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 