in \partial Q_{r}\) then \(|t| + |x| + |v| \ge r^{3}\). Therefore, for \(3 \alpha = \alpha _{0}\) and \(r \le R/16\), by (6.2),
$$\begin{aligned} \sup _{(s,y,w)\in Q_{R/16}} \frac{|f(s,y,w)-f(0,0,0)|}{(|s|+|y|+|w|)^{\alpha }} & = \sup _{(s,y,w)\in \partial Q_{r}, r\in (0, R/16)} \frac{|f(s,y,w)-f(0,0,0)|}{(|s|+|y|+|w|)^{\alpha }} \nonumber \\ & \le \sup _{(s,y,w)\in Q_{r}, r\in (0, R/16)} \frac{|f(s,y,w)-f(0,0,0)|}{r^{\alpha _{0}}} \nonumber \\ & \le \sup _{r \in (0,R/16)} \varphi (r) \nonumber \\ & \le \frac{C}{R^{\alpha _{0}}} \sup _{(t,x,v) \in Q_{R}}|f(t,x,v)|. \end{aligned}$$
(6.3)
Now we consider the general case. For any \((t_{*},x_{*},v_{*}) \in Q_{R/32}(t_{0}, x_{0}, 0)\), define the translated function
$$\begin{aligned} F(T,X,V) & = f(t,x,v),\\ T & = t-t_{*},\quad X = x-x_{*} - Tv_{*}, \quad V = v-v_{*}. \end{aligned}$$
Then F satisfies
$$\begin{aligned} \partial _{T} F + V \cdot \nabla _{X} F = \nabla _{V} \cdot (\Sigma _{G}(t,x,v) \nabla _{V} F) + a_{g}(t,x,v) \cdot \nabla _{V} F. \end{aligned}$$
Therefore, by (6.3),
$$\begin{aligned} \sup _{(s,y,w)\in Q_{R_{1}/16}} \frac{|F(s,y,w)-F(0,0,0)|}{(|s|+|y|+|w|)^{\alpha }} \le \frac{C}{{R_{1}}^{\alpha _{0}}} \sup _{(t,x,v) \in Q_{R_{1}}}|F(t,x,v)| \end{aligned}$$
for every \(R_{1} < 1\). Since \(|v_{*}| \le R/128\),
$$\begin{aligned} (t,x,v) \in Q_{R/64}(t_{*},x_{*},v_{*}) \text { implies } (T,X,V) \in Q_{R/32} \end{aligned}$$
and
$$\begin{aligned} (T,X,V) \in Q_{R/2} \text { implies } (t,x,v) \in Q_{R}(t_{*},x_{*},v_{*}). \end{aligned}$$
Therefore, by (6.3)
$$\begin{aligned}&\sup _{(t,x,v)\in Q_{R/64}(t_{*},x_{*},v_{*})}\frac{|f(t,x,v)-f(t_{*} ,x_{*},v_{*})|}{(|t-t_{*}|+|x-x_{*}|+|v-v_{*}|)^{\alpha }}\\&\qquad \le (1+|v_{*}|)^{\alpha } \sup _{(t,x,v)\in Q_{R/64}(t_{*},x_{*},v_{*} )}\frac{|f(t,x,v)-f(t_{*},x_{*},v_{*})|}{((1+|v_{*}|)|t-t_{*}|+|x-x_{*} |+|v-v_{*}|)^{\alpha }}\\&\qquad \le C \sup _{(T,X,V)\in Q_{R/32}} \frac{|F(T,X,V)-F(0,0,0)|}{((1+|v_{*}|)|T|+|X+v_{*} T|+|V|)^{\alpha }}\\&\qquad \le C \sup _{(T,X,V)\in Q_{R/32}} \frac{|F(T,X,V)-F(0,0,0)|}{(|T|+|X|+|V|)^{\alpha }}\\&\qquad \le \frac{C}{{R}^{\alpha _{0}}} \sup _{(T,X,V) \in Q_{R/2}}|F(T,X,V)|\\&\qquad \le \frac{C}{{R}^{\alpha _{0}}} \sup _{(t,x,v) \in Q_{R}}|f(t,x,v)|. \end{aligned}$$
So the proof is complete. \(\square \)

6.2 Global Hölder Estimate

In this subsection, we will derive a Hölder continuity for the solution of (1.22). Let f(txv) be a weak solution of (1.22) in the sense of Definition 4.1. Then
$$\begin{aligned} {{\tilde{f}}}(t,x,v) \, {:=} \, {\left\{ \begin{array}{ll} f(t,x,v), & \text { if }t\ge 0,\\ f_{0}(x,v), & \text { if }-1 \le t< 0 \end{array}\right. } \end{aligned}$$
satisfies
$$\begin{aligned} {{\tilde{f}}}_{t} + v \cdot \nabla _{x} {{\tilde{f}}} - {{\bar{A}}}_{g} {{\tilde{f}}} = \tilde{S}(t,x,v), \end{aligned}$$
where \({{\bar{A}}}_g\) and \({{\bar{K}}}_g\) is defined as in (3.2), and (3.3),
$$\begin{aligned} {{\tilde{S}}}(t,x,v) = {\left\{ \begin{array}{ll} (v\cdot \nabla _{x} - {{\bar{A}}}_{f_{0}})f_{0}(x,v), & \text { if }t\le 0,\\ {{\bar{K}}}_{g} f(t,x,v), & \text { if }t> 0. \end{array}\right. } \end{aligned}$$
Since U(ts) is the solution operator of (4.5). Then f satisfies
$$\begin{aligned} f(t) = U(t, -1)f_{0} + \int _{-1}^{t} U(t, s){{\tilde{S}}}(s)\mathrm{d}s. \end{aligned}$$
Fisrt, we will obtain a uniform Hölder continuity of U(ts)f. Finally, we will derive a uniform Hölder continuity of f(t).

As a starting point, we introduce a technical lemma to obtain a uniform Hölder continuity of U(ts)f.

Lemma 6.6

Let\((t_{*}, x_{*}, v_{*})\in {\mathbb {R}}_{+} \times {\mathbb {R}}^{3} \times {\mathbb {R}}^{3}\), \(N -1/2 \le |v_{*}| \le N+1/2\), \(m>9\)andObe an orthonormal matrix. Define
$$\begin{aligned} D := \begin{bmatrix} (1+|v_{*}|)^{-3/2}&0&0\\ 0&(1+|v_{*}|)^{-1/2}&0\\ 0&0&(1+|v_{*}|)^{-1/2} \end{bmatrix} , \end{aligned}$$
(6.4)
$$\begin{aligned} X := D^{-1}O^{T}(x-v_{*}(t-t_{*})), \quad X_{*} := D^{-1}O^{T}x_{*}, \quad V := D^{-1}O^{T}(v-v_{*}),\\ r_{0} := (2+N)^{-m}, \quad r_{1} := (2+N)^{-\frac{2m}{3}+\frac{5}{6}}, \quad r_{2} := (2+N)^{-\frac{4m}{9}+ \frac{13}{18}}. \end{aligned}$$
Then if\((t,x,v)\in Q_{r_{0}}(t_{*}, x_{*}, v_{*})\), then\((t,X,V) \in Q_{r_{1}}(t_{*}, X_{*}, 0)\). Moreover, if\((t,X,V) \in Q_{128 r_{1}} (t,X_{*},0)\), then\((t,x,v) \in Q_{128r_{2}}(t_{*}, x_{*}, v_{*})\).

Proof

If \((t,x,v) \in Q_{r_{0}}(t_{*}, x_{*}, v_{*})\), then
$$\begin{aligned} |t-t_{*}| & \le r_{0}^2 \le r_{1}^2,\\ |X-X_{*}| & = |D^{-1}O^{T}(x-x_{*} - v_{*}(t-t_{*}))|\\ & \le (2 + N)^{3/2}(r_{0}^3 + (N+1/2)r_{0}^2)\\ & \le (2+N)^{3/2}\left( (2+N)^{-3m} + (N+1/2)(2+N)^{-2m} \right) \\ & \le (2+N)^{3/2}(2+N)^{1-2m}\\ & \le r_{1}^{3}, \end{aligned}$$
and
$$\begin{aligned} |V| & = |D^{-1}O^{T}(v-v_{*})|\\ & \le (2+N)^{3/2}r_{0}\\ & \le r_{1}. \end{aligned}$$
Conversely, if \((t,X,V) \in Q_{128r_{1}}(t,X_{*},0)\), then
$$\begin{aligned} |t-t_{*}| \le (128 r_{1})^2 \le (128 r_{2})^2 \end{aligned}$$
and
$$\begin{aligned} |v-v*| = |ODV| \le (1/2+N)^{-1/2}r_{1} \le 128 r_{2}. \end{aligned}$$
Since \(128r_{1}\le 1\) and \((1/2 +N)^{-1/2}(1+N) \le 128(2+N)^{1/2}\), we have
$$\begin{aligned} |x-x_{*}| & = |OD(X-X_{*}) + v_{*}(t-t_{*})|\\ & \le (1/2+N)^{-1/2}(r_{1}^3 + N r_{1}^2)\\ & \le (1/2+N)^{-1/2}(1+N)r_{1}^2\\ & \le (128)^3(2+N)^{\frac{1}{2}-\frac{4m}{3} + \frac{5}{3}}\\ & \le (128 r_{2})^3. \end{aligned}$$
So the proof is complete. \(\square \)

Lemma 6.7

(Uniform Hölder for (4.5)) Assume (2.6). Letfbe a solution of (4.5) in\(Q_{1}(t_{0},x_{0},v_{0})\). Then there exist\(\vartheta >0\), \(\vartheta _{0}>0\), \(C_{\vartheta }\), and\(\alpha \in (0,1)\)depending only on dimension such that
$$\begin{aligned} \sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime }) \in Q_{1}(t_{0},x_{0},v_{0})} \frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{(|t-t^{\prime }| + |x-x^{\prime }| +|v-v^{\prime }|)^{\alpha }}\le C \Vert\; f\Vert _{\infty , \vartheta }\le C_{\vartheta } \Vert\, f_{0}\Vert _{\infty , \vartheta + \vartheta _{0}}. \end{aligned}$$
(6.5)

Proof

By the integration by parts,
$$\begin{aligned} a_{g}\cdot \nabla _{v} f & = -\left\{ \phi ^{ij}*[v_{i} \mu ^{1/2}g]\right\} \partial _{j} f - \left\{ \phi ^{ij}*[\mu ^{1/2}\partial _{j} g]\right\} \partial _{i} f\\ & = -\left( \left\{ \phi ^{ij}*[v_{i} \mu ^{1/2}g]\right\} + \left\{ \phi ^{ij}*[\mu ^{1/2}\partial _{i} g]\right\} \right) \partial _{j} f\\ & = -\left( 2\left\{ \phi ^{ij}*[v_{i} \mu ^{1/2}g]\right\} + \left\{ \phi ^{ij}*[\partial _{i}(\mu ^{1/2} g)]\right\} \right) \partial _{j} f\\ & = -\left( 2\left\{ \phi ^{ij}*[v_{i} \mu ^{1/2}g]\right\} + \left\{ \partial _{i} \phi ^{ij}*[\mu ^{1/2} g]\right\} \right) \partial _{j} f\\ & = -2 v\cdot (\sigma _{\sqrt{\mu }g} \nabla _{v} f) - \partial _{i} \sigma _{\sqrt{\mu }g}^{ij} \partial _{j} f. \end{aligned}$$
Let \(N:= |v_{0}|\).

To obtain (6.5), we split the proof in two cases; \(|(t,x,v)-(t',x',v')| \le (2+N)^{-3m}\) or \(|(t,x,v)-(t',x',v')| > (2+N)^{-3m}\) for some \(m>0\) to be determined later. For the first case, we will consider a new center \((t_{*}, x_{*}, v_{*})\in Q_{1}\), such that \((t,x,v), (t',x',v') \in Q_{(2+N)^{-m}}(t_{*},x_{*},v_{*})\). Note that \(N-1/2 \le |v_{*}| \le N+1/2\).

Therefore, it is enough to prove that for every \((t_{*},x_{*},v_{*})\in Q_{1}(t_{0},x_{0},v_{0} )\),
$$\begin{aligned} \sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{(2+N)^{-m}}(t_{*} ,x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{(|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{\alpha } }\le C\Vert\; f\Vert _{\infty ,\vartheta } \end{aligned}$$
(6.6)
and
$$\begin{aligned} \sup _{\begin{subarray}{c} (t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{1} (t_{0},x_{0},v_{0}),\\ |t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }| > (2+N)^{-3m} \end{subarray} }\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{(|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{\alpha }}\le C\Vert\; f\Vert _{\infty ,\vartheta }. \end{aligned}$$
(6.7)
We first focus on (6.6). Consider the the following translation
$$\begin{aligned} {\bar{f}}(t,y,w):=f(t,x,v), \end{aligned}$$
where \(x=y+v_{*}(t-t_{*})\), \(v=v_{*}+w\). Then it is easy to check that \({\bar{f}}\) satisfies
$$\begin{aligned} \partial _{t}{\bar{f}}+w\cdot \nabla _{y}{\bar{f}}=\nabla _{w}\cdot ({\bar{\sigma }} _{G}\nabla _{w}{\bar{f}})+(v_{*}+w)\cdot ({\bar{\sigma }}_{\sqrt{\mu }g}\nabla _{w} {\bar{f}})-\sum _{ij}\partial _{i}{\bar{\sigma }}_{\sqrt{\mu }g}^{ij}\partial _{j} {\bar{f}}, \end{aligned}$$
where \({\bar{\sigma }}_{G}(t,y,w):=\sigma _{G}(t,x,v)\), \({\bar{\sigma }}_{\mu }(t,y,w):=\sigma _{\mu }(t,x,v)\), and \({\bar{\sigma }}_{\sqrt{\mu }g}(t,y,w):=\sigma _{\sqrt{\mu }g}(t,x,v)\). Let O be an orthonormal constant matrix which will be determined later. Next consider
$$\begin{aligned} {\tilde{f}}(t,\xi ,\nu ):={\bar{f}}(t,y,w), \end{aligned}$$
where \(y=O\xi \), \(w=O\nu \). Then we have
$$\begin{aligned} \partial _{t}{\bar{f}}(t,y,w) & = \partial _{t}{\tilde{f}}(t,\xi ,\nu ),\nonumber \\ w\cdot \nabla _{y}{\bar{f}}(t,y,w) & = \sum _{i}w_{i}\partial _{y_{i}}(\tilde{f}(t,\xi ,\nu ))\nonumber \\ & = \sum _{i,k}w_{i}\partial _{\xi _{k}}{\tilde{f}}(t,\xi ,\nu )\frac{\partial \xi _{k}}{\partial {y_{i}}}\nonumber \\ & = \sum _{i,k}O_{ik}w_{i}\partial _{\xi _{k}}{\tilde{f}}(t,\xi ,\nu ) \nonumber \\ & = \sum _{k}(O^{T}w)_{k}\cdot \partial _{\xi _{k}}{\tilde{f}}(t,\xi ,\nu )\nonumber \\ & = \nu \cdot \nabla _{\xi }{\tilde{f}}(t,\xi ,\nu ), \end{aligned}$$
(6.8)
where \(O_{ik}\) is the (ik) component of O. We used the following formula to derive the third equality in (6.8),
$$\begin{aligned} \frac{\partial \xi _{k}}{\partial {y_{i}}}=\frac{\partial \sum _{l}O_{lk}y_{l} }{\partial {y_{i}}}=O_{ik}. \end{aligned}$$
Similarly,
$$\begin{aligned} {{\bar{\sigma }}}_{G}(t,y,w) \nabla _{w} {{\bar{f}}}(t,y,w) & = \sum _{j}{{\bar{\sigma }}} _{G}^{ij}(t,y,w)\partial _{w_{j}} {{\tilde{f}}}(t,\xi ,\nu ) \nonumber \\ & = \sum _{j,k} {{\bar{\sigma }}}_{G}^{ij}(t,y,w) O_{jk}\partial _{\nu _{k}} \tilde{f}(t,\xi ,\nu ). \end{aligned}$$
(6.9)
Define \({\tilde{\sigma }}_{G}(t,\xi ,\nu ):=O^{T}{\bar{\sigma }}_{G}(t,y,w)O\). Note that
$$\begin{aligned} O{\tilde{\sigma }}_{G}(t,\xi ,\nu )={\bar{\sigma }}_{G} (t,y,w)O. \end{aligned}$$
(6.10)
Then by (6.9) and (6.10), we have
$$\begin{aligned} \nabla _{w}\cdot ({\bar{\sigma }}_{G}(t,y,w)\nabla _{w}{\bar{f}}(t,y,w)) & = \sum _{i,j}\partial _{w_{i}}({\bar{\sigma }}_{G}^{ij}(t,y,w)\partial _{w_{j}} {\tilde{f}}(t,\xi ,\nu ))\\ & = \sum _{i,j,k}\partial _{w_{i}}(O_{jk}{\bar{\sigma }}_{G}^{ij}(t,y,w)\partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu ))\\ & = \sum _{i,j,k}\partial _{w_{i}}(O_{ij}{\tilde{\sigma }}_{G}^{jk}(t,\xi ,\nu )\partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu ))\\ & = \sum _{i,j,k,l}O_{il}O_{ij}\partial _{\nu _{*}}({\tilde{\sigma }}_{G}^{jk} (t,\xi ,\nu )\partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu ))\\ & = \sum _{k,l}\partial _{\nu _{*}}({\tilde{\sigma }}_{G}^{lk}(t,\xi ,\nu )\partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu ))\\ & = \nabla _{\nu }\cdot ({\tilde{\sigma }}_{G}(t,\xi ,\nu )\nabla _{\nu }{\bar{f}} (t,\xi ,\nu )). \end{aligned}$$
In the last equality, we used \(O^{T}O=I\). Similarly, define \({\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu ):=O^{T}{\bar{\sigma }}_{\sqrt{\mu } g}(t,y,w)O\), then
$$\begin{aligned} v_{*}\cdot ({\bar{\sigma }}_{\sqrt{\mu }g}(t,y,w)\nabla _{w}{\bar{f}}(t,y,w)) & = v_{*}\cdot (O{\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu )\nabla _{\nu }\tilde{f}(t,\xi ,\nu ))\\ & = (O^{T}v_{*})\cdot ({\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu )\nabla _{\nu }{\tilde{f}}(t,\xi ,\nu ))\\ & = \nu _{*}\cdot ({\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu )\nabla _{\nu }\tilde{f}(t,\xi ,\nu )), \end{aligned}$$
where \(\nu _{*} = O^{T} v_{*}\),
$$\begin{aligned} w\cdot ({\bar{\sigma }}_{\sqrt{\mu }g}(t,y,w)\nabla _{w}{\bar{f}}(t,y,w)) & = w\cdot (O{\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu )\nabla _{\nu }{\tilde{f}} (t,\xi ,\nu ))\\ & = (O^{T}w)\cdot ({\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu )\nabla _{\nu }\tilde{f}(t,\xi ,\nu ))\\ & = \nu \cdot ({\tilde{\sigma }}_{\sqrt{\mu }g}(t,\xi ,\nu )\nabla _{\nu }\tilde{f}(t,\xi ,\nu )), \end{aligned}$$
and
$$\begin{aligned} \sum _{i,j}\partial _{w_{i}}{\bar{\sigma }}_{\sqrt{\mu }g}^{ij}(t,y,w)\partial _{w_{j}}{\bar{f}}(t,y,w) & = \sum _{i,j,k}\partial _{w_{i}}{\bar{\sigma }}_{\sqrt{\mu }g}^{ij}(t,y,w)\partial _{w_{j}}{\tilde{f}}(t,\xi ,\nu )\\ & = \sum _{i,j,k}\partial _{w_{i}}{\bar{\sigma }}_{\sqrt{\mu }g}^{ij}(t,y,w)O_{jk} \partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu )\\ & = \sum _{i,j,k,l}O_{il}\partial _{\nu _{*}}{\bar{\sigma }}_{\sqrt{\mu }g} ^{ij}(t,y,w)O_{jk}\partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu )\\ & = \sum _{k,l}\partial _{\nu _{*}}{\tilde{\sigma }}_{\sqrt{\mu }g}^{lk}(t,\xi ,\nu )\partial _{\nu _{k}}{\tilde{f}}(t,\xi ,\nu ). \end{aligned}$$
Therefore \({\tilde{f}}\) satisfies
$$\begin{aligned} \partial _{t}{\tilde{f}}+\nu \cdot \nabla _{\xi }{\tilde{f}}=\nabla _{\nu }\cdot ({\tilde{\sigma }}_{G}\nabla _{\nu }{\tilde{f}})+(\nu _{*}+\nu )\cdot (\tilde{\sigma }_{\sqrt{\mu }g}\nabla _{\nu }{\tilde{f}})-\sum _{k,l}\partial _{l}\tilde{\sigma }_{\sqrt{\mu }g}^{lk}\partial _{k}{\tilde{f}}. \end{aligned}$$
We split \({{\tilde{\sigma }}}_{G}(t,\xi ,\nu )\) in three parts.
$$\begin{aligned} {{\tilde{\sigma }}}_{G}(t,\xi ,\nu ) & = O^{T} {{\bar{\sigma }}}_{\mu }(0) O\\& \quad + O^{T}\left( {{\bar{\sigma }}}_{\mu }(w) - {{\bar{\sigma }}}_{\mu }(0)\right) O\\& \quad + O^{T} {{\bar{\sigma }}}_{\sqrt{\mu }g} (t,y,w) O\\ & = {{\tilde{\sigma }}}_{1} + {{\tilde{\sigma }}}_{2} + {{\tilde{\sigma }}}_{3}. \end{aligned}$$
Choose orthonormal vectors \(o_{1}=v_{*}/|v_{*}|\), \(o_{2}\), \(o_{3}\) and
$$\begin{aligned} O := \begin{bmatrix} o_{1}&o_{2}&o_{3} \end{bmatrix} . \end{aligned}$$
Note that
$$\begin{aligned} \nu _{*} = O^{T} v_{*} = \begin{bmatrix} |v_{*}|\\ 0\\ 0 \end{bmatrix} . \end{aligned}$$
Moreover \({{\bar{\sigma }}}_{\mu }(0)\) has a simple eigenvalue \(\lambda _{1}(v_{*})\) associated with the vector \(v_{*}\), and a double eigenvalue \(\lambda _{2} (v_{*})\) associated with \(v^{\perp }\). Therefore,
$$\begin{aligned} {{\tilde{\sigma }}}_{1} = \begin{bmatrix} \lambda _{1}(v_{*})&0&0\\ 0&\lambda _{2}(v_{*})&0\\ 0&0&\lambda _{2}(v_{*}) \end{bmatrix} . \end{aligned}$$
Note that \(\lambda _{1}(v_{*})\) and \(\lambda _{2}(v_{*})\) satisfy
$$\begin{aligned} \frac{1}{C}(1+N)^{-3} & \le \lambda _{1}(v_{*}) \le C (1+N)^{-3}, \nonumber \\ \frac{1}{C}(1+N)^{-1} & \le \lambda _{2}(v_{*}) \le C (1+N)^{-1}. \end{aligned}$$
(6.11)
Since \(\partial _{v_{k}} (\sigma _{\mu })^{ij}(v) \le C(1+|v|)^{-2}\), by the mean value theorem,
$$\begin{aligned} |({{\bar{\sigma }}}_{\mu })^{ij}(w)-({{\bar{\sigma }}}_{\mu }(0))| \le C(1+N)^{-3 +1}(2+N)^{-m}. \end{aligned}$$
Therefore,
$$\begin{aligned} |({{\tilde{\sigma }}}_{2})^{ij}| \le C(1+N)^{-2}(2+N)^{-m}. \end{aligned}$$
Define
$$\begin{aligned} D_{u}(\nu ,\nu ^{\prime };v):=\nu ^{T}\sigma _{u}(v)\nu ^{\prime }. \end{aligned}$$
Then we can easily check that
$$\begin{aligned} |D_{u}(\nu ,\nu ^{\prime };v)|\le |D_{u}(\nu ,\nu ;v)|^{1/2}|D_{u}(\nu ^{\prime } ,\nu ^{\prime };v)|^{1/2}. \end{aligned}$$
Since \(|v-v_{*}|<(2+N)^{-m}\) and \(o_{1}=v_{*}/|v_{*}|\), we have
$$\begin{aligned} |(I-P_{v})o_{1}| & = \frac{|(I-P_{v})v_{*}|}{|v_{*}|} \nonumber \\ & = \frac{|-v+v_{*}+v-P_{v}v_{*}|}{|v_{*}|} \nonumber \\ & \le \frac{|v-v_{*}|+|v-P_{v}v_{*}|}{|v_{*}|} \nonumber \\ & = \frac{|v-v_{*}|+|P_{v}(v-v_{*})|}{|v_{*}|} \nonumber \\ & = 2\frac{|v-v_{*}|}{|v_{*}|}\le C(2+N)^{-m}. \end{aligned}$$
(6.12)
Note that
$$\begin{aligned} ({\tilde{\sigma }}_{3})^{ij}=o_{i}^{T}\sigma _{\sqrt{\mu }g}(v)o_{j} . \end{aligned}$$
(6.13)
Therefore, by (2.7),
$$\begin{aligned} |({\tilde{\sigma }}_{3})^{11}| & = |D_{\sqrt{\mu }g}(o_{1};v)| \nonumber \\ & \le C\Vert g\Vert _{\infty }\left( (1+|v|)^{-3}|P_{v}o_{1} |^{2}+(1+|v|)^{-1}|(I-P_{v})o_{1}|^{2}\right) \nonumber \\ & \le C\Vert g\Vert _{\infty }\left( (1+N)^{-3}+(1+N)^{-1}(2+N)^{-2m}\right) \nonumber \\ & \le C\Vert g\Vert _{\infty }(1+N)^{-3}, \end{aligned}$$
(6.14)
and for \((i,j)\in \{(1,2),(1,3),(2,1),(3,1)\}\),
$$\begin{aligned} |({\tilde{\sigma }}_{3})^{ij}| & \le |D_{\sqrt{\mu }g}(o_{1};v)|^{1/2} |D_{\sqrt{\mu }g}(o_{k};v)|^{1/2}\\ & \le C\Vert g\Vert _{\infty }(1+N)^{-3/2}\left( (1+|v|)^{-3} |P_{v}o_{k}|^{2}+(1+|v|)^{-1}|(I-P_{v})o_{k}|^{2}\right) ^{1/2}\\ & \le C\Vert g\Vert _{\infty }(1+N)^{-3/2}(1+N)^{-1/2}\\ & = C\Vert g\Vert _{\infty }(1+N)^{-2}, \end{aligned}$$
where \(k=2\) or 3. Finally, for \(i,j=2\) or 3,
$$\begin{aligned} |({\tilde{\sigma }}_{3})^{ij}| & \le |D_{\sqrt{\mu }g}(o_{i};v)|^{1/2} |D_{\sqrt{\mu }g}(o_{j};v)|^{1/2}\\ & = C\Vert g\Vert _{\infty }\left( (1+|v|)^{-3}|P_{v}o_{i}|^{2} +(1+|v|)^{-1}|(I-P_{v})o_{i}|^{2}\right) ^{1/2}\\&\quad \times \left( (1+|v|)^{-3}|P_{v}o_{j}|^{2}+(1+|v|)^{-1}|(I-P_{v})o_{j}|^{2}\right) ^{1/2}\\&\le C\Vert g\Vert _{\infty }(1+N)^{-1}. \end{aligned}$$
Finally, consider the dilation matrix D as in (6.4) and the dilated function
$$\begin{aligned} F(t,X,V):={\tilde{f}}(t,\xi ,\nu ), \end{aligned}$$
where \(\xi =DX\), \(\nu =DV\). Then we can easily check that F satisfies
$$\begin{aligned} \partial _{t}F+V\cdot \nabla _{X}F=\nabla _{V}\cdot (\Sigma \nabla _{V}F)+(\nu _{l}+\nu )^{T}D\Sigma _{3}\nabla _{V}F+\sum _{k,l}\partial _{V_{l}}\Sigma _{3} ^{lk}\partial _{V_{k}}F, \end{aligned}$$
where \(\Sigma =\Sigma _{1}+\Sigma _{2}+\Sigma _{3}\), \(\Sigma _{i}(t,X,V)=D^{-1} {\tilde{\sigma }}_{i}(t,\xi ,\nu )D^{-1}\) for \(i=1,2,3\). Then by (6.11), we have
$$\begin{aligned} \frac{1}{C}\le (\Sigma _{1})^{ii}\le C,\quad (\Sigma _{1})^{ij}=0\text { for }i\ne j \end{aligned}$$
and
$$\begin{aligned} |(\Sigma _{2})^{ij}| & \le C(1+N)(2+N)^{-m},\\ |(\Sigma _{3})^{ij}| & \le C\Vert g\Vert _{\infty }. \end{aligned}$$
Moreover, since \(|\nu |\le (2+N)^{-m}\),
$$\begin{aligned} |D(\nu _{*}+\nu )| & \le |D\nu _{*}|+|D\nu |\\ & \le (1+N)^{-1/2}+C(1+N)^{-1/2}(2+N)^{-m}, \end{aligned}$$
and since \(\partial _{\nu _{k}}{\tilde{\sigma }}_{3}(\nu )\le C\Vert g\Vert _{\infty }(1+N)^{-2}\), we have
$$\begin{aligned} |\partial _{V_{l}}\Sigma _{3}^{lk}| & = |d_{l}^{-1}d_{k}^{-1}\partial _{V_{l} }{\tilde{\sigma }}_{3}(\nu )|\\ & \le d_{k}^{-1}|\partial _{\nu _{l}}{\tilde{\sigma }}_{3}(\nu )|\\ & \le \Vert g\Vert _{\infty }(1+N)^{-1/2}, \end{aligned}$$
where \(d_{k}\)s are the kth diagonal element of D. Choose \(m>4\) such that \(|(\Sigma _{2})^{ij}|,|(\nu _{*}+\nu )^{T}D\Sigma _{3}|\le \varepsilon \ll 1\). If \(\Vert g\Vert _{\infty }\le \varepsilon \), then any eigenvalue of \(\Sigma \) is bounded above and below uniformly in N. Therefore, by Lemma 6.5, there exist a constant \(C>0\) uniformly in N and a constant \(\alpha \in (0,1)\), depending only on dimension such that
$$\begin{aligned}&\sup _{(t,X,V),(t^{\prime },X^{\prime },V^{\prime })\in Q_{r_{1}}(t_{*} ,X_{*},0)}\frac{|F(t,X,V)-F(t^{\prime },X^{\prime },V^{\prime })|}{(|t-t^{\prime }|+|X-X^{\prime }|+|V-V^{\prime }|)^{\alpha }} \nonumber \\&\quad \le \frac{C}{(r_{1})^{3\alpha }}\Vert\; f\Vert _{L^{\infty }(Q_{128r_{1} }(t_{*},X_{*},0))}, \end{aligned}$$
(6.15)
where \(X_{*}=D^{-1}\xi _{*}\), \(\xi _{*}=O^{T}y_{l}\), \(y_{l}=x_{*}-v_{*}t_{*}\) and \(r_{1}\) is defined as in Lemma 6.5. Note that
$$\begin{aligned}&\frac{1}{(1+N)^{\alpha }}\sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{r_{0}}(t_{*},x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{(|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{\alpha }} \nonumber \\&\quad \le \sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{r_{0} }(t_{*},x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{((2+N)|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{ \alpha }}, \end{aligned}$$
(6.16)
where \(r_{0}\) and \(r_{1}\) are defined as in Lemma 6.6. Moreover, we have
$$\begin{aligned}&(2+N)|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }| \nonumber \\&\quad =(2+N)|t-t^{\prime }|+|ODX+v_{*}(t-t_{*})-(ODX^{\prime }+v_{*} (t^{\prime }-t_{*}))|+|ODV+v_{*}-(ODV^{\prime }+v_{*})| \nonumber \\&\quad \ge (2+N)|t-t^{\prime }|+|OD(X-X^{\prime })|-|v_{*}||t-t^{\prime }|+|OD(V-V^{\prime })| \nonumber \\&\quad \ge |t-t^{\prime }|+|OD(X-X^{\prime })|+|OD(V-V^{\prime })| \nonumber \\&\quad \ge (1+N)^{-3/2}(|t-t^{\prime }|+|(X-X^{\prime })|+|(V-V^{\prime })|). \end{aligned}$$
(6.17)
By (6.15), (6.16), (6.17), and Lemma 6.6, we have
$$\begin{aligned}&\sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{r_{0}}(t_{*} ,x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{((2+N)|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{ \alpha }} \nonumber \\&\quad \le \sup _{(t,X,V),(t^{\prime },X^{\prime },V^{\prime })\in Q_{r_{1} }(t_{*},X_{*},0)}\frac{|F(t,X,V)-f(t^{\prime },X^{\prime },V^{\prime } )|}{(1+N)^{-3 \alpha /2 }(|t-t^{\prime }|+|(X-X^{\prime })|+|(V-V^{\prime }|)^{ \alpha }}. \end{aligned}$$
(6.18)
Combine (6.15), (6.16), and (6.18). Then we have
$$\begin{aligned}&\sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{r_{0}}(t_{*} ,x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{(|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{\alpha } } \nonumber \\&\quad \le (2+N)^{\alpha }\sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{r_{0}}(t_{*},x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{((2+N)|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{\alpha }} \nonumber \\&\quad \le \frac{(2+N)^{\alpha }}{(1+N)^{-3 \alpha /2}}\sup _{(t,X,V),(t^{\prime },X^{\prime },V^{\prime })\in Q_{r_{1}}(t_{*},X_{*},0)} \frac{|F(t,X,V)-f(t^{\prime },X^{\prime },V^{\prime })|}{(|t-t^{\prime }|+|X-X^{\prime }|+|(V-V^{\prime }|)^{\alpha }} \nonumber \\&\quad \le \frac{C(2+N)^{\alpha }}{(1+N)^{-3 \alpha /2}(r_{1})^{3\alpha } }\Vert\; f\Vert _{L^{\infty }(Q_{128r_{1}}(t_{*},X_{*},0))}. \end{aligned}$$
(6.19)
By Lemma 6.6,
$$\begin{aligned} \Vert\; f\Vert _{L^{\infty }(Q_{128r_{1}}(t_{*},X_{*},0))}\le \Vert\; f\Vert _{L^{\infty }(Q_{128r_{2}}(t_{*},x_{*},v_{*}))}. \end{aligned}$$
(6.20)
Choose \(m>9\) such that \(128r_{2}<1\). Then we have
$$\begin{aligned} \Vert\; f\Vert _{L^{\infty }(Q_{128r_{2}}(t_{*},x_{*},v_{*}))} & \le C(1+N)^{-\vartheta }\Vert (1+|v|)^{\vartheta }f\Vert _{L^{\infty }(Q_{128r_{2}}(t_{*},x_{*},v_{*}))} \nonumber \\ & \le C(1+N)^{-\vartheta }\Vert (1+|v|)^{\vartheta }f\Vert _{L^{\infty }}, \end{aligned}$$
(6.21)
for every \(\vartheta >0\). Finally combining (6.19)–(6.21), we have
$$\begin{aligned}&\sup _{(t,x,v),(t^{\prime },x^{\prime },v^{\prime })\in Q_{r_{0}}(t_{*},x_{*},v_{*})\cap Q_{1}(t_{0},x_{0},v_{0})}\frac{|f(t,x,v)-f(t^{\prime },x^{\prime },v^{\prime })|}{(|t-t^{\prime }|+|x-x^{\prime }|+|v-v^{\prime }|)^{\alpha }} \nonumber \\&\quad \le \frac{C(1+N)^{\alpha -\vartheta }}{(1+N)^{-3 \alpha /2}(r_{1})^{3\alpha }}\Vert (1+|v|)^{\vartheta }f\Vert _{L^{\infty }} \nonumber \\&\quad \le C(1+N)^{\alpha -\vartheta +\frac{3 \alpha }{2}+2m\alpha -\frac{5\alpha }{2}}\Vert (1+|v|)^{\vartheta }f\Vert _{L^{\infty }} \nonumber \\&\quad =C(1+N)^{-\vartheta + 2m\alpha }\Vert (1+|v|)^{\vartheta }f\Vert _{L^{\infty }}, \end{aligned}$$
(6.22)
where \(\vartheta >0\) will be determined later.
To prove (6.7),
$$\begin{aligned}&\sup _{\begin{subarray}{c} (t,x,v),(t',x',v') \in Q_1(t_0,x_0,v_0)\\ |t-t'| + |x-x'| +|v-v'| > (2+N)^{-3m} \end{subarray} }\frac{|f(t,x,v)-f(t',x',v')|}{(|t-t'| + |x-x'| +|v-v'|)^\alpha } \nonumber \\&\quad \le 2 (2+N)^{3\alpha m }\Vert\; f\Vert _{L^\infty (Q_1(t_0,x_0,v_0))} \nonumber \\&\quad \le 2 C(1+N)^{3 \alpha m - \vartheta }\Vert (1+|v|)^\vartheta f\Vert _{L^\infty (Q(t_0,x_0,v_0;1))} \nonumber \\&\quad \le 2 C(1+N)^{3 \alpha m - \vartheta }\Vert (1+|v|)^\vartheta f\Vert _{L^\infty }. \end{aligned}$$
(6.23)
Now choose
$$\begin{aligned} \vartheta > 3 \alpha m. \end{aligned}$$
(6.24)
Then from (6.22) to (6.24), we prove (6.6) and (6.7). Therefore, we have
$$\begin{aligned} \sup _{(t,x,v),(t',x',v') \in Q_1(t_0,x_0,v_1)} \frac{|f(t,x,v)-f(t',x',v')|}{(|t-t'| + |x-x'| +|v-v'|)^\alpha } \le C\Vert\; f\Vert _{\infty , \vartheta }. \end{aligned}$$
By Theorem 5.14, we have (6.5). \(\square \)

Now we will prove Theorem 1.4.

Proof of Theorem 1.4

Since f satisfies (1.22), we have
$$\begin{aligned} f_t + v\cdot \nabla _x f - {{\bar{A}}}_g f = {{\bar{K}}}_g f. \end{aligned}$$
(6.25)
Define
$$\begin{aligned} {{\tilde{f}}}(t,x,v) = {\left\{ \begin{array}{ll} f(t,x,v), & \text { if }t\ge 0,\\ f_0(x,v), & \text { if }-1 \le t< 0. \end{array}\right. } \end{aligned}$$
Consider \({{\tilde{S}}}(t,x,v) = (\partial _t + v\cdot \nabla _x - {{\bar{A}}}_g){{\tilde{f}}}\). Then for \(t\le 0\),
$$\begin{aligned} {{\tilde{S}}}(t,x,v) & = (\partial _t + v\cdot \nabla _x - {{\bar{A}}}_g){{\tilde{f}}}\\ & = (v\cdot \nabla _x - {{\bar{A}}}_{f_{0}})f_{0}(x,v)\\ & = -f_{0t}, \end{aligned}$$
where \(f_{0t}\) was defined in Theorem 1.1. Since f is a weak solution of (1.22) in the sense of Definition 4.1, for \(t>0\), \({{\tilde{S}}}(t,x,v) = {{\bar{K}}}_g f(t,x,v)\). Thus, \({{\tilde{f}}}\) satisfies
$$\begin{aligned} {{\tilde{f}}}_t + v \cdot \nabla _x {{\tilde{f}}} - {{\bar{A}}}_g {{\tilde{f}}} = {{\tilde{S}}}(t,x,v). \end{aligned}$$
Since U(ts) is the solution operator for \(\partial _t + v \cdot \nabla _x - {{\tilde{A}}}_g = 0\). Then \({{\tilde{f}}}\) satisfies
$$\begin{aligned} {{\tilde{f}}}(t) = U(t,-1)f_0 + \int _{-1}^{t} U(t,s) {{\tilde{S}}}(s) \mathrm{d}s. \end{aligned}$$
Let \(0<{{\bar{\varepsilon }}}\ll 1\) be given. Note that by Lemma 6.5, there exists \(\alpha >0\) such that \(U(t,-1){{\tilde{f}}}(-1)\) is uniformly Hölder continuous on \((0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3\).
For every \(0 \le t_2 \le t_1\), \(|t_1 - t_2| + |x_2 - x_1| + |v_2-v_1| = {{\bar{\varepsilon }}}\),
$$\begin{aligned}&|f(t_1 ,x_1,v_1)- f(t_2,x_2,v_2)|\\&\quad \le \left| (U(t_1,-1)f_0)(x_1,v_1) - (U(t_2,-1)f_0)(x_2,v_2)\right| \\&\qquad + \left| \int _{t_2}^{t_1} (U(t_1,s){{\tilde{S}}}(s))(x_1,v_1)\mathrm{d}s\right| \\&\qquad + \left| \int _{t_2-{{\bar{\varepsilon }}}^\alpha }^{t_2}\left( (U(t_1,s){{\tilde{S}}}(s))(x_1,v_1) - (U(t_2,s){{\tilde{S}}}(s))(x_2,v_2)\right) \mathrm{d}s\right| \\&\qquad + \left| \int _{-1}^{t_2-{{\bar{\varepsilon }}}^\alpha }\left( (U(t_1,s){{\tilde{S}}}(s))(x_1,v_1) - (U(t_2,s){{\tilde{S}}}(s))(x_2,v_2)\right) \mathrm{d}s\right| \\&\quad \le (I) + (II) +(III) + (IV). \end{aligned}$$
By Lemma 6.7, there exist \(\vartheta \), l, C and \(C_{\vartheta , l}\) such that
$$\begin{aligned} (I) \le (C \Vert\, f_0\Vert _{\infty , \vartheta } + C_{\vartheta ,l}\Vert\, f_0\Vert _{2,\vartheta +l}) {{\bar{\varepsilon }}}^{\alpha }. \end{aligned}$$
By Lemma 3.8, (1.25), and (5.16), there exists \(l_0\) such that
$$\begin{aligned} (II)\le {{\bar{\varepsilon }}} \sup _{s>0}\Vert {{\tilde{K}}}_g f(s)\Vert _\infty \le C{{\bar{\varepsilon }}} \sup _{s>0} \Vert\; f(s)\Vert \le C {{\bar{\varepsilon }}}(\Vert\, f_0\Vert _\infty +\Vert\, f_0\Vert _{2,l_0}). \end{aligned}$$
Note that for \(s\le 0\),
$$\begin{aligned} \Vert {{\tilde{S}}}(s)\Vert _{\infty , \vartheta } \le C\Vert\, f_{0t}\Vert _{\infty , \vartheta } , \end{aligned}$$
(6.26)
$$\begin{aligned} \Vert {{\tilde{S}}}(s)\Vert _{2, \vartheta } \le C\Vert\, f_{0t}\Vert _{2, \vartheta }, \end{aligned}$$
(6.27)
and for \(s\ge 0\), by Lemma 5.9 and Theorems 1.2 and 1.3 there exists \(l_0\) such that
$$\begin{aligned} \Vert {{\tilde{S}}}(s)\Vert _{\infty , \vartheta } \le C(1+s)^{-2}(\Vert\, f_0\Vert _{2,\vartheta +l_{0}} + \Vert\, f_0\Vert _{\infty ,\vartheta +l_{0}}) \end{aligned}$$
(6.28)
and
$$\begin{aligned} \Vert {{\tilde{S}}}(s)\Vert _{2,\vartheta +l} \le C(1+s)^{-2}(\Vert\, f_0\Vert _{2, \vartheta +l_{0}} + \Vert\, f_0\Vert _{\infty , \vartheta +l_{0} }). \end{aligned}$$
(6.29)
Therefore, by (6.26) and (6.29), we have
$$\begin{aligned} (III) & \le {{\bar{\varepsilon }}}^\alpha \sup _{s\in (t_2 - {{\bar{\varepsilon }}}^\alpha , t_2)}\Vert {{\tilde{S}}}(s)\Vert _{\infty }\\ & \le C{{\bar{\varepsilon }}}^\alpha (\Vert\, f_{0t}\Vert _{\infty }+\Vert\, f_0\Vert _{2, l_{0}} + \Vert\, f_0\Vert _{\infty , l_{0} }). \end{aligned}$$
For \(-1 \le s \le t_2 - {{\bar{\varepsilon }}}^\alpha \), we have
$$\begin{aligned}&|t_1 - (t_2 +1 - {{\bar{\varepsilon }}}^\alpha )| + |x_1 - x_2| + |v_1 - v_2|\\&\quad = |t_1 - t_2 -1 + {{\bar{\varepsilon }}}^\alpha | + |x_1 - x_2| + |v_1 - v_2|\\&\quad \le 1 - {{\bar{\varepsilon }}}^\alpha + |t_1 - t_2| + |x_1 - x_2| + |v_1 - v_2|\\&\quad \le 1, \\&|t_2 - (t_2 + 1 - {{\bar{\varepsilon }}}^\alpha )| + |x_2 - x_2| + |v_2 - v_2| = 1- {{\bar{\varepsilon }}}^\alpha \le 1. \end{aligned}$$
Therefore,
$$\begin{aligned} (t_i , x_i, v_i ) \in Q_1(t_2 + 1 - {{\bar{\varepsilon }}}^\alpha , x_2, v_2) \subset (s,\infty ) \times {\mathbb {T}}^3 \times {\mathbb {R}}^3, \end{aligned}$$
for each \(i = 1,2\). Therefore by Lemma 6.7, there exist \(\vartheta \), l, C and \(C_{\vartheta , l}\) such that
$$\begin{aligned}&(U(t_1,s){{\tilde{S}}}(s))(x_1,v_1) - (U(t_2,s){{\tilde{S}}}(s))(x_2,v_2)\\&\le (C \Vert {{\tilde{S}}}(s)\Vert _{\infty , \vartheta } + C_{\vartheta ,l}\Vert {{\tilde{S}}}(s)\Vert _{2,\vartheta +l}) {{\bar{\varepsilon }}}^{\alpha }, \end{aligned}$$
for \(-1\le s\le t_{2}- {{\bar{\varepsilon }}}^{\alpha }\). Therefore, by Lemma 6.7, (6.26), (6.28), and (6.27), (6.29), we have
$$\begin{aligned} (IV) & \le \left| \int _{-1}^{0}\left( (U(t_1,s){{\tilde{S}}}(s))(x_1,v_1) - (U(t_2,s){{\tilde{S}}}(s))(x_2,v_2)\right) \mathrm{d}s\right| \\& \quad + \left| \int _{0}^{t_2-{{\bar{\varepsilon }}}^\alpha }\left( (U(t_1,s){{\tilde{S}}}(s))(x_1,v_1) - (U(t_2,s){{\tilde{S}}}(s))(x_2,v_2)\right) \mathrm{d}s\right| \\ & \le C {{\bar{\varepsilon }}}^{\alpha } \left( \sup _{s\in [-1,0]} (\Vert {{\tilde{S}}}(s)\Vert _{\infty ,\vartheta } + \Vert {{\tilde{S}}}(s)\Vert _{2, \vartheta +l}) + \int _{0}^{\infty } (\Vert {{\tilde{S}}}(s)\Vert _{\infty , \vartheta } + \Vert {{\tilde{S}}}(s)\Vert _{2, \vartheta + l}) \mathrm{d}s \right) \\ & \le C {{\bar{\varepsilon }}}^\alpha \left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert\, f_{0t}\Vert _{2, \vartheta + l} + (\Vert\, f_0\Vert _{2, \vartheta +l_{0}} + \Vert\, f_0\Vert _{\infty , \vartheta +l_{0} })\int _{0}^{\infty }(1+s)^{-2}\mathrm{d}s\right) \\ & \le C {{\bar{\varepsilon }}}^\alpha \left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert\, f_{0t}\Vert _{2, \vartheta + l} + \Vert\, f_0\Vert _{2, \vartheta +l_{0}} + \Vert\, f_0\Vert _{\infty , \vartheta +l_{0} }\right) . \end{aligned}$$
Now we update \(\vartheta \) to \(\vartheta + 2 + \max \{l, l_{0}\}\). Then, by Proposition 2.1, we have
$$\begin{aligned} |f(t_1 ,x_1,v_1)- f(t_2,x_2,v_2)| & \le (I) + (II) + (III) + (IV)\\ & \le C {{\bar{\varepsilon }}}^\alpha \left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert\, f_0\Vert _{\infty , \vartheta }\right) . \end{aligned}$$
Thus we complete the proof. \(\square \)

7 Hölder Estimate and \(S^{p}\) Bound

Let f be a weak solution of (1.22) in the sense of Definition 4.1. Define
$$\begin{aligned} {{\bar{f}}}(t,x,v) = {\left\{ \begin{array}{ll} f(t,x,v), & \text { if }t\ge 0,\\ f_0(x,v), & \text { if }-1 \le t< 0. \end{array}\right. } \end{aligned}$$
Then \({{\bar{f}}}\) satisfies
$$\begin{aligned} \begin{aligned} \partial _t {{\bar{f}}} + v \cdot \nabla _x {{\bar{f}}} - \sigma _G^{ij} \partial _{v_i,v_j} {{\bar{f}}}&= {\left\{ \begin{array}{ll} - \partial _{v_i} \sigma _G^{ij}\partial _{v_j} f + a_g \cdot \nabla _v f + K_1 f + J_g f, & \text { if }t\ge 0,\\ (v\cdot \nabla _x - \sigma _{\mu + \mu ^{1/2}f_0}^{ij} \partial _{v_i,v_j}) f_0, & \text { if }-1 \le t< 0 \end{array}\right. }\\&= {\left\{ \begin{array}{ll} - \partial _{v_i} \sigma _G^{ij}\partial _{v_j} f + a_g \cdot \nabla _v f + K_1 f + J_g f, & \text { if }t\ge 0,\\ - f_{0t} + \partial _{v_{i}}\sigma _{\mu + \mu ^{1/2}f_0}^{ij} \partial _{v_{j}}f_{0} , & \text { if }-1 \le t< 0, \end{array}\right. } \end{aligned} \end{aligned}$$
(7.1)
where \(\sigma _G\) is defined as in (2.1) with \(G = \mu + \mu ^{1/2}g\) and
$$\begin{aligned} K_1 f & = - \mu ^{-1/2}\partial _i \left\{ \mu \left[ \phi ^{ij}* \left\{ \mu ^{1/2}[\partial _j f + v_j f] \right\} \right] \right\} \nonumber \\ & = 2v_i \mu \left[ \phi ^{ij}* \left\{ \mu ^{1/2}[\partial _j f + v_j f] \right\} \right] - \mu ^{1/2}\left[ \partial _i \phi ^{ij}* \left\{ \mu ^{1/2}[\partial _j f + v_j f] \right\} \right] , \end{aligned}$$
(7.2)
and
$$\begin{aligned} J_g f = - v\cdot \sigma v f -\partial _i\left\{ \phi ^{ij}*[\mu ^{1/2}\partial _j g] \right\} f + \left\{ \phi ^{ij}*[v_i \mu ^{1/2} \partial _j g] \right\} f + \partial _{i}\sigma ^{i}f. \end{aligned}$$
(7.3)

Lemma 7.1

For every\(\beta >0\)and\(p>3\),
$$\begin{aligned} \Vert K_1f\Vert _{L^p(n \le |v| \le n+1)} \le \frac{C_{p,\beta }}{n^ \beta }\left( \Vert\; f\Vert _{L^p} + \Vert D_v f\Vert _{L^p}\right) , \end{aligned}$$
(7.4)
where\(K_1\)is defined as in (7.2).

Proof

Since \(K_1\) is defined as in (7.2), it is enough to show that
$$\begin{aligned} \left\| \mu (v)\int _{{\mathbb {R}}^3}|v-v'|^\vartheta \mu (v')h(v') \mathrm{d}v'\right\| _{L^{p}}\le \Vert h\Vert _{L^p}. \end{aligned}$$
By the Hölder inequality,
$$\begin{aligned}&\int _{(0,\infty )\times {\mathbb {T}}^3\times {\mathbb {R}}^3} \mu ^p(v)\left( \int _{{\mathbb {R}}^3}|v-v'|^\vartheta \mu (v') h(v') \mathrm{d}v'\right) ^p \mathrm{d}v\mathrm{d}x\mathrm{d}t\\&\quad \le \int _{(0,\infty )\times {\mathbb {T}}^3\times {\mathbb {R}}^3} \mu ^p(v)\left( \int _{{\mathbb {R}}^3}|v-v'|^{\vartheta p'} \mu ^{p'}(v')\mathrm{d}v'\right) ^{p/p'} \left( \int _{{\mathbb {R}}^3}h^p(v') \mathrm{d}v'\right) \mathrm{d}v\mathrm{d}x\mathrm{d}t\\&\quad \le \int _{{\mathbb {R}}^3} \mu ^p(v) (1+|v|)^{\vartheta p} \mathrm{d}v \int _{(0,\infty )\times {\mathbb {T}}^3} \left( \int _{{\mathbb {R}}^3}h^p(v') \mathrm{d}v'\right) \mathrm{d}x\mathrm{d}t\\&\quad =C_p \Vert h\Vert _{L^p}^p. \end{aligned}$$
\(\square \)
Clearly, we have
$$\begin{aligned} \Vert J_g f\Vert _{L^p(n \le |v| \le n+1)} \le (C+\Vert g\Vert _\infty ) \Vert\; f\Vert _{L^p(n \le |v| \le n+1)} \le Cn^{- \beta } \Vert (1+|v|)^{\beta }f\Vert _{L^p(n \le |v| \le n+1)}. \end{aligned}$$
(7.5)

Theorem 7.2

(Theorem 3.3 in [5]) Let\(\Omega \)be a bounded open set in\({\mathbb {R}}^7\)and letfbe a strong solution in\(\Omega \)to the equation
$$\begin{aligned} \sum _{i,j=1}^3 \sigma ^{ij}(t,x,v)\partial _{v_i,v_j} f + Yf = h, \end{aligned}$$
where\(Y = -\partial _t - v\cdot \nabla _x\). Suppose that\(\sigma \)is uniformly elliptic,
$$\begin{aligned} \Vert \sigma ^{ij}\Vert _{C^{\alpha }(\Omega )} \le C, \end{aligned}$$
(7.6)
and\(f, h\in L^p\). Then\(\partial _{v_i,v_j} f \in L^p_{\text {loc}}\), \(Yf \in L^p_{\text {loc}}\)and for every open set\(\Omega '\subset \subset \Omega \)there exists a positive constant\(c_1\)depending only on\(p, \Omega ', \Omega , \alpha , C\)and elliptic constant of\(\sigma \)such that
$$\begin{aligned} \Vert \partial _{v_i,v_j} f\Vert _{L^p(\Omega ')} & \le c_1 (\Vert\; f\Vert _{L^p(\Omega )} + \Vert h\Vert _{L^p(\Omega )}),\\ \Vert Y f\Vert _{L^p(\Omega ')} & \le c_1 (\Vert\; f\Vert _{L^p(\Omega )} + \Vert h\Vert _{L^p(\Omega )}). \end{aligned}$$

Remark 7.3

Especially, from [5]
$$\begin{aligned} c_1 = c_2(\lambda _1, \lambda _2)c_3(\text {dist}(\Omega ',\Omega ),\alpha ,C,p), \end{aligned}$$
where \(\lambda _1\), \(\lambda _2\) are the smallest and the largest eigenvalues, respectively. More precisely, there exist \(C>0\) and \(\alpha '>0\) such that
$$\begin{aligned} c_2(\lambda _1,\lambda _2) := \max _{|x|^2+|v|^2+|t|^2 =1} |\Gamma ^0_{v_i,v_j}(x,v,t)| \le C(\lambda _1^{-1} + \lambda _2)^{\alpha '}, \end{aligned}$$
where \(\Gamma ^0(\zeta ^{-1}\circ z)\) is a fundamental solution of
$$\begin{aligned} \sum _{i,j=1}^3 \sigma ^{ij}(\tau ,\xi ,\nu )\partial _{v_i,v_j} f + Yf = 0 \end{aligned}$$
(7.7)
and \(\circ \) is a Lie group operation corresponding to (7.7) for some \(C>0\).

Remark 7.4

Since \(f(t,\cdot ,v)\) is a periodic function on \({\mathbb {T}}^3\), we extend it to a periodic function on \((3{\mathbb {T}})^3\). Note that
$$\begin{aligned} \Vert\; f\Vert _{L^p(3{\mathbb {T}}^3)}^p = 27\Vert\; f\Vert _{L^p({\mathbb {T}}^3)}^p. \end{aligned}$$
Define
$$\begin{aligned} \Vert\; f\Vert _{S^p(\Omega )} := \Vert\; f\Vert _{L^p(\Omega )} + \Vert D_v f\Vert _{L^p(\Omega )} + \Vert D_{vv}f\Vert _{L^p(\Omega )} + \Vert Yf\Vert _{L^p(\Omega )}, \end{aligned}$$
where \(Y = -\partial _t -v\cdot \nabla _x\).

Lemma 7.5

Assume (2.6). Letfbe a weak solution of (1.6), (1.7), and (1.22) in the sense of Definition 4.1. Suppose thatgsatisfies\(\Vert g\Vert _{C^{\alpha }((0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3)} \le C'\)for some\(0<\alpha <1\)and\(C>0\). Then there exist\(\vartheta >0\), \(p>3\), \(C_{\vartheta ,\alpha , C', p}\)such that
$$\begin{aligned} \Vert\; f\Vert _{S^p((0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3)} \le C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert D_v f_0\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) . \end{aligned}$$
(7.8)

Proof

Since \(\Vert g\Vert _{C^{\alpha }((0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3)} \le C\), \(\sigma _G\) satisfies (7.6). Apply Theorem 7.2 to (7.1) with \(\Omega '=\Omega '_n\) and \(\Omega = \Omega _n\), where
$$\begin{aligned} \Omega ' & = \Omega '_n:=\{t\ge 0, x\in {\mathbb {T}}^3, n \le |v| \le n+1\},\\ \Omega & = \Omega _n:=\{t\ge -1, x\in 3{\mathbb {T}}^3, n-1/2 \le |v| \le n+3/2\}. \end{aligned}$$
Let \(\sigma _G\), \(K_1\), and \(J_g\) be defined as in (2.1), (7.2), (7.3). Then, we have
$$\begin{aligned} \Vert\; f\Vert _{S^p(\Omega '_n)}^p & \le (c_1)^p \left( \Vert \partial _{v_i}\sigma _G^{ij}\partial _{v_j} f\Vert _{L^p(\Omega '_n)}^p + \Vert a_g \cdot \nabla _v f\Vert _{L^p(\Omega '_n)}^p + \Vert K_1 f + J_g f\Vert _{L^p(\Omega '_n)}^p\right. \\& \left. \quad + \Vert (v\cdot \nabla _x - \sigma _G^{ij} \partial _{v_i,v_j}) f_0\Vert _{L^p\left( (-1,0)\times 3{\mathbb {T}}^3 \times B(0;n-1/2,n+3/2)\right) }^p\right) , \end{aligned}$$
where \(B(z;r_1,r_2) := B(z;r_2){\setminus} B(z;r_1)\). By Lemma 2.3 and Remark 7.3,
$$\begin{aligned} c_2(\lambda _1, \lambda _2) = C n^a \end{aligned}$$
for some \(a>0\) and
$$\begin{aligned} c_3(\text {dist}(\Omega ', \Omega ), \alpha , C, p) = C_{\alpha ,C,p}. \end{aligned}$$
Therefore,
$$\begin{aligned} c_1 = C_{\alpha ,C,p} n^a \end{aligned}$$
for some \(a>0\). Let
$$\begin{aligned} {{\tilde{\Omega }}}'_n = \{t\ge 0, x\in {\mathbb {T}}^3, n-1/2\le |v| \le n+3/2\}. \end{aligned}$$
Then by Remark 7.4,
$$\begin{aligned} \Vert\; f\Vert _{S^p(\Omega '_n)}^p & \le (C_{\alpha ,C,p}n^a)^p \left( \Vert \partial _{v_i}\sigma _G^{ij}\partial _{v_j} f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p + \Vert a_g \cdot \nabla _v f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p + \Vert K_1 f + J_g f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p\right. \nonumber \\&\left. \quad + \Vert (v\cdot \nabla _x - \sigma _G^{ij} \partial _{v_i,v_j}) f_0\Vert _{L^p\left( (-1,0)\times {\mathbb {T}}^3 \times B(0;n-1/2,n+3/2)\right) }^p\right) , \end{aligned}$$
(7.9)
where \({{\tilde{\Omega }}}'_n = \{t\ge 0, x\in {\mathbb {T}}^3, n-1/2\le |v| \le n+3/2\}\). Note that by the standard interpolation, we have
$$\begin{aligned} \Vert D_v f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p \le \varepsilon ' \Vert D_{vv} f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p + \frac{C}{\varepsilon '}\Vert\; f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p. \end{aligned}$$
(7.10)
Let \(\beta > 2a+4\) and \(\varepsilon ' = \varepsilon _0\) which can be determined later. Then by (7.4), (7.5), and (7.10), we have
$$\begin{aligned}&\sum (C_{\alpha ,C,p}n^a)^p \Vert K_1 f + J_g f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p \nonumber \\&\quad \le \sum (C_{\beta , \alpha , C, p})^p n^{p(a - \beta )}\left( \Vert\; f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3\right) }^p + \Vert D_v f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p + \Vert (1+|v|)^\beta f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p\right) \nonumber \\&\quad \le (C_{\beta , \alpha , C, p})^p\left( \Vert (1+|v|)^{\beta } f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p + \Vert D_v f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p\right) \nonumber \\&\quad \le (C_{\beta , \alpha , C, p, \varepsilon _0})^p\Vert (1+|v|)^{\beta } f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p + \varepsilon _0 \Vert D_{vv} f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p \end{aligned}$$
(7.11)
for some \(\varepsilon _0\). Similarly,
$$\begin{aligned}&\sum (C_{\alpha ,C,p}n^a)^p \Vert (v\cdot \nabla _x - \sigma _G^{ij} \partial _{v_i,v_j})f_0(x,v)\Vert _{L^p\left( (-1,0)\times {\mathbb {T}}^3 \times B(0;n-1/2,n+3/2)\right) }^p \nonumber \\&\quad \le \sum (C_{\alpha ,C,p})^pn^{p(a- \beta )} \Vert (1+|v|)^\beta (v\cdot \nabla _x - \sigma _G^{ij} \partial _{v_i,v_j})f_0(x,v)\Vert _{L^p\left( {\mathbb {T}}^3 \times B(0;n-1/2,n+3/2)\right) }^p \nonumber \\&\quad \le (C_{\alpha ,C,p})^p\Vert (v\cdot \nabla _x - \sigma _G^{ij} \partial _{v_i,v_j})f_0\Vert _{p, \beta }^p. \end{aligned}$$
(7.12)
Choose \(\varepsilon ' = \varepsilon n^{- p (a+2)}\) small enough. Since \(\Vert \partial _{v_i} \sigma _G^{ij}\Vert _\infty <C\), \(\Vert a_g\Vert _\infty \le C\), by (7.10) we have
$$\begin{aligned}&\sum (C_{\alpha ,C,p}n^a)^p \left( \Vert \partial _{v_i} \sigma _G^{ij} \partial _{v_j}f)\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p + \Vert a_g \cdot \nabla _v f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p\right) \nonumber \\&\quad \le \sum (C_{\alpha ,C,p})^p \left( \varepsilon n^{-2p} \Vert D_{vv} f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p + \varepsilon ^{-1} n^{2p (a + 1)}\Vert\; f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p\right) \nonumber \\&\quad \le \sum (C_{\alpha ,C,p})^p \left( \varepsilon n^{-2p} \Vert D_{vv} f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p + n^{p (2a + 2 - \beta )}\Vert (1+|v|)^{\beta }f\Vert _{L^p({{\tilde{\Omega }}}'_n)}^p\right) \nonumber \\&\quad \le (C_{\beta ,\alpha ,C,p})^p \left( \varepsilon \Vert D_{vv} f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p + C_{\varepsilon } \Vert (1+|v|)^\beta f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p\right) . \end{aligned}$$
(7.13)
Combining (7.9)–(7.13) and absorbing \(\Vert D_{vv}f\Vert \) term on RHS to the LHS, we have
$$\begin{aligned}&\Vert\; f\Vert _{S^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p\\&\quad \le (C_{\beta , \alpha , C, p})^p\left( \Vert (1+|v|)^\beta f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p+ \Vert (v\cdot \nabla _x - \sigma _F^{ij} \partial _{v_i,v_j})f_0\Vert _{p, \beta }^p\right) \\&\quad \le (C_{\beta , \alpha , C, p})^p\left( \Vert (1+|v|)^\beta f\Vert _{L^p\left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) }^p+ \Vert\, f_{0t}\Vert _{p, \beta }^p + \Vert D_v f_0\Vert _{p, \beta -2}^p\right) . \end{aligned}$$
Finally, by Lemmas 2.4 and 5.11, the standard interpolation, and Proposition 2.1, we have (7.8). \(\square \)

Here we introduce some regularity results in \(S^p\) norm.

Theorem 7.6

(Theorem 2.1 in [24]) Let\(f\in S^p({\mathbb {R}}^7)\), with\(1<p<\infty \).
  1. (1)

    If \(2p>14\) and \(p<14\), then \(f\in C^{\gamma }({\mathbb {R}}^7)\), with \(\gamma = \frac{2p-14}{p}\);

     
  2. (2)

    if \(p>14\), then \(\partial _{v_i}f \in C^\delta ({\mathbb {R}}^7)\), with \(\delta = \frac{p-14}{p}\).

     
Define \({\left| \!\left| \!\left| (t,x,v) \right| \!\right| \!\right| } = \rho \), where \(\rho \) is a unique positive solution to the equation
$$\begin{aligned} \frac{t^{2}}{\rho ^{4}}+\frac{|x|^{2}}{\rho ^{6}}+\frac{|v|^{2}}{\rho ^{2}} = 1 \end{aligned}$$
and
$$\begin{aligned} (\tau ,\xi ,\nu )^{-1}\circ (t,x,v) = (t- \tau , x- \xi + (t- \tau )\nu , v- \nu ). \end{aligned}$$
Now we can deduce the following lemma.

Lemma 7.7

Assume (2.6). Letfbe a weak solution of (1.6), (1.7), and (1.22) in the sense of Definition 4.1. Suppose thatgsatisfies\(\Vert g\Vert _{C^{\alpha }((0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3)} \le C\)for some\(0<\alpha <1\)and\(C>0\).

If\(2p > 14\)then, letting\(\alpha _1 = \min \left\{ 1, \frac{2p - 14}{p}\right\} \), there exist\(\vartheta >0\)and\(C=C_{\vartheta ,\alpha , C, p}\)such that
$$\begin{aligned} \frac{|f(t,x,v)-f(\tau ,\xi ,\nu )|}{{\left| \!\left| \!\left| (\tau ,\xi ,\nu )^{-1}\circ (t,x,v) \right| \!\right| \!\right| }^{\alpha _1}}\le C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert D_v f_0\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) \end{aligned}$$
for every\((t,x,v),(\tau ,\xi ,\nu )\in (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3\), \((t,x,v)\ne (\tau ,\xi ,\nu )\).
If\(p > 14\)then, letting\(\alpha _2 = \frac{p - 14}{p}\), there exist\(\vartheta >0\)and\(C=C_{\vartheta ,\alpha , C, p}\)such that
$$\begin{aligned} \frac{|\partial _{v_i}f(t,x,v)-\partial _{v_i}f(\tau ,\xi ,\nu )|}{{\left| \!\left| \!\left| (\tau ,\xi ,\nu )^{-1}\circ (t,x,v) \right| \!\right| \!\right| }^{\alpha _2}}\le C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert D_v f_0\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) \end{aligned}$$
for every\((t,x,v),(\tau ,\xi ,\nu )\in (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3\), \((t,x,v)\ne (\tau ,\xi ,\nu )\).

Proof

It immediately follows from Lemma 7.5 and Theorem 7.6. \(\square \)

Remark 7.8

If \(\tau =t\) and \(\xi = x\), then \({\left| \!\left| \!\left| (t,x,\nu )^{-1}\circ (t,x,v) \right| \!\right| \!\right| } = |v- \nu |\).

By Remark 7.8 and Lemma 7.7, we have

Lemma 7.9

Assume (2.6). Letfbe a weak solution of (1.6), (1.7), and (1.22) in the sense of Definition 4.1. Suppose thatgsatisfies\(\Vert g\Vert _{C^{\alpha }((0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3)} \le C\)for some\(0<\alpha <1\)and\(C>0\). Let\(p=14\), then there exist\(\vartheta >0\)and\(C=C_{\vartheta ,\alpha , C, p}\)such that
$$\begin{aligned} \Vert D_v f\Vert _{L^\infty \left( (0,\infty )\times {\mathbb {T}}^3 \times {\mathbb {R}}^3 \right) } \le C\left( \Vert\, f_{0t}\Vert _{\infty , \vartheta } + \Vert D_v f_0\Vert _{\infty , \vartheta } + \Vert\, f_{0}\Vert _{\infty , \vartheta }\right) . \end{aligned}$$

8 Proof of Theorem 1.1

In this section we will use an iteration argument to prove the existence and the uniqueness of the weak solution of (1.5)–(1.7) in the sense of Definition 4.1. We first construct the function sequence as follows. Define \(f^{(0)}(t,x,v) := f_{0}(x,v)\). Since \(f_{0}(x,v)\) satisfies (1.16), by Lemma 4.2, we can define \(f^{(1)}\) as a solution of (1.6), (1.7), and (1.22) with \(g=f^{(0)}\). Moreover, by Theorem 1.3, \(f^{(1)}\) satisfies the assumption in Lemma 2.4. Thus we can also define \(f^{(2)}\) as a solution of (1.6), (1.7), and (1.22) with \(g=f^{(1)}\). Inductively we can define a function sequence \(f^{(n)}\) for \(n\ge 0\).

Lemma 8.1

There exist\(C, \vartheta _0>0\), and\(0<\varepsilon _0\ll 1\)such that if\(f_{0}\)satisfies
$$\begin{aligned} \Vert\, f_{0}\Vert _{\infty , \vartheta } \le \varepsilon _{0}, \end{aligned}$$
then
$$\begin{aligned} \sup _{n\in {\mathbb {N}}, t\ge 0}\Vert\; f^{(n)}(t)\Vert _{\infty , \vartheta } \le C\Vert\, f_{0}\Vert _{\infty , \vartheta + \vartheta _{0}}. \end{aligned}$$

Sketch of proof

It is clear by applying Theorem 1.3 to \(f^{(n)}\) inductively on n. \(\square \)

Lemma 8.2

Let\(f_0\)be a given function satisfying (1.16). Let\(f_1\)and\(f_2\)be weak solutions of (1.6), (1.7), and (1.22) in the sense of Definition 4.1with\(g=g_1\), \(g=g_2\)respectively. Suppose that\(g_1\)and\(g_2\)are uniformly Hölder continuous functions satisfying (2.6). Then we have
$$\begin{aligned}&\frac{1}{2}\Vert (f_1-f_2)(t)\Vert _{2, {{\bar{\vartheta }}}}^2 + \left( \frac{1}{2}-C \varepsilon \right) \int _0^t \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2 \mathrm{d}s \nonumber \\&\quad \le \int _0^t C(\Vert\, f_2\Vert _{\infty } + \Vert \nabla _{v}f_2\Vert _{\infty }) \min \left\{ \Vert (g_1-g_2)(s)\Vert _{2, {{\bar{\vartheta }}}}^2,\Vert (g_1-g_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2\right\} \mathrm{d}s + C\int _0^t \Vert (f_1-f_2)\Vert _{2, {{\bar{\vartheta }}}}^2 \mathrm{d}s, \end{aligned}$$
(8.1)
where\({{\bar{\vartheta }}}<0\)is a constant defined as in Theorem 2.8. Therefore by the Gronwall inequality for every\(t_0>0\),
$$\begin{aligned} \sup _{t\in (0, t_0)} \frac{1}{2}\Vert (f^{(n+1)}-f^{(n+2)})(t)\Vert _{2, {{\bar{\vartheta }}}} \le e^{Ct_0}C_{\varepsilon }t_0 \sup _{s\in (0,t_0)} \Vert (f^{(n)}-f^{(n+1)})(s)\Vert _{2, {{\bar{\vartheta }}}}. \end{aligned}$$
(8.2)

Proof

Note that \(f_1\), \(f_2\), \(g_1\), and \(g_2\) satisfy
$$\begin{aligned} \partial _t (f_1-f_2) + v\cdot \nabla _x(f_1-f_2) + L(f_1-f_2) = \Gamma (g_1,f_1-f_2) + \Gamma (g_1-g_2,f_2). \end{aligned}$$
Multiplying the above equation by \(w^{2 {{\bar{\vartheta }}}}(f_1-f_2)\) and integrating both sides of the resulting equation yields
$$\begin{aligned}&\frac{1}{2}\Vert (f_1-f_2)(t)\Vert _{2,{{\bar{\vartheta }}}} + \int _0^t \left( w^{2 {{\bar{\vartheta }}}}L(f_1-f_2)(s),(f_1-f_2)(s) \right) \mathrm{d}s \nonumber \\&\quad = \int _0^t \left( w^{2 {{\bar{\vartheta }}}} \Gamma (g_1(s),(f_1-f_2)(s)),(f_1-f_2)(s) \right) \mathrm{d}s + \int _0^t \left( w^{2 {{\bar{\vartheta }}}} \Gamma ((g_1-g_2)(s),f_2(s)),(f_1-f_2)(s) \right) \mathrm{d}s \nonumber \\&\quad = \int _0^t ((I) + (II)) \mathrm{d}s. \end{aligned}$$
(8.3)
By Lemma 2.7, we have
$$\begin{aligned} \left( w^{2 \vartheta _0} L(f_1-f_2)(s),(f_1-f_2)(s) \right) \ge \frac{1}{2} \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2 - C\Vert (f_1-f_2)(s)\Vert _{2, {{\bar{\vartheta }}}}^2. \end{aligned}$$
Since \(g_1\) satisfies (2.6), by Theorem 2.8, we have
$$\begin{aligned} (I) & \le C\Vert g_1(s)\Vert _\infty \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2\\ & \le C \varepsilon \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2. \end{aligned}$$
Since we want to control (II) in terms of \(\Vert g_{1} - g_{2}\Vert _{2, {{\bar{\vartheta }}}}\), \(\Vert g_{1} - g_{2}\Vert _{\sigma , {{\bar{\vartheta }}}}\), and \(\Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}\), we have to show that
$$\begin{aligned} \Vert \nabla _{v} f_2\Vert _{\infty } < \infty . \end{aligned}$$
Since \(g_2\) is uniformly Hölder continuous by Lemma 7.9, \(D_v f_2\) is uniformly bounded. Therefore by Theorems 2.8 and 1.3, Lemma 7.9, and Young’s inequality, we have
$$\begin{aligned} (II) & \le C\left( \Vert\, f_2\Vert _{\infty } + \Vert \nabla _{v} f_2\Vert _{\infty } \right) \min \{ \Vert (g_1-g_2)(s)\Vert _{2, {{\bar{\vartheta }}}},\Vert (g_1-g_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}\} \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}\\ & \le C_{\varepsilon }\left( \Vert\, f_2\Vert _{\infty } + \Vert \nabla _{v} f_2\Vert _{\infty } \right) \left( \min \left\{ \Vert (g_1-g_2)(s)\Vert _{2, {{\bar{\vartheta }}}}^2,\Vert (g_1-g_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2\right\} + C\varepsilon \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2\right) . \end{aligned}$$
Then we have
$$\begin{aligned}&\frac{1}{2}\Vert (f_1-f_2)(t)\Vert _{2, {{\bar{\vartheta }}}}^2 + \frac{1}{2}\int _0^t \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2 \mathrm{d}s - C\int _0^t \Vert (f_1-f_2)(s)\Vert _{2, {{\bar{\vartheta }}}}^2 \mathrm{d}s\\&\quad \le \int _0^t \Big( C\varepsilon \Vert (f_1-f_2)(s)\Vert _{\sigma , {{\bar{\vartheta }}}}^2+ C_{\varepsilon }\left( \Vert\, f_2\Vert _{\infty } + \Vert \nabla _{v} f_2\Vert _{\infty } \right) \times \min \left\{ \Vert (g_1-g_2)(s)\Vert _{2, \vartheta _0}^2,\Vert (g_1-g_2)(s)\Vert _{\sigma , \vartheta _0}^2\right\}\Big) \mathrm{d}s. \end{aligned}$$
Therefore we have (8.1). \(\square \)

Proof of Theorem 1.1

Proof of (1)
  • Existence Let \(t_0\) be a given positive constant and \({{\bar{\vartheta }}}<0\) be a constant defined as in Theorem 2.8. Since \(f_0\) satisfies (1.16), by Theorem 1.4, \(f^{(n)}\) is Hölder continuous uniformly in n. Therefore, by (8.1) we have
    $$\begin{aligned}&\Vert (f^{(n+1)}-f^{(n)})(t)\Vert _{2, \vartheta _0}^2\\&\quad \le C'\int _0^t \Vert (f^{(n)} - f^{(n-1)})(s)\Vert _{2, \vartheta _0}^2 \mathrm{d}s + C\int _0^t\Vert (f^{(n+1)} - f^{(n)})(s)\Vert _{2,\vartheta _0}^2 \mathrm{d}s. \end{aligned}$$
    for some C and \(C'\). Then we will show that
    $$\begin{aligned} \Vert (f^{(n)} - f^{(n-1)})(t)\Vert _{2, \vartheta _0}^2 \le \frac{e^{Cn t_0}(C't)^n}{n!} \end{aligned}$$
    (8.4)
    for every \(t\in (0,t_0)\) and \(n\ge 1\) by the induction on n. Suppose that (8.4) holds for \(n=k\), then
    $$\begin{aligned} C'\int _0^t \Vert (f^{(k)} - f^{(k-1)})(s)\Vert _{2, \vartheta _0}^2 \mathrm{d}s \le \frac{e^{Ck t_0}(C't)^{k+1}}{(k+1)!} \end{aligned}$$
    for \(t\in (0,t_0)\). Therefore, we have
    $$\begin{aligned} \Vert (f^{(k+1)}-f^{(k)})(t)\Vert _{2, \vartheta _0}^2 \le \frac{e^{Ck t_0}(C't)^{k+1}}{(k+1)!} + C\int _0^t\Vert (f^{(k+1)} - f^{(k)})(s)\Vert _{2,\vartheta _0}^2 \mathrm{d}s \end{aligned}$$
    for every \(t\in (0,t_0)\). Then by the Gronwall inequality, we have
    $$\begin{aligned} \Vert (f^{(k+1)}-f^{(k)})(t)\Vert _{2, \vartheta _0}^2 \le \frac{e^{Ck t_0}(C't)^{k+1}}{(k+1)!} e^{Ct} \le \frac{e^{C(k+1) t_0}(C't)^{k+1}}{(k+1)!} \end{aligned}$$
    for every \(t\in (0,t_0)\). Thus we have (8.4) for every \(n\in {\mathbb {N}}\). Moreover, we have
    $$\begin{aligned} \lim _{N \rightarrow \infty } \sum _{n>N} \sup _{0\le t\le t_0} \Vert (f^{(n)} - f^{(n-1)})(t)\Vert _{2, \vartheta _0} = 0. \end{aligned}$$
    Thus \(f^{(n)}\) is a Cauchy sequence in \(L^2([0,t_{0}]\times {\mathbb {T}}^3 \times {\mathbb {R}}^3, w^{\vartheta _{0}}\mathrm{d}t\mathrm{d}x\mathrm{d}v)\). Let \(f = \lim _{n \rightarrow \infty } f^{(n)}\). Then by (8.2), f is a weak solution of (1.5)–(1.7) in the sense of Definition 4.1.
  • Uniqueness Suppose that f and g are weak solutions of (1.5)–(1.7) in the sense of Definition 4.1. Then by (8.1), we have
    $$\begin{aligned} \frac{1}{2}\Vert (f-g)(t)\Vert _{2, \vartheta _0}^2 + \left( \frac{1}{2}-C \varepsilon \right) \int _{0}^{t} \Vert (f-g)(s)\Vert _{\sigma , \vartheta _0}^2 \mathrm{d}s \le C\int _{0}^{t} \Vert (f-g)(s)\Vert _{2, \vartheta _0}^2 \mathrm{d}s. \end{aligned}$$
    Since \(C \varepsilon ' < 1/4\), we have
    $$\begin{aligned} \frac{1}{2}\Vert (f-g)(t)\Vert _{2, \vartheta _0}^2 \le C\int _0^t \Vert (f-g)(s)\Vert _{2, \vartheta _0}^2 \mathrm{d}s. \end{aligned}$$
    Therefore, by the Gronwall inequality, we have
    $$\begin{aligned} \Vert (f-g)(t)\Vert _{2, \vartheta _0}^2 = 0 \end{aligned}$$
    for every \(t\in (0,t_0)\).
Since \(t_0\) is arbitrary, we conclude that the weak solution of (1.5)–(1.7) in the sense of Definition 4.1 uniquely exists globally in time.

Proof of (3) We can apply f to Theorems 1.21.4, and Lemma 7.9. Then we have (1.17)–(1.21).

Proof of (2) Let \(F = \mu + \sqrt{\mu }f\), where f is the weak solution of (1.5)–(1.7) in the sense of Definition 4.1. Consider
$$\begin{aligned} \partial _{t} F + v \cdot \nabla _{x} F = Q(F,F) = \sigma _{F}^{ij} \partial _{v_{i}v_{j}}F + 8 \pi F^2. \end{aligned}$$
(8.5)
Similar to Definition 4.1, we can define a weak solution to (8.5) and we can easily check that F is a weak solution to (8.5). Since f satisfies (1.19), by Lemma 2.4, \(\sigma _{F}\) is a non-negative definite matrix. Therefore, in a similar manner to Sect. 3, we can obtain a weak minimum principle for (8.5). Thus, if \(F(0) \ge 0\), then \(F(t)\ge 0\). \(\square \)

Notes

Acknowledgements

Yan Guo is supported in part by NSF grant #DMS-1810868, Chinese NSF Grant #10828103, as well as a Simon Fellowship. Hyung Ju Hwang was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF-2017R1E1A1A03070105, NRF-2019R1A5A1028324).

Compliance with Ethical Standards

Conflict of Interest

The authors declare that they have no conflict of interest.

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Copyright information

© Peking University 2020

Authors and Affiliations

  1. 1.Department of MathematicsPohang University of Science and TechnologyPohangSouth Korea
  2. 2.Division of Applied MathmaticsBrown UniversityProvidenceUSA

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