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 