---
abstract: 'In this paper we study the following stochastic Hamiltonian system in $\mR^{2d}$ (a second order stochastic differential equation), $$\dif \dot X_t=b(X_t,\dot X_t)\dif t+\sigma(X_t,\dot X_t)\dif W_t,\ \ (X_0,\dot X_0)=(x,\v)\in\mR^{2d},$$ where $b(x,\v):\mR^{2d}\to\mR^d$ and $\sigma(x,\v):\mR^{2d}\to\mR^d\otimes\mR^d$ are two Borel measurable functions. We show that if $\sigma$ is bounded and uniformly non-degenerate, and $b\in H^{2/3,0}_p$ and $\nabla\sigma\in L^p$ for some $p>2(2d+1)$, where $H^{\alpha,\beta}_p$ is the Bessel potential space with differentiability indices $\alpha$ in $x$ and $\beta$ in $\v$, then the above stochastic equation admits a unique strong solution so that $(x,\v)\mapsto Z_t(x,\v):=(X_t,\dot X_t)(x,\v)$ forms a stochastic homeomorphism flow, and $(x,\v)\mapsto Z_t(x,\v)$ is weakly differentiable with ess.$\sup_{x,\v}\mE\left(\sup_{t\in[0,T]}|\nabla Z_t(x,\v)|^q\right)<\infty$ for all $q\geq 1$ and $T\geq 0$. Moreover, we also show the uniqueness of probability measure-valued solutions for kinetic Fokker-Planck equations with rough coefficients by showing the well-posedness of the associated martingale problem and using the superposition principle established by Figalli [@Fi] and Trevisan [@Tre].'
address: |
     School of Mathematics and Statistics, Wuhan University, Wuhan, Hubei 430072, P.R.China\
    Email: XichengZhang@gmail.com 
author:
- |
    XICHENG ZHANG\
    \
    [*Dedicated to the 60th birthday of Professor Michael Röckner*]{}
title: Stochastic Hamiltonian flows with singular coefficients
---

[^1]

Introduction
============

Consider the following [*second order*]{} time dependent stochastic differential equation (abbreviated as SDE): $$\begin{aligned}
\dif \dot X_t=b_t(X_t,\dot X_t)\dif t+\sigma_t(X_t,\dot X_t)\dif W_t,\ \ (X_0,\dot X_0)=(x,\v)\in\mR^{2d},\end{aligned}$$ where $b_t(x,\v):\mR_+\times\mR^{2d}\to\mR^d$ and $\sigma_t(x,\v):\mR_+\times\mR^{2d}\to\mR^d\otimes\mR^d$ are two Borel measurable functions, $\dot X_t$ denotes the first order derivative of $X_t$ with respect to $t$, and $W_t$ is a $d$-dimensional standard Brownian motion on some probability space $(\Omega,\sF,\mP)$. When $\sigma=0$, the above equation is the classical Newtonian mechanic equation, which describes the motion of a particle. When $\sigma\not=0$, it means that the motion is perturbed by some random external force. More backgrounds about the above stochastic Hamiltonian system are referred to [@So; @Ta], etc.

It is noticed that if we let $Z_t:=(X_t,\dot X_t)$, then $Z_t$ solves the following [*one order*]{} (degenerate) SDE: $$\begin{aligned}
\label{SDE0}
\dif Z_t=(\dot X_t,b_t(Z_t))\dif t+(0,\sigma_t(Z_t)\dif W_t),\ \ Z_0=z=(x,\v)\in\mR^{2d},\end{aligned}$$ and whose time-dependent infinitesimal generator is given by $$\begin{aligned}
\label{LL}
\sL^{a,b}_t f(x,\v):=\tr(a_t\cdot\nabla^2_\v f)(x,\v)+(\v\cdot\nabla_x f)(x,\v)+(b_t\cdot\nabla_\v f)(x,\v).\end{aligned}$$ Here $a_t(x,\v):=\tfrac{1}{2}(\sigma_t\sigma^*_t)(x,\v)$, $\nabla^2_\v f(x,\v)$ stands for the Hessian matrix, the asterisk and tr($\cdot$) denote the transpose and the trace of a matrix respectively. Moreover, let $\mu_t$ be the probability distributional measure of $Z_t$ in $\mR^{2d}$. By Itô’s formula, one knows that $\mu_t$ solves the following Fokker-Planck equation in the distributional sense: $$\begin{aligned}
\label{FP}
\p_t\mu_t=(\sL^{a,b}_t)^*\mu_t,\ \ \mu_0=\delta_z,\end{aligned}$$ where $\delta_z$ is the Dirac measure at $z$. More precisely, for any $f\in C^2_c(\mR^{2d})$, $$\p_t\mu_t(f)=\mu_t(\sL^{a,b}_t f),\ \mu_0(f)=f(z),$$ where $\mu_t(f)=\int f\dif\mu_t=\mE f(Z_t)$. In the literature $\sL^{a,b}_t$ is also called kinetic Fokker-Planck or Kolmogorov’s operator.

During the past decade, there is an increasing interest in the study of SDEs with singular or rough coefficients. In the non-degenerate case, Krylov and Röckner [@Kr-Ro] showed the strong uniqueness to the following SDE in $\mR^d$: $$\dif X_t=b_t(X_t)\dif t+\dif W_t, \ X_0=x,$$ where $b\in L^q_{lock}(\mR^+;L^p(\mR^d))$ with $\frac{d}{p}+\frac{2}{q}<1$. The argument in [@Kr-Ro] is based on Girsanov’s theorem and some estimates from the theory of PDE. In this framework, Fedrizzi and Flandoli [@Fe-Fl; @Fe-Fl0] studied the well-posedness of stochastic transport equations with rough coefficients. When $b$ is bounded measurable, the Malliavin differentiability of $X_t$ with respect to sample path $\omega$ and the weak differentiability of $X_t$ with respect to starting point $x$ were recently studied in [@Me] and [@Mo-Ni-Pr] respectively. We also mention that weak uniqueness was studied in [@Ba-Ch] and [@Ji] under rather weak assumptions on $b$ (belonging to some Kato’s class). Moreover, the multiplicative noise case was studied in [@Zh0; @Zh2; @Zh4] by using Zvonkin’s transformation [@Zv] and some careful estimates of second order parabolic equations.

In the degenerate case, Chaudru de Raynal [@Ch] firstly showed the strong well-posedness for SDE under the assumptions that $\sigma$ is Lipschitz continuous and $b$ is $\alpha$-Hölder continuous in $x$ and $\beta$-Hölder continuous in $\v$ with $\alpha\in (\frac{2}{3},1)$ and $\beta\in(0,1)$. The proofs in [@Ch] strongly depend on some explicit estimates for Kolmogorov operator with constant coefficients and Zvonkin’s transformation. In a recent joint work [@Wa-Zh] with F.Y. Wang, we also showed the strong uniqueness and homeomorphism property for under weaker Hölder-Dini’s continuity assumption on $b$. The proofs in [@Wa-Zh] rely on a characterization of Hölder-Dini’s spaces and gradient estimates for the semigroup associated with the kinetic operator. Notice that in [@Ch] and [@Wa-Zh], more general degenerate SDEs are considered, while, the case with critical differentiability indices $\alpha=\frac{2}{3}$ and $\beta=0$ is left open. The purpose of this work is to establish a similar theory for degenerate SDE as in Krylov and Röckner’s paper [@Kr-Ro] (see also [@Zh4]). In particular, the critical indices $\alpha=\frac{2}{3}$ and $\beta=0$ are covered. More precisely, we aim to prove that \[Main\] Suppose that for some $K\geq 1$ and all $(t,x,\v)\in\mR_+\times\mR^{2d}$, $$\begin{aligned}
K^{-1}|\xi|\leq |\sigma^*_t(x,\v)\xi|\leq K|\xi|,\ \  \forall\xi\in\mR^d,\tag{{\bf UE}}\end{aligned}$$ where $\sigma^*$ denotes the transpose of matrix $\sigma$, and for some $p>2(2d+1)$, $$\kappa_0:=\sup_{s\geq 0}\|\nabla\sigma_s\|^p_p+\int_0^\infty\|(\mI-\Delta_x)^{\frac{1}{3}} b_s\|_p^p\dif s<\infty.$$ Then for any $z=(x,\v)\in\mR^{2d}$, SDE admits a unique strong solution $Z_t(z)=(X_t,\dot X_t)$ so that $(t,z)\mapsto Z_t(z)$ has a bi-continuous version. Moreover,

1.  There is a null set $N$ such that for all $\omega\notin N$ and for each $t\geq 0$, the map $z\mapsto Z_t(z,\omega)$ is a homeomorphism on $\mR^{2d}$.

2.  For each $t\geq 0$, the map $z\mapsto Z_t(z)$ is weakly differentiable a.s., and for any $q\geq 1$ and $T>0$, $$\begin{aligned}
    \label{YN1}
    \mathrm{ess.}\sup_{z}\mE\left(\sup_{t\in[0,T]}|\nabla Z_t(z)|^q\right)<\infty,\end{aligned}$$ where $\nabla$ denotes the generalized gradient.

3.  Let $\sigma^n$ and $b^n$ be the regularized approximations of $\sigma$ and $b$ (see below for definitions). Let $Z^n$ be the corresponding solution of SDE associated with $(\sigma^n,b^n)$. For any $q\geq 1$ and $T>0$, there exits a constant $C>0$ only depending on $T,K,\kappa_0,d,p,q$ such that $$\mE\left(\sup_{t\in[0,T]}|Z^n_t-Z_t|^q\right)\leq C\left(\|b^n-b\|^q_{\mL^p(T)}+n^{(\frac{2d}{p}-1)q}\right),\ \ n\in\mN.$$

As a corollary, we have the following local well-posedness result by a standard localization argument. Suppose that for any $T, R>0$, there exists a constant $K_{T,R}\geq 1$ such that for all $(t,x,\v)\in[0,T]\times B_R$ and $\xi\in\mR^d$, $$\begin{aligned}
K_{T,R}^{-1}|\xi|\leq |\sigma^*_t(x,\v)\xi|\leq K_{T,R}|\xi|,\label{Con22}\end{aligned}$$ where $B_R:=\{(x,\v): |(x,\v)|\leq R\}$, and for some $p>2(2d+1)$, $$\sup_{t\in[0,T]}\|\nabla(\sigma_t\chi_R)\|^p_{p}+\int_0^T\|(\mI-\Delta_x)^{\frac{1}{3}} (b_s\chi_R)\|_p^p\dif s\leq K_{T,R},$$ where $\chi_R:\mR^{2d}\to[0,1]$ is a smooth function with $\chi_R(z)=1$ for $|z|\leq R$ and $\chi_R(z)=0$ for $|z|>2R$. Then for any fixed $(x,\v)\in\mR^{2d}$, SDE admits a unique local strong solution $(X_t,\dot X_t)$ up to the explosion time $\zeta$.

Let $$\sigma^R_t(z):=\sigma_t(z\chi_R(z)),\ \ b^R_t(z):=b_t(z)\chi_R(z).$$ By the assumptions, one sees that $(\sigma^R, b^R)$ satisfies the conditions of Theorem \[Main\]. Hence, there exists a unique solution to the following SDE: $$\dif Z^R_t=(\dot X^R_t,b^R_t(Z^R_t))\dif t+(0,\sigma^R_t(Z^R_t)\dif W_t),\ \ Z^R_0=z=(x,\v)\in\mR^{2d},$$ where $Z^R_t=(X^R_t,\dot X^R_t)$. Define $$\zeta_R:=\inf\Big\{t\geq 0: |Z^R_t|\geq R\Big\},\ \ Z_t:=Z^R_t,\ t\in[0,\zeta_R].$$ Since $Z^{R'}_t|_{[0,\zeta_R]}=Z^R_t|_{[0,\zeta_R]}$ for $R'>R$, one sees that $R\mapsto \zeta_R$ is increasing and the above $Z_t$ is well-defined. Clearly, $\zeta=\lim_{R\to\infty}\zeta_R$ is the explosion time of $Z_t$, and $Z_t$ uniquely solves before $\zeta$.

The strategy of proving Theorem \[Main\] is still based on Zvonkin’s transformation. As in the non-degenerate case [@Zh4], we need to establish the $L^p$-maximal regularity estimate to the following degenerate parabolic equation (see Theorem \[Th32\] below): $$\p_t u=\sL^{a,b}_t u+f,\ \ u_0=0.$$ Here we shall use the freezing coefficient argument and the $L^p$-estimate established in [@Br-Cu-La-Pr] and [@Bo] for degenerate operators with constant coefficients (see also [@Ch-Zh] for the case of nonlocal operators). Compared with [@Br-Cu-La-Pr] and [@Pr], we not only consider the optimal regularity of $u$ along the nondegenerate $\v$-direction, but also the optimal regularity of $u$ along the degenerate $x$-direction.

On the other hand, from the viewpoint of PDE, the well-posedness of Fokker-Planck equation (especially uniqueness) with rough coefficients is a quite involved problem. Since $a$ and $b$ possess less regularities and $\sL^{a,b}_t$ is a degenerate operator, the direct analytical approach seems not work (cf. [@Bo-Kr-Ro1; @Bo-Kr-Ro2]). Let $\cP(\mR^{2d})$ be the set of all probability measures on $\mR^{2d}$. We shall use a probabilistic method to prove the following result. \[Main2\] Suppose that $\sigma$ satisfies [**(UE)**]{} and for any $T>0$, $$\lim_{|z-z'|\to 0}\sup_{t\in[0,T]}\|\sigma_t(z)-\sigma_t(z')\|=0,$$ and $b\in L^q_{loc}(\mR_+; L^q(\mR^{2d}))$ for some $q\in(2(2d+1),\infty]$. Then for any $\nu\in\cP(\mR^{2d})$, there exists a unique probability measure-valued solution $\mu_t\in\cP(\mR^{2d})$ to in the distributional sense in the class that $t\mapsto\mu_t$ is weakly continuous with $\mu_0=\nu$ and $$\int^t_0\!\!\int_{\mR^{2d}}(|\v|+|b_s(x,\v)|)\mu_s(\dif x,\dif\v)\dif s<\infty,\ \ t>0.$$

The proof of this result is based on Figalli and Trevisan’s superposition characterization for the solutions of Fokker-Planck equation in terms of martingale problem associated with $\sigma$ and $b$. More precisely, Figalli [@Fi] and Trevisan [@Tre] showed that for any weakly continuous probability measure-valued solution $\mu_t$ of with initial value $\nu\in\cP(\mR^{2d})$, there exists a martingale solution for operator $\sL^{a,b}_t$ (a probability measure $\mP_\nu$ over the space of all continuous functions from $\mR_+$ to $\mR^{2d}$ denoted by $\Omega$) such that for all $t\in\mR_+$, $$\int_{\mR^{2d}}\varphi(z)\mu_t(\dif z)=\int_{\Omega}\varphi(\omega_t)\mP_\nu(\dif \omega),$$ where $t\mapsto\omega_t$ is the coordinate process over $\Omega$. Hence, in order to prove Theorem \[Main2\], it suffices to show the well-posedness of martingale problem for $\sL^{a,b}_t$ in the sense of Stroock and Varadhan [@St-Va]. This will be achieved by proving some Krylov’s type estimate (see Theorem \[Th41\] below), which is also a key tool for proving Theorem \[Main\]. It is remarked that in [@Ro-Zh], we have already used this technique to show the uniqueness of measure-valued solutions and $L^p$-solutions to possibly degenerate second order Fokker-Planck equations under some weak conditions on the coefficients (but not the case of Theorem \[Main2\]).

This paper is organized as follows: in Section 2, we introduce some anisotropic fractional Bessel potential spaces, and prepare some useful estimates for later use. In Section 3, we show the $L^p$-maximal regularity estimate for kinetic Fokker-Planck equations. In Section 4, we study the martingale problem associated with $(\sigma,b)$ under the same assumptions as in Theorem \[Main2\] by showing the basic Krylov’s type estimate. In particular, we first prove Theorem \[Main2\]. In Section 5, we then prove Theorem \[Main\] by using Zvonkin’s transformation and Krylov’s estimate obtained in the previous section. In Appendix, a stochastic Gronwall’s type lemma used in Section 5 is given.

Convention: The letter $C$ with or without subscripts will denote an unimportant constant, whose value may change in different places. Moreover, $A\preceq B$ means that $A\leq CB$ for some constant $C>0$, and $A\asymp B$ means that $C^{-1}B\leq A\leq CB$ for some $C>1$.

After this work was finished, I was informed by Professor Enrico Priola during “The 8th International Conference on Stochastic Analysis and its Applications" held at BIT that, very recently, Fedrizzi, Flandoli, Priola and Vovelle [@Fe-Fl-Pr-Vo] also obtained the strong well-posedness together with their flow property of SDE under the conditions $\sigma_t(z)=\mI$ and $b_t(z)=b(z)$ possessing the following regularity $$\|(\mI-\Delta_x)^{s/2}b\|_p<\infty$$ for some $s>2/3$ and $p>6d$.

Preliminaries
=============

For $\alpha\geq 0$ and $p\in(1,\infty)$, let $H^\alpha_p:=H^\alpha_p(\mR^d):=(\mI-\Delta)^{-\frac{\alpha}{2}}(L^p(\mR^d))$ be the usual Bessel potential space with norm $$\|f\|_{\alpha,p}:=\|(\mI-\Delta)^{\frac{\alpha}{2}} f\|_p,$$ where $\|\cdot\|_p$ is the usual $L^p$-norm, and $\Delta$ is the Laplacian. For $\alpha\in(0,2)$, let $\Delta^{\frac{\alpha}{2}}:=-(-\Delta)^{\frac{\alpha}{2}}$ be the usual fractional Laplacian. Notice that up to a constant $C(\alpha,d)>0$, an alternative definition of $\Delta^{\frac{\alpha}{2}}$ is given by $$\begin{aligned}
\label{Def1}
\Delta^{\frac{\alpha}{2}} f(x):=\lim_{\eps\downarrow 0}\int_{|y|\geq\eps}\delta_yf(x)|y|^{-d-\alpha}\dif y,\ \ \delta_yf(x):=f(x+y)-f(x).\end{aligned}$$ We will frequently use such a definition below. It is well-known that by the boundedness of Riesz’s transformation (cf. [@St]), $$\begin{aligned}
\label{Riesz}
\|\Delta^{\frac{1}{2}}f\|_p\asymp\|\nabla f\|_p,\ p>1,\end{aligned}$$ and an equivalent norm in $H^\alpha_p$ is given by $$\begin{aligned}
\label{Equiv}
\|f\|_{\alpha,p}\asymp\|f\|_{p}+\|\Delta^{\frac{\alpha-[\alpha]}{2}}\nabla^{[\alpha]} f\|_p,\end{aligned}$$ where $[\alpha]$ is the integer part of real number $\alpha$, and we have used the convention $\Delta^0:=\mI$. Notice that for $\alpha\in(0,1]$ and $p>1$, $$\begin{aligned}
\label{LI0}
\|f(\cdot+x)-f(\cdot)\|_p\preceq \|\Delta^{\frac{\alpha}{2}}f\|_p|x|^\alpha,\end{aligned}$$ and in particular, $$\begin{aligned}
\label{LI}
\|f(\cdot+x)-f(\cdot)\|_p\preceq \|f\|_{\alpha,p}(|x|^\alpha\wedge 1).\end{aligned}$$ Moreover, we also have the following interpolation inequality: for any $0\leq\alpha<\beta<\infty$, $$\begin{aligned}
\label{Inter}
\|f\|_{\alpha,p}\leq C(p,d,\alpha,\beta)\|f\|^{\frac{\beta-\alpha}{\beta}}_p\|f\|^{\frac{\alpha}{\beta}}_{\beta,p},\end{aligned}$$ and the following Sobolev embedding results hold: for any $\alpha\in(0,1)$, if $p\alpha>d$, then $$\begin{aligned}
\label{Em0}
\|f\|_\infty+\sup_{x\not=y}\frac{|f(x)-f(y)|}{|x-y|^\gamma}\leq C(p,d,\alpha,\gamma)\|f\|_{\alpha,p},\ \gamma\in(0,\alpha-\tfrac{d}{p});\end{aligned}$$ if $p\alpha<d$, then $$\begin{aligned}
\label{Em1}
\|f\|_{q}\leq C(p,d,\alpha,q)\|f\|_{\alpha,p},\ \ q\in[p,\tfrac{pd}{d-p\alpha}].\end{aligned}$$ All the above facts are standard and can be found in [@Be-Lo] and [@St].

To treat the kinetic Fokker-Planck operator, we introduce the following anisotropic Bessel potential spaces. Let $C^\infty_c(\mR^{2d})$ be the space of all smooth functions on $\mR^{2d}$ with compact supports. For $\alpha,\beta\geq 0$, we define the Bessel potential space $H^{\alpha,\beta}_p:=H^{\alpha,\beta}_p(\mR^{2d})$ as the completion of $C^\infty_c(\mR^{2d})$ with respect to norm: $$\|f\|_{\alpha,\beta; p}:=\|(\mI-\Delta_x)^{\frac{\alpha}{2}}f\|_p+\|(\mI-\Delta_\v)^{\frac{\beta}{2}}f\|_p.$$ Notice that by the Mihlin multiplier theorem (cf. [@Be-Lo]), $$\|f\|_{\alpha,\beta;p}\asymp \|f\|_p+\|\Delta_x^{\frac{\alpha}{2}}f\|_p+\|\Delta_\v^{\frac{\beta}{2}}f\|_p
\asymp \|((\mI-\Delta_x)^{\frac{\alpha}{2}}+(\mI-\Delta_\v)^{\frac{\beta}{2}})f\|_p.$$ In the following, we simply write $$H^{\infty,\infty}_p:=H^{\infty,\infty}_p(\mR^{2d}):=\cap_{\alpha,\beta\geq 0}H^{\alpha,\beta}_p(\mR^{2d}).$$ (i) For any $\alpha,\beta\geq 0$, $\theta\in[0,1]$ and $p>1$, there is a constant $C=C(\alpha,\beta,\theta,p,d)>0$ such that $$\begin{aligned}
&\|(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}(\mI-\Delta_\v)^{\frac{(1-\theta)\beta}{2}}f\|_p\leq C\|f\|_{\alpha,\beta;p},\label{Em8}\\
&\quad \|\Delta_x^{\frac{\theta\alpha}{2}}\Delta_\v^{\frac{(1-\theta)\alpha}{2}}f\|_p\leq C\|(\Delta_x+\Delta_\v)^{\frac{\alpha}{2}}f\|_p.\label{Em88}\end{aligned}$$ In particular, for any $\alpha\geq 0$, $\beta\geq 1$ and $p>1$, we have $$\begin{aligned}
\label{E28}
\|\nabla_\v f\|_{\alpha(\beta-1)/\beta,\beta-1;p}\leq C\|f\|_{\alpha,\beta;p}.\end{aligned}$$ (ii) Let $\alpha,\beta\geq 0$ and $p>1$ with $d\not=\frac{p\alpha\beta}{\alpha+\beta}$. Set $$p^*:=
\left\{
\begin{aligned}
&dp/(d-\tfrac{p\alpha\beta}{\alpha+\beta}),&\ \ d>\tfrac{p\alpha\beta}{\alpha+\beta};\\
&\infty,&\ \ d<\tfrac{p\alpha\beta}{\alpha+\beta}. 
\end{aligned}
\right.$$ For any $q\in[p,p^*]$, there is a constant $C=C(\alpha,\beta,p,q,d)>0$ such that $$\begin{aligned}
\label{E29}
\|f\|_{q}\leq C\|f\|_{\alpha,\beta;p}.\end{aligned}$$

\(i) It follows by the Mihlin multiplier theorem and .

\(ii) For , by , and with $\theta=\frac{\beta}{\alpha+\beta}$, we have $$\begin{aligned}
\|f\|^q_q&=\int_{\mR^d}\|f(\cdot,\v)\|_q^q\dif \v \preceq
\int_{\mR^d}\|(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}f(\cdot,\v)\|_p^q\dif \v\\
&\leq\left(\int_{\mR^d}\left(\int_{\mR^d}|(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}f(x,\v)|^q\dif\v\right)^{p/q}\dif x\right)^{q/p}\\
&\preceq\left(\int_{\mR^d}\|(\mI-\Delta_\v)^{\frac{(1-\theta)\beta}{2}}(\mI-\Delta_x)^{\frac{\theta\alpha}{2}}f(x,\cdot)\|^p_{p}\dif x\right)^{q/p}
\preceq\|f\|^q_{\alpha,\beta;p},\end{aligned}$$ where the second inequality is due to Minkovskii’s inequality.

Let $a:\mR^{2d}\to\mR^d\otimes\mR^d$ be a measurable function. Write $$\sL^a u:=\mathrm{tr}(a\cdot\nabla^2_\v u)+\v\cdot\nabla_x u.$$ We have \[Le22\] Let $\alpha\in(0,1)$ and $p>d/\alpha$. Suppose that $$\kappa_0:=\sup_{\v}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|_p+\|a\|_\infty<\infty.$$ For any $\eps\in(0,1)$, there is a constant $C_\eps=C_\eps(p,d,\alpha,\kappa_0)>0$ such that for all $u\in H^{\infty,\infty}_p$, $$\begin{aligned}
\label{E210}
\|[\Delta_x^{\frac{\alpha}{2}},\sL^a]u\|_p\leq\eps\|\Delta_x^{\frac{\alpha}{2}}\nabla^2_\v u\|_{p}+C_\eps \|\nabla^2_\v u\|_{p},\end{aligned}$$ where $[\Delta_x^{\frac{\alpha}{2}},\sL^a]u:=\Delta_x^{\frac{\alpha}{2}}(\sL^a u)-\sL^a(\Delta_x^{\frac{\alpha}{2}}u)$.

Notice that by definition , $$[\Delta_x^{\frac{\alpha}{2}},\sL^a]u=\tr\big(\Delta_x^{\frac{\alpha}{2}} a\cdot\nabla^2_\v u\big)
+\int_{\mR^d}\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)|y|^{-d-\alpha}\dif y.$$ Hence, $$\|[\Delta_x^{\frac{\alpha}{2}},\sL^a]u\|_p\leq\|\tr(\Delta_x^{\frac{\alpha}{2}} a\cdot\nabla^2_\v u)\|_p
+\int_{\mR^d}\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p|y|^{-d-\alpha}\dif y.$$ Let $\beta\in(\frac{d}{p},\alpha)$. By , we have $$\begin{aligned}
\|\tr(\Delta_x^{\frac{\alpha}{2}} a\cdot\nabla^2_\v u)\|_p^p
&\preceq\int_{\mR^d}\|\Delta_x^{\frac{\alpha}{2}} a(\cdot,\v)\|^p_{p}\|\nabla^2_\v u(\cdot,\v)\|_\infty^p\dif \v\\
&\preceq\sup_{\v}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|^p_p\|\nabla^2_\v u\|_{\beta,0; p}^p,\end{aligned}$$ and for $\gamma\in(0,\beta-\frac{d}{p})$, $$\begin{aligned}
&\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p^p
\preceq\int_{\mR^d}\|\delta_y a(\cdot,\v)\|_{p}^p\cdot\|\nabla^2_\v\delta_y u(\cdot,\v)\|_{\infty}^p\dif \v\\
&\qquad\stackrel{\eqref{LI0}}{\preceq}|y|^{(\alpha+\gamma)p}\int_{\mR^d}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|_p^p\|\nabla^2_\v u(\cdot,\v)\|_{\beta,p}^p\dif\v\\
&\qquad\ \preceq |y|^{(\alpha+\gamma)p}\sup_{\v}\|\Delta_x^{\frac{\alpha}{2}}a(\cdot,\v)\|^p_p\|\nabla^2_\v u\|_{\beta,0;p}^p.\end{aligned}$$ Moreover, it is easy to see that $$\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p\preceq\|a\|_\infty\|\nabla^2_\v u\|_p.$$ Therefore, $$\big\|\tr\big(\delta_{(y,0)}a\cdot\nabla^2_\v\delta_{(y,0)}u\big)\big\|_p\preceq \kappa_0(|y|^{\alpha+\gamma}\wedge 1)\|\nabla^2_\v u\|_{\beta,0;p}.$$ Combining the above calculations, we get for some $C=C(p,d,\alpha,\beta)>0$, $$\begin{aligned}
\label{LJ1}
\|[\Delta_x^{\frac{\alpha}{2}},\sL^a]u\|_p\leq C\kappa_0\|\nabla^2_\v u\|_{\beta,0;p},\end{aligned}$$ On the other hand, by the interpolation inequality and Young’s inequality, we have for any $\eps\in(0,1)$, $$\|\nabla^2_\v u\|_{\beta,0;p}\preceq \|\nabla^2_\v u\|^\frac{\beta}{\alpha}_{\alpha,0;p}\|\nabla^2_\v u\|^\frac{\alpha-\beta}{\alpha}_{p}
\leq\eps\|\nabla^2_\v u\|_{\alpha,0;p}+C_\eps\|\nabla^2_\v u\|_{p}.$$ Estimate now follows by .

\[Le23\] For any $\alpha,\beta\in(0,1)$ and $p>(\alpha+\beta)d/(\alpha\beta)$, there is a constant $C=C(\alpha,\beta,p,d)>0$ such that for all $b\in H^{\alpha,0}_p$ and $u\in H^{\infty,\infty}_p$, $$\|b\cdot\nabla_\v u\|_{\alpha,0; p}\leq C\|b\|_{\alpha,0;p}\Big(\|\Delta_x^{\frac{\alpha}{2}}\nabla_\v u\|_{0,\beta;p}+\|\nabla_\v u\|_{0,\beta; p}\Big).$$

Notice that by , $$\begin{aligned}
\|b\cdot\nabla_\v u\|_{\alpha,0;p}\preceq\|b\cdot\nabla_\v u\|_p+\|\Delta_x^{\frac{\alpha}{2}}(b\cdot\nabla_\v u)\|_p.\end{aligned}$$ By definition , we have $$\begin{aligned}
&\|\Delta_x^{\frac{\alpha}{2}}(b\cdot\nabla_\v u)\|_p\leq\|(\Delta_x^{\frac{\alpha}{2}} b)\cdot\nabla_\v u\|_p
+\|b\cdot\nabla_\v \Delta_x^{\frac{\alpha}{2}}u\|_p\\
&\qquad+\int_{\mR^d}\big\|\delta_{(y,0)}b\cdot\nabla_\v\delta_{(y,0)}u\big\|_p|y|^{-d-\alpha}\dif y=:I_1+I_2+I_3.\end{aligned}$$ For $I_1$, since $p>(\alpha+\beta)d/(\alpha\beta)$, by with $q=\infty$, we have $$\begin{aligned}
I_1\preceq \|\Delta_x^{\frac{\alpha}{2}} b\|_p\|\nabla_\v u\|_\infty\preceq\|b\|_{\alpha,0; p}\|\nabla_\v u\|_{\alpha,\beta;p}.\end{aligned}$$ For $I_2$, since $p\alpha>d$, by we have $$\begin{aligned}
I_2^p&=\int_{\mR^d}\|b(\cdot,\v)\cdot\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p\dif \v\\
&\preceq\int_{\mR^d}\|b(\cdot,\v)\|_\infty^p\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p\dif \v\\
&\preceq\int_{\mR^d}\|b(\cdot,\v)\|_{\alpha,p}^p\dif \v\sup_\v\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p.\end{aligned}$$ For $I_3$, by and again, we have for any $\gamma\in(0,\alpha-\frac{d}{p})$, $$\begin{aligned}
&\int_{\mR^d}\|\delta_yb(\cdot,\v)\cdot\nabla_\v\delta_yu(\cdot,\v)\|_p^p\dif \v
\preceq\int_{\mR^d}\|\delta_y b(\cdot,\v)\|_p^p\|\delta_y\nabla_\v u(\cdot,\v)\|_\infty^p\dif \v\\
&\qquad\preceq\int_{\mR^d}\|b(\cdot,\v)\|_{\alpha,p}^p(|y|^{\alpha p}\wedge 1)\|\nabla_\v u(\cdot,\v)\|_{\alpha,p}^p(|y|^{\gamma p}\wedge 1)\dif \v\\
&\qquad\preceq\left(\int_{\mR^d}\|b(\cdot,\v)\|_{\alpha,p}^p\dif \v\right)\sup_\v\|\nabla_\v u(\cdot,\v)\|_{\alpha,p}^p(|y|^{(\alpha+\gamma)p}\wedge 1).\end{aligned}$$ On the other hand, notice that by $p\beta>d$ and , $$\begin{aligned}
\sup_\v\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(\cdot,\v)\|_p^p
\leq\int_{\mR^d}\sup_\v |\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(x,\v)|^p\dif x
\preceq\int_{\mR^d}\|\nabla_\v\Delta_x^{\frac{\alpha}{2}} u(x,\cdot)\|^p_{\beta, p}\dif x,\end{aligned}$$ and similarly, $$\|b\cdot\nabla_\v u\|^p_p\preceq\int_{\mR^d} \|b(\cdot,\v)\|^p_{\alpha,p}\|\nabla_\v u(\cdot,\v)\|^p_{p}\dif\v\preceq\|b\|^p_{\alpha,0;p}\|\nabla_\v u\|^p_{0,\beta;p}.$$ Combining the above calculations, we obtain the desired estimate.

Let $\varrho:\mR^{2d}\to[0,\infty)$ be a smooth function with support in the unit ball and $\int\varrho=1$. Define $$\begin{aligned}
\label{Rho}
\varrho_\eps(z):=\eps^{-2d}\varrho(\eps^{-1}z),\ \ \eps\in(0,1),\end{aligned}$$ and for a locally integrable function $u:\mR^{2d}\to\mR$, $$u_{\eps}(z):=u*\varrho_\eps(z)=\int_{\mR^{2d}}u(z')\varrho_\eps(z-z')\dif z'.$$ Let $\sP$ be an operator on the space of locally integrable functions. We define $$\begin{aligned}
\label{Def0}
[\varrho_\eps, \sP]u:=(\sP u)*\varrho_\eps-\sP(u*\varrho_\eps).\end{aligned}$$ We need the following commutator estimate results. \[Le24\]

(i) Let $p\in[1,\infty)$ and $q,r\in[p,\infty]$ with $\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$. For any $b\in L^q(\mR^{2d})$ and $u\in H^{0,1}_r$, we have $$\begin{aligned}
    \label{ES9}
    \lim_{\eps\to 0}\|[\varrho_\eps, b\cdot\nabla_\v]u\|_p=0.\end{aligned}$$

(ii) Let $a:\mR^{2d}\to\mR^d\otimes\mR^d$ be a bounded measurable function. For any $p\in[1,\infty)$ and $u\in H^{0,2}_p$, we have $$\begin{aligned}
    \label{ES8}
    \lim_{\eps\to 0}\|[\varrho_\eps, \sL^a]u\|_p=0.\end{aligned}$$

\(i) It follows by [@Zh22 Lemma 4.2].

\(ii) By definition, we can write for $z=(x,\v)$, $$\begin{aligned}
[\varrho_\eps, \sL^a]u(z)&=(\sL^au)*\varrho_\eps(z)-\sL^a(u*\varrho_\eps)(z)\\
&=\int_{\mR^{2d}}\tr((a(z')-a(z))\cdot\nabla^2_\v u(z'))\varrho_\eps(z-z')\dif z'\\
&\quad+\int_{\mR^{2d}}(\v'-\v) \cdot\nabla_x u(z')\varrho_\eps(z-z')\dif z'\\
&=:I^\eps_1(t,z)+I^\eps_2(t,z).\end{aligned}$$ For $I^\eps_1(t,z)$, by Jensen’s inequality and the assumption, we have $$\begin{aligned}
|I^\eps_1(t,z)|^p&\leq \int_{\mR^{2d}}|\tr((a(z')-a(z))\cdot\nabla^2_\v u(z'))|^p\varrho_\eps(z-z')\dif z'\\
&\leq(2\|a\|_\infty)^p\int_{\mR^{2d}}\|\nabla^2_\v u(z')\|^p\varrho_\eps(z-z')\dif z'.\end{aligned}$$ For $I^\eps_2(t,z)$, by the integration by parts and Hölder’s inequality, we have $$\begin{aligned}
|I^\eps_2(t,z)|^p&=\left|\int_{\mR^{2d}}(\v'-\v) \cdot \nabla_x\varrho_\eps(z-z')u(z')\dif z'\right|^p\\
&\leq \eps^{p}\left(\int_{\mR^{2d}}|\nabla_x\varrho_\eps(z-z')|\cdot|u(z')|\dif z'\right)^p\\
&\leq \eps^{p}\left(\int_{\mR^{2d}}|\nabla_x\varrho_\eps(z)|\dif z\right)^{p-1}
\int_{\mR^{2d}}|\nabla_x\varrho_\eps(z-z')|\cdot|u(z')|^p\dif z'\\
&=\left(\int_{\mR^{2d}}|\nabla_x\varrho(z)|\dif z\right)^{p-1}
\int_{\mR^{2d}}|(\nabla_x\varrho)_\eps(z-z')|\cdot|u(z')|^p\dif z'.\end{aligned}$$ Combining the above calculations, we get $$\|[\varrho_\eps, \sL^a]u\|_p\leq C\|u\|_{0,2;p}.$$ Hence, for any $u\in H^{0,2}_p$, it is easy to see that $$\begin{aligned}
\label{GH0}
\lim_{\eps'\to 0}\sup_{\eps\in(0,1)}\|[\varrho_\eps, \sL^a](u_{\eps'}-u)\|_p\leq C\lim_{\eps'\to 0}\|u_{\eps'}-u\|_{0,2;p}=0.\end{aligned}$$ Moreover, for fixed $\eps'\in(0,1)$, since $u_{\eps'}\in H^{\infty,\infty}_p$, by [@Zh22 Lemma 4.2] we have $$\lim_{\eps\to 0}\|[\varrho_\eps, \sL^a]u_{\eps'}\|_p=0,$$ which together with implies .

Let $\sigma_t(x,\v)=\sigma_t$ be independent of $(x,\v)$. Define for $t<s$, $$\begin{aligned}
\label{PP}
P_{t,s}f(x,\v)=\mE f(x+(s-t)\v+X_{t,s}, \v+V_{t,s}),\end{aligned}$$ where $$(X_{t,s},V_{t,s})=\left(\int^s_tV_{t,r}\dif r, \int^s_t\sigma_r\dif W_r\right).$$

We need the following basic $L^p$-regularity estimates related to $P_{t,s}$, which plays a basic role in the next section. \[Th25\] Let $T>0$. Suppose that for some $K>0$ and all $t\in[0,T]$, $$K^{-1}|\xi|\leq |\sigma^*_t\xi|\leq K|\xi|,\ \ \xi\in\mR^d.$$

(i) For any $\alpha,\beta\geq 0$ and $p>1$, there exists a positive constant $C=C(K,T,p,d,\alpha,\beta)$ such that for all $f\in L^p(\mR^{2d})$ and $0\leq t<s\leq T$, $$\begin{aligned}
    \label{Gr}
    \begin{split}
    \|P_{t,s}f\|_{\alpha,0;p}\leq C(s-t)^{-\frac{3\alpha}{2}}\|f\|_p,\\
    \|P_{t,s}f\|_{0,\beta;p}\leq C(s-t)^{-\frac{\beta}{2}}\|f\|_p.
    \end{split}\end{aligned}$$

(ii) For any $p>1$, there exists a positive constant $C_p=C_p(K,d)$ such that for all $\lambda\geq 0$ and $f\in\mL^p(T)=L^p([0,T]\times\mR^{2d})$, $$\begin{aligned}
    \label{Pri22}
    \|\nabla^2_\v u^\lambda\|_{\mL^P(T)}+\|\Delta^{\frac{1}{3}}_x u^\lambda\|_{\mL^P(T)}\leq C_p\|f\|_{\mL^P(T)},\end{aligned}$$ where $u^\lambda_t(x,\v):=\int^T_t\e^{\lambda(t-s)}P_{t,s}f_s(x,\v)\dif s$ satisfies $$\p_t u^\lambda+\sL^{a,0}_tu^\lambda-\lambda u^\lambda+f=0$$ in the distributional sense.

\(i) It follows by the following gradient estimate and the interpolation theorem (see [@Wa-Zh Theorem 2.10]), $$\|\nabla^k_x\nabla^m_\v P_{t,s}f\|_p\leq C(s-t)^{-\frac{3k+m}{2}}\|f\|_p,\ k,m\in\mN_0.$$ (ii) It is a consequence of [@Br-Cu-La-Pr] and [@Bo Theorem 2.1] (see also [@Ch-Zh Theorem 3.3]).

Notice that in the references [@Wa-Zh] and [@Ch-Zh], the positions of $t$ and $s$ are exchanged.

Maximal $L^p$-solutions of kinetic Fokker-Planck equations
==========================================================

Throughout this section, we fix $T>0$. Let $p\in(1,\infty)$ and $\alpha,\beta\geq 0$. For $t\in[0,T]$, we introduce the following Banach spaces with natural norms: $$\mL^p(t,T):=L^p([t,T];L^p(\mR^{2d})),\ \ \mH^{\alpha,\beta}_p(t,T):=L^p([t,T];H^{\alpha,\beta}_p(\mR^{2d})).$$ For simplicity of notation, we write $$\mL^p(T):=\mL^p(0,T),\ \ \mH^{\alpha,\beta}_p(T):=\mH^{\alpha,\beta}_p(0,T).$$

We assume that $a:[0,T]\times\mR^{2d}\to\mR^d\otimes\mR^d$ is symmetric and satisfies that for some $K\geq 1$ and $\delta\in(0,1)$, $$\begin{aligned}
\tag{\bf H$^{\delta,p}_K$}
\left\{
\begin{aligned}
&K^{-1}\cdot\mI\leq a_t(z)\leq K\cdot\mI,\,\,(t,z)\in[0,T]\times\mR^{2d}\\
&\omega_a(\delta):=\sup_{|z-z'|\leq\delta}\sup_{t\in[0,T]}\|a_t(z)-a_t(z')\|\leq\tfrac{1}{2(C_p+1)}
\end{aligned}
\right\},\end{aligned}$$ where $C_p$ is the same as in . Here and in the remainder of this paper, $\|\cdot\|$ denotes the Hilbert-Schmidt norm. For $\lambda\geq 0$, consider the following backward kinetic Fokker-Planck equation $$\begin{aligned}
\label{PDE}
\p_t u+\sL^{a,b}_t u-\lambda u+f=0, \ \ u_T=0,\end{aligned}$$ where $f_t(x,\v):[0,T]\times\mR^{2d}\to\mR$ is a Borel function. We first introduce the following notion of solutions to the above equation. \[Def31\] Let $p\in(1,\infty)$ and $f\in\mL^p(T)$. A Borel function $u\in \mH^{0,2}_p(T)$ is called a solution of if for any $\varphi\in C^\infty_c(\mR^{2d})$ and all $t\in[0,T]$, $$\begin{aligned}
\label{Sol}
\begin{split}
\<u_t,\varphi\>&=\int^T_t\<\tr(a_s\cdot\nabla^2_\v u_s),\varphi\>\dif s-\int^T_t\<\v\cdot\nabla_x\varphi, u_s\>\dif s\\
&\quad+\int^T_t\<b_s\cdot\nabla_\v u_s,\varphi\>\dif s-\lambda\int^T_t\<u_s,\varphi\>\dif s+\int^T_t\<f_s,\varphi\>\dif s,
\end{split}\end{aligned}$$ where $\<u_t,\varphi\>:=\int_{\mR^{2d}}u_t(z)\varphi(z)\dif z$. The main aim of this section is to show that \[Th32\] Let $\alpha\in[0,\frac{2}{3})$, $\beta\in(1,2)$ and $p>\frac{2}{(2-3\alpha)\wedge(2-\beta)}$ be not equal to $\frac{d(\alpha+\beta)}{\alpha(\beta-1)}$. Suppose that $a$ satisfies [**(H$^{\delta,p}_K$)**]{}, and for some $q\in[p\vee \frac{d(\alpha+\beta)}{\alpha(\beta-1)},\infty]$, $$\kappa_0:=\|b\|_{L^p([0,T]; L^q(\mR^{2d}))}<\infty.$$

(i) For any $f\in \mL^p(T)$, there exists a unique solution $u=u^\lambda$ to in the sense of Definition \[Def31\] with $$\begin{aligned}
    \label{Pri0}
    \|u^\lambda\|_{\mH^{2/3,2}_p(T)}\leq C\|f\|_{\mL^p(T)}, \ \end{aligned}$$ and for all $t\in[0,T]$, $$\begin{aligned}
    \label{Pri00}
    \|u^\lambda_t\|_{\alpha,\beta;p}\leq C((T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{(3\alpha)\vee\beta}{2}}\|f\|_{\mL^p(t,T)},\end{aligned}$$ where the constant $C$ only depends on $d,\delta,K,\alpha,\beta,p, q,T$ and $\kappa_0$.

(ii) If in addition, we also assume that $p>\frac{d(3\beta-1)}{2(\beta-1)}$ and $$\kappa_1:=\sup_{t,\v}\|\Delta_x^{\frac{1}{3}}\sigma_t(\cdot,\v)\|_p+\|b\|_{\mH^{2/3,0}_p(T)}<\infty,$$ then for any $f\in \mH^{2/3,0}_p(T)$, the unique solution $u$ also satisfies $$\begin{aligned}
    \label{Pr1}
    \|\nabla_x\nabla_\v u^\lambda\|_{\mL^p(T)}+\|\Delta^{\frac{1}{3}}_x\nabla^2_\v u^\lambda\|_{\mL^p(T)}\leq C\|f\|_{\mH^{2/3,0}_p(T)},\end{aligned}$$ and for all $t\in[0,T]$, $$\begin{aligned}
    \label{Pr2}
    \|\Delta_x^{\frac{1}{3}}  u^\lambda_t\|_{\alpha,\beta;p}\leq C((T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{(3\alpha)\vee\beta}{2}}\|f\|_{\mH^{2/3,0}_p(t,T)},\end{aligned}$$ where the constant $C$ only depends on $d,\delta,K,\alpha,\beta,p, T$ and $\kappa_1$.

In order to emphasize the dependence of the unique solution $u$ on $a,b$ and $T,\lambda, f$, we sometimes denote $u=\cR^{a,b}_{\lambda,T}(f)$.

Case $b=0$
----------

In this subsection we first consider the case of $b=0$ by using the freezing coefficient argument, and show the following basic existence and uniqueness result for equation . \[Th33\] Let $p>1$ and $\alpha\in[0,\frac{2}{3})$, $\beta\in[0,2)$. Suppose [**(H$^{\delta,p}_K$)**]{} holds.

(i) For any $f\in \mL^p(T)$, there exists a unique solution $u=u^\lambda$ to in the sense of Definition \[Def31\] so that $$\begin{aligned}
    \label{Pri}
    \|u^\lambda\|_{\mH^{2/3,2}_p(T)}\leq C_1\|f\|_{\mL^p(T)}.\end{aligned}$$

(ii) If $p>\frac{2}{(2-3\alpha)\wedge(2-\beta)}$, then for all $t\in[0,T]$, $$\begin{aligned}
    \label{Pri2}
    \|u^\lambda_t\|_{\alpha,\beta;p}\leq C_2((T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{(3\alpha)\vee\beta}{2}}\|f\|_{\mL^p(t,T)}.\end{aligned}$$

Here $C_1=C_1(d,\delta,K,p,T)$ and $C_2=C_2(d,\delta,K,\alpha,\beta,p,T)$ are increasing with respect to $T$.

We show the apriori estimates and by the freezing coefficient argument. The existence of a solution follows by the standard continuity argument. We divide the proof into five steps.

[**(a)**]{} First of all, we assume that $u\in C([0,T]; H^{\infty,\infty}_p)$ satisfies for Lebesgue almost all $t\in[0,T]$. For given $p\geq 1$, let $\phi$ be a nonnegative symmetric smooth function on $\mR^{2d}$ with support in the unit ball and $$\int_{\mR^{2d}}|\phi(z)|^p\dif z=1.$$ Let $\delta\in(0,1)$ be as in [**(H$^{\delta,p}_K$)**]{} and set $$\phi_\delta(z):=\delta^{-2d/p}\phi(z/\delta),$$ and for $z^o=(x^o,\v^o)$ and $t\in[0,T]$, define $$z^o_t:=(x^o-t\v^o,\v^o), \  \phi^{z^o_t}_\delta(z):=\phi_\delta(z^o_t-z).$$ By definition, it is easy to see that $$\begin{aligned}
\label{LK1}
\int_{\mR^{2d}}|\phi^{z^o_t}_\delta(z)|^p\dif z^o=1,\ \ t\in[0,T],\ z\in\mR^{2d},\end{aligned}$$ and for $j=1,2$, $$\begin{aligned}
\label{LK2}
\sup_{t\in[0,T]}\sup_{z\in\mR^{2d}}\int_{\mR^{2d}}|\nabla^j_\v\phi^{z^o_t}_\delta(z)|^p\dif z^o\leq C_\delta.\end{aligned}$$ Define the freezing functions at point $z^o=(x^o,\v^o)$ as follows: $$a^{z^o}_t:=a_{t}(z^o_t),\ \ u^{z^o}_{\delta,t}(z):=u_t(z)\phi^{z^o_t}_\delta(z).$$ By and easy calculations, one sees that $$\begin{aligned}
\label{LK88}
\p_t u^{z^o}_\delta+\mathrm{tr}(a^{z^o}_t\cdot\nabla^2_{\v} u^{z^o}_\delta)+\v\cdot \nabla_x u^{z^o}_\delta-\lambda u^{z^o}_\delta=g^{z^o}_{\delta},\end{aligned}$$ where for $z=(x,\v)$, $$\begin{aligned}
g^{z^o}_{\delta,t}(z)&:=\mathrm{tr}(a^{z^o}_t\cdot\nabla^2_{\v} u^{z^o}_{\delta,t})(z)-\mathrm{tr}(a_t\cdot\nabla^2_\v u_t)(z)\phi_\delta^{z^o_t}(z)\\
&\quad+(\v-\v^o)\cdot \nabla_x \phi_\delta^{z^o_t}(z) u_t(z)+f_t(z)\phi_\delta^{z^o_t}(z).\end{aligned}$$ We have the following [*claim:*]{} $$\begin{aligned}
\label{LK3}
\begin{split}
&\left(\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}
\leq \omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}\\
&\quad+C_\delta\Big(\|u\|_{\mL^p(t,T)}+\|f\|_{\mL^p(t,T)}\Big),\ \ t\in[0,T].
\end{split}\end{aligned}$$ [*Proof of the claim:*]{} Observe that $$\begin{aligned}
g^{z^o}_{\delta,t}(z)&=\mathrm{tr}((a^{z^o}_t-a_t)\cdot\nabla^2_\v u_t)(z)\phi_\delta^{z^o_t}(z)
+\mathrm{tr}(a^{z^o}_t\cdot(\nabla_\v u_t\otimes\nabla_\v\phi^{z^o_t}_\delta))(z)\\
&\quad+[\mathrm{tr}(a^{z^o}_t\cdot\nabla^2_\v\phi^{z^o_t}_\delta)(z)+(\v-\v^o)\cdot \nabla_x \phi_\delta^{z^o_t}(z)] u_t(z)
+f_t(z)\phi_\delta^{z^o_t}(z)\\
&=:I^\delta_1(t,z,z^o)+I^\delta_2(t,z,z^o)+I^\delta_3(t,z,z^o)+I^\delta_4(t,z,z^o).\end{aligned}$$ For $I^\delta_1(t,z,z^o)$, since the support of $\phi_\delta$ is in $B_\delta:=\{z\in\mR^{2d}: |z|\leq\delta\}$, by the definition of $\omega_a(\delta)$ and , we have $$\begin{aligned}
\left(\int_{\mR^{2d}}\|I^\delta_1(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}
&\leq \omega_a(\delta)\left(\int_{\mR^{2d}}\|\nabla^2_\v u\cdot\phi^{z^o_\cdot}_\delta\|^p_{\mL^p(t,T)}\dif z^o\right)^{1/p}\\
&=\omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}.\end{aligned}$$ For $I^\delta_2(t,z,z^o)$, by we have $$\left(\int_{\mR^{2d}}\|I^\delta_2(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\leq C_\delta\|\nabla_\v u\|_{\mL^p(t,T)}.$$ For $I^\delta_3(t,z,z^o)$, we similarly have $$\left(\int_{\mR^{2d}}\|I^\delta_3(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\leq C_\delta\|u\|_{\mL^p(t,T)}.$$ For $I^\delta_4(t,z,z^o)$, by we have $$\left(\int_{\mR^{2d}}\|I^\delta_4(\cdot,\cdot,z^o)\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}=\|f\|_{\mL^p(t,T)}.$$ Combining the above calculations, and by the interpolation inequality and Young’s inequality, we get the claim.

Now by , we have $$\begin{aligned}
\label{ES6}
\begin{split}
&\|\nabla^2_\v u\|_{\mL^p(t,T)}
=\left(\int_{\mR^{2d}}\|(\nabla^2_\v u)\phi^{z^o_\cdot}_\delta\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\\
&\quad\leq \left(\int_{\mR^{2d}}\|\nabla^2_\v (u\phi^{z^o_\cdot}_\delta)-(\nabla^2_\v u)\phi^{z^o_\cdot}_\delta\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}\\
&\qquad+\left(\int_{\mR^{2d}}\|\nabla^2_\v u^{z^o}_\delta\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}=:I_1+I_2.
\end{split}\end{aligned}$$ For $I_1$, by and the interpolation inequality, we have $$I_1\leq C_\delta\Big(\|\nabla_\v u\|_{\mL^p(t,T)}+\|u\|_{\mL^p(t,T)}\Big)\leq \omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}+C_{\delta}\|u\|_{\mL^p(t,T)}.$$ For $I_2$, noticing that by and Duhamel’s formula, $$\begin{aligned}
\label{EQ1}
u^{z^o}_{\delta,t}(z)=\int^T_t\e^{\lambda(t-s)} P^{z^o}_{t,s}g^{z^o}_{\delta,s}(z)\dif s,\end{aligned}$$ where $P^{z^o}_{t,s}$ is defined by in terms of $\sigma^{z^o}_t:=(a^{z^o}_t)^{1/2}$, by and , we have $$\begin{aligned}
\begin{split}
I_2\leq C_0&\left(\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\right)^{1/p}
\leq C_0\omega_a(\delta)\|\nabla^2_\v u\|_{\mL^p(t,T)}\\
&\quad+C_\delta\Big(\|u\|_{\mL^p(t,T)}+C\|f\|_{\mL^p(t,T)}\Big).
\end{split}\end{aligned}$$ Substituting these two estimates into and by $\omega_a(\delta)\leq\frac{1}{2(C_0+1)}$, we get $$\begin{aligned}
\label{LK9}
\|\nabla^2_\v u\|_{\mL^p(t,T)}\leq C\Big(\|u\|_{\mL^p(t,T)}+\|f\|_{\mL^p(t,T)}\Big).\end{aligned}$$ Similarly, one can show that (see also step (c) below) $$\begin{aligned}
\label{LK8}
\|\Delta_x^{\frac{1}{3}}u\|_{\mL^p(t,T)}\leq C\Big(\|u\|_{\mL^p(t,T)}+\|f\|_{\mL^p(t,T)}\Big).\end{aligned}$$

[**(b)**]{} By and the contraction of operator $P^{z^o}_{t,s}$ in $L^p(\mR^{2d})$, we have $$\begin{aligned}
\|u_t\|_p^p&\stackrel{\eqref{LK1}}{=}\int_{\mR^{2d}}\|u^{z^o}_{\delta,t}\|^p_p\dif z^o\leq\int_{\mR^{2d}}\left(\int^T_t \|g^{z^o}_{\delta,s}\|_{p}\dif s\right)^p\dif z^o\\
&\leq \left(\int^T_t\e^{p\lambda(t-s)/(p-1)}\dif s\right)^{p-1}\int_{\mR^{2d}}\int^T_t\|g^{z^o}_{\delta,s}\|^p_{p}\dif s\dif z^o\\
&\stackrel{\eqref{LK3}}{\preceq} ((T-t)\wedge\lambda^{-1})^{p-1}\Big(\|\nabla^2_\v u\|_{\mL^p(t,T)}^p+\|u\|_{\mL^p(t,T)}^p+\|f\|_{\mL^p(t,T)}^p\Big)\\
&\stackrel{\eqref{LK9}}{\preceq}((T-t)\wedge\lambda^{-1})^{p-1}\left(\int^T_t\|u_s\|_{p}^p\dif s+\|f\|_{\mL^p(t,T)}^p\right),\end{aligned}$$ which yields by Gronwall’s inequality that $$\begin{aligned}
\label{ES77}
\|u_t\|^p_p\leq C((T-t)\wedge\lambda^{-1})^{p-1}\|f\|_{\mL^p(t,T)}^p.\end{aligned}$$ Substituting it into and , we obtain , and also by , $$\begin{aligned}
\label{LP2}
\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\leq C\|f\|_{\mL^p(t,T)}^p,\ \ t\in[0,T].\end{aligned}$$

[**(c)**]{} Let $\alpha\in(0,\frac{2}{3})$. By , and Hölder’s inequality, we have $$\begin{aligned}
\|u^{z^o}_{\delta,t}\|_{\alpha,0; p}&\leq \int^T_t\e^{\lambda(t-s)} \|P^{z^o}_{t,s}g^{z^o}_{\delta,s}\|_{\alpha,0;p}\dif s
\preceq\int^T_t\e^{\lambda(t-s)}(s-t)^{-\frac{3\alpha}{2}}\|g^{z^o}_{\delta,s}\|_{p}\dif s\no\\
&\preceq \left(\int^T_t\e^{\lambda(t-s)} (s-t)^{-\frac{3p\alpha}{2(p-1)}}\dif s\right)^{1-\frac{1}{p}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}\no\\
&\preceq( (T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{3\alpha}{2}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}.\label{LP1}\end{aligned}$$ By again, we have $$\begin{aligned}
\label{LP4}
\begin{split}
\|\Delta_x^{\frac{\alpha}{2}}u_t\|^p_{p}&=\int_{\mR^{2d}}\|(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta\|^p_{p}\dif z^o 
\preceq\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}} u^{z^o}_{\delta,t}\|^p_{p}\dif z^o \\
&\qquad+\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}} (u_t\phi^{z^o_t}_\delta)-(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta\|^p_{p}\dif z^o.
\end{split}\end{aligned}$$ By and , we have $$\begin{aligned}
\begin{split}
\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}}u^{z^o}_{\delta,t}\|^p_{p}\dif z^o 
&\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{3p\alpha}{2}}\int_{\mR^{2d}}\|g^{z^o}_{\delta}\|_{\mL^p(t,T)}^p\dif z^o\\
&\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{3p\alpha}{2}}\|f\|^p_{\mL^p(t,T)}.
\end{split}\label{LP5}\end{aligned}$$ On the other hand, noticing that by definition , $$\Delta_x^{\frac{\alpha}{2}} (u_t\phi^{z^o_t}_\delta)-(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta
=u_t\cdot\Delta_x^{\frac{\alpha}{2}}\phi^{z^o_t}_\delta+\int_{\mR^d}\delta_{(y,0)}u_t\cdot\delta_{(y,0)}\phi^{z^o_t}_\delta|y|^{-d-\alpha}\dif y,$$ and $$\sup_z\int_{\mR^{2d}}|\Delta_x^{\frac{\alpha}{2}}\phi_\delta(z^o_t-z)|^p\dif z^o\leq C_\delta,$$ $$\sup_z\left(\int_{\mR^{2d}}|\delta_{(y,0)}\phi^{z^o_t}_\delta(z)|^p\dif z^o\right)^{1/p}\leq C_\delta(|y|\wedge 1),$$ by Minkovskii’s inequality, we have $$\begin{aligned}
\label{LP6}
\begin{split}
&\int_{\mR^{2d}}\|\Delta_x^{\frac{\alpha}{2}} (u_t\phi^{z^o_t}_\delta)-(\Delta_x^{\frac{\alpha}{2}} u_t)\phi^{z^o_t}_\delta\|_p^p\dif z^o
\preceq\int_{\mR^{2d}}\|u_t\cdot\Delta_x^{\frac{\alpha}{2}}\phi^{z^o_t}_\delta\|_p^p\dif z^o\\
&\qquad+\int_{\mR^{2d}}\left(\int_{\mR^d}|\delta_{(y,0)}u_t(z)|\cdot\left(\int_{\mR^{2d}}|\delta_{(y,0)}\phi^{z^o_t}_\delta(z)|^p\dif z^o\right)^{1/p}|y|^{-d-\alpha}\dif y\right)^p\dif z\\
&\qquad\preceq \|u_t\|^p_p+\int_{\mR^{2d}}\left(\int_{\mR^d}|\delta_{(y,0)}u_t(z)|\cdot(|y|\wedge 1)|y|^{-d-\alpha}\dif y\right)^p\dif z\\
&\qquad\preceq \|u_t\|^p_p+\left(\int_{\mR^d}\|\delta_{(y,0)}u_t\|_p\cdot(|y|\wedge 1)|y|^{-d-\alpha}\dif y\right)^p\preceq\|u_t\|^p_p.
\end{split}\end{aligned}$$ Combining , and with , we arrive at $$\|u_t\|^p_{\alpha,0; p}\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{3p\alpha}{2}}\|f\|^p_{\mL^p(t,T)}.$$ Similarly, for any $\beta\in(0,2)$, one can show that $$\|u_t\|^p_{0,\beta; p}\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{p\beta}{2}}\|f\|^p_{\mL^p(t,T)}.$$ Combining the above two estimates, we obtain .

Below we assume that $u\in\mH^{0,2}_p(T)$ is a solution of in the sense of Definition \[Def31\]. Let $\varrho_\eps$ be defined by and set $$u_\eps:=u*\varrho_\eps,\ \ f_\eps:=f*\varrho_\eps.$$ Taking $\varphi=\varrho_\eps(z-\cdot)$ in , we obtain $$\p_t u_\eps+\sL^a_t u_\eps-\lambda u_\eps+[\varrho_\eps,\sL^a_t]u_t+f_\eps=0,\ \ u_{\eps,T}=0,$$ where $\sL^a_t:=\sL^{a,0}_t$ and $[\varrho_\eps,\sL^a_t]$ is defined by . By what we have proved, it holds that $$\|u_\eps\|_{\mH^{2/3,2}_p(T)}\leq C_1(\|f_\eps\|_{\mL^p(T)}+\|[\varrho_\eps,\sL^{a}]\|_{\mL^p(T)}).$$ Since $\nabla^2_\v u\in\mL^p(T)$, by the property of convolutions and (ii) of Lemma \[Le24\], we get by taking limits. Similarly, we also have .

Finally, we use the standard continuity argument to show the existence of a solution (see [@Kr0]). Consider the following parametrized equation: $$\begin{aligned}
\label{PPE}
\p_t u+\sL^{a_\tau}_t u-\lambda u+f=0, \ \ u_T=0,\end{aligned}$$ where $\tau\in[0,1]$ and $a_\tau:=K(1-\tau)\mI+\tau a$. Since $K^{-1}\cdot\mI\leq a\leq K\cdot\mI$, we obviously have $$K^{-1}\cdot\mI\leq a_\tau\leq K\cdot \mI,\ \ \omega_{a_\tau}(\delta)=\omega_a(\delta).$$ Hence the apriori estimate holds for with constant $C_1$ independent of $\tau\in[0,1]$. Suppose that is solvable for some $\tau_0\in[0,1)$. We want to show that is also solvable for any $\tau\in[\tau_0,\tau_0+\frac{1}{4KC_1})$, where $C_1$ is the constant in . Let $u^0=0$ and for $n\in\mN$, define $u^n$ recursively by $$\p_t u^{n}+\sL^{a_{\tau_0}}_t u^{n}-\lambda u^n+\tr((a_\tau-a_{\tau_0})\cdot\nabla^2_\v u^{n-1})+f=0, \ \ u^{n}_T=0.$$ By the apriori estimate , we have $$\begin{aligned}
\|u^n\|_{\mH^{2/3,2}_p(T)}&\leq C_1\|\tr((a_\tau-a_{\tau_0})\cdot\nabla^2_\v u^{n-1})+f\|_{\mL^p(T)}\\
&\leq 2KC_1(\tau-\tau_0)\|u^{n-1}\|_{\mH^{2/3,2}_p(T)}+C_1\|f\|_{\mL^p(T)}\\
&\leq \tfrac{1}{2}\|u^{n-1}\|_{\mH^{2/3,2}_p(T)}+C_1\|f\|_{\mL^p(T)},\end{aligned}$$ and similarly, $$\|u^n-u^m\|_{\mH^{2/3,2}_p(T)}\leq \tfrac{1}{2}\|u^{n-1}-u^{m-1}\|_{\mH^{2/3,2}_p(T)},$$ which imply that $u^n$ is a Cauchy sequence in $\mH^{2/3,2}_p(T)$. It is easy to see that the limit $u$ of $u^n$ satisfies . Since is solvable for $\tau=0$ by (ii) of Theorem \[Th25\], by repeatedly using what we have proved finitely many times, we get the solvability of for $\tau=1$.

Next we show further regularity of the solution under extra assumption. Suppose that for some $p>3d/2$, [**(H$^{\delta,p}_K$)**]{} holds and $$\kappa_1:=\sup_{(t,\v)\in[0,T]\times\mR^d}\|\Delta_x^{\frac{1}{3}}\sigma_t(\cdot,\v)\|_p<\infty.$$

(i) For any $f\in \mH^{2/3,0}_p(T)$, the unique solution $u$ of PDE also satisfies $$\begin{aligned}
    \label{E35}
    \|\nabla_x\nabla_\v u\|_{\mL^p(T)}+\|\Delta^{\frac{1}{3}}_x\nabla^2_\v u\|_{\mL^p(T)}\leq C_3\|f\|_{\mH^{2/3,0}_p(T)}.\end{aligned}$$

(ii) Let $\alpha\in[0,\frac{2}{3})$ and $\beta\in[0,2)$. If $p>\frac{2}{(2-3\alpha)(2-\beta)}\vee\frac{3d}{2}$, then $$\begin{aligned}
    \label{E36}
    \|\Delta_x^{\frac{1}{3}} u_t\|_{\alpha,\beta;p}\leq C_4((T-t)\wedge\lambda^{-1})^{1-\frac{1}{p}-\frac{(3\alpha)\vee\beta}{2}}\|f\|_{\mH^{2/3,0}_p(T)}.\end{aligned}$$

Here $C_3=C_3(d,\delta,K,\kappa_1, p,T)$ and $C_4=C_4(d,\delta,K,\kappa_1, \alpha,\beta,p,T)$ are increasing with respect to $T$.

As in the proof of Theorem \[Th33\], it suffices to show the apriori estimates and . Notice that using with $\alpha=\frac{2}{3}$, $\beta=2$ and by , $$\begin{aligned}
\label{ES00}
\|\Delta_x^{\frac{1}{6}}\nabla_\v u\|_{\mL^p(t,T)}\preceq\|u\|_{\mH^{2/3,2}_p(t,T)}\preceq\|f\|_{\mL^p(t,T)}.\end{aligned}$$ Assume $u\in C([0,T];H^{\infty,\infty}_p)$ and let $w_t(x,\v):=\Delta_x^{\frac{1}{3}}u_t(x,\v)$. By we have $$\p_t w+\sL^a_t w+[\Delta_x^{\frac{1}{3}},\sL^a_t]u+\Delta_x^{\frac{1}{3}}f=0, w_T=0.$$ By definition and the assumptions, it is easy to see that $$\sup_{(t,\v)\in[0,T]\times\mR^d}\|\Delta_x^{\frac{1}{3}}a_t(\cdot,\v)\|_p<\infty.$$ Hence, by , and , we have for any $\eps>0$, $$\begin{aligned}
&\|\nabla_x\nabla_\v u\|_{\mL^p(T)}+\|\Delta_x^{\frac{2}{3}}u\|_{\mL^p(T)}+\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}\\
&\quad\stackrel{\eqref{Riesz}}{\preceq}\|\Delta_x^{\frac{1}{6}}\nabla_\v w\|_{\mL^p(T)}+\|\Delta_x^{\frac{1}{3}} w\|_{\mL^p(T)}+\|\nabla^2_\v w\|_{\mL^p(T)}\\
&\quad\preceq \|[\Delta_x^{\frac{1}{3}},\sL^a_t]u+\Delta_x^{\frac{1}{3}}f\|_{\mL^p(T)}\\
&\quad\leq\|[\Delta_x^{\frac{1}{3}},\sL^a_t]u\|_{\mL^p(T)}+\|f\|_{\mH^{2/3,0}_p(T)}\\
&\quad\preceq\eps\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}+C_\eps\big(\|\nabla^2_\v u\|_{\mL^p(T)}+\|f\|_{\mH^{2/3,0}_p(T)}\big)\\
&\quad\preceq\eps\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}+C_\eps\|f\|_{\mH^{2/3,0}_p(T)},\end{aligned}$$ which implies the desired estimate by letting $\eps$ be small enough. As for , it follows by applying to $w$ and using the above estimate.

Proof of Theorem \[Th32\]
-------------------------

By a standard fixed point argument or Picard’s iteration, it suffices to prove the apriori estimate . Let $\alpha\in(0,\frac{2}{3})$, $\beta\in(1,2)$ and $p>\frac{2}{(2-3\alpha)\wedge(2-\beta)}$ be not equal to $\frac{d(\alpha+\beta)}{\alpha(\beta-1)}$.

\(i) Let $q\in[p\vee \frac{d(\alpha+\beta)}{\alpha(\beta-1)},\infty]$ and $r\in[p,\infty]$ with $\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$. Notice that by Hölder’s inequality and , and using $(\alpha(\beta-1)/\beta, \beta-1)$ in place of $(\alpha,\beta)$ in , $$\|b\cdot\nabla_\v u\|^p_{\mL^p(t,T)}\leq \int^T_t\|b_s\|^p_{q}\|\nabla_\v u_s\|^p_{r}\dif s\preceq \int^T_t\|b_s\|^p_{q}\|u_s\|^p_{\alpha,\beta; p}\dif s.$$ Thus, by we have $$\begin{aligned}
\|u_t\|_{\alpha,\beta;p}^p&\preceq ((T-t)\wedge\lambda^{-1})^{p-1-\frac{p((3\alpha)\vee\beta)}{2}}\|b\cdot\nabla_\v u+f\|^p_{\mL^p(t,T)}\\
&\preceq((T-t)\wedge\lambda^{-1})^{p-1-\frac{p((3\alpha)\vee\beta)}{2}}\|f\|^p_{\mL^p(t,T)}+\int^T_t\|b_s\|^p_{q}\|u_s\|^p_{\alpha,\beta;p}\dif s,\end{aligned}$$ which implies by Gronwall’s inequality.

On the other hand, by we have $$\|u\|_{\mH^{2/3,0}_p(T)}^p\preceq \|b\cdot\nabla_\v u+f\|^p_{\mL^p(T)}\preceq
\int^T_0\|b_s\|^p_{q}\|u_s\|^p_{\alpha,\beta;p}\dif s+\|f\|^p_{\mL^p(T)},$$ which in turn implies by .

\(ii) Let $p>\frac{d(3\beta-1)}{2(\beta-1)}$. By Lemma \[Le23\] with $\alpha=\frac{2}{3}$, we have $$\begin{aligned}
\|b\cdot\nabla_\v u\|^p_{\mH^{2/3,0}_p(t,T)}
&\preceq \int^T_t\|b_s\|^p_{2/3,0;p}\Big(\|\Delta_x^{\frac{1}{3}}\nabla_\v u_s\|_{0,\beta-1; p}^p+\|\nabla_\v u_s\|_{0,\beta-1;p}^p\Big)\dif s\\
&\preceq \int^T_t\|b_s\|^p_{2/3,0;p}\Big(\|\Delta_x^{\frac{1}{3}}u_s\|_{0,\beta; p}^p+\|u_s\|_{0,\beta;p}^p\Big)\dif s.\end{aligned}$$ Thus, by and , we have $$\begin{aligned}
\|\Delta_x^{\frac{1}{3}} u_t\|^p_{\alpha,\beta;p}&\preceq ((T-t)\wedge\lambda^{-1})^{p-1-\frac{p((3\alpha)\vee\beta)}{2}}\|b\cdot\nabla_\v u+f\|^p_{\mH^{2/3,0}_p(t,T)}\\
&\preceq ((T-t)\wedge\lambda^{-1})^{p-1-\frac{p((3\alpha)\vee\beta)}{2}}\|f\|^p_{\mH^{2/3,0}_p(t,T)}\\
&\quad+\int^T_t\|b_s\|^p_{2/3,0;p}\|\Delta_x^{\frac{1}{3}}u_s\|_{0,\beta; p}^p\dif s,\end{aligned}$$ which yields by Gronwall’s inequality.

Moreover, by we have $$\begin{aligned}
\|\nabla_x\nabla_\v u\|_{\mL^p(T)}+\|\Delta_x^{\frac{1}{3}}\nabla^2_\v u\|_{\mL^p(T)}\leq 
C\|b\cdot\nabla_\v u+f\|_{\mH^{2/3,0}_p(T)}\leq C\|f\|_{\mH^{2/3,0}_p(T)},\end{aligned}$$ which gives . The proof is complete.

Well-posedness of martingale problem
====================================

Let $\Omega=C(\mR_+;\mR^{2d})$ be the space of all continuous functions from $\mR_+$ to $\mR^{2d}$, which is endowed with the locally uniform convergence topology. Let $$Z_t(\omega):=\omega_t$$ be the coordinate process on $\Omega$. For $t\geq 0$, let $$\sF_t:=\sigma\Big\{Z_s: s\in[0,t]\Big\}, \ \ \sF:=\vee_{t\geq 0}\sF_t.$$ We first recall the following notions of martingale solution and weak solution (see [@St-Va]). Let $\sigma:\mR_+\times\mR^{2d}\to\mR^d\otimes\mR^d$ and $b:\mR_+\times\mR^{2d}\to\mR^d$ be Borel measurable functions and $\sL^{a,b}_t$ be defined by with $a=\frac{1}{2}\sigma\sigma^*$.

(i) (Martingale solution) For given $(r,z)\in\mR_+\times\mR^{2d}$, a probability measure $\mP=\mP_{r,z}$ on $(\Omega,\sF)$ is said a solution to the martingale problem for $\sL^{a,b}_t$ starting from $(r,z)$ if $$\mP\left(Z_t=z,t\in[0,r]\ \mathrm{ and } \int^t_r(\|a_s(Z_s)\|+|b_s(Z_s)|)\dif s<\infty,t\geq r\right)=1,$$ and for all $\varphi\in C^\infty_c(\mR^d)$, $$\begin{aligned}
    \label{MM4}
    [r,\infty)\ni t\mapsto \varphi(Z_t)-\int^t_r\sL^{a,b}_s \varphi(Z_s)\dif s=:M^{r,\varphi}_t\end{aligned}$$ is an $\sF_t$-martingale with respect to $\mP$ after time $r$. We denote by $\sP^{\sigma,b}_{r,z}$ the set of all martingale solutions associated with $(\sigma,b)$ and starting from $(r,z)$.

(ii) (Well-posedness) One says that the martingale problem for $\sL^{a,b}_t$ is well-posed if for each $(r,z)\in\mR_+\times\mR^{2d}$, there is exactly one solution $\mP_{r,z}$ to the martingale problem for $\sL^{a,b}_t$ starting from $(r,z)$.

(iii) (Weak solution) A triple $((\tilde Z,\tilde W); (\tilde \Omega,\tilde \sF,\tilde \mP);(\tilde \sF_t)_{t\geq 0})$ is called a weak solution of SDE with starting point $(r,z)\in\mR_+\times\mR^{2d}$ if $(\tilde \sF_t)_{t\geq 0}$ satisfies the usual conditions, $\tilde W$ is a $d$-dimensional $\tilde \sF_t$-Brownian motion, and $\tilde Z=(\tilde X,\dot{\tilde  X})$ is an $\tilde \sF_t$-adapted $\mR^{2d}$-valued process satisfying that $$\int^t_r(\|a_s(\tilde Z_s)\|+|b_s(\tilde Z_s)|)\dif s<\infty, \ t\geq r,\ \  \tilde\mP-a.s.,$$ and $$\tilde Z_t=z+\int^t_r(\dot{\tilde  X}_s,b_s(\tilde Z_s))\dif s+\int^t_r(0,\sigma_s(\tilde Z_s)\dif\tilde  W_s).$$

It is well known that the martingale solutions and weak solutions are equivalent, for example, see [@Ka-Sh p.318, Proposition 4.11].

Krylov’s type estimate
----------------------

In this subsection we first show the following important estimate of Krylov’s type for weak solutions. \[Th41\] Suppose that $\sigma$ satisfies [**(UE)**]{} and $$\begin{aligned}
\label{EJ1}
\lim_{|z-z'|\to 0}\sup_{t\in[0,T]}\|\sigma_t(z)-\sigma_t(z')\|=0,\  T>0,\end{aligned}$$ and $b\in \cap_T\mL^q(T)$ for some $q>(2(2d+1),\infty]$. Then for any $p>2d+1$ and $T>0$, there is a constant $C>0$ depending on $T,\sigma,p,q,d, \|b\|_{\mL^q(T)}$ such that for any $(r,z)\in[0,T)\times\mR^{2d}$, any weak solution $((\tilde Z,\tilde W); (\tilde \Omega,\tilde \sF,\tilde \mP);(\tilde \sF_t)_{t\geq 0})$ of SDE with starting point $(r,z)$, and $r\leq t_0<t_1\leq T$ and $f\in\mL^p(T)$, $$\begin{aligned}
\label{BB9}
\tilde\mE\left(\int^{t_1}_{t_0} f_s(\tilde Z_s)\dif s\Big|\tilde \sF_{t_0}\right)
\leq C(t_1-t_0)^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)},\end{aligned}$$ where $C>0$ is increasing in $T,\omega_\sigma(\delta)$ and $\|b\|_{\mL^q(T)}$.

Without loss of generality, we assume $(r,z)=(0,0)$ and drop the tilde in the definition of weak solutions for simplicity. We divide the proof into four steps.

[**(a)**]{} Let $a_t(z):=\frac{1}{2}(\sigma_t\sigma^*_t)(z)$ and $p\in(2(2d+1),q]$. For any $0\leq t_0<t_1\leq T$, $\lambda\geq 1$ and $f\in\mL^p(t_0,t_1)$, let $u=\cR^{a,b}_{\lambda,t_1}(f)$ be the solution of PDE with terminal time $T=t_1$. By and with $\alpha=\frac{4d}{3(2d+1)}$ and $\beta=\frac{4d}{2d+1}$, $$\begin{aligned}
\label{44}
\begin{split}
&\|\cR^{a,b}_{\lambda,t_1}(f)\|_{\mL^\infty(t_0,t_1)}
\preceq \big\|\cR^{a,b}_{\lambda,t_1}(f)\big\|_{L^\infty([t_0,t_1];H^{\alpha,\beta}_p)}\\
&\quad\preceq ((t_1-t_0)\wedge\lambda^{-1})^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)},
\end{split}\end{aligned}$$ and by , , with $\alpha=\frac{4d+1}{3(2d+1)}$ and $\beta=\frac{4d+1}{2d+1}$, $$\begin{aligned}
\label{404}
\begin{split}
&\|\cR^{a,0}_{\lambda,t_1}(f)\|_{\mL^\infty(t_0,t_1)}\preceq 
\big\|\cR^{a,0}_{\lambda,t_1}(f)\big\|_{L^\infty([t_0,t_1];H^{\alpha,\beta}_p)}\\
&\quad\preceq ((t_1-t_0)\wedge\lambda^{-1})^{\frac{1}{2(2d+1)}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)},
\end{split}\end{aligned}$$ where the $C$ in the above $\preceq$ only depend on $d,p,q,K,T,\omega_a(\delta)$ and $\|b\|_{\mL^q(T)}$.

Now, for any $R>0$, define a stopping time $$\begin{aligned}
\label{TR}
\tau_R:=\Bigg\{t\geq 0: \int^t_0 |b_s(Z_s)|\dif s>R\Bigg\}.\end{aligned}$$ Let $\varrho_\eps$ be as in . We introduce a $d\times d$ matrix-valued function, which is crucial for us below. For $t\geq 0$ and $z\in\mR^{2d}$, let $$a^{\eps,R}_{t}(z):=
\left\{
\begin{array}{ll}
\tfrac{\mE\big(a_t(Z_t)\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}{\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)},\ 
& \mathrm{ if }\ \ \mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)\not=0,\\
a_t(z),\ & \mathrm{ if }\ \ \mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)=0.
\end{array}
\right.$$ By [**(UE)**]{} and the definitions, we have $$K^{-2}\cdot\mI\leq a^{\eps,R}_{t}(z)\leq K^2\cdot\mI,$$ and for all $|z-z'|\leq\eps$, $$\begin{aligned}
\|a^{\eps,R}_{t}(z)-a^{\eps,R}_{t}(z')\|&\leq\|a_{t}(z)-a_{t}(z')\|+\left\|\tfrac{\mE\big((a_t(Z_t)-a_t(z))\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}
{\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}\right\|\\
&+\left\|\tfrac{\mE\big((a_t(Z_t)-a_t(z'))\varrho_\eps(Z_t-z')1_{t<\tau_R}\big)}{\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}\right\|\leq 3\omega_a(\eps).\end{aligned}$$ Let $C_0$ be the same as in . By , one may choose $\delta_0$ small enough such that for all $\eps\in(0,\delta_0)$ and $R>0$, $$\begin{aligned}
\label{App}
\omega_{a^{\eps,R}}(\eps)\leq 3\omega_a(\delta_0)\leq\tfrac{1}{2(C_0+1)}.\end{aligned}$$

[**(b)**]{} In this step we show that for any $p>2(2d+1)$ and $f\in\mL^p(T)$, $$\begin{aligned}
\label{BB90}
\mE\left(\int^{T}_0 f_s(Z_s)\dif s\right)\leq C\|f\|_{\mL^p(T)}.\end{aligned}$$ By a standard density argument, we may assume $f\in C_c([0,T]\times\mR^{2d})$. Let $$u^{\eps,R}:=\cR^{a^{\eps,R},0}_{\lambda, T}(f),\ \ u^{\eps,R}_{\eps}:=u^{\eps,R}*\varrho_\eps,\ \ f_\eps:=f*\varrho_\eps.$$ By Itô’s formula, we have $$\begin{aligned}
\label{EY1}
\begin{split}
&\mE\Big(u^{\eps,R}_{\eps,T\wedge\tau_R}(Z_{T\wedge\tau_R})-u_{\eps,0}(Z_0)\Big)
=\mE\left(\int^{T\wedge\tau_R}_0\!\!\! (\p_s u^{\eps,R}_{\eps,s}+\sL^{a,b}_su^{\eps,R}_{\eps,s})(Z_s)\dif s\right).
\end{split}\end{aligned}$$ Noticing that by Definition \[Def31\], $$\p_su^{\eps,R}_{\eps, s}+(\sL^{a^{\eps,R},0}_su^{\eps,R}_s)*\varrho_\eps-\lambda u^{\eps,R}_{\eps, s}+f_\eps=0,$$ and by the definitions of $u^{\eps,R}_\eps$ and $a^{\eps,R}$, $$\begin{aligned}
&\mE\big(\tr(a_s\cdot\nabla^2_\v u^{\eps,R}_{\eps,t})(Z_s)1_{s<\tau_R}\big)\\
&=\int_{\mR^{2d}}\mE\big(\tr(a_s(Z_s)\cdot\nabla^2_\v u^{\eps,R}_s(z))\varrho_\eps(Z_s-z)1_{s<\tau_R}\big)\dif z\\
&=\int_{\mR^{2d}}\tr(a^{\eps,R}_s(z)\cdot\nabla^2_\v u^{\eps,R}_s(z))\mE\big(\varrho_\eps(Z_s-z)1_{s<\tau_R}\big)\dif z\\
&=\mE\Big((\tr(a^{\eps,R}_s\cdot\nabla^2_\v u^{\eps,R}_s)*\varrho_\eps)(Z_s)1_{s<\tau_R}\Big),\end{aligned}$$ by easy calculations, one sees that $$\begin{aligned}
&\mE\Big((\p_s u^{\eps,R}_{\eps,t}+\sL^{a,b}_su^{\eps,R}_{\eps,t})(Z_s)1_{s<\tau_R}\Big)\\
&=\int_{\mR^{2d}}\!\!\!\mE\Big((\dot X_s-\v)\cdot\nabla_x\varrho_\eps(Z_s-z)1_{s<\tau_R}\Big)u^{\eps,R}_s(z)\dif z\\
&\quad+\mE\Big((b_s\cdot\nabla_\v u^{\eps,R}_{\eps,t}+\lambda u^{\eps,R}_s-f_\eps)(Z_s)1_{s<\tau_R}\Big)\\
&\leq\|u^{\eps,R}_s\|_\infty\left(\lambda+\int_{\mR^{2d}}|\v|~|\nabla_x\varrho_\eps|(x,\v)\dif x\dif\v\right)\\
&\quad+\|\nabla_\v u^{\eps,R}_s\|_\infty\mE\big(|b_s|(Z_s)1_{s<\tau_R}\big)
-\mE\big( f_\eps(Z_s)1_{s<\tau_R}\big).\end{aligned}$$ Substituting this into , we obtain $$\begin{aligned}
&\mE\left(\int^{T\wedge\tau_R}_0f_\eps(Z_s)\dif s\right)\leq 
(\|\nabla_x\varrho\|_1+2+\lambda)\|u^{\eps,R}\|_{\mL^\infty(T)}\\
&\qquad+\|\nabla_\v u^{\eps,R}\|_{\mL^\infty(T)}\mE\left(\int^{T\wedge\tau_R}_0|b_s|(Z_s)\dif s\right).\end{aligned}$$ By , and with $b\equiv0$, there is a $C=C(d,p,K,T,\omega_a(\delta_0))>0$ such that for all $\eps\in(0,\delta_0)$ and $\lambda>1$, $$\begin{aligned}
&\mE\left(\int^{T\wedge\tau_R}_0f_{\eps,s}(Z_s)\dif s\right)\leq 
C(1+\lambda)T^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(T)}\\
&\qquad+C\lambda^{\frac{1}{p}-\frac{1}{2(2d+1)}}\|f\|_{\mL^p(T)}\mE\left(\int^{T\wedge\tau_R}_0|b_s|(Z_s)\dif s\right),\end{aligned}$$ which implies, by letting $\eps\to 0$ and $\lambda$ large enough, that for any $\delta>0$, there is a $C_\delta>0$ such that for all $f\in\mL^p(T)$, $$\begin{aligned}
\label{BB090}
\mE\left(\int^{T\wedge\tau_R}_0 f_s(Z_s)\dif s\right)\leq \left(C_\delta+\delta\mE\left(\int^{T\wedge\tau_R}_0|b_s|(Z_s)\dif s\right)\right)\|f\|_{\mL^p(T)}.\end{aligned}$$ In particular, choosing $f_s=|b_s|$ and $\delta\leq1/(2\|b\|_{\mL^q(T)})$, we get $$\mE\left(\int^{T\wedge\tau_R}_0|b_s|(Z_s)\dif s\right)\leq C\|b\|_{\mL^q(T)}.$$ Substituting this into and letting $R\to\infty$, we get . [**(c)**]{} In this step we show that holds for $p=q>2(2d+1)$. Let $0\leq t_0<t_1\leq T$ and $f\in \mL^q(t_0,t_1)$, and write $$u:=-\cR^{a,b}_{0,t_1}(f)\in\mH^{0,2}_q(t_1),\ \ u_\eps:=u*\varrho_\eps.$$ Noticing that by definitions, $$\begin{aligned}
\p_t u_{\eps,t}+\sL^{a,b}_tu_{\eps,t}=f_{\eps,t}+[\varrho_\eps,\sL^{a,b}_t]u_t,\ \ u_{\eps,t_1}=0,\end{aligned}$$ by Itô’s formula, we have $$\begin{aligned}
\mE^{\sF_{t_0}}\Big(u_{\eps,t_1}(Z_{t_1})-u_{\eps,t_0}(Z_{t_0})\Big)
=\mE^{\sF_{t_0}}\!\left(\int^{t_1}_{t_0} \Big(f_{\eps,s}+[\varrho_\eps,\sL^{a,b}_s]u_s\Big)(Z_s)\dif s\right),\label{BB2}\end{aligned}$$ where $\mE^{\sF_{t_0}}(\cdot)=\mE(\cdot|^{\sF_{t_0}})$. Since by , and Lemma \[Le24\], $$\lim_{\eps\to 0}\mE\left(\int^{t_1}_{t_0}|[\varrho_\eps,\sL^{a,b}_s]u_s|(Z_s)\dif s\right)\preceq
\lim_{\eps\to 0}\|[\varrho_\eps,\sL^{a,b}_\cdot]u\|_{\mL^q(t_0,t_1)}=0$$ and $$\lim_{\eps\to 0}\mE\left(\int^{t_1}_{t_0}|f_{\eps,s}-f_s|(Z_s)\dif s\right)\preceq
\lim_{\eps\to 0}\|f_\eps-f\|_{\mL^q(t_0,t_1)}=0,$$ taking limits $\eps\to 0$ for both sides of (\[BB2\]) and by , we obtain $$\begin{aligned}
\label{KJ0}
\mE^{\sF_{t_0}}\left(\int^{t_1}_{t_0} f_s(Z_s)\dif s\right)
\leq 2\|u\|_{\mL^\infty(t_0,t_1)}\leq C(t_1-t_0)^{\frac{1}{2d+1}-\frac{1}{q}}\|f\|_{\mL^q(t_0,t_1)}.\end{aligned}$$

[**(d)**]{} Let $0\leq t_0<t_1\leq T$. By with $f_s=|b_s|$ and Corollary \[Cor1\] below, we have for any $\lambda>0$, $$\begin{aligned}
\label{KJ1}
\mE^{\sF_{t_0}}\exp\left(\lambda\int^{t_1}_{t_0}|b_s(Z_s)|\dif s\right)\leq C(\lambda,\|b\|_{\mL^q(T)})<\infty.\end{aligned}$$ Define $$\cE_{t_0,t_1}:=\exp\left\{\int^{t_1}_{t_0}(\sigma^{-1}_sb_s)(Z_s)\dif W_s-\frac{1}{2}\int^{t_1}_{t_0}|(\sigma^{-1}_sb_s)(Z_s)|^2\dif s\right\}.$$ By Novikov’s criterion, $\mE(\cE_{t_0,t_1})=1$, and for any $\gamma\in\mR$, by and Hölder’s inequality, $$\begin{aligned}
\label{KJ2}
\mE\Big(\cE_{t_0,t_1}^\gamma|\sF_{t_0}\Big)<\infty.\end{aligned}$$ Define a new probability $\mQ_{t_0, t_1}:=\cE_{t_0,t_1}\mP$. By Girsanov’s theorem, $\bar W_\cdot:=W_\cdot+\int^\cdot_{t_0}b_s(Z_s)\dif s$ is still a Brownian motion under $\mQ_{t_0,t_1}$. Moreover, $Z_t$ satisfies $$Z_t=Z_{t_0}+\int^t_{t_0}(\dot{X}_s,0)\dif s+\int^t_{t_0}(0,\sigma_s(Z_s)\dif\bar  W_s).$$ By the same argument as used in [**(c)**]{} with $b\equiv 0$, since in this case, we only need to control $\|u\|_{\mL^\infty(t_0,t_1)}$ and holds for any $p>2d+1$, we obtain that there is a constant $C>0$ such that for all $p>2d+1$ and $f\in\mL^p(T)$, $$\mE^{\mQ_{t_0,t_1}}\left(\int^{t_1}_{t_0} f_s(Z_s)\dif s\Big|\sF_{t_0}\right)
\leq C(t_1-t_0)^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(t_0,t_1)}.$$ Finally, by and Hölder’s inequality, we obtain .

We have the following useful corollary. (Khasminskii’s type estimate)\[Cor1\] In the same framework of Theorem \[Th41\], letting $\beta:=\frac{1}{2d+1}-\frac{1}{p}$ and $C$ be the same as in , we have

(i) For each $m\in\mN$ and $0\leq t_0<t_1\leq T$, it holds that $$\begin{aligned}
    \tilde\mE^{\tilde\sF_{t_0}}\left(\int^{t_1}_{t_0}f_s(\tilde Z_s)\dif s\right)^m\leq m!(C\|f\|_{\mL^p(T)}(t_1-t_0)^\beta)^m,\end{aligned}$$ where $\tilde\mE^{\tilde\sF_{t_0}}(\cdot)=\tilde\mE(\cdot|^{\tilde\sF_{t_0}})$.

(ii) For any $\lambda>0$ and $0\leq t_0<t_1\leq T$, it holds that $$\begin{aligned}
    \tilde\mE^{\tilde\sF_{t_0}}\exp\left(\lambda\int^{t_1}_{t_0}f_s(\tilde Z_s)\dif s\right)\leq 2^{T (2C\lambda\|f\|_{\mL^p(T)})^{1/\beta}}.\end{aligned}$$

\(i) Still we drop the tilde below. For $m\in\mN$, noticing that $$\left(\int^{t_1}_{t_0}g(s)\dif s\right)^m=m!\int\!\!\!\cdots\!\!\!\int_{\Delta^m}g(s_1)\cdots g(s_m)\dif s_1\cdots\dif s_m,$$ where $$\Delta^m:=\Big\{(s_1,\cdots, s_m):  t_0\leq s_1\leq s_2\leq\cdots\leq s_m\leq t_1\Big\},$$ by , we have $$\begin{aligned}
\mE^{\sF_{t_0}}\left(\int^{t_1}_{t_0}f_s(Z_s)\dif s\right)^m
&=m!\mE^{\sF_{t_0}}\left(\int\!\!\!\cdots\!\!\!\int_{\Delta^m}f_{s_1}(Z_{s_1})\cdots f_{s_m}(Z_{s_m})\dif s_1\cdots\dif s_m\right)\\
&=m!\mE^{\sF_{t_0}}\Bigg(\int\!\!\!\cdots\!\!\!\int_{\Delta^{m-1}}f_{s_1}(Z_{s_1})\cdots f_{s_{m-1}}(Z_{s_{m-1}})\\
&\quad\times\mE\left( \int^{t_1}_{s_{m-1}}f_{s_m}(Z_{s_m})\dif s_m\Big|_{\sF_{s_{m-1}}}\right)\dif s_1\cdots\dif s_{m-1}\Bigg)\\
&\leq m!\mE^{\sF_{t_0}}\int\!\!\!\cdots\!\!\!\int_{\Delta^{m-1}}f_{s_1}(Z_{s_1})\cdots f_{s_{m-1}}(Z_{s_{m-1}})\\
&\quad \times C(t_1-s_{m-1})^\beta\|f\|_{\mL^p(T)}\dif s_1\cdots\dif s_{m-1}\leq\cdots\\
&\leq m!(C\|f\|_{\mL^p(T)}(t_1-t_0)^\beta)^m.\end{aligned}$$ (ii) For $\lambda>0$, let us choose $n$ such that for $s_j=t_0+\frac{j(t_1-t_0)}{n}$, $$\lambda C\|f\|_{\mL^p(T)}(s_{j+1}-s_{j})^\beta\leq1/2.$$ Then by (i) we have $$\begin{aligned}
\mE^{\sF_{s_j}}\exp\left(\lambda\int^{s_{j+1}}_{s_j}f_s(Z_s)\dif s\right)
&=\sum_{m}\frac{1}{m!}\mE^{\sF_{s_j}}\left(\lambda\int^{s_{j+1}}_{s_j}f_s(Z_s)\dif s\right)^m\leq 2.\end{aligned}$$ Hence, $$\begin{aligned}
&\mE^{\sF_{t_0}}\exp\left(\lambda\int^{t_1}_{t_0}f_s(Z_s)\dif s\right)=\mE^{\sF_{t_0}}\left(\prod_{j=0}^{n-1}\exp\left(\lambda\int^{s_{j+1}}_{s_j}f_s(Z_s)\dif s\right)\right)\\
&=\mE^{\sF_{t_0}}\left(\prod_{j=0}^{n-2}\exp\left(\lambda\int^{s_{j+1}}_{s_j}f_s(Z_s)\dif s\right)
\mE^{\sF_{s_{n-1}}}\exp\left(\lambda\int^{s_{n}}_{s_{n-1}}f_s(Z_s)\dif s\right)\right)\\
&\leq 2\mE^{\sF_{t_0}}\left(\prod_{j=0}^{n-2}\exp\left(\lambda\int^{s_{j+1}}_{s_j}f_s(Z_s)\dif s\right)\right)\leq\cdots\leq 2^{n}.\end{aligned}$$ The proof is complete.

Well-posedness of martingale problem
------------------------------------

In this subsection we show the well-posedness of martingale problem for $\sL^{a,b}_t$. More precisely, \[Th62\] Suppose that [**(UE)**]{} holds, and for any $T>0$, $$\begin{aligned}
\label{CC9}
\lim_{|z-z'|\to 0}\sup_{t\in[0,T]}\|\sigma_t(z)-\sigma_t(z')\|=0,\end{aligned}$$ and $b\in\mL^q(T)$ for some $q\in(2(2d+1),\infty]$. For each $(r,z)\in\mR_+\times\mR^{2d}$, the set $\sP^{\sigma,b}_{r,z}$ has one and only one point. In particular, the martingale problem for $\sL^{a,b}_t$ is well-posed.

Below, we shall fix starting point $(r,z)\in\mR_+\times\mR^{2d}$ and divide the proof into three steps.

[**(a)**]{} We first show the uniqueness. For $\varphi\in C^\infty_c(\mR^{2d})$ and $t_1>r$, let $u=\cR^{a,b}_{0,t_1}(\varphi)\in\mH^{2/3,2}_p(t_1)$, which satisfies $$\begin{aligned}
\label{KJ4}
\p_t u+\sL^{a,b}_t u+\varphi=0,\ \ u_{t_1}=0.\end{aligned}$$ Let $u^\eps_t(z):=u_t*\varrho_\eps(z)$ and $\mP\in\sP^{\sigma,b}_{r,z}$. By , and a standard approximation for the time variable, one sees that $$t\mapsto u^\eps(t,Z_{t})-\int^{t}_r(\p_s+\sL^{a,b}_s)u^\eps(s,Z_s)\dif s$$ is an $\sF_t$-martingale under $\mP$ after time $r$. Thus, by , we have $$\begin{aligned}
u^\eps_r(z)&=-\mE\left(\int^{t_1}_r(\p_s u+\sL^{a,b}_s)u^\eps_s(Z_s)\dif s\right)\\
&=\mE\left(\int^{t_1}_r\big([\varrho_\eps, \sL^{a,b}_s]u_s+\varphi^\eps\big)(Z_s)\dif s\right).\end{aligned}$$ By Theorem \[Th41\] and Lemma \[Le24\], taking limits $\eps\to 0$ for both sides yields $$u_r(z)=\mE\left(\int^{t_1}_r\varphi(Z_s)\dif s\right),\ t_1\geq r.$$ In particular, we have for any $\mP_1,\mP_2\in\sP^{\sigma,b}_{r,z}$ and $t\geq r$, $$\mE_{\mP_1}\varphi(Z_t)=\mE_{\mP_2}\varphi(Z_t).$$ By [@St-Va Theorem 6.2.3], we get the uniqueness.

[**(b)**]{} For $n\in\mN$, let $\varrho_{1/n}$ be defined by with $\eps=1/n$, and define $$b^n_t:=b_t*\varrho_{1/n}, \ \sigma^n_t:=\sigma_t*\varrho_{1/n}.$$ Clearly, $$b^n\in L^q([0,T]; C^\infty_b(\mR^{2d})), \ \sigma^n\in L^\infty([0,T]; C^\infty_b(\mR^{2d}))$$ and $$\|b^n\|_{\mL^q(T)}\leq\|b\|_{\mL^q(T)},\ \omega_{a^n}(\delta)\leq\omega_{a}(\delta).$$ By the classical theory of SDEs, the following SDE admits a unique strong solution $Z^n_t=(X^n_t,\dot X^n_t)$ $$\dif Z^n_t=(\dot X^n_t,b^n_t(Z^n_t))\dif t+(0,\sigma^n_t(Z^n_t)\dif W_t),\ Z^n_t|_{[0,r]}=z.$$ By Corollary \[Cor1\], for any $m\in\mN$ and $T>0$, there is a constant $C=C(m,T)>0$ such that for all $r\leq t_0<t_1\leq T$ and $n\in\mN$, $$\begin{aligned}
\mE\left|\int^{t_1}_{t_0}b^n_s(Z^n_s)\dif s\right|^m\leq C\|b\|_{\mL^q(T)}^m(t_1-t_0)^{m(\frac{1}{2d+1}-\frac{1}{p})}.\end{aligned}$$ Let $\mP_n$ be the probability distribution of $Z^n$ in $\Omega$. By the above moment estimate, it is by now standard to show that $(\mP_n)_{n\in\mN}$ is tight.

[**(c)**]{} By extracting a subsequence, without loss of generality, we may assume that $\mP_n$ weakly converges to some probability measure $\mP$. To see that $\mP\in\sP^{\sigma,b}_{r,z}$, it suffices to show that for any $\varphi\in C^\infty_c(\mR^{2d})$, $M^{r,\varphi}_t$ defined by is an $\sF_t$-martingale under $\mP$. Equivalently, for any $r\leq t_0\leq t_1$ and any bounded continuous $\sF_{t_0}$-measurable $G$, $$\begin{aligned}
\label{MM5}
\mE_{\mP}(M^{r,\varphi}_{t_1}G)=\mE_{\mP}(M^{r,\varphi}_{t_0}G).\end{aligned}$$ Notice that $$\begin{aligned}
\label{Lim3}
\mE_{\mP_n}(M^{r,\varphi}_{n,t_1}G)=\mE_{\mP_n}(M^{r,\varphi}_{n,t_0}G),\end{aligned}$$ where $$M^{r,\varphi}_{n,t_i}:=\varphi(Z_{t_i})-\int^{t_i}_r\sL^{a^n,b^n}_s \varphi(Z_s)\dif s,\ i=0,1.$$ Let us prove the following limit: for $i=0,1$, $$\begin{aligned}
\label{Lim0}
\lim_{n\to\infty}\mE_{\mP_n}\left(G\int^{t_i}_r(b^n_s\cdot\nabla_\v\varphi)(Z_s)\dif s\right)=\mE_{\mP}\left(G\int^{t_i}_r(b_s\cdot\nabla_\v\varphi)(Z_s)\dif s\right).\end{aligned}$$ For any $p\in(2d+1,q)$ and $T>0$, by Theorem \[Th41\], there is a constant $C>0$ such that for all $n\in\mN$ and $f\in \mL^p(T)$, $$\begin{aligned}
\mE_{\mP_n}\left(\int^T_r f_s(Z_s)\dif s\right)\leq C\|f\|_{\mL^p(T)}.\end{aligned}$$ Let $f\in C_c([0,T]\times\mR^{2d})$. By taking weak limits, we have $$\begin{aligned}
\mE_{\mP}\left(\int^T_r f_s(Z_s)\dif s\right)&=\lim_{n\to\infty}\mE_{\mP_n}\left(\int^T_r f_s(Z_s)\dif s\right)\leq C\|f\|_{\mL^p(T)}.\end{aligned}$$ By a monotone class argument, the above estimate still holds for all $f\in\mL^p(T)$. Let the support of $\varphi$ be contained in $B_R$. Thus, if we let $\mP_\infty=\mP$, then $$\begin{aligned}
\label{Lim1}
\begin{split}
&\lim_{m\to\infty}\sup_{n\in\mN\cup\{\infty\}}\mE_{\mP_n}\left(G\int^{t_i}_r|(b^m_s-b_s)\cdot\nabla_\v\varphi|(Z_s)\dif s\right)\\
&\quad\leq C\|G\|_\infty\|\nabla_\v\varphi\|_\infty\lim_{m\to\infty}\|(b^m-b)1_{B_R}\|_{\mL^p(r,t_i)}=0.
\end{split}\end{aligned}$$ On the other hand, for each $m\in\mN$, since $\omega\mapsto G(\omega)\int^{t_i}_r(b^m_s\cdot\nabla_\v\varphi)(Z_s(\omega))\dif s$ is a continuous and bounded functional, we have $$\lim_{n\to\infty}\mE_{\mP_n}\left(G\int^{t_i}_r(b^m_s\cdot\nabla_\v\varphi)(Z_s)\dif s\right)=\mE_{\mP}\left(G\int^{t_i}_r(b^m_s\cdot\nabla_\v\varphi)(Z_s)\dif s\right).$$ Combining this with , we get . Similarly, one can show $$\begin{aligned}
\label{Lim2}
\lim_{n\to\infty}\mE_{\mP_n}\left(G\int^{t_i}_r\sL^{a^n,0}_r\varphi(Z_s)\dif s\right)=\mE_{\mP}\left(G\int^{t_i}_r\sL^{a,0}_r\varphi(Z_s)\dif s\right).\end{aligned}$$ Finally, by taking weak limits for both sides of and using and , we get . The proof is complete.

When $b$ is bounded measurable ($q=\infty$), this result has been proven in [@Pr] and [@Me]. Therein, more general equations were considered. However, by comparing with the original proof of Stroock and Varadhan [@St-Va Chapter 7], our proof is quite different from [@Pr; @Me] as our starting point is a global apriori Krylov’s estimate (see Theorem \[Th41\]). In principle, it is reasonable to believe that our argument is applicable for more general equations as studied in [@Pr; @Me]. By suitable localization techniques as developed in [@Pr], it is possible to weaken the global assumptions in Theorem \[Th62\] as local ones together with some non-explosion conditions.

Proof of Theorem \[Main2\]
--------------------------

Let $\nu\in\cP(\mR^{2d})$. By Theorem \[Th62\], the probability measure $\mP_\nu(A):=\int_{\mR^{2d}}\mP_{0,z}(A)\nu(\dif z)$ is the unique martingale solution for $\sL^{a,b}_t$ starting from $\nu$ at time $0$. The conclusion of Theorem \[Main2\] now follows by [@Tre Theorem 2.5].

Proof of Theorem \[Main\]
=========================

In this section we assume that $\sigma$ satisfies [**(UE)**]{} and for some $p>2(2d+1)$, $$\begin{aligned}
\label{Con5}
\kappa_0:=\|b\|_{L^p(\mR_+; H^{2/3,0}_p)}+\|\nabla\sigma\|_{L^\infty(\mR_+; L^p)}<\infty.\end{aligned}$$ For $n\in\mN$, let $\varrho_{1/n}$ be defined by with $\eps=1/n$, as in the previous, define $$\begin{aligned}
\label{Reg}
b^n_t:=b_t*\varrho_{1/n}, \ \sigma^n_t:=\sigma_t*\varrho_{1/n},\ \ b^\infty_t:=b_t, \ \sigma^\infty_t:=\sigma_t.\end{aligned}$$ \[Le51\] Assume [**(UE)**]{} and . Then [**(H$^{\delta,p}_K$)**]{} holds for $a^n:=\frac{1}{2}\sigma^n(\sigma^n)^*$ uniformly with respect to $n$, and there exists a constant $C=C(d,p)>0$ such that for all $s$, $$\|\sigma^n_s-\sigma_s\|_\infty\leq C\|\nabla \sigma_s\|_p n^{\frac{2d}{p}-1},\ \ \sup_{\v}\|\Delta_x^{\frac{1}{3}}\sigma^n_s(\cdot,\v)\|_p\leq C\|\nabla\sigma_s\|_p.$$

Since $p>2d$, by Morrey’s inequality, there is a constant $C=C(d,p)>0$ such that $$|\sigma_s(z)-\sigma_s(z')|\leq C|z-z'|^{1-\frac{2d}{p}}\|\nabla \sigma_s\|_p,\ \ z,z'\in\mR^{2d}.$$ From this, it is easy to see that [**(H$^{\delta,p}_K$)**]{} holds for $a^n=\frac{1}{2}\sigma^n(\sigma^n)^*$ uniformly with respect to $n$, and $$\begin{aligned}
\|\sigma^n_s-\sigma_s\|_\infty&\leq\int_{\mR^{2d}}\|\sigma_s(\cdot+z)-\sigma_s(\cdot)\|_\infty\varrho_{1/n}(z)\dif z\\
&\leq C\|\nabla \sigma_s\|_p\int_{\mR^{2d}}|z|^{1-\frac{2d}{p}}\varrho_{1/n}(z)\dif z\leq C\|\nabla \sigma_s\|_p n^{\frac{2d}{p}-1}.\end{aligned}$$ Moreover, since $p>4d$, by and , we have $$\begin{aligned}
\sup_{\v}\|\Delta_x^{\frac{1}{3}}\sigma^n_s(\cdot,\v)\|^p_p
&\leq\int_{\mR^d}\sup_\v |\Delta_x^{\frac{1}{3}}\sigma^n_s(x,\v)|^p_p\dif x
\preceq\|\Delta_x^{\frac{1}{3}}\sigma^n_s\|^p_p+\|\Delta_x^{\frac{1}{3}}\Delta_\v^{\frac{1}{8}}\sigma^n_s\|^p_{p}\\
&\preceq\|\Delta^{\frac{11}{24}}\sigma^n_s\|^p_{p}
\preceq\|\Delta^{\frac{1}{2}}\sigma^n_s\|^p_{p}\stackrel{\eqref{Riesz}}{\preceq}\|\nabla\sigma^n_s\|^p_p.\end{aligned}$$ Here $\Delta=\Delta_x+\Delta_\v$. The proof is complete.

Let $T>0$ and $\lambda\geq 1$. For $n\in\mN\cup\{\infty\}$, let $\u^n_\lambda\in \mH^{0,2}_p(T)$ uniquely solve the following PDE: $$\p_t\u^n_\lambda+\sL^{a^n,b^n}_t\u^n_\lambda-\lambda \u^n_\lambda+b^n=0,\ \ \u^n_{_\lambda,T}=0,$$ where $a^n:=\frac{1}{2}\sigma^n(\sigma^n)^*$. By Lemma \[Le51\] and Theorem \[Th32\], there is a constant $C=C(d,p,\kappa_0,K)>0$ such that for all $n\in\mN\cup\{\infty\}$, $$\begin{aligned}
\label{PP1}
\|\nabla\nabla_\v\u^n_\lambda\|_{\mL^p(T)}\leq C\|b\|_{\mH^{2/3,0}_p(T)},\end{aligned}$$ and by , , with $\alpha=\frac{4d+1}{3(2d+1)}$ and $\beta=\frac{4d+1}{2d+1}$, $$\begin{aligned}
\|\nabla_\v\u^n_\lambda\|_{\mL^\infty(T)}\leq C\|\u^n_\lambda\|_{L^\infty(0,T; \mH^{\alpha,\beta}_p)}
\leq C(T\wedge\lambda^{-1})^{\frac{1}{2(2d+1)}-\frac{1}{p}}\|b\|_{\mL^p(T)}.\label{PP2}\end{aligned}$$ \[Le52\] Under [**(UE)**]{} and , there is a constant $C>0$ depending only on $d,p,\kappa_0,K, T,\lambda$ such that for all $n,m\in\mN$, $$\|\u^n_\lambda-\u^m_\lambda\|_{\mL^\infty(T)}+\|\nabla_\v\u^n_\lambda-\nabla_\v\u^m_\lambda\|_{\mL^\infty(T)}
\leq C\Big(\|b^n-b^m\|_{\mL^p(T)}+(n\wedge m)^{\frac{2d}{p}-1}\Big).$$

Let $\w^{n,m}:=\u^n_\lambda-\u^m_\lambda$. By equation , we have $$\p_t\w^{n,m}+(\sL^{a^n,b^n}_t-\lambda)\w^{n,m}+(\sL^{a^n,b^n}_t-\sL^{a^m,b^m}_t)\u^m_\lambda+b^n-b^m=0.$$ Noticing that $$g^{n,m}_{\lambda,t}:=(\sL^{a^n,b^n}_t-\sL^{\sigma^m,b^m}_t)\u^m_{\lambda,t}=\tr((a^n_t-a^m_t)\cdot\nabla^2_\v \u^m_{\lambda,t})
+(b^n_t-b^m_t)\cdot\nabla_\v \u^m_{\lambda,t},$$ by , and Lemma \[Le51\], we have $$\begin{aligned}
\|g^{n,m}_\lambda\|_{\mL^p(T)}&\preceq\|a^n-a^m\|_{\mL^\infty(T)}\|\nabla^2_\v \u^m_\lambda\|_{\mL^p(T)}
+\|b^n-b^m\|_{\mL^p(T)}\|\nabla_\v \u^m_\lambda\|_{\mL^\infty(T)}\\
&\preceq (n\wedge m)^{\frac{2d}{p}-1}+\|b^n-b^m\|_{\mL^p(T)}.\end{aligned}$$ By and with $\alpha=\frac{4d}{3(2d+1)}$ and $\beta=\frac{4d}{2d+1}$, we have $$\begin{aligned}
\|\w^{n,m}_t\|_\infty\preceq\|\w^{n,m}_t\|_{\alpha,\beta;p}&\preceq \|g^{n,m}_\lambda\|_{\mL^p(T)}+\|b^n-b^m\|_{\mL^p(T)}\\
&\preceq (n\wedge m)^{\frac{2d}{p}-1}+\|b^n-b^m\|_{\mL^p(T)},\end{aligned}$$ and by , , with $\alpha=\frac{4d+1}{3(2d+1)}$ and $\beta=\frac{4d+1}{2d+1}$, $$\begin{aligned}
\|\nabla_\v\w^{n,m}_t\|_\infty&\preceq \|\nabla_\v\w^{n,m}_t\|_{\alpha(\beta-1)/\beta,\beta-1;p}
\preceq \|\w^{n,m}_t\|_{\alpha,\beta;p}\\
&\preceq (n\wedge m)^{\frac{2d}{p}-1}+\|b^n-b^m\|_{\mL^p(T)}.\end{aligned}$$ The proof is complete.

For $n\in\mN\cup\{\infty\}$, let $$\begin{aligned}
\label{HH}
H^n_t(x,\v):=\v+\u^n_{\lambda,t}(x,\v).\end{aligned}$$ By , one can choose $\lambda$ large enough (being independent of $n$ and fixed below) so that $$\begin{aligned}
\label{T0}
\|\nabla_\v\u^n_\lambda\|_{\mL^\infty(T)}\leq\tfrac{1}{2},\end{aligned}$$ and thus, $$\begin{aligned}
\label{DIF}
\tfrac{1}{2}|\v-\v'|\leq |H^n_t(x,\v)-H^n_t(x,\v')|\leq \tfrac{3}{2}|\v-\v'|.\end{aligned}$$ Observing that $$\p_tH^n+\sL^{a^n,b^n}_tH^n-\lambda \u^n_\lambda=0,$$ by Itô’s formula, we have $$\begin{aligned}
\label{SDE1}
H^n_t(Z^n_t)=H^n_0(Z^n_0)+\lambda\int^t_0\u^n_{\lambda,s}(Z^n_s)\dif s+\int^t_0\Theta^n_s(Z^n_s)\dif W_s,\end{aligned}$$ where $\Theta^n_s(z):=(\nabla_\v H^n_s\cdot \sigma^n_s)(z)$ satisfies by and that $$\|\Theta^n\|_\infty\leq 2\|\sigma\|_\infty,$$ and for the given $p>2(2d+1)$, $$\begin{aligned}
\label{633}
\begin{split}
\|\nabla\Theta^n\|_{\mL^p(T_0)}&\leq 2\|\nabla\sigma^n\|_{\mL^p(T_0)}+\|\sigma^n\|_\infty\|\nabla\nabla_\v H^n\|_{\mL^p(T_0)}\\
&\leq 2\|\nabla\sigma\|_{\mL^p(T_0)}+C\|\sigma\|_\infty\|b\|_{\mH^{2/3,0}_p(T_0)}.
\end{split}\end{aligned}$$ For $n\in\mN$, since $b^n\in L^p([0,T]; C^\infty_b(\mR^{2d}))$ and $\sigma^n\in L^\infty([0,T]; C^\infty_b(\mR^{2d}))$, the following SDE admits a unique solution $Z^n_t=(X^n_t,\dot X^n_t)$ $$\begin{aligned}
\label{App0}
\dif Z^n_t=(\dot X^n_t,b^n_t(Z^n_t))\dif t+(0,\sigma^n_t(Z^n_t)\dif W_t),\ \ Z^n_0=z=(x,\v)\in\mR^{2d}.\end{aligned}$$ We have \[Le53\] Under [**(UE)**]{} and , for any $q\geq 2$, there is a constant $C>0$ such that for all $n,m\in\mN$, $$\begin{aligned}
\label{PP4}
\left\|\sup_{t\in[0,T_0]}|Z^n_t-Z^m_t|\right\|_{L^q(\Omega)}\leq 
C\left(\|b^n-b^m\|_{\mL^p(T_0)}+(n\wedge m)^{\frac{2d}{p}-1}\right).\end{aligned}$$

By and Itô’s formula, we have $$\begin{aligned}
&|H^n_t(Z^n_t)-H^m_t(Z^m_t)|^2=|H^n_0(z)-H^m_0(z)|^2+\int^t_0\|\Theta^n_s(Z^n_s)-\Theta^m_s(Z^m_s)\|^2\dif s\\
&\qquad\qquad+2\lambda\int^t_0\<H^n_s(Z^n_s)-H^m_s(Z^m_s), \u^n_{\lambda,s}(Z^n_s)-\u^m_{\lambda,s}(Z^m_s)\>\dif s\\
&\qquad\qquad+2\int^t_0\<H^n_s(Z^n_s)-H^m_s(Z^m_s), (\Theta^n_s(Z^n_s)-\Theta^m_s(Z^m_s))\dif W_s\>,\end{aligned}$$ and also, $$|X^n_t-X^m_t|^2=2\int^t_0\<X^n_s-X^m_s,\dot X^n_s-\dot X^m_s\>\dif s.$$ If we set $$\xi_t:=|H^n_t(Z^n_t)-H^n_t(Z^m_t)|^2+|X^n_t-X^m_t|^2,$$ then $$\begin{aligned}
\xi_t&\leq 2\|H^n_t-H^m_t\|^2_\infty+2|H^n_t(Z^n_t)-H^m_t(Z^m_t)|^2+|X^n_t-X^m_t|^2\\
&\leq\xi_0+\int^t_0\zeta^{(1)}_s\dif s+\int^t_0\zeta^{(2)}_s\dif W_s+\int^t_0\xi_s\beta_s\dif s+\int^t_0\xi_s\alpha_s\dif W_s,\end{aligned}$$ where $\xi_0:=3\|\u^n-\u^m\|^2_{\mL^\infty(T_0)}$ and $$\begin{aligned}
\zeta^{(1)}_s&:=4\|\Theta^n_s(Z^m_s)-\Theta^m_s(Z^m_s)\|^2+6\lambda|H^n_s(Z^m_s)-H^m_s(Z^m_s)|^2,\\
\zeta^{(2)}_s&:=4(\Theta^n_s(Z^n_s)-\Theta^m_s(Z^m_s))^*(H^n_s(Z^n_s)-H^m_s(Z^m_s))\\
&\quad-4(\Theta^n_s(Z^n_s)-\Theta^n_s(Z^m_s))^*(H^n_s(Z^n_s)-H^n_s(Z^m_s)),\\
\beta_s&:=4\|\Theta^n_s(Z^n_s)-\Theta^n_s(Z^m_s)\|^2/\xi_s+6\lambda+\lambda|\dot X^n_s-\dot X^m_s|^2/\xi_s\\
&\quad+2\<X^n_s-X^m_s,\dot X^n_s-\dot X^m_s\>/\xi_s,\\
\alpha_s&:=4(\Theta^n_s(Z^n_s)-\Theta^n_s(Z^m_s))^*(H^n_s(Z^n_s)-H^n_s(Z^m_s))/\xi_s.\end{aligned}$$ By , one has $$\begin{aligned}
\label{Es31}
\xi_t\asymp|Z^n_t-Z^m_t|^2=|X^n_t-X^m_t|^2+|\dot X^n_t-\dot X^m_t|^2,\end{aligned}$$ which implies by that $$\begin{aligned}
\label{PP3}
|\beta_s|+|\alpha_s|^2\preceq 1+\big(\cM|\nabla \Theta^n_s|(Z^n_s)\big)^2+\big(\cM|\nabla \Theta^n_s|(Z^m_s)\big)^2=:G^{n,m}_s.\end{aligned}$$ In view of $p>2(2d+1)$ and $\sup_n\|\cM|\nabla \Theta^n|\|_{\mL^p(T)}<\infty$, by (ii) of Corollary \[Cor1\], we have for any $\gamma>0$, $$\sup_{n,m}\mE\exp\left\{\gamma\int^{T}_0G^{n,m}_s\dif s\right\}<\infty.$$ Therefore, by Lemma \[Le35\] below with $q_0=q$, $q_1=q_2=q_3=3q/2$, we have $$\begin{aligned}
\label{513}
\|\xi^*_{T}\|_q\preceq \|\u^n-\u^m\|^2_{\mL^\infty(T)}+\left\|\int^{T}_0|\zeta^{(1)}_s|\dif s\right\|_{3q/2}
+\left\|\int^{T}_0|\zeta^{(2)}_s|^2\dif s\right\|^{1/2}_{3q/4},\end{aligned}$$ where $\xi^*_{T}:=\sup_{t\in[0,T]}\xi_t$.

Now, letting $$\ell_{n,m}:=\|\sigma^n-\sigma^m\|_{\mL^\infty(T)}+\|\u^n-\u^m\|_{\mL^\infty(T)}+\|\nabla\u^n-\nabla\u^m\|_{\mL^\infty(T)},$$ by definitions, we have $$\begin{aligned}
|\zeta_s^{(1)}|\preceq\ell^2_{n,m},\end{aligned}$$ and by and , $$\begin{aligned}
|\zeta^{(2)}_s|^2&\preceq |(\Theta^n_s(Z^n_s)-\Theta^m_s(Z^m_s))^*(H^n_s(Z^m_s)-H^m_s(Z^m_s))|^2\\
&\quad+|(\Theta^n_s(Z^m_s)-\Theta^m_s(Z^m_s))^*(H^n_s(Z^n_s)-H^n_s(Z^m_s))|^2\\
&\preceq \Big(\|\Theta^n_s(Z^m_s)-\Theta^m_s(Z^m_s)\|^2+\|\Theta^n_s(Z^n_s)-\Theta^n_s(Z^m_s)\|^2\Big)\\
&\quad\times|\u^n_s(Z^m_s)-\u^m_s(Z^m_s)|^2+\|\Theta^n_s(Z^m_s)-\Theta^m_s(Z^m_s)\|^2\xi_s\\
&\preceq \ell^4_{n,m}+\ell^2_{n,m}G^{n,m}_s\xi_s,\end{aligned}$$ where $G^{n,m}_s$ is defined by . Substituting these estimates into and by Hölder’s inequlity and Young’s inequality, we have for any $\eps\in(0,1)$, $$\begin{aligned}
\|\xi^*_{T}\|_q
&\preceq\ell^2_{n,m}+\ell_{n,m}\left\|\int^{T}_0G^{n,m}_s\xi_s\dif s\right\|^{1/2}_{3q/4}\no\\
&\preceq\ell^2_{n,m}+\ell_{n,m}\left\|\xi^*_{T}\int^{T}_0G^{n,m}_s\dif s\right\|^{1/2}_{3q/4}\no\\
&\preceq\ell^2_{n,m}+\ell_{n,m}\left\|\int^{T}_0G^{n,m}_s\dif s\right\|^{1/2}_{3q}\|\xi^*_{T}\|_q^{\frac{1}{2}}\no\\
&\preceq C_\eps\ell^2_{n,m}+\eps\|\xi^*_{T}\|_q.\label{ERQ}\end{aligned}$$ Letting $\eps$ be small enough and noticing that by Lemmas \[Le51\] and \[Le52\], $$\ell_{n,m}\leq C\left(\|b^n-b^m\|_{\mL^p(T)}+(n\wedge m)^{\frac{2d}{p}-1}\right),$$ we obtain the desired estimate from .

(Existence of strong solutions)\[Th54\] Under [**(UE)**]{} and , for any $q\geq 2$, there is a continuous $\sF_t$-adapted process $Z_t=(X_t,\dot X_t)$ solving equation , and such that $$\begin{aligned}
\label{PP7}
\left\|\sup_{t\in[0,T]}|Z^n_t-Z_t|\right\|_{L^q(\Omega)}\leq C\left(\|b^n-b\|_{\mL^p(T)}+n^{\frac{2d}{p}-1}\right).\end{aligned}$$

First of all, by , there exists a continuous $\sF_t$-adapted process $Z^\infty_t:=Z_t=(X_t,\dot X_t)$ satisfying . Moreover, for any $p>2d+1$, there is a positive constant $C=C(T,K,\kappa_0,d,p)$ such that for any $f\in\mL^p(T)$, $$\begin{aligned}
\label{Est1}
\mE\left(\int^{T}_0f_s(Z^\infty_s)\dif s\right)\leq C\|f\|_{\mL^p(T)}.\end{aligned}$$ In fact, for any $f\in C_c([0,T]\times\mR^{2d})$, by we have $$\begin{aligned}
\label{Est2}
\mE\left(\int^{T}_0f_s(Z^n_s)\dif s\right)\leq C\|f\|_{\mL^p(T)},\end{aligned}$$ which gives by taking limit $n\to\infty$ and a monotone class argument.

Below we show that $Z_t$ solves SDE by taking limits for . For this, we only need to show the following two limits: $$\begin{aligned}
&\lim_{n\to\infty}\mE\left|\int^t_0 b^n_s(Z^n_s)\dif s-\int^t_0 b_s(Z_s)\dif s\right|=0,\label{Lim7}\\
&\lim_{n\to\infty}\mE\left|\int^t_0 \sigma^n_s(Z^n_s)\dif W_s-\int^t_0 \sigma_s(Z_s)\dif W_s\right|=0.\label{Lim8}\end{aligned}$$ For , by and we have $$\lim_{n\to\infty}\sup_{m\in\mN\cup\{\infty\}}\mE\left(\int^t_0 |b^n_s(Z^m_s)-b_s(Z^m_s)|\dif s\right)
\leq C\lim_{n\to\infty}\|b^n-b\|_{\mL^p(t)}=0,$$ and for each $n\in\mN$, by the dominated convergence theorem, $$\lim_{m\to\infty}\mE\left(\int^t_0 |b^n_s(Z^m_s)-b^n_s(Z^\infty_s)|\dif s\right)=0.$$ Limit is similar. The proof is complete.

Now, by a result of Cherny [@Che], the existence of strong solutions together with the weak uniqueness (see Theorem \[Th62\]) implies the pathwise uniqueness. However, to show the homeomorphism property of $z\mapsto Z_t(z)$, one needs the following $q$-order moment estimate for all $q\in\mR$, which clearly implies the pathwise uniqueness. \[Le55\] For any $q\in\mR$, there is a constant $C>0$ such that for any two solutions $Z_t$ and $Z_t'$ of SDE with starting points $z=(x,\v)$ and $z'=(x',\v')$ respectively, $$\begin{aligned}
\label{Es7}
\mE\left(\sup_{t\in[0,T]}|Z_t-Z_t'|^{2q}\right)\leq C|z-z'|^{2q}.\end{aligned}$$ Moreover, we also have $$\begin{aligned}
\label{Es8}
\mE\left(\sup_{t\in[0,T]}(1+|Z_t|^2)^{q}\right)\leq C(1+|z|^2)^{q}.\end{aligned}$$

Let $H=H^\infty$ be defined by . By and Itô’s formula, we have $$\begin{aligned}
|H_t(Z_t)-H_t(Z'_t)|^2&=|H_0(Z_0)-H_0(Z'_0)|^2+\int^t_0\|\Theta_s(Z_s)-\Theta_s(Z'_s)\|^2\dif s\\
&+2\lambda\int^t_0\<H_s(Z_s)-H_s(Z'_s), \u_{\lambda,s}(Z_s)-\u_{\lambda,s}(Z'_s)\>\dif s\\
&+2\int^t_0\<H_s(Z_s)-H_s(Z'_s),(\Theta_s(Z_s)-\Theta_s(Z'_s))\dif W_s\>.\end{aligned}$$ As in the proof of Lemma \[Le53\], if we set $$\xi_t:=|H_t(Z_t)-H_t(Z'_t)|^2+|X_t-X'_t|^2,$$ then $$\xi_t=\xi_0+\int^t_0\xi_s\beta_s\dif s+\int^t_0\xi_s\alpha_s\dif W_s,$$ where $$\begin{aligned}
\beta_s&:=[\|\Theta_s(Z_s)-\Theta_s(Z'_s)\|^2+2\<X_s-X'_s,\dot X_s-\dot X'_s\>]/\xi_s\\
&\quad+2\lambda\<H_s(Z_s)-H_s(Z'_s), \u_{\lambda,s}(Z_s)-\u_{\lambda,s}(Z'_s)\>/\xi_s\end{aligned}$$ and $$\alpha_s:=2(\Theta_s(Z_s)-\Theta_s(Z'_s))^*(H_s(Z_s)-H_s(Z'_s))/\xi_s.$$ By Dolèans-Dade’s exponential formula, we have $$\begin{aligned}
\label{H6}
\xi_t=\xi_0\exp\left\{\int^t_0\alpha_s\dif W_s+\int^t_0\Big[\beta_s-\tfrac{1}{2}|\alpha_s|^2\Big]\dif s\right\}\end{aligned}$$ By , one has $$\xi_t\asymp|Z_t-Z'_t|^2=|X_t-X'_t|^2+|\dot X_t-\dot X'_t|^2.$$ Hence, by and below, $$|\beta(s)|+|\alpha(s)|^2\preceq 1+(\cM|\nabla \Theta_s|(Z_s))^2+(\cM|\nabla \Theta_s|(Z'_s))^2.$$ Since $\|\cM|\nabla \Theta|\|_{\mL^p(T)}<\infty$ and $p>2(2d+1)$, by Corollary \[Cor1\], we have for any $\gamma>0$, $$\begin{aligned}
\label{Exp}
\mE\exp\left\{\gamma\int^{T}_0\big(|\beta_s|+|\alpha_s|^2\big)\dif s\right\}<\infty.\end{aligned}$$ For $q\in\mR$, let $$\sE^q_t:=\exp\left\{q\int^t_0\alpha_s\dif W_s-\tfrac{q^2}{2}\int^t_0|\alpha_s|^2\dif s\right\}.$$ By and Novikov’s criterion, $t\mapsto \sE^q_t$ is a continuous martingale. Therefore, by , Hölder’s inequality and Doob’s maximal inequality, $$\begin{aligned}
&\mE\left(\sup_{t\in[0,T]}|Z_t-Z'_t|^{2q}\right)\\
&\preceq|z-z'|^{2q}
\mE\left(\sup_{t\in[0,T]}\exp\left\{q\int^t_0\alpha_s\dif W_s+q\int^t_0[\beta_s-\tfrac{1}{2}|\alpha_s|^2]\dif s\right\}\right)\\
&\preceq|z-z'|^{2q}
\left(\mE\left((\sE^q)^*_{T}\right)^2\right)^{\frac{1}{2}}
\left(\mE\exp\left\{\int^{T}_0\big((q^2+q)|\alpha_s|^2+q\beta_s\big)\dif s\right\}\right)^{\frac{1}{2}}\\
&\preceq|z-z'|^{2q}\left(\mE(\sE^{q}_{T})^2\right)^{\frac{1}{2}}
\preceq|z-z'|^{2q},\end{aligned}$$ which gives .

On the other hand, if we let $$\xi_t:=1+|H_t(Z_t)|^2+|X_t|^2,$$ then for any $q\in\mR$, by Itô’s formula, we have $$\begin{aligned}
&\xi_t^q=\xi_0^q+q\int^t_0\xi_s^{q-1}(2\<X_s,\dot X_s\>+\|\Theta_s(Z_s)\|^2-\lambda\<H_s(Z_s),\u_{\lambda,s}(Z_s)\>)\dif s\\
&+2q\int^t_0\xi_s^{q-1}\<H_s(Z_s), \Theta_s(Z_s)\dif W_s\>+2q(q-1)\int^t_0\xi_s^{q-2}|\Theta^*_s(Z_s)H_s(Z_s)|^2\dif s.\end{aligned}$$ Noticing that $$|H_t(Z_t)|\asymp1+|\dot X_t|,\ \ \xi_t\asymp 1+|Z_t|^2=1+|X_t|^2+|\dot X_t|^2,$$ by Burkholder’s inequality, we have $$\begin{aligned}
\mE\left(\sup_{s\in[0,t]}\xi_t^{2q}\right)\preceq \xi_0^{2q}+\mE\int^t_0\xi_s^{2q}\dif s,\end{aligned}$$ which in turn gives by Gronwall’s inequality.

Now we can give

The existence and uniqueness of a strong solution in time interval $[0,T]$ follows by Theorem \[Th54\] and Lemma \[Le55\]. The bi-continuous version of $(t,z)\mapsto Z_t(z)$ follows by and Kolmogorov’s continuity criterion.

[**(A)**]{} As for the homeomorphism property, it follows by Lemma \[Le55\] and Kunita’s argument (see [@Ku; @Zh2]).

[**(B)**]{} The weak differentiability of $z\mapsto Z_t(z)$ and estimate follow by and [@Xi-Zh Theorem 1.1].

[**(C)**]{} It follows by .

Appendix
========

The following stochastic Gronwall’s type lemma is probably well-known. Since we can not find it in the literature, a proof is provided here for the reader’s convenience. \[Le35\] For given $T>0$, let $(\xi_t)_{t\in [0, T]}$ and $(\beta_t)_{t\in [0, T]}$ (resp. $(\alpha_t)_{t\in [0, T]}$) be two real-valued (resp. $\mR^d$-valued) measurable $\sF_t$-adapted processes. Let $\zeta_t$ be an Itô process with the form: $$\zeta_t=\zeta_0+\int^t_0\zeta^{(1)}_s\dif s+\int^t_0\zeta^{(2)}_s\dif W_s.$$ Suppose that for any $\gamma>0$, $$\begin{aligned}
\kappa_\gamma:=\mE\exp\left\{\gamma\int^T_0\big(|\beta_s|+|\alpha_s|^2\big)\dif s\right\}<\infty,\label{C1}\end{aligned}$$ and $$\begin{aligned}
\label{C2}
0\leq\xi_t\leq\zeta_t+\int^t_0\xi_s\beta_s\dif s+\int^t_0\xi_s\alpha_s\dif W_s.\end{aligned}$$ Then for any $q_0\in[1,\infty)$ and $q_1,q_2,q_3>q_0$, there is a constant $C>0$ only depending on $q_i,\kappa_{\gamma}, i=0,1,2,3$ such that $$\begin{aligned}
\|\xi^*_T\|_{q_0}\leq C\left(\|\zeta_0\|_{q_1}+\left\|\int^T_0|\zeta_s^{(1)}|\dif s\right\|_{q_2}
+\left\|\int^T_0|\zeta_s^{(2)}|^2\dif s\right\|^{1/2}_{q_3/2}\right),\label{HG8}\end{aligned}$$ where $\xi^*_T:=\sup_{t\in[0,T]}\xi_t$ and $\|\cdot\|_{q_i}$ denotes the norm in $L^{q_i}(\Omega)$.

Write $$\begin{aligned}
\eta_t&:=\zeta_t+\int^t_0\xi_s\beta_s\dif s+\int^t_0\xi_s\alpha_s\dif W_s\\
&=\zeta_t+\int^t_0\eta_s\bar\beta_s\dif s+\int^t_0\eta_s\bar\alpha_s\dif W_s,\end{aligned}$$ where $\bar\beta_s:=\xi_s\beta_s/\eta_s$ and $\bar\alpha_s:=\xi_s\alpha_s/\eta_s$. Here we use the convention $\frac{0}{0}:=0$.

Define $$M_t:=\exp\left\{\int^t_0\bar\alpha_s\dif W_s+\int^t_0(\bar\beta_s-\tfrac{1}{2}|\bar\alpha_s|^2)\dif s\right\}.$$ By Itô’s formula, we have $$M_t=1+\int^t_0M_s\bar\beta_s\dif s+\int^t_0M_s\bar\alpha_s\dif W_s$$ and $$\begin{aligned}
\eta_t=M_t\left[\zeta_0+\int^t_0 M^{-1}_s(\zeta^{(1)}_s-\<\bar\alpha_s,\zeta_s^{(2)}\>)\dif s+\int^t_0M^{-1}_s \zeta^{(2)}_s\dif W_s\right].\end{aligned}$$ Hence, $$\begin{aligned}
\|\eta^*_T\|_{q_0}&\leq\|M^*_T\zeta_0\|_{q_0}+\left\|M^*_T(M^{-1})^*_T\int^T_0|\zeta^{(1)}_s|\dif s\right\|_{q_0}\\
&\quad+\left\|M^*_T(M^{-1})^*_T\int^T_0|\bar\alpha_s|\cdot|\zeta_s^{(2)}|\dif s\right\|_{q_0}\\
&\quad+\left\|M^*_T\sup_{t\in[0,T]}\left|\int^t_0M^{-1}_s\zeta_s^{(2)}\dif W_s\right|\right\|_{q_0}\\
&=:I_1+I_2+I_3+I_4.\end{aligned}$$ Noticing that by , $$\begin{aligned}
\label{C3}
|\bar\beta_s|\leq|\beta_s|,\ \ |\bar\alpha_s|\leq|\alpha_s|,\end{aligned}$$ for any $p\in\mR$, by , Hölder’s inequality and Doob’s maximal inequality, we have $$\begin{aligned}
\label{Expo}
\mE\left(\sup_{t\in[0,T]}M_t^p\right)<\infty.\end{aligned}$$ Thus, by Hölder’s inequality and , we have $$I_1\leq C\|\zeta_0\|_{q_1},\ \ 
I_2\leq C\left\|\int^T_0|\zeta^{(1)}_s|\dif s\right\|_{q_2},$$ and by , $$\begin{aligned}
I_3&\leq\left\|M^*_T(M^{-1})^*_T\left(\int^T_0|\alpha_s|^2\dif s\right)^{1/2}\left(\int^T_0|\zeta_s^{(2)}|^2\dif s\right)^{1/2}\right\|_{q_0}\\
&\leq C\left\|\left(\int^T_0|\zeta_s^{(2)}|^2\dif s\right)^{1/2}\right\|_{q_3}=C\left\|\int^T_0|\zeta_s^{(2)}|^2\dif s\right\|^{1/2}_{q_3/2}.\end{aligned}$$ Similarly, by Hölder and Burkholder’s inequalities, we also have $$I_4\leq C\left\|\int^T_0|\zeta_s^{(2)}|^2\dif s\right\|^{1/2}_{q_3/2}.$$ Combining the above estimates, we obtain .

Let $f$ be a locally integrable function on $\mR^d$. The Hardy-Littlewood maximal function is defined by $$\cM f(x):=\sup_{0<r<\infty}\frac{1}{|B_r|}\int_{B_r}f(x+y)\dif y,$$ where $B_r:=\{x\in\mR^{d}: |x|<r\}$. The following result can be found in [@Cr-De Appendix A]. \[Le2\] (i) There exists a constant $C_d>0$ such that for all $f\in C^\infty(\mR^{d})$ and $x,y\in \mR^{d}$, $$\begin{aligned}
|f(x)-f(y)|\leq C_d |x-y|(\cM|\nabla f|(x)+\cM|\nabla f|(y)).\label{Es2}\end{aligned}$$ (ii) For any $p>1$, there exists a constant $C_{d,p}$ such that for all $f\in L^p(\mR^d)$, $$\begin{aligned}
\|\cM f\|_p\leq C_{d,p}\|f\|_p.\label{Es30}\end{aligned}$$

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[^1]: This work is supported by NNSFs of China (Nos. 11271294, 11325105)
