\section{Introduction and formulation of the problem}\label{intro}

We study slightly compressible fluid flows in rotating porous media.
The fluid has density $\rho$, velocity $v$, and pressure $p$.
The porous media is rotated with a constant angular velocity $\tilde\Omega \vec k$, with the unit vector $\vec k$ being the axis of rotation, and
$\tilde\Omega\ge 0$ being the angular speed. Let $x$ be the coordinate vector of a position in the rotating frame.

The equation for fluid flows written in the rotating frame, see e.g. \cite{VadaszBook}, is
\begin{equation}\label{Drot}
\frac{\mu}{k}v+ c\rho \tilde\Omega \vec k\times v + \rho \tilde\Omega^2 \vec k\times (\vec k\times x)=-\nabla p+\rho \vec g,
\end{equation}
where $\mu$ is the dynamic viscosity, $k$ the permeability,
 $c\rho \tilde\Omega \vec k\times v$ the Coriolis acceleration,
$\tilde\Omega^2 \vec k\times (\vec k\times x)$ the centripetal acceleration,
and $\vec g$ the gravitational acceleration.

The basic assumption for equation \eqref{Drot} is that the flows obey the Darcy's law
\be
\frac{\mu}{k}v=-\nabla p+\rho \vec g.
\end{equation}

However, in many situations, for instance, when the Reynolds number is large, this assumption is invalid. Instead, Forchheimer equations \cite{Forchh1901,Forchheimerbook} are usually used to model the flows in these cases. For example, the two-term Forchheimer's law states that
\begin{equation}
av+b|v|v=-\nabla +\rho \vec g,
\end{equation}
where  $a, b>0$ are some physical parameters.
(See also Forchheimer's three-term and power laws in, e.g., \cite{MuskatBook,BearBook,NieldBook}.)

A general form of the Forchheimer equations, taking into account Muskat's dimension analysis \cite{MuskatBook}, is
\begin{equation}\label{Fnr}
 \sum_{i=0}^N a_i \rho^{\alpha_i} |v|^{\alpha_i} v=-\nabla p +\rho \vec g.
\end{equation}
Here we focus on the explicit dependence on the density, leaving the dependence on the dynamic viscosity and permeability encoded in the coefficients $a_i$'s.

The interested reader is referred to the books \cite{BearBook,NieldBook,StraughanBook} for more information about the Forchheimer equations and a larger family of Brinkman-Darcy-Forchheimer equations. For their mathematical analysis in the case of incompressible fluids, see e.g. \cite{Zabensky2015a,ChadamQin,Payne1999a,Payne1999b,MTT2016,CKU2006,KR2017} and references therein. For the treatments of compressible fluids, see \cite{ABHI1,HI1,HI2,HIKS1,HKP1,HK1,HK2,CHK1,CHK2,CH1,CH2}. It is noted that the Forchheimer flows have drawn much less attention of mathematical research compared to the Darcy flows, and among the papers devoted to them, the number of those on compressible fluids is much smaller than that on the incompressible one.

Then the  equation for the rotating flows corresponding to \eqref{Fnr}, written in the rotating frame, is
 \begin{equation}\label{FM}
 \sum_{i=0}^N a_i \rho^{\alpha_i} |v|^{\alpha_i} v+ \frac{2\rho \tilde\Omega}{\tilde\phi} \vec k\times v+\rho \tilde\Omega^2 \vec k\times (\vec k\times x)=-\nabla p +\rho \vec g,
 \end{equation}
where $\tilde \phi\in(0,1)$ is the constant porosity.
In particular, when $N=1$, the specific Forchheimer's two-term law for rotating fluids \cite{Ward64,VadaszBook} is
 \begin{equation}
\frac{\mu}{k} v+ \frac{c_F \rho}{\sqrt{k}}|v|v + \frac{2\rho \tilde\Omega}{\tilde\phi} \vec k\times v+\rho \tilde\Omega^2 \vec k\times (\vec k\times x)=-\nabla p +\rho \vec g,
 \end{equation*}
where $c_F$ is the Forchheimer constant.
Even in this case, there is no mathematical analysis for compressible fluids in literature.

We make one simplification  in \eqref{FM}: replacing $\displaystyle \frac{2\rho \tilde\Omega}{\tilde\phi} \vec k\times v$  with
\begin{equation}\label{sim}
 \frac{2\rho_* \tilde\Omega}{\tilde\phi} \vec k\times v, \quad \rho_*=const.\ge 0.
\end{equation}

We then approximate equation \eqref{FM} by
 \begin{equation}\label{FMR}
 \sum_{i=0}^N a_i \rho^{\alpha_i} |v|^{\alpha_i} v+ \mathcal R \vec k\times v+\rho \tilde\Omega^2 \vec k\times (\vec k\times x)=-\nabla p +\rho \vec g,
 \end{equation}
where
\begin{equation}\label{Rdef}
\mathcal R=\frac{2\rho_* \tilde\Omega}{\tilde\phi}=const.\ge 0.
\end{equation}

Let $g:\mathbb{R}^+\rightarrow\mathbb{R}^+$ be a generalized polynomial  defined by
\begin{equation}\label{eq2}
g(s)=a_0 + a_1s^{\alpha_1}+\cdots +a_Ns^{\alpha_N}=\sum_{i=0}^N a_i s^{\alpha_i}\quad\text{for } s\ge 0,
\end{equation}
where $N\ge 1$ is an integer, the powers $\alpha_0=0<\alpha_1<\alpha_2<\ldots<\alpha_N$ are real numbers,
and the coefficients $a_0,a_N>0$ and $a_1,a_2,\ldots,a_{N-1}\ge 0$ are constants.

Then equation \eqref{FMR} can be rewritten as
\begin{equation}\label{FM1}
 g(\rho|v|) v+ \mathcal R \vec k\times v=-\nabla p +\rho \vec g - \rho \tilde\Omega^2 \vec k\times (\vec k\times x).
\end{equation}

Multiplying both sides of  \eqref{FM1}  by $\rho$ gives
 \begin{equation}\label{neweq1}
 g(|\rho v|) \rho v + \mathcal R \vec k\times (\rho v)  =-\rho\nabla p+ \rho^2 \vec g- \rho^2 \tilde\Omega^2 \vec k\times (\vec k\times x).
 \end{equation}

We will solve  for $\rho v$ from \eqref{neweq1}, which is possible thanks to the following lemma.

\begin{lemma}\label{Fsolve}
Given any vector $k\in\mathbb R^3$, the function $F_0(v)\stackrel{\rm def}{=} g(|v|)v + k\times v$ is a bijection from $\mathbb R^3$ to $\mathbb R^3$.
\end{lemma}
\begin{proof}
Note that $F_0$ is a continuous function on $\mathbb R^3$ and
$$ \frac{F_0(v)\cdot v}{|v|}=\frac{g(|v|)|v|^2}{|v|}\to\infty \text{ as }|v|\to\infty.$$
Then it is well-known that $F_0(\mathbb R^3)=\mathbb R^3$, see e.g. \cite[Theorem 3.3]{Deimling2010}.

It remains to prove that $F_0$ is one-to-one.
Let $v, w\in \mathbb R^3$. We have
\begin{equation*}
\begin{split}
(F_0(v)-F_0(w))\cdot (v-w)
&= (g(|v|)v-g(|w|))\cdot (v-w)+ [k\times (v-w)]\cdot(v-w)\\
&=(g(|v|)v-g(|w|)w)\cdot (v-w) \\
&= a_0|v-w|^2+\sum_{i=1}^N a_i (|v|^{\alpha_i}v-|w|^{\alpha_i} w)\cdot (v-w).
\end{split}
\end{equation*}

By applying \cite[Lemma 4.4, p.~13]{DiDegenerateBook} to each
$(|v|^{\alpha_i}v-|w|^{\alpha_i} w)\cdot (v-w)$, for $i\ge 1$,
we then obtain
\begin{equation}\label{Fmono}
(F_0(v)-F_0(w))\cdot (v-w)\ge a_0|v-w|^2+\sum_{i=1}^N C_i a_i |v - w|^{\alpha_i+2},
\end{equation}
where $C_i>0$ depends on $\alpha_i$ and $a_i$, for $i=1,\ldots, N.$

If $F_0(v)=F_0(w)$, it follows the monotonicity \eqref{Fmono} that
\begin{equation*}
0=(F_0(v)-F_0(w))\cdot (v-w)\ge a_0 |v - w|^2,
\end{equation*}
which implies $v=w$.
\end{proof}

Let vector $\vec k=(k_1,k_2,k_3)$  be fixed now with $k_1^2+k_2^2+k_3^2=1$.
We denote by ${\mathbf J} $ the $3\times 3$ matrix representing the rotation $\vec k\times$, that is, ${\mathbf J} x=\vec k\times x$ for all $x\in\mathbb R^3$. Explicitly,
we have
\begin{equation}\label{JR}
{\mathbf J} =\begin{pmatrix}   0 & -k_3 & k_2\\ k_3 & 0 & -k_1 \\ -k_2& k_1 & 0 \end{pmatrix}
\text{ and }
{\mathbf J}^2=\begin{pmatrix}
     k_1^2-1&k_1 k_2 & k_1 k_3\\
     k_1 k_2&k_2^2-1&k_2 k_3\\
     k_1 k_3& k_2 k_3&k_3^2-1
    \end{pmatrix}.
\end{equation}

\begin{definition}\label{FX} Throughout the paper, the function $g$ in \eqref{eq2} is fixed. We define the function $F:\mathbb R^3\to\mathbb R^3$ by
\begin{equation}\label{Fdef}
F(v)= g(|v|)v + \mathcal R  {\mathbf J} v \quad \text{ for }v\in\mathbb R^3,
\end{equation}
and denote its inverse function, which exists thanks to Lemma \ref{Fsolve}, by
\begin{equation} \label{Xdef}
X=F^{-1}.
\end{equation}
\end{definition}

Since $F$ is odd, then so is $X$.
Returning to equation \eqref{neweq1}, we can invert
 \begin{equation}\label{new2}
\rho v= -X(\rho\nabla p - \rho^2 \vec g + \rho^2 \tilde\Omega^2 {\mathbf J}^2 x).
\end{equation}

The equation of state for (isothermal)  slightly compressible fluids is
   \begin{equation}\label{slight}
\frac{1}{\rho}   \frac{d\rho}{dp}=\kappa,
\quad \text{where the constant compressibility } \kappa>0 \text { is small}.
   \end{equation}

Using \eqref{slight}, we write \eqref{new2} as
\begin{equation}\label{rveq}
\rho v= -X(\kappa^{-1} \nabla \rho - \rho^2 \vec g+ \rho^2 \tilde\Omega^2 {\mathbf J}^2 x).
\end{equation}

Consider the equation of continuity
\begin{equation}\label{eq5}
\tilde \phi\frac{\partial \rho}{\partial t} +\nabla\cdot(\rho v)=0,
\end{equation}
where $t$ is the time variable.
Combining \eqref{eq5} with \eqref{rveq} gives
\begin{equation}\label{eq0}
\tilde \phi\frac{\partial \rho}{\partial t}=\nabla\cdot(X(\kappa^{-1} \nabla \rho - \rho^2 \vec g+ \rho^2 \tilde \Omega^2 {\mathbf J}^2 x)).
\end{equation}

In the rotating frame, the gravitational field $\vec g$ becomes $\vec g(t)=\tilde{\mathcal G} e_0(t)$,
with  the unit vector
\begin{equation}
e_0(t)=(- \sin \theta \cos(\tilde\Omega t+\omega_0), -\sin\theta \sin(\tilde\Omega t+\omega_0), \cos \theta) ,
\end{equation}
where $\theta\in[0,\pi]$ is the fixed angle between $\vec g$ and $\vec k$, the number $\omega_0=const.$,
and $\tilde{\mathcal G}>0$ is the gravitational constant.

We make a simple change of variable  $u=\rho/\kappa$. Then we obtain from \eqref{eq0} the partial differential equation (PDE)
\begin{equation}\label{ueq}
\phi u_t=  \nabla \cdot\Big (X\big(\nabla u+u^2 \mathcal Z(x,t)\big)\Big),
\end{equation}
where
\begin{equation}\label{Zx}
\mathcal Z(x,t)=-\mathcal G e_0(t)+ \Omega^2  {\mathbf J}^2 x,
\end{equation}
\begin{equation}
\phi =\kappa\tilde \phi,\quad
\mathcal G=\kappa^2\tilde{\mathcal G},\quad \Omega=\kappa \tilde \Omega.
\end{equation}

To reduce the complexity in our mathematical treatment, hereafter, we consider the involved parameters and all equations to be non-dimensional. This is allowed by using appropriate scalings.

In this paper, we study the initial and boundary value problem (IBVP) for equation \eqref{ueq}.
More specifically, let $U$ be an open, bounded  set in $\mathbb R^3$ with $C^1$ boundary $\Gamma=\partial U$.
We study the following problem
\begin{equation}\label{ibvpg}
\begin{aligned}
\begin{cases}
\phi u_t=\nabla\cdot \Big(X\big(\nabla u+u^2 \mathcal Z(x,t)\big)\Big)\quad &\text{in}\quad U\times (0,\infty)\\
u(x,0)=u_0(x) \quad &\text{in}\quad U\\
u(x,t)= \psi(x,t)\quad &\text{in}\quad \Gamma\times (0,\infty),
\end{cases}
\end{aligned}
\end{equation}
where the initial data $u_0(x)$ and the Dirichlet boundary data $\psi(x,t)$ are given.

We will focus on the mathematical analysis of problem \eqref{ibvpg}.
We obtain various estimates of the solution in terms of the initial and boundary data.
These estimates show how the solutions, in space and time, can be controlled by the initial and boundary data.
We emphasize that the dependence on the problem's key parameters, including the angular speed of rotation, are expressed explicitly in our results.

The paper is organized as follows.
In section \ref{Prem}, we establish basic properties of the function $X$ which are crucial to our understanding of problem \eqref{ibvpg}.
They reveal the nonlinear structure and the degeneracy of the nonlinear parabolic equation \eqref{ueq}. Moreover, they have explicit dependence on the physical parameters, which, as stated above, is an important goal of this paper.
In section \ref{maxsec}, we prove the maximum principle for non-negative solutions of equation \eqref{ueq} in Theorem \ref{maxpr}. Using this, we derive the maximum estimates for non-negative solutions of the IBVP \eqref{ibvpg} in Corollary \ref{maxcor}.
Section \ref{Gradprep} contains the Lady\v{z}enskaja-Ural{$'$}\def\cprime{$'$} \def\cprime{$'$}ceva-typed embedding, Theorem \ref{LUembed}, with the weight $K[w,Q]$ which is related to the type of degeneracy of the nonlinear PDE \eqref{ueq}. This is one of the key tools in obtaining higher integrability for the gradient later.
In section \ref{L2asub}, we establish the estimate for the $L^{2-a}_{x,t}$-norm of the gradient in Theorem \ref{L2apos}. It was done through the $\mathcal K$-weighted $L^2$-norm first, see Proposition \ref{L2a}, and then by the interpretation of the weight $\mathcal K$.
In section \ref{gradsec}, we estimate the $L^{s}_{x,t}$-norms of the gradient, which is interior in the spatial variables,  for any finite number $s>2-a$. Specifically, we obtain estimates for $2-a<s\le 4-a$ in subsection \ref{L4ma}, and for $s>4-a$ in subsection \ref{Lhigher}.
We use the iteration method by Lady\v{z}enskaja and Ural{$'$}\def\cprime{$'$} \def\cprime{$'$}ceva \cite{LadyParaBook68}. This is a classical technique but, with suitable modifications based on the structure of equation \eqref{ueq}, applies well to our complicated nonlinear PDE. Moreover, it is sufficiently explicit to allow us to track all the necessary constants.
Section \ref{maxintime} is devoted to the estimates for the the gradient's $L^\infty_tL^s_x$-norms, which, of course, are stronger norms than those in the previous section.

It is worth mentioning that the derived estimates in this paper are already complicated, therefore, we strive to make them coherent, and hence more digestible, rather than sharp.

Concerning the simplification \eqref{sim}, it is a common strategy when encountering a new nonlinear problem. As presented above, it allows us to formulate the whole fluid dynamics system as a scalar parabolic equation \eqref{ueq}.
Such approximation, usually with some average density $\rho_*$, makes the problem much more accessible, while still gives insights on the flows' behaviors.
More importantly, this approach prompts the way to analyze the full model. Indeed, in the general case, the $\mathcal R$ in \eqref{Rdef} becomes $\mathcal R(u)$, and the PDE \eqref{ueq} becomes
\begin{equation}\label{unext}
\phi u_t=  \nabla \cdot\Big (X\big(u,\nabla u+u^2 \mathcal Z(x,t)\big)\Big),
\end{equation}
with $X(u, y)$ defined in the same way as \eqref{Xdef}.
Therefore, we can reduce the fluid dynamics system to a scalar PDE again. Furthermore, the properties of function $X$ established in subsection \ref{Xpropsec}  with explicit dependence on $\mathcal R=\mathcal R(u)$,  and other $X$-related results in section \ref{Gradprep} will play fundamental roles in understating the structure of the PDE \eqref{unext}.
This will be pursued and reported in a sequel (part II) of this paper.

Finally, in spite of our focus on the slightly compressible fluids, the techniques developed in the current paper can be combined with those in our previous work \cite{CHK1,CHK2} to model and analyze other types of compressible fluid flows such as the rotating isentropic flows for gases.

\section{Preliminaries}\label{Prem}

This section contains prerequisites and basic results on function $X$.

\subsection{Notation and elementary inequalities}
A vector $x\in\mathbb R^3$ is denoted by a $3$-tuple $(x_1,x_2,x_3)$ and considered as a column vector, i.e., a $3\times 1$ matrix. Hence $x^{\rm T}$ is the $1\times 3$ matrix $(x_1\ x_2\ x_3)$.

For a function $f=(f_1,f_2,\ldots,f_m):\mathbb R^n\to\mathbb R^m$, its derivative is the $m\times n$ matrix
\begin{align*}
f'=Df=\Big(\frac{\partial f_i}{\partial x_j}\Big)_{1\le i\le m,\,1\le j\le n}.
\end{align*}

In particular, when $n=3$ and $m=1$, i.e.,  $f:\mathbb R^3\to \mathbb R$, the derivative is
$$f'=Df=(f_{x_1}\ f_{x_2}\ f_{x_3}),$$
while its gradient vector is
$$\nabla f=(f_{x_1},f_{x_2},f_{x_3})=(Df)^{\rm T}.$$
The Hessian matrix is $$D^2 f\stackrel{\rm def}{=} D(\nabla f)=  \Big(\frac{\partial^2 f}{\partial x_j \partial x_i} \Big)_{i,j=1,2,3}.$$

Interpolation inequality for integrals:
\begin{equation}\label{Lpinter}
\int |f|^sd\mu\le \Big(\int |f|^p d\mu\Big)^\frac{q-s}{q-p}\Big(\int |f|^q d\mu\Big)^\frac{s-p}{q-p}\text{ for }0<p<s<q.
\end{equation}

\medskip
For two vectors $x,y\in \mathbb R^3$, their dot product is $x\cdot y=x^{\rm T}y=y^{\rm T}x$, while $xy^{\rm T}$ is the $3\times 3$ matrix $(x_iy_j)_{i,j=1,2,3}$.

Let $\mathbf A=(a_{ij})$ and $\mathbf B=(b_{ij})$ be any $3\times 3$ matrices of real numbers. Their inner product is
\begin{equation*}
\mathbf A:\mathbf B\stackrel{\rm def}{=} {\rm trace}(\mathbf A\mathbf B^{\rm T})=\sum_{i,j=1}^3 a_{ij}b_{ij}.
\end{equation*}

The Euclidean norm of the matrix $\mathbf A$ is $$|\mathbf A|=(\mathbf A:\mathbf A)^{1/2}=(\sum_{i,j=1}^3 a_{ij}^2)^{1/2}.$$
(Note that we do not use $|\mathbf A|$ to denote the determinant in this paper.)

When $\mathbf A$ is considered as a linear operator, another norm is defined by
\begin{equation}\label{opnorm}
|\mathbf A|_{\rm op}=\max\Big\{ \frac{|Ax|}{|x|}:x\in\mathbb R^3,x\ne 0\Big\} = \max\{ |Ax|:x\in\mathbb R^3,|x|=1\}.
\end{equation}

We have the following inequalities
\begin{align}
\label{mm0}
|\mathbf A y|&\le |\mathbf A|\cdot |y| \quad\text{ for any  }y\in\mathbb R^3,\\
\label{mop}
|\mathbf A y|&\le |\mathbf A|_{\rm op}\cdot |y|\quad  \text{ for any  }y\in\mathbb R^3,\\
\label{mmi}
|\mathbf A \mathbf B|&\le |\mathbf A|\cdot |\mathbf B|
\end{align}

It is also known that
\begin{equation}\label{nnorms}
|\mathbf A|_{\rm op}\le |\mathbf A|\le c_*|\mathbf A|_{\rm op},
\end{equation}
where $c_*$ is a positive constant independent of $\mathbf A$.

In particular, for matrix ${\mathbf J}$, we observe, for any $x\in\mathbb R^3$, that
\begin{equation}\label{Jxineq}
|{\mathbf J} x|\le |\vec k|\,|x|=|x|\quad\text{ and }\quad |{\mathbf J}^2x|\le |{\mathbf J} x|\le |x|.
\end{equation}
By choosing $x\ne 0$ perpendicular to $\vec k$, we conclude, for the norm \eqref{opnorm}, that
$$|{\mathbf J}|_{\rm op}=|{\mathbf J}^2|_{\rm op}=1.$$
For the Euclidean norm,  we have, from explicit formulas in \eqref{JR}, that
\begin{equation}\label{Jnorm}
|{\mathbf J}|^2=2|\vec k|^2=2,
\end{equation}
\begin{equation}\label{J2norm}
|{\mathbf J}^2|^2=3-2|\vec k|^2+|\vec k|^4=2.
\end{equation}

\medskip
We recall below some more elementary inequalities that will be used in this paper. First,
\begin{equation}\label{ee3}
 \frac{x^p+y^p}2\le (x+y)^p\le 2^{(p-1)^+}(x^p+y^p)\quad  \text{for all } x,y\ge 0,\quad  p>0,
\end{equation}
where $z^+=\max\{z, 0\}$ for any $z\in\mathbb R$.
Particularly,
\begin{equation}\label{ee2}
(x+y)^p\le 2^p(x^p+y^p)\quad  \text{for all } x,y\ge 0,\quad p>0,
\end{equation}
\begin{equation}\label{ee4}
x^\beta \le x^\alpha+x^\gamma\quad \text{for all } 0\le \alpha\le \beta\le\gamma.
\end{equation}
\begin{equation}\label{ee5}
x^\beta \le 1+x^\gamma \quad \text{for all } 0\le \beta\le\gamma.
\end{equation}

By the triangle inequality and the second inequality of \eqref{ee3}, we have
\begin{equation}\label{ee6}
|x-y|^p\ge 2^{-(p-1)^+}|x|^p-|y|^p \quad \text{for all } x,y\in\mathbb R^n,\quad p>0.
\end{equation}

\subsection{Properties of the function $X$}\label{Xpropsec}
It is obvious that the structure of the parabolic equation \eqref{ueq} depends greatly on the properties of the function $X$.
Thus, this subsection is devoted to studying $X$.

Recall that the functions $F$ and $X$ are defined in Definition \ref{FX}.
Throughout the paper, we denote
\begin{equation} \label{adef}
a=\frac{\alpha_N}{1+\alpha_N}\in(0,1),
\end{equation}
\begin{equation}\label{defxione}
\chi_0=g(1)=\sum_{i=0}^N a_i\text{ and }\chi_1=g(1)+\mathcal R=\chi_0+\mathcal R.
\end{equation}

\begin{lemma}\label{lem21}

{\rm (i)} One has
\begin{equation}\label{X0}
\frac{c_1\chi_1^{-1}|y|}{(1+|y|)^a}\le |X(y)|\le \frac{c_2\chi_1^a|y|}{(1+|y|)^a} \text{ for all }y\in \mathbb R^3,
\end{equation}
where
$c_1=\min\{1,\chi_0\}^a $ and $c_2=2^a c_1^{-1}\min\{a_0,a_N\}^{-1}$.
Alternatively,
\begin{equation}\label{X1}
 \chi_1^{-(1-a)}|y|^{1-a}-1\le |X(y)|\le c_3|y|^{1-a} \text{ for all }y\in \mathbb R^3,
 \end{equation}
where $c_3=(a_N)^{a-1}$.

{\rm (ii)} One has
\begin{equation}\label{X2}
\frac{c_4 \chi_1^{-2} |y|^2}{(1+|y|)^a}  \le X(y)\cdot y\le \frac{c_2 \chi_1^a |y|^2}{(1+|y|)^a} \text{ for all }y\in \mathbb R^3,
\end{equation}
where $c_4=(\min\{1,a_0,a_N\}/2^{\alpha_N})^{1+a}$.
Alternatively,
\begin{equation}\label{X3}
c_5\chi_1^{-2} (|y|^{2-a}-1)  \le X(y)\cdot y\le c_3|y|^{2-a}\text{ for all }y\in \mathbb R^3,
\end{equation}
where $c_5=2^{-a}c_4$.
\end{lemma}
\begin{proof}
Let $y\in\mathbb R^3$ and $v=X(y)$. Then, by \eqref{Xdef},
\begin{equation}\label{Fvy} F(v)=y.\end{equation}

(i) Since $g(|v|)v$ and $\vec k\times v$ are orthogonal, we have from \eqref{Fdef} and \eqref{Fvy} that
\begin{equation}
 |y|^2 = g(|v|)^2|v|^2+\mathcal R^2|{\mathbf J} v|^2.
\end{equation*}
This and \eqref{Jxineq} show that
\begin{equation}
g(|v|)^2|v|^2\le |y|^2\le (g(|v|)^2+\mathcal R^2)|v|^2.
\end{equation*}
Thus,
\begin{equation}\label{gf1}
g(|v|)|v|\le |y|\le (g(|v|)+\mathcal R)|v|.
\end{equation}

\textit{Proof of \eqref{X1}.}
From the first inequality in \eqref{gf1},
\begin{equation}\label{vf}
|y| \ge g(|v|)|v|\ge a_N |v|^{\alpha_N+1}, \text{ which implies }|v|\le (a_N^{-1} |y|)^\frac{1}{\alpha_N+1}=c_3|y|^{1-a}.
\end{equation}
So we obtain the second inequality of \eqref{X1}.
From \eqref{gf1},
\begin{equation}\label{fov}
|y|\le \chi_1(1+|v|)^{\alpha_N+1}.
\end{equation}
Then we obtain the first inequality in \eqref{X1}.

\textit{Proof of \eqref{X0}.}
Since $X(0)=0$, we consider only $y,v\ne 0$.

\textit{Case $|v|>1$.} By \eqref{gf1}, $|y|> g(1)\cdot 1=\chi_0$.
Furthermore, by \eqref{vf},
\begin{equation}\label{v1}
|v|\le (a_N^{-1}|y|)^{1-a}=\frac{2^a a_N^{a-1}|y|}{(|y|+|y|)^a}\le \frac{2^a \chi_1^a a_N^{-1}|y|}{(|y|+\chi_0)^a}\le \frac{2^a a_N^{-1}\chi_1^a|y|}{(\min\{1,\chi_0\})^a(|y|+1)^a}.
\end{equation}

On the other hand, we have from \eqref{gf1} that $|y|\le (g(1)|v|^{\alpha_N}+\mathcal R)|v|\le \chi_1 |v|^{\alpha_N+1}$. Then
\begin{equation}\label{v2}
|v|\ge (\chi_1^{-1}|y|)^{1-a}\ge \frac{\chi_1^{a-1}|y|}{|y|^a}\ge \frac{\chi_0^a \chi_1^{-1}|y|}{(|y|+1)^a}.
\end{equation}

\textit{Case $0<|v|\le 1$.} It follows \eqref{gf1} that
$a_0 |v|\le  |y|\le \chi_1 |v|\le \chi_1$.
Thus,
\begin{align}\label{v3}
|v|&\le a_0^{-1} |y|=\frac{a_0^{-1}|y|(|y|+\chi_0)^a}{(|y|+\chi_0)^a}\le \frac{a_0^{-1}|y|(\chi_1+\chi_0)^a}{(|y|+\chi_0)^a}
\le \frac{ a_0^{-1}(2\chi_1)^a|y|}{(\min\{1,\chi_0\})^a(|y|+1)^a},\\
\label{v4}
|v|&\ge \chi_1^{-1} |y|\ge \chi_1^{-1} |y| \frac{1}{(|y|+1)^a}.
\end{align}

From \eqref{v1} and \eqref{v3}, we obtain the second inequality in \eqref{X0}.
From \eqref{v2} and \eqref{v4}, we obtain the first inequality in \eqref{X0}.

(ii) Note that the second inequality of \eqref{X2}, respectively \eqref{X3}, follows the Cauchy-Schwarz inequality and the second inequality of \eqref{X0}, respectively \eqref{X1}. Thus, we focus on proving the first inequalities of \eqref{X2} and \eqref{X3}.
We have from \eqref{Fdef} and \eqref{gf1} that
\begin{equation*}
X(y)\cdot y=v\cdot F(v)=g(|v|)|v|^2
\ge  \frac{g(|v|)|y|^2}{(g(|v|)+\mathcal R)^2}  .
\end{equation*}

We estimate
\begin{align}
\label{g2}
g(|v|)+\mathcal R
&\le \chi_1(1+|v|)^{\alpha_N},\\
\intertext{and, by using \eqref{ee2},}
\label{g1}
g(|v|)
&\ge  \min\{a_0,a_N\} (1+|v|^{\alpha_N})
\ge \min\{a_0,a_N\} 2^{-\alpha_N}(1+|v|)^{\alpha_N}.
\end{align}
Hence,
\begin{align*}
\frac{g(|v|)}{(g(|v|)+\mathcal R)^2}
\ge   \frac{\min\{a_0,a_N\} 2^{-\alpha_N}}{\chi_1^2(1+|v|)^{\alpha_N}}.
\end{align*}
Note, by \eqref{ee3}, that we also have
\begin{equation*}
1+|y|\ge 1+g(|v|)|v|\ge \min\{1,a_N\}(1+|v|^{\alpha_N+1})
\ge \min\{1,a_N\}2^{-\alpha_N}(1+|v|)^{\alpha_N+1}.
\end{equation*}
Then,
\begin{equation}\label{vy}
(1+|v|)^{\alpha_N} \le  (1+|y|)^a (2^{\alpha_N}/\min\{1,a_N\})^a.
\end{equation}
Therefore,
\begin{equation}\label{gg}
\frac{g(|v|)}{(g(|v|)+\mathcal R)^2}
\ge \frac{\min\{a_0,a_N\} 2^{-\alpha_N}}{\chi_1^2(1+|y|)^a}(\min\{1,a_N\}2^{-\alpha_N})^a\ge \frac{c_4 \chi_1^{-2}}{(1+|y|)^a}.
\end{equation}
Hence we obtain the first inequality in \eqref{X2}.
Then the first inequality of \eqref{X3} follows this by considering $|y|\le 1$ and $|y|>1$ separately.
\end{proof}

\begin{remark}
Compared to \eqref{X1}, inequality in \eqref{X0} indicates $X(y)\to0$ as $|y|\to0$ at the rate $|y|$, while $X(y)\to\infty$ as $|y|\to\infty$ at a different rate $|y|^{1-a}$.
This refined form \eqref{X0} and its proof originate from \cite[Lemma 2.1]{CHIK1}.
\end{remark}

\begin{lemma}\label{Fpinv}
The function $F$ belongs to $C^1(\mathbb R^3)$, and the derivative matrix $F'(v)$ is invertible for each $v\in\mathbb R^3$.
Consequently, $X\in C^1(\mathbb R^3)$ and $X'(y)$ is invertible for each $y\in\mathbb R^3$.
\end{lemma}
\begin{proof}
Elementary calculations show that
\begin{align}
 \label{Fprime}
F'(v)&=g'(|v|)\frac{v v^{\rm T}}{|v|}+g(|v|)\mathbf I_3+ \mathcal R {\mathbf J}\text{ for }v\ne 0,\\
\label{Fp0}
F'(0)&=g(0)\mathbf I_3+\mathcal R {\mathbf J}.
\end{align}
Clearly, $F'(v)$ is continuous at $v\ne 0$, and $F'(v)\to F'(0)$ as $v\to 0$. Therefore, $F\in C^1(\mathbb R^3)$.

For $z\in \mathbb R^3$, we have
\begin{align*}
z^{\rm T} F'(v) z
&= \Big(g'(|v|)\frac{(z^{\rm T}v)^2}{|v|}+g(|v|)|z|^2\Big)\text{ for }v\ne 0,\\
z^{\rm T} F'(0) z&=g(0)|z|^2.
\end{align*}
Since $g'(s)>0$ for $s>0$, it follows that
\begin{equation}\label{Fzz}
z^{\rm T} F'(v) z \ge g(|v|)|z|^2 \text{ for all } v,z\in\mathbb R^3.
\end{equation}

Let $v\in\mathbb R^3$. If $F'(v)z=0$, then $0=z^{\rm T} F'(v)z\ge g(|v|)|z|^2$, which implies that $z=0$. Hence, $F'(v)$ is invertible.

By the Inverse Function Theorem, the statements for $X$ follow those for $F$.
\end{proof}

\begin{lemma}\label{Xder}
For any $y\in\mathbb R^3$, the derivative matrix $X'(y)$ satisfies
\begin{equation}\label{Xprime}
c_6\chi_1^{-1}(1+|y|)^{-a}\le |X'(y)|\le  c_7(1+\chi_1)^a (1+|y|)^{-a},
\end{equation}
\begin{equation}\label{hXh}
\xi^{\rm T} X'(y)\xi \ge c_8 \chi_1^{-2}(1+|y|)^{-a} |\xi|^2\text{ for all }\xi\in \mathbb R^3,
\end{equation}
where
\begin{align*}
c_6= \sqrt3(2^{-\alpha_N}\min\{1,a_N\})^a/(\alpha_N+2),\quad c_7=c_* 2^{\alpha_N}/\min\{a_0,a_N\},\quad
c_8=c_4/(\alpha_N+2)^2.
\end{align*}
\end{lemma}
\begin{proof}
Let $y\in\mathbb R^3$, then, by \eqref{Xdef},  $X'(y)=(F'(v))^{-1}$, with $F(v)=y.$
We first claim that
\begin{equation}\label{claim}
\frac{c_9}{g(|v|)+\mathcal R}\le |X'(y)|\le \frac{c_*}{g(|v|)},
\end{equation}
where $c_9=\sqrt3/(\alpha_N+2)$, and $c_*>0$ is the positive constant in \eqref{nnorms}.

Accepting \eqref{claim} for a moment, we prove the inequality \eqref{Xprime}.
Observe, by  \eqref{fov}, that
\begin{equation}\label{g5}
1+|y|\le (\chi_1+1)(1+|v|)^{\alpha_N}.
\end{equation}
On the one hand, \eqref{g1} and \eqref{g5} yield
\begin{equation}\label{g3}
g(|v|)
\ge \min\{a_0,a_N\} 2^{-\alpha_N} (1+|y|)^a/(1+\chi_1)^a.
\end{equation}
On the other hand, \eqref{g2} and \eqref{vy} give
\begin{equation}\label{g4}
g(|v|)+\mathcal R\le \chi_1(1+|y|)^a (2^{\alpha_N}/\min\{1,a_N\})^a.
\end{equation}
Then, by combining \eqref{claim}, \eqref{g3} and \eqref{g4}, we obtain \eqref{Xprime}.

We now prove the claim \eqref{claim}.

\medskip
\noindent\textit{Proof of the first inequality in \eqref{claim}.}
First, we consider $y,v\ne 0$. In \eqref{Fprime}, the matrix $g'(|v|)|v|^{-1}v v^{\rm T}+g(|v|)\mathbf I_3$ is symmetric, while $\mathcal R {\mathbf J}$ is anti-symmetric. Hence they are orthogonal, and, together with \eqref{Jnorm}, we have
\begin{equation*}
\begin{split}
|F'(v)|^2
&=\Big|g'(|v|)\frac{v v^{\rm T}}{|v|}+g(|v|)\mathbf I_3\Big|^2+\mathcal R^2 |{\mathbf J}|^2\\
&={\rm trace}\Big\{\big(g'(|v|)\frac{v v^{\rm T}}{|v|}+g(|v|)\mathbf I_3\big)^2\Big\}+2\mathcal R^2\\
&={\rm trace}\Big( (g'(|v|))^2v v^{\rm T} +2g'(|v|)g(|v|)\frac{v v^{\rm T}}{|v|} +(g(|v|))^2I_3\Big)+2\mathcal R^2.
\end{split}
\end{equation*}
Since trace$(vv^{\rm T})=|v|^2$, we have
\begin{align*}
|F'(v)|^2
&=(g'(|v|))^2 |v|^2 +2g'(|v|)g(|v|)|v| +3(g(|v|))^2+2\mathcal R^2\\
&=(g'(|v|) |v| +g(|v|))^2 +2g(|v|)^2+2\mathcal R^2.
\end{align*}

Note that
\begin{equation*}
g'(|v|)>0 \quad \text{ and }\quad
g'(|v|)|v|\le \alpha_N g(|v|).
\end{equation*}

Then
\begin{equation}\label{i1}
|F'(v)|^2\le ((\alpha_N+1)^2+2)g(|v|)^2 +2\mathcal R^2
\le (\alpha_N+2)^2 (g(|v|)+\mathcal R)^2.
\end{equation}

Similarly, we have from \eqref{Fp0} that
\begin{equation}\label{i2}
|F'(0)|^2=3g(0)^2+ 2\mathcal R^2\le 3 (g(0)+\mathcal R)^2 \le (\alpha_N+2)^2(g(0)+\mathcal R)^2 .
\end{equation}

From \eqref{i1} and \eqref{i2}, it follows
\begin{equation}\label{FpU}
|F'(v)|\le (\alpha_N+2)(g(|v|)+\mathcal R)\text{ for all }v\in \mathbb R^3.
\end{equation}

By \eqref{mmi}, $$\sqrt 3 = |\mathbf I_3| =| F'(v)X'(y) | \le |F'(v)|\cdot |X'(y)|,$$ which gives
\begin{align*}
|X'(y)|\ge \frac{\sqrt 3}{ |F'(v)|} \ge \frac{\sqrt3}{(\alpha_N+2)(g(|v|)+\mathcal R) }=\frac{c_9}{g(|v|)+\mathcal R}.
\end{align*}

\medskip
\noindent\textit{Proof of  the second inequality in \eqref{claim}.}
For $z\ne 0$, by Cauchy-Schwarz inequality and \eqref{Fzz}, we have
\begin{equation}\label{Fpz}
|F'(v)z|\ge \frac{|z^{\rm T} F'(v)z|}{|z|} \ge g(|v|)|z|.
\end{equation}

For any $\xi\in\mathbb R^3\setminus \{0\}$, applying \eqref{Fpz}  to $z=X'(y)\xi$, which is non-zero thanks to $X'(y)$ being invertible, gives
\begin{align*}
|\xi|\ge g(|v|)|X'(y)\xi|.
\end{align*}
Thus, the operator norm of $X'(y)$ is bounded by $|X'(y)|_{\rm op}\le 1/g(|v|)$,  and then, by relation \eqref{nnorms},
\begin{equation*}
|X'(y)|\le c_*|X'(y)|_{\rm op}\le c_*/g(|v|).
\end{equation*}

\medskip
\noindent\textit{Proof of \eqref{hXh}.}
Let $u=X'(y)\xi=(F'(v))^{-1}\xi$, which gives $\xi=F'(v)u$.
Rewriting $[X'(y)\xi]\cdot \xi$ in terms of $u,v$ and using property \eqref{Fzz}, we have
\begin{equation}\label{Xh1}
[X'(y)\xi]\cdot \xi= u\cdot [ F'(v)u ]\ge |u|^2 g(|v|)=|X'(y)\xi|^2 g(|v|).
\end{equation}
From \eqref{mm0} and \eqref{FpU}, for any $\xi\in\mathbb R^3$:
\begin{equation*}
|\xi|=|F'(v)X'(y)\xi|\le |F'(v)| \cdot |X'(y)\xi|\le (\alpha_N+2) (g(|v|)+\mathcal R)|X'(y)\xi|,
\end{equation*}
thus
\begin{equation}\label{Xh2}
|X'(y)\xi|\ge \frac{|\xi|}{(\alpha_N+2)(g(|v|)+\mathcal R)}.
\end{equation}
Combining \eqref{Xh1}, \eqref{Xh2} and \eqref{gg} yields
\begin{align*}
\xi^{\rm T}X'(y)\xi&\ge \frac{g(|v|)}{(\alpha_N+2)^2(g(|v|)+\mathcal R)^2} |\xi|^2\ge \frac{c_4\chi_1^{-2} |\xi|^2}{(\alpha_N+2)^2(1+|y|)^a}.
\end{align*}
This proves \eqref{hXh}.
\end{proof}

\section{Maximum estimates}\label{maxsec}

We will estimate the solutions of \eqref{ibvpg} by the maximum principle.
Denote \begin{equation}\label{Phi0} \Phi(x,t)=\nabla u(x,t)+u^2(x,t)\mathcal Z(x,t).\end{equation}

We re-write the PDE in \eqref{ibvpg} in the non-divergence form as
\begin{align*}
\phi u_t=X'(\Phi):(D\Phi)^{\rm T}
&=X'(\Phi):\big(D^2 u + u^2 D\mathcal Z(x,t)+ 2u \mathcal Z(x,t) Du\big)^{\rm T}\\
&=\mathbf A:\big (D^2 u + u^2 \Omega^2 {\mathbf J}^2 + 2u (\mathcal Z(x,t) Du)^{\rm T}\big ),
\end{align*}
where $\mathbf A=\mathbf A(x,t)= X'(\Phi(x,t))$.

We write $\mathbf A=\mathbf S+\mathbf T,$ where $\mathbf S$ and $\mathbf T$ are the symmetric and anti-symmetric parts of $\mathbf A$, i.e.,
$$\mathbf S=(s_{ij})_{i,j=1,2,3}=\frac12(\mathbf A+\mathbf A^{\rm T})\quad \text{and}\quad
\mathbf T=\frac12(\mathbf A-\mathbf A^{\rm T}).$$

Sine $D^2u$ is symmetric, we have
$\mathbf T:D^2u=0$, hence $\mathbf A:D^2u=\mathbf S :D^2u$.

Similarly, ${\mathbf J}^2$ is symmetric, and we have $\mathbf A:{\mathbf J}^2=\mathbf S:{\mathbf J}^2$.

Therefore,
\begin{equation}\label{usym}
\phi u_t=\mathbf S:D^2 u+u^2 \Omega^2 \mathbf S:{\mathbf J}^2 +  2u \mathbf A:(\mathcal Z(x,t) Du)^{\rm T}.
\end{equation}

Equation \eqref{usym} turns out to possess a maximum principle.
Recall that the parabolic boundary of $U\times(0,T]$ is
\begin{equation*}
\partial_p (U\times(0,T])=\bar U\times[0,T]\setminus U\times (0,T].
\end{equation*}

\begin{theorem}[Maximum principle]\label{maxpr}
Suppose $T>0$, and $u\in C(\bar U\times[0,T])\cap C_{x,t}^{2,1}(U\times (0,T])$ with $u\ge 0$ is a solution of \eqref{ueq} on $U\times (0,T]$. Then
 \begin{equation}\label{max-u}
 \max_{\bar U\times [0,T]} u =\max_{\partial_p (U\times(0,T])} u.
 \end{equation}
\end{theorem}
\begin{proof}
We make use of equation \eqref{usym} which is equivalent to \eqref{ueq}.
We examine the second term on the right-hand side of \eqref{usym}.
Direct calculations using the formula of ${\mathbf J}^2$  in \eqref{JR} give
$$\mathbf S:{\mathbf J}^2
=\sum_{i,j=1}^3 k_ik_j s_{ij} -\sum_{i=1}^3 s_{ii}=\vec k^{\rm T}\mathbf S\vec k -\rm{trace}(\mathbf S).$$

By \eqref{hXh}, we have
$\xi^{\rm T}\mathbf A\xi \ge 0$ for all $\xi\in\mathbb R^3$,
and, hence,
\begin{equation}\label{Axx}
\xi^{\rm T}\mathbf S\xi \ge 0\text{ for all }\xi\in\mathbb R^3.
\end{equation}

By  \eqref{Axx} and the fact $\mathbf S$ is symmetric, we have $\mathbf S\ge 0$ with eigenvalues $\lambda_1,\lambda_2,\lambda_3\ge 0$. Then, applying Cauchy-Schwarz's inequality and \eqref{mop} to $\vec k^{\rm T}\mathbf S \vec k$, we obtain
\begin{equation} \label{newpos}
\mathbf S:{\mathbf J}^2 \le |\mathbf S|_{\rm op} |\vec k|^2 -\rm{trace}(\mathbf S)=\max\{\lambda_1,\lambda_2,\lambda_3\}\cdot 1 - (\lambda_1+\lambda_2+\lambda_3)\le 0.
\end{equation}

Let $\varepsilon>0$. Set  $u^\varepsilon(x,t)=e^{-\varepsilon t}u(x,t)$ and
$\displaystyle M_\varepsilon=\max_{\bar U\times [0,T]} u^\varepsilon$.
We prove that
 \begin{equation}\label{maxe}
\max_{\bar U\times [0,T]} u^\varepsilon = \max_{\partial_p (U\times (0,T))} u^\varepsilon.
 \end{equation}

Suppose \eqref{maxe} is false. Then $M_\varepsilon>0$ and there exists a point $(x_0,t_0)\in U\times (0,T]$ for such that
$u^\varepsilon(x_0,t_0)=M_\varepsilon$. At this maximum point $(x_0,t_0)$ we have
\begin{equation}\label{vt1}
u^\varepsilon_t (x_0,t_0)\ge 0,
\quad \nabla u^\varepsilon(x_0,t_0)=0,\quad   D^2 u^\varepsilon(x_0,t_0)\le 0.
\end{equation}

We observe the followings:

(a) The second property of \eqref{vt1} implies $\nabla u(x_0,t_0)=0$.

(b) On the one hand, we have from \eqref{Axx} that $\mathbf S(x,t)\ge 0$ on $U\times(0,T]$.
On the other hand, the last property of \eqref{vt1} implies $D^2 u(x_0,t_0)\le 0$.
Then it is well-known that $\mathbf S(x_0,t_0): D^2u(x_0,t_0)\le 0$, see e.g. \cite[Chapter 2, Lemma 1]{FriedmanPara2008}.

From \eqref{usym}, \eqref{newpos}, and (a), (b), we obtain $u_t(x_0,t_0)\le 0$.
Therefore,
\begin{equation*}
u^\varepsilon_t(x_0,t_0)=-\varepsilon e^{-\varepsilon t_0} u(x_0,t_0) +e^{-\varepsilon t_0} u_t(x_0,t_0)
\le -\varepsilon  u^\varepsilon(x_0,t_0)=-\varepsilon M_\varepsilon <0.
\end{equation*}
This contradicts the first inequality in \eqref{vt1}, hence,  \eqref{maxe} holds true.
 Note that
 \begin{equation*}
e^{-\varepsilon T}\max_{\bar U\times [0,T]} u\le \max_{\bar U\times [0,T]} u^\varepsilon = \max_{\partial_p (U\times (0,T))} u^\varepsilon\le \max_{\partial_p (U\times (0,T))} u
\le \max_{\bar U\times [0,T]} u.
 \end{equation*}
Then letting $\varepsilon\to 0$ yields \eqref{max-u}.
\end{proof}

\begin{corollary}\label{maxcor}
 Let $u\in C(\bar U\times[0,T])\cap C_{x,t}^{2,1}(U\times (0,T])$ with $u\ge 0$ solve problem \eqref{ibvpg} on a time interval $[0,T]$ for some $T>0$. Then it holds for all $t\in[0,T]$ that
\begin{equation}\label{uM}
\max_{x\in\bar U}u(x,t)\le M_0(t)\stackrel{\rm def}{=} \sup_{x\in U} |u_0(x)|+\sup_{(x,\tau)\in \partial U \times (0,t]} |\psi(x,\tau)|.
\end{equation}
 \end{corollary}
 \begin{proof}
  Because of the continuity of $u$ on $\bar U\times [0,t]$, the quantity $M_0(t)$, in fact, equals the maximum of $u$ on $\partial_p(U\times(0,t])$. Then applying inequality \eqref{max-u} to $T=t$ yields estimate \eqref{uM}.
 \end{proof}

\section{Preparations for the gradient estimates}\label{Gradprep}

This section contains technical preparations for the estimates for different norms of the gradient $\nabla u$ in the next three  sections.

Given two mappings $Q:U\to\mathbb R^3$ and  $w:U\to \mathbb R$, we define a function $K[w,Q]$ on $U$ by
\begin{equation}\label{Ku}
K[w,Q](x)=(1+|\nabla w(x) + w^2(x) Q(x) |)^{-a}\quad\text{ for } x\in U.
\end{equation}

This function will be conveniently used in comparisons with $X(\nabla w(x) + w^2(x) Q(x))$ arising in the PDE \eqref{ueq}.

\begin{lemma}\label{kuglem}
If $s\ge a$, then the following inequalities hold on $U$:
\begin{align}
\label{kug1}
 K[w,Q]|\nabla w|^s
 &\le 2^{2s-a}|\nabla w|^{s-a}+2^{2s+1-a}(1+|w^2 Q|^s),\\
\label{kug2}
K[w,Q]|\nabla w|^s&\ge 3^{-1}|\nabla w|^{s-a} - 3^{-1} (1+|w^2 Q|^s).
\end{align}
\end{lemma}
\begin{proof}
We denote, in this proof, $K=K[w,Q]$.
Let $s\ge a$, by the triangle inequality and inequalities \eqref{ee2}, \eqref{ee5}, we have
\begin{align*}
 K|\nabla w|^s
 &\le 2^s K\cdot (|\nabla w +w^2 Q|^s+|w^2 Q|^s)\\
 &\le 2^s |\nabla w +w^2 Q|^{s-a}+ 2^s K|w^2 Q|^s\\
 &\le 2^s\cdot 2^{s-a}( |\nabla w|^{s-a} + |w^2 Q|^{s-a})+2^s K|w^2 Q|^s.
\end{align*}
Using inequality \eqref{ee5} and the fact $K\le 1$, we estimate
\begin{equation*}
|w^2 Q|^{s-a}\le 1+|w^2 Q|^s, \quad K|w^2 Q|^s\le 1+|w^2 Q|^s.
\end{equation*}
Hence,
\begin{equation}\label{Kw1}
 K|\nabla w|^s
\le  2^{2s-a}|\nabla w|^{s-a} +(2^{2s-a}+2^s)(1+|w^2 Q|^s).
\end{equation}
Noticing that $2^{2s-a}+2^s\le 2\cdot 2^{2s-a}$, we obtain \eqref{kug1} from \eqref{Kw1}.

To prove \eqref{kug2}, we write $ |\nabla w|^{s-a} = K|\nabla w|^{s-a} (1+|\nabla w+w^2 Q| )^a$, and apply
inequality \eqref{ee3} to have
\begin{align*}
 |\nabla w|^{s-a} &\le K|\nabla w|^{s-a} (1+|\nabla w|^a+|w^2 Q|^a )
 =K\cdot (|\nabla w|^{s-a} +|\nabla w|^s+|\nabla w|^{s-a}|w^2 Q|^a ).
\end{align*}

Concerning the last sum between the parentheses, applying inequality \eqref{ee5} to its first term gives
\begin{align*}
|\nabla w|^{s-a}\le 1 + |\nabla w|^s,
\end{align*}
and applying Young's inequality to its last term with the conjugate powers $s/(s-a)$ and $s/a$, when $s>a$, gives
\begin{align*}
|\nabla w|^{s-a}|w^2 Q|^a\le |\nabla w|^s+|w^2 Q|^s.
\end{align*}
Obviously, this inequality also holds when $s=a$. Thus,
\begin{equation}\label{pre-kug}
|\nabla w|^{s-a}
\le 3K|\nabla w|^s+K\cdot(1+|w^2 Q|^s)
\le 3K|\nabla w|^s+(1+|w^2 Q|^s).
\end{equation}
We obtain \eqref{kug2}.
\end{proof}

For our later convenience, we rewrite inequality \eqref{pre-kug}, by replacing $s-a$ with $s$,  as
\begin{equation}\label{kugs}
|\nabla w|^s\le 3K[w,Q]|\nabla w|^{s+a} + (1+|w^2 Q|^{s+a}) \text{ for all }s\ge 0.
\end{equation}

\begin{theorem}\label{LUembed}
For each $s\geq1$, there exists a constant $C>0$ depending only on $s$ and number $a$ in \eqref{adef} such that  for any function $w\in C^2(U)$ and non-negative function
$\zeta\in C_c^1(U)$, the following inequality holds
\begin{equation}\label{LUineq}
\begin{split}
\int_U K[w,Q] |\nabla w|^{2s+2} \zeta^2 dx
&\le
 C \sup_{\text{\rm supp} \zeta }(w^2) \Big \{
 \int_U K[w,Q] |D^2 w|^2 (|\nabla w|^{2s-2} +1)\zeta^2 dx\\
&\quad + \int_U K[w,Q] |\nabla w|^{2s}  \Big(|\nabla \zeta|^2 +(|w||Q'|+|Q|)^2\zeta^2 w^2 \Big) dx\Big\} \\
&\quad +C\int_U \Big( 1+  | w^2 Q|^{2s}\Big)  | w^2Q |^{2s+2}  \zeta^2 dx.
\end{split}
\end{equation}

Assume, in addition, that $Q$ and $Q'$ are bounded on $U$. Then,
\begin{equation}\label{ibt}
\begin{split}
&\int_U K[w,Q] |\nabla w|^{2s+2} \zeta^2 dx
\le
 C M_w^2 \int_U K[w,Q] |D^2 w|^2 |\nabla w|^{2s-2}  \zeta^2 dx\\
&\quad + CM_w^2 \int_U K[w,Q]|\nabla w|^{2s}  (|\nabla \zeta|^2 + (M_w\mu_Q+M_Q)^2 M_w^2 \zeta^2 ) dx \\
&\quad +  C{\rm sgn}(s-1) M_w^2 \int_U K[w,Q] |D^2 w|^2 \zeta^2 dx
+  C |U|(M_w^2 M_Q)^{2s+2}(1+M_w^2 M_Q)^{2s},
\end{split}
\end{equation}
where
\begin{align*}
M_w=\sup_{x\in\text{\rm supp} \zeta } |w(x)|,\quad M_Q=\sup_{x\in U} |Q(x)|,\quad \mu_Q=\sup_{x\in U} |Q'(x)|.
\end{align*}
\end{theorem}
\begin{proof}
For convenience in computing the derivatives, we will first establish \eqref{LUineq} with $K[w,Q]$ being replaced by the following function
\begin{equation}\label{Kstar}
K_*(x)= (1+ |\nabla w(x) + w^2(x)Q(x)|^2 )^{-a/2}.
\end{equation}

In this proof, the symbol $C$ denotes a generic positive constant depending only on $s$ and number $a$ in \eqref{adef}, while $C_\varepsilon>0$ depends on $s$, $a$ and $\varepsilon$.

We use Einstein's summation convention in our calculations.
Let
$$I =  \int_U K_*(x) |\nabla w(x)|^{2s+2} \zeta^2(x) dx.$$
By integration by parts and direct calculations, we see that
\begin{align*}
I
&=\int_U (K_*  |\nabla w|^{2s} \partial_i w \zeta^2)\cdot \partial_i w dx
 = - \int_U \partial_i (K_*  |\nabla w|^{2s} \partial_i w \zeta^2)\cdot  w dx\\
 &=I_1+I_2+I_3+I_4,
\end{align*}
where
\begin{align*}
I_1& = a \int_U \frac{(\nabla w + w^2Q) \cdot (\partial_i \nabla w +  2w\partial_i w Q+ w^2  \partial_i Q)   }{(1+ |\nabla w + w^2Q|^2)^{a/2+1}} |\nabla w|^{2s} \partial_i w\cdot  \zeta^2 \cdot w dx,\\
I_2&= - \int_U K_* |\nabla w|^{2s} \Delta w \cdot \zeta^2  \cdot w dx,\\
I_3&= - 2s \int_U K_* \Big(|\nabla w|^{2s-2} \partial_i\partial_j w \partial_j w\Big)\cdot  \partial_i w \cdot \zeta^2  \cdot w dx,\\
I_4&= -2 \int_U K_* |\nabla w|^{2s} \partial_i w \zeta \partial_i  \zeta \cdot w dx.
\end{align*}

Above, we used $\partial_i(|\nabla w|^{2s})=\partial_i(|\nabla w|^2)^s=s(|\nabla w|^2)^{s-1}\partial_i (|\nabla w|^2)$ to avoid possible singularities when $\nabla w=0$.

We estimate $I_1$ first. Observe that
\begin{equation*}
\frac{1}{(1+ |\nabla w + w^2Q|^2)^{a/2+1}}
= \frac{K_*}{1+|\nabla w + w^2Q|^2}\le\frac{2K_*}{(1+|\nabla w + w^2Q|)^2} .
\end{equation*}

By Cauchy-Schwarz and triangle inequalities, we have
\begin{align*}
I_1
&\le C \int_U \frac{K_* }{(1+|\nabla w + w^2Q|)^2} \cdot |\nabla w + w^2Q|\\
&\qquad \cdot \Big (|D^2 w| |\nabla w|^{2s+1}  +  |w||\nabla w|^{2s+2}|Q|+ w^2 |\nabla w|^{2s+1}|Q'| \Big)   \zeta^2 \cdot |w| dx \\
&\le  J_1+J_2+J_3,
\end{align*}
where
\begin{align*}
 J_1&= C\int_U \frac{K_* }{1+|\nabla w + w^2Q|}(|D^2 w| |\nabla w|^{2s+1}  )   \zeta^2 \cdot |w| dx,\\
 J_2&= C\int_U \frac{K_* }{1+|\nabla w + w^2Q|}( |wQ||\nabla w|^{2s+2})   \zeta^2 \cdot |w| dx,\\
 J_3&= C\int_U \frac{K_* }{1+|\nabla w + w^2Q|}(w^2 |Q'||\nabla w|^{2s+1} )   \zeta^2 \cdot |w| dx.
\end{align*}

For $J_1$, by using triangle inequality
$|\nabla w|\le  |\nabla w+ w^2  Q |+ |w^2  Q|,$
we have
\begin{equation*}
\begin{split}
J_1 &\le  C\int_U \frac{K_* }{1+|\nabla w + w^2  Q|}(|\nabla w+ w^2  Q |^{2s+1} + |w^2  Q|^{2s+1})  |D^2 w|    \zeta^2 |w| dx\\
&\le C  \int_U K_*  \Big( |\nabla w+ w^2  Q |^{2s}+    |w^2  Q|^{2s+1}\Big)  |D^2 w|    \zeta^2 |w| dx.
\end{split}
\end{equation*}
Applying triangle inequality and \eqref{ee2} gives
$$ |\nabla w+ w^2  Q |^{2s}\le C( |\nabla w|^{2s}+ |w^2  Q|^{2s}).$$
Thus,
\begin{equation*}
J_1 \le C \int_U K_*  |\nabla w|^{2s}  |D^2 w|    \zeta^2 |w| dx + C \int_U K_*  \Big( |w^2  Q |^{2s}+  |w^2  Q|^{2s+1}\Big)  |D^2 w|    \zeta^2 |w| dx.
\end{equation*}

Let $\varepsilon>0$. Denote
\begin{equation*}
I_5=\int_U K_* |D^2 w|^2|\nabla w|^{2s-2}\zeta^2 w^2 dx,\quad
I_6=\int_U K_* |D^2 w|^2\zeta^2 w^2 dx.
\end{equation*}

In the last inequality for $J_1$, we apply Cauchy's inequality to obtain
\begin{equation*}
\begin{split}
J_1 &\le C \int_U K_*  |\nabla w|^{2s}  |D^2 w|    \zeta^2 |w| dx + C \int_U K_*  \Big( |w^2  Q |^{2s}+  |w^2  Q|^{2s+1}\Big)  |D^2 w|    \zeta^2 |w| dx\\
&\le \varepsilon I+ C_\varepsilon I_5 +  C \int_U K_* \Big( |w^2  Q |^{4s}+  |w^2  Q|^{4s+2}\Big)    \zeta^2 dx
+CI_6.
\end{split}
\end{equation*}

Similarly,
\begin{equation*}
\begin{split}
J_2 &\le C \int_U \frac{K_* }{1+|\nabla w + w^2  Q|}(|\nabla w+ w^2  Q |^{2s+2} + |w^2  Q|^{2s+2})  \zeta^2 w^2 |Q|dx\\
&\le C  \int_U K_*  \Big( |\nabla w+ w^2  Q |^{2s+1}+    |w^2  Q|^{2s+2}\Big)    \zeta^2 w^2|Q| dx\\
&\le  C \int_U K_*  \Big( |\nabla w|^{2s+1}+ |w^2  Q |^{2s+1}+  |w^2  Q|^{2s+2}\Big)   \zeta^2 w^2|Q| dx.
\end{split}
\end{equation*}

For $J_3$, neglecting the denominator in the integrand gives
\begin{equation*}
J_3 \le  C\int_U K_* |\nabla w|^{2s+1}  \zeta^2 |w|^3|Q'| dx.
\end{equation*}

Combining the above estimates for $J_1$, $J_2$ and $J_3$ yields
\begin{align*}
 I_1\le \varepsilon I+ C_\varepsilon I_5 +CI_6+ J_4+J_5,
\end{align*}
where
\begin{align*}
 J_4&= C\int_U K_* |\nabla w|^{2s+1}  (|Q|+|w||Q'|) \zeta^2 w^2 dx,\\
J_5&=  C \int_U K_* \Big( |w^2  Q |^{4s}+  |w^2  Q|^{4s+2}\Big)    \zeta^2 dx
+ C\int_U K_*  \Big( |w^2  Q |^{2s+1}+  |w^2  Q|^{2s+2}\Big)   \zeta^2 w^2 |Q|dx.
\end{align*}

We estimate, by using Cauchy's inequality,
\begin{equation*}
J_4
\le \varepsilon I+C_\varepsilon\int_U K_* |\nabla w|^{2s}   (|Q|+|w||Q'|)^2 \zeta^2 w^4 dx,
\end{equation*}
and, with $K_*\le 1$,
\begin{align*}
 J_5&\le  C \int_U \Big( |w^2  Q |^{4s}+  |w^2  Q|^{4s+2}\Big)    \zeta^2 dx
+ C\int_U \Big( |w^2  Q |^{2s+1}+  |w^2  Q|^{2s+2}\Big)   \zeta^2 w^2 |Q|dx\\
&=C\int_U ( |w^2 Q |^{4s}+  |w^2 Q|^{4s+2}+ |w^2 Q |^{2s+2}+  |w^2 Q|^{2s+3}) \zeta^2 dx.
\end{align*}
Applying inequality \eqref{ee4} gives
\begin{equation*}
J_5\le    C\int_U \Big( | w^2Q |^{2s+2}+  | w^2 Q|^{4s+2}\Big)    \zeta^2 dx
= C\int_U \Big( 1+  | w^2 Q|^{2s}\Big)  | w^2Q |^{2s+2}  \zeta^2 dx.
\end{equation*}

Therefore,
\begin{align*}
 I_1&\le 2\varepsilon I+ C_\varepsilon I_5 +CI_6+C_\varepsilon\int_U K_* |\nabla w|^{2s} (|Q|+|w||Q'|)^2  \zeta^2  w^4 dx\\
&\quad + C\int_U \Big( 1+  | w^2 Q|^{2s}\Big)  | w^2Q |^{2s+2}  \zeta^2 dx.
\end{align*}

The terms $I_2$, $I_3$, $I_4$ can be bounded simply  by
\begin{align*}
I_2+I_3+I_4
&\le \int_U K_* |\nabla w|^{2s} |\Delta w| \zeta^2  |w| dx
 + 2 s \int_U K_* |\nabla w|^{2s} |D^2  w|  \zeta^2  |w| dx\\
&\quad +2 \int_U K_* |\nabla w|^{2s+1}  \zeta |\nabla \zeta| |w|dx\\
&\le C\int_U K_*  |\nabla w|^{2s} |D^2 w | \zeta^2 |w| dx
+2\int_U K_*  |\nabla w|^{2s+1}  \zeta|\nabla \zeta|\, |w| dx.
\end{align*}
Applying Cauchy's inequality to each integral gives
\begin{align*}
I_2+I_3+I_4
& \le (\varepsilon I + C_\varepsilon I_5)+\Big(\varepsilon I +C_\varepsilon \int_U K_*  |\nabla w|^{2s}  |\nabla \zeta|^2 w^2 dx\Big).
\end{align*}

Combining the estimates of $I_1$ and $I_2+I_3+I_4$, we have
\begin{equation*}
I \le  4\varepsilon I +C_\varepsilon I_5+CI_6
+  C_\varepsilon I_7 + C\int_U \Big( 1+  | w^2 Q|^{2s}\Big)  | w^2Q |^{2s+2}  \zeta^2 dx,
\end{equation*}
where
\begin{align*}
I_7=\int_U K_* |\nabla w|^{2s}  \Big(|\nabla \zeta|^2 +(|w||Q'|+|Q|)^2\zeta^2 w^2 \Big)w^2 dx.
\end{align*}

Selecting $\varepsilon=1/8$, we obtain
\begin{equation}\label{Ks}
I \le  C(I_5+I_6+I_7)
 +C\int_U \Big( 1+  | w^2 Q|^{2s}\Big)  | w^2Q |^{2s+2}  \zeta^2 dx.
\end{equation}

In each integral of $I_5$, $I_6$ and $I_7$, we bound
\begin{align*}
\zeta^2 w^2\le \zeta^2\sup_{\rm supp \zeta}(w^2),\quad |\nabla \zeta|^2 w^2\le |\nabla \zeta|^2 \sup_{\rm supp \zeta}(w^2).
\end{align*}
It then follows \eqref{Ks} that
\begin{equation}\label{LU0}
\begin{split}
\int_U K_* |\nabla w|^{2s+2} \zeta^2 dx
&\le
 C \sup_{\text{\rm supp} \zeta }(w^2) \Big \{
 \int_U K_* |D^2 w|^2 (|\nabla w|^{2s-2} +1)\zeta^2 dx\\
&\quad + \int_U K_* |\nabla w|^{2s}  \Big(|\nabla \zeta|^2 +(|w||Q'|+|Q|)^2\zeta^2 w^2 \Big) dx\Big\} \\
&\quad +C\int_U \Big( 1+  | w^2 Q|^{2s}\Big)  | w^2Q |^{2s+2}  \zeta^2 dx.
\end{split}
\end{equation}

We compare $K_*$ in \eqref{Kstar} with $K[w,Q]$ in \eqref{Ku}.
Because
\begin{equation*}
2^{-1}(1+ |\nabla w + w^2 Q|)^2 \le 1+ |\nabla w+ w^2 Q|^2 \le (1+ |\nabla w + w^2 Q|)^2,
\end{equation*}
then
\begin{equation}\label{KK}
K[w,Q](x)\le K_*(x)\le 2^{a/2} K[w,Q](x)\quad\forall x\in U.
\end{equation}

Applying the first, respectively second, inequality of \eqref{KK} to the left-hand side, respectively right-hand side, of \eqref{LU0}, we obtain inequality \eqref{LUineq}.

Now, consider the case  $w$, $Q$ and $Q'$ are bounded. By simple estimates of the last two integrals of \eqref{LUineq} using the numbers $M_w$, $M_Q$, $\mu_Q$, and by using the following estimate
$$|\nabla w|^{2s-2}+1\le 2|\nabla w|^{2s-2}+{\rm sgn}(s-1)$$
for the first integral on the right-hand side of \eqref{LUineq},
we obtain inequality  \eqref{ibt}.
\end{proof}

\section{Gradient estimates (I) }\label{L2asub}

This section is focused on \emph{a priori} estimates for the gradient of a solution $u(x,t)$ of the IBVP \eqref{ibvpg}.

Hereafter, we fix $T>0$. Let $u(x,t)$ be a $C_{x,t}^{2,1}(\bar U\times(0,T])\times C(\bar U\times[0,T])$ solution of \eqref{ibvpg}, not necessarily non-negative.

In the estimates of the Lebesgue norms below, we will use the energy method. For that, it is convenient to shift the solution to a function vanishing at the boundary.

Let $\Psi(x,t)$ be the extension of $\psi(x,t)$ from $\Gamma\times(0,T]$ to $\bar U\times[0,T]$. It is assumed to have necessary regularity needed for calculations below. All our following estimates, as far as the boundary data is concerned, will be expressed in terms of $\Psi$. This will not lose the generality since we can always translate them into $\psi$-dependence estimates, see e.g. \cite{HI1,JerisonKenig1995}.

Define $\bar u=u-\Psi$ and $\bar u_0=u_0-\Psi$.
We derive from \eqref{ibvpg} the equations for $\bar u$:
\begin{equation}\label{ubar}
\begin{aligned}
\begin{cases}
\phi \bar u_t=\nabla\cdot (X(\Phi(x,t))-\phi \Psi_t\quad &\text{on}\quad U\times (0,T],\\
\bar u(x,0)=\bar u_0(x) \quad &\text{on}\quad U,\\
\bar u(x,t)= 0\quad &\text{on}\quad \Gamma\times (0,T],
\end{cases}
\end{aligned}
\end{equation}
where $\Phi(x,t)$ is the same notation as \eqref{Phi0}, i.e.,
$$\Phi(x,t)=\nabla u(x,t)+u^2(x,t)\mathcal Z(x,t)=\nabla u(x,t)+W(x,t),$$
with $W(x,t)=u^2(x,t)\mathcal Z(x,t).$

The following ``weight'' function will play important roles in our statements and proofs:
$$\mathcal K(x,t)=K[u(\cdot,t),\mathcal Z(\cdot,t)](x)= (1+|\Phi(x,t)|)^{-a}.$$

We estimate the $L^{2-a}_{x,t}$-norm for $\nabla u$ in terms of the initial and boundary data.
Define
\begin{equation}\label{MZ}
M_{\mathcal Z}=\sup_{\bar U\times[0,T]}|\mathcal Z(x,t)|,
\end{equation}
\begin{equation}\label{Estar}
{\mathcal E}_0=\inttx{0}{T}{U} (\chi_1^{2(2+a)}|\nabla \Psi|^2+ \phi\chi_1^2 (|\Psi_t|^2+\Psi^2))dxdt.
\end{equation}

\begin{xnotation
In the remaining of this paper, the constant $C$ is positive and generic with varying values in different places.
It depends on number $N$, the coefficients $a_i$'s and powers $\alpha_i$'s of the function $g$ in \eqref{eq2}, the number $c_*$ in \eqref{nnorms}, and the set $U$.
In sections \ref{gradsec} and \ref{maxintime}, it further depends on number $s$, the subsets $U'$, $V$, $U_k$'s of $U$.
However, it does not depend on the initial and boundary data of $u$, the functions $\Phi(x,t)$, $\mathcal Z(x,t)$, and numbers $T$, $T_0$, $t_0$, $\phi$, $\mathcal R$, $\chi_*$, $M_*$, whenever these are introduced. In particular, it is independent of the cut-off function $\zeta$ in Lemmas \ref{lem61}, \ref{lem62}, \ref{lem67}, Proposition \ref{prop63}, and inequality \eqref{iterate2}.
\end{xnotation}

\begin{proposition}\label{L2a} One has
 \begin{align}
 \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt
 &\le  C\Big\{ \chi_1^2 \phi (\|\bar u_0\|_{L^2}^2+\|u\|_{L^2(U\times (0,T))}^2)\notag\\
 &\quad\qquad+(1+\chi_1)^{2(2+a)}M_{\mathcal Z}^2\|u\|_{L^4(U\times (0,T))}^4+{\mathcal E}_0\Big\},\label{gradu0}\\
 \inttx{0}{T}{U} |\nabla u|^{2-a} dxdt
 &\le  C \Big\{\chi_1^2 \phi (\|\bar u_0\|_{L^2}^2+\|u\|_{L^2(U\times (0,T))}^2) \notag\\
 &\quad\qquad +(1+\chi_1)^{2(2+a)}(T+M_{\mathcal Z}^2\|u\|_{L^4(U\times (0,T))}^4)+{\mathcal E}_0\Big\}. \label{gradu1}
 \end{align}
\end{proposition}
\begin{proof}
In this proof, we denote $J= \int_U \Psi_t \bar u dx$.
Multiplying the PDE \eqref{ubar} by $\bar u$, integrating over domain $U$ and using  integration by parts, we have
 \begin{align*}
\frac{\phi }{2} \frac{d}{dt} \int_U \bar u^2 dx
&=- \int_U X(\Phi)\cdot  \nabla \bar u dx-\phi J\\
&=-\int_UX(\Phi)\cdot \Phi dx
+\int_UX(\Phi)\cdot (\nabla \Psi + W) dx-\phi  J.
\end{align*}

By \eqref{X0} and \eqref{X2}, we have
\begin{align*}
\frac\phi 2  \frac{d}{dt} \int_U \bar u^2   dx&\le -c_4\chi_1^{-2}\int_U \mathcal K |\Phi|^2 dx
+c_2\chi_1^a \int_U \mathcal K |\Phi|(|\nabla \Psi|+|W|)dx-\phi  J.
\end{align*}
For the second integral on the right-hand side, applying Cauchy's inequality gives
\begin{align*}
\frac\phi 2  \frac{d}{dt} \int_U \bar u^2   dx&\le -c_4\chi_1^{-2}\int_U \mathcal K |\Phi|^2 dx\\
&\quad+\int_U \Big[\varepsilon\mathcal K|\Phi|^2 +C\varepsilon^{-1}\chi_1^{2a}\mathcal K(|\nabla \Psi|+|W|)^2\Big]dx-\phi  J.
\end{align*}

Choosing $\varepsilon=c_4\chi_1^{-2}/2$, we obtain
\begin{align}\label{ewJ}
\frac\phi 2  \frac{d}{dt} \int_U \bar u^2   dx&\le -\frac{c_4\chi_1^{-2}}{2}\int_U \mathcal K |\Phi|^2 dx+C \chi_1^{2(1+a)}\int_U  (|\nabla \Psi|^2+|W|^2)dx-\phi  J.
\end{align}

Note that
\begin{equation}\label{Wbound}
|W|\le M_{\mathcal Z} u^2.
\end{equation}

Writing $\bar u$ in $J$ as $u-\Psi$, and using Cauchy's inequality, we have
\begin{equation}\label{pJ}
\phi |J|\le \int_U \phi |\Psi_t| (|u|+|\Psi|) dx\le \phi \int_U (u^2+|\Psi_t|^2+|\Psi|^2)dx.
\end{equation}

Utilizing estimates \eqref{Wbound} and \eqref{pJ} in \eqref{ewJ}, we derive
\begin{align*}
\frac\phi 2  \frac{d}{dt} \int_U \bar u^2   dx
+\frac{c_4\chi_1^{-2}}{2}\int_U \mathcal K|\Phi|^2 dx
&\le C\chi_1^{2(1+a)}M_{\mathcal Z}^2\int_U u^4 dx\\
&\quad+C \chi_1^{2(1+a)}\int_U  |\nabla \Psi|^2dx
+\phi\int_U (u^2 + |\Psi_t|^2+\Psi^2)dx.
\end{align*}

Then integrating from $0$ to $T$ gives
\begin{equation}
\begin{aligned}\label{KP}
 \inttx{0}{T}{U} \mathcal K|\Phi|^2 dx dt
 &\le  C \chi_1^2 \phi \|\bar u_0\|_{L^2}^2 +C\chi_1^{2(2+a)}M_{\mathcal Z}^2\inttx{0}{T}{U} u^4 dxdt+C\phi \chi_1^2\inttx{0}{T}{U} u^2 dxdt\\
&\quad  +C{\mathcal E}_0.
 \end{aligned}
\end{equation}

For the left-hand side of \eqref{KP}, we use inequalities \eqref{ee6} and \eqref{Wbound} to have
\begin{align*}
\mathcal K|\Phi|^{2}=\mathcal K|\nabla u+W|^2\ge 2^{-1}\mathcal K |\nabla u|^2-\mathcal K|W|^2\ge 2^{-1} \mathcal K|\nabla u|^2-M_{\mathcal Z}^2 u^4.
\end{align*}

Combining this with \eqref{KP} yields
\begin{align*}
 \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dx
 &\le  2\inttx{0}{T}{U} \mathcal K|\Phi|^2 dx dt + 2M_{\mathcal Z}^2\inttx{0}{T}{U} u^4 dxdt\\
 &\le  C \chi_1^2 \phi \|\bar u_0\|_{L^2}^2 +C(1+\chi_1)^{2(2+a)}M_{\mathcal Z}^2\inttx{0}{T}{U} u^4 dxdt\\
 &\quad+C\phi\chi_1^2\inttx{0}{T}{U} u^2 dxdt
  +C{\mathcal E}_0,
 \end{align*}
which proves \eqref{gradu0}.

Using \eqref{kug2} and \eqref{Wbound}, we estimate the integrand on the left-hand side of \eqref{gradu0} by
  \begin{align*}
  \mathcal K|\nabla u|^2\ge 3^{-1}|\nabla u|^{2-a} -(1+|W|^2)\ge 3^{-1}|\nabla u|^{2-a} -(1+(u^2M_{\mathcal Z})^2).
  \end{align*}
  It results in
\begin{align*}
 \inttx{0}{T}{U} |\nabla u|^{2-a} dxdt
 &\le  3\inttx{0}{T}{U} \mathcal K|\nabla u|^2 dx + 3\inttx{0}{T}{U} (1+M_{\mathcal Z}^2 u^4) dxdt\\
 &\le  C \chi_1^2 \phi (\|\bar u_0\|_{L^2}^2+\|u\|_{L^2(U\times (0,T))}^2) \\
 &\quad+C(1+\chi_1)^{2(2+a)}\inttx{0}{T}{U} (1+M_{\mathcal Z}^2 u^4) dxdt+ C{\mathcal E}_0,
 \end{align*}
which proves  \eqref{gradu1}.
 \end{proof}

To have more specific estimates, we examine the bounds for the constituents of the PDE in \eqref{ubar}.
Note, from \eqref{Zx} and \eqref{Jxineq},  that
\begin{equation*}
|\mathcal Z(x,t)|
\le \mathcal G + \Omega^2 |{\mathbf J}^2 x|
\le \mathcal G + \Omega^2 r_0
=\mathcal G(1+\Omega_*^2),
\end{equation*}
where
\begin{equation} \label{r0}
r_0=\max\{|x|: x\in \bar U \}
\text{ and }
\Omega_*=\Omega \sqrt{r_0/\mathcal G}=\tilde\Omega\sqrt{ r_0/\tilde{\mathcal G}}.
\end{equation}
Thus, the number $M_{\mathcal Z}$ in \eqref{MZ} can be bounded by
\begin{equation}\label{MZ1}
 M_{\mathcal Z}\le \mathcal G(1+\Omega_*)^2.
\end{equation}

Next, it is obvious that
$D\mathcal Z(x,t)=\Omega^2 {\mathbf J}^2$,
hence, by \eqref{J2norm},
\begin{equation}\label{DZ}
|D\mathcal Z(x,t)|=\mu_{\mathcal Z}\stackrel{\rm def}{=} \sqrt 2 \Omega^2=\sqrt 2(\mathcal G/r_0) \Omega_*^2.
\end{equation}

We rewrite $\mathcal R$ in \eqref{Rdef} as
\begin{equation}\label{RR1}
\mathcal R=\frac{2\rho_* \Omega}{\phi}
=\frac{2\rho_* \sqrt{\mathcal G/r_0}}{\phi} \Omega_*
=\frac{2\rho_* \sqrt{ \tilde{\mathcal G}/r_0}}{\tilde\phi} \Omega_*.
\end{equation}

From \eqref{MZ1}, \eqref{DZ} and \eqref{RR1}, we conveniently relate the upper bounds of $\mathcal R$, $M_{\mathcal Z}$, $\mu_{\mathcal Z}$ to a single parameter $\chi_*$ as follows
\begin{equation}\label{RMm}
\mathcal R\le \chi_*\stackrel{\rm def}{=} \max\{1,d_*(1+\Omega_*)\},\quad M_{\mathcal Z}\le \chi_*^2,\quad \mu_{\mathcal Z}\le \chi_*^2,
\end{equation}
where
\begin{equation}
d_*
=\sqrt{\mathcal G/r_0}\max\Big\{\frac{2\rho_*}{\phi},\sqrt{r_0},2^{1/4}\Big\}.
\end{equation*}

The reason for \eqref{RMm}, with the choice of $\chi_*\ge 1$, is to simplify many estimates that will be obtained later, and specify the dependence of those estimates on the parameters $d_*$ and $\Omega_*$.

\begin{remark}
A very common situation is that the rotation is about the vertical axis, then $\vec k=\pm (0,0,1)$, and
\begin{equation}\label{Jmatrix}
{\mathbf J}=\pm \begin{pmatrix}   0 & -1 & 0\\ 1 & 0 & 0 \\ 0& 0 & 0 \end{pmatrix},\quad
{\mathbf J}^2=-\begin{pmatrix}
     1&0 & 0\\
     0&1&0\\
     0& 0&0
    \end{pmatrix}.
\end{equation}
Thanks to \eqref{Jmatrix}, we can, in this case, replace the $r_0$ in \eqref{r0} with a smaller number, namely,
$$r_0=\max\{(x_1^2+x_2^2)^{1/2}: x=(x_1,x_2,x_3)\in \bar U \}.$$
\end{remark}

\begin{definition}\label{BEN}
 We will use the following quantities in our estimates throughout the paper:
\begin{align*}

  M_*&=\sup_{ U\times[0,T]} |u|, \quad
{\mathcal E}_*=\inttx{0}{T}{U} (|\nabla \Psi|^2+\phi |\Psi_t|^2+\phi \Psi^2)dxdt,\\
{\mathcal N}_0&=\phi \|\bar u_0\|_{L^2}^2 + TM_*^2+{\mathcal E}_*,\quad
{\mathcal N}_*=\phi\|\bar u_0\|_{L^2}^2+T+ {\mathcal E}_*,\\
{\mathcal N}_2&=\phi( \|\bar u_0\|_{L^2}^2+\|\nabla u_0\|_{L^2}^2)+T+ {\mathcal E}_*,\\
{\mathcal N}_s&=\phi( \|\bar u_0\|_{L^2}^2+\|\nabla u_0\|_{L^2}^2+\|\nabla u_0\|_{L^s}^s)+T+ {\mathcal E}_* \text{ for $s>2$.}
\end{align*}
\end{definition}

The estimates obtained in the rest of this paper will depend on the quantities in Definition \ref{BEN}. Among those, only $M_*$ still depends on the solution $u$.
However, this quantity can be bounded in terms of initial and boundary data by using different techniques.
For instance, in our original problem, $u=\rho/\kappa\ge 0$, hence we have from Corollary \ref{maxcor} that
$M_*\le C M_0(T)$, see \eqref{uM}.
Therefore, in the following, we say ``the estimates are expressed in terms of the initial and boundary data'' even when they contain $M_*$.

\medskip
As stated in section \ref{intro}, we will keep tracks of the dependence on certain physical parameters, particularly, the angular speed. Note, by \eqref{defxione} and \eqref{RMm}, that
\begin{equation}\label{xichi}
\chi_1\le C\chi_*.
\end{equation}
With this and the fact $\chi_*\ge 1$, we compare ${\mathcal E}_0$ in \eqref{Estar} with ${\mathcal E}_*$ by
\begin{equation}\label{EEs}
{\mathcal E}_0\le C\chi_*^{2(2+a)}{\mathcal E}_*.
\end{equation}
It is clear that
\begin{align}
\label{NNN0}
{\mathcal E}_*&\le {\mathcal N}_0\le (M_*+1)^2 {\mathcal N}_*,\\
\label{NNNs}
{\mathcal E}_*&\le {\mathcal N}_*\le {\mathcal N}_2\le {\mathcal N}_s \text{ for $s>2$.}
\end{align}

\begin{theorem}\label{L2apos}
One has
\begin{align}
\label{gradu2}
 \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt&\le C \Big\{\chi_*^2\phi \|\bar u_0\|_{L^2}^2
 + \chi_*^{2(4+a)}TM_*^2(M_*+1)^2+\chi_*^{2(2+a)}{\mathcal E}_*\Big\},\\
 \label{gradu3}
 \inttx{0}{T}{U} |\nabla u(x,t)|^{2-a}dxdt &\le C\Big\{ \chi_*^2\phi \|\bar u_0\|_{L^2}^2
 + \chi_*^{2(4+a)}T(M_*+1)^4+\chi_*^{2(2+a)}{\mathcal E}_*\Big\}.
\end{align}
Consequently, the following more concise estimates hold
\begin{equation}
\label{gradu4}
 \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt\le C\chi_*^{2(4+a)}(M_*+1)^2 {\mathcal N}_0,
\end{equation}
\begin{equation}
\label{gradu6a}
 \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt \le C\chi_*^{2(4+a)}(M_*+1)^4 {\mathcal N}_*,
\end{equation}
\begin{equation}
\label{gradu6b}
 \inttx{0}{T}{U} |\nabla u(x,t)|^{2-a}dxdt \le C\chi_*^{2(4+a)}(M_*+1)^4 {\mathcal N}_*.
\end{equation}
\end{theorem}
\begin{proof}
 By the definition of $M_*$, we obviously have
\be
|u|\le M_*\text{ on } \bar U\times[0,T],
\end{equation}
hence,
\begin{equation}\label{uu}
\|u\|_{L^2(U\times (0,T))}^2\le CTM_*^2\quad\text{and}\quad
\|u\|_{L^4(U\times (0,T))}^4\le CTM_*^4.
\end{equation}
Using \eqref{uu}, the relations \eqref{xichi}, \eqref{EEs}, and estimate of $M_{\mathcal Z}$ in \eqref{RMm} for the right-hand side of \eqref{gradu0}, we have
\begin{align*}
 \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt
 &\le C\chi_*^2 \phi (\|\bar u_0\|_{L^2}^2 +TM_*^2) +C(1+\chi_*)^{2(2+a)}\cdot \chi_*^4\cdot  T M_*^4 +C\chi_*^{2(2+a)}{\mathcal E}_*\\
 &\le C\chi_*^2 \phi \|\bar u_0\|_{L^2}^2 +C\chi_*^{2(2+a)+4} T M_*^2  (1+M_*^2) +C\chi_*^{2(2+a)}{\mathcal E}_*.
\end{align*}
Then \eqref{gradu2} and \eqref{gradu4} follow. From \eqref{gradu4}, we infer \eqref{gradu6a} thanks to the last relation in \eqref{NNN0}.

Similarly, we have from  \eqref{gradu1} that
\begin{equation*}
\inttx{0}{T}{U} |\nabla u(x,t)|^{2-a}dxdt \le C\Big\{ \chi_*^2\phi (\|\bar u_0\|_{L^2}^2+TM_*^2)+\chi_*^{2(2+a)}(T+ \chi_*^4 \cdot T M_*^4) +\chi_*^{2(2+a)}{\mathcal E}_*\Big\}.
\end{equation*}
Then \eqref{gradu3} and \eqref{gradu6b} follow.
\end{proof}

We remark that while \eqref{gradu3} gives a direct estimate for the $L^{2-a}$-norm, the alternative estimate in form of \eqref{gradu2}  prepares for the iterations in section \ref{gradsec} below.

\begin{remark}\label{smlrmk1}
From the point of view of pure PDE estimates, the right-hand side of \eqref{gradu4} can be small, for a fixed $T>0$, while the right-hand side of \eqref{gradu6a}
and  \eqref{gradu6b} cannot. It is because $|u|$, $|\bar u_0|$, and ${\mathcal E}_*$ being small will make ${\mathcal N}_0$, but not ${\mathcal N}_*$, small.
\end{remark}

\section{Gradient estimates (II)} \label{gradsec}

In this section, we establish the interior
$L^s$-estimate of $\nabla u $ for all $s > 0$.

For the remainder of this paper, $\zeta$ always denotes a function in $C^2_c(U\times [0,T])$ that satisfies
$0\le \zeta(x,t)\le 1$ for all $(x,t)$.
When such a function is specified, the quantity $\mathcal D_s$ is defined, for $s\ge 0$, by
\begin{equation}\label{Ds}
\mathcal D_s= \int_U|\nabla u_0(x)|^{2s+2}\zeta^2(x,0) dx.
\end{equation}

The next two lemmas \ref{lem61} and \ref{lem62} are the main technical steps for the later iterative estimates of the gradient.

\begin{lemma} \label{lem61}
For any $s\ge 0$, one has
\begin{equation}\label{iterate1}
\begin{aligned}
&\phi \sup_{t\in[0,T]}\int_U |\nabla u(x,t)|^{2s+2} \zeta^2(x,t) dx,\\
&(s+1)c_8\chi_1^{-2}\inttx{0}{T}{U} \mathcal K  |D^2 u|^2   |\nabla u|^{2s}  \zeta^2  dx dt\le \phi \mathcal D_s +CI_0,
\end{aligned}
\end{equation}
where
\begin{align*}
I_0&=\mu_{\mathcal Z}^2 (1+\chi_1)^{2(1+a)}\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s} u^4   \zeta^2 dx dt \\
&\quad + (1+\chi_1)^{2(1+a)}\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} (M_{\mathcal Z}^2 u^2\zeta^2 + |\nabla \zeta|^2)   dx dt + \inttx{0}{T}{U} |\nabla u|^{2s+2} \zeta|\zeta_t| dx dt.
\end{align*}
Consequently,
\begin{align}
 \label{irat0}
\phi \sup_{t\in[0,T]}\int_U |\nabla u(x,t)|^{2s+2} \zeta^2(x,t) dx
&\le \phi \mathcal D_s +C J_0,\\
\label{iterate4}
\inttx{0}{T}{U} \mathcal K  |D^2 u|^2   |\nabla u|^{2s}  \zeta^2  dx dt
&\le C \chi_*^2 (\phi \mathcal D_s+J_0),
\end{align}
where
\begin{align*}
J_0
&=\chi_*^{2(3+a)}\widetilde M^4\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s}  \zeta^2 dx dt \\
&\quad +  \chi_*^{2(3+a)}(\widetilde M+1)^2\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} (\zeta^2 + |\nabla \zeta|^2)   dx dt
 + \inttx{0}{T}{U} |\nabla u|^{2s+2} \zeta|\zeta_t| dx dt,
\end{align*}
with
\begin{equation}
\widetilde M=\sup\{ |u(x,t)|:(x,t)\in \operatorname{supp}\zeta\}.
\end{equation}
\end{lemma}
\begin{proof}
For each $j=1,2,3$ and $t\in(0,T)$, let $\{\varphi_n(x)\}_{n=1}^\infty$ be a sequence in $C_c^\infty(U)$ that approximates $|\nabla u(x,t)|^{2s} \partial_j u(x,t)\zeta^2(x,t)$ in $W_0^{1,2}(U)$. Multiplying equation \eqref{ueq} by $-\partial_j \varphi_n(x)$, integrating the resulting equation over $U$, and using the integration by parts twice for the right-hand side give
\begin{align*}
-\phi\int_U \frac{\partial u}{\partial t} \partial_j\varphi_n dx
&=  -\int_U \frac{\partial}{\partial x_i}\Big[ X_i(\Phi(x,t)) \Big] \partial_j \varphi_n dx
=\int_U X_i(\Phi(x,t))  \partial_i \partial_j \varphi_n dx\\
&=  -\int_U \frac{\partial}{\partial x_j}\Big[ X_i(\Phi(x,t)) \Big] \partial_i \varphi_n dx.
\end{align*}

Passing $n\to\infty$ and summing in $j$ yield
\begin{align*}
-\phi\int_U \frac{\partial u}{\partial t}\nabla\cdot ( |\nabla u|^{2s} \nabla u\zeta ^2)dx
&=  -\int_U \frac{\partial}{\partial x_j}\Big[ X_i(\Phi(x,t)) \Big] \partial_i (|\nabla u|^{2s} \partial_j u \zeta^2) dx.
\end{align*}

Performing integration by parts again for the left-hand side, we obtain
\begin{equation}\label{esti}
\begin{aligned}
& \frac{\phi}{2s+2} \frac{d}{dt} \int_U |\nabla u|^{2s+2} \zeta^2 dx\\
&= -\int_U \frac{\partial}{\partial x_j}\Big[ X_i(\Phi(x,t)) \Big] \partial_i (|\nabla u|^{2s} \partial_j u \zeta^2) dx+\frac{1}{s+1}\int_U |\nabla u|^{2s+2} \zeta\zeta_t dx\\
&=I_1+I_2+I_3+I_4,
\end{aligned}
\end{equation}
where
\begin{align*}
 I_1&= -\int_U \frac{\partial}{\partial x_j}\Big[ X_i(\Phi(x,t)) \Big]  \partial_j \partial_i u  |\nabla u|^{2s}  \zeta^2 dx,\\
I_2&= -2\int_U \frac{\partial}{\partial x_j}\Big[ X_i(\Phi(x,t)) \Big]  \partial_j u \, |\nabla u|^{2s} \, \zeta \partial_i  \zeta dx,\\
I_3&= -2s\int_U \frac{\partial}{\partial x_j}\Big[ X_i(\Phi(x,t)) \Big] \partial_j u   \, (|\nabla u|^{2s-2} \partial_i\partial_l u\partial_l u)\, \zeta^2 dx,\\
I_4&=\frac{1}{s+1} \int_U|\nabla u|^{2s+2} \zeta\zeta_t dx.
\end{align*}

For $i,j=1,2,3$, denote $X'_{ij}(y)=\partial X_i(y)/\partial y_j$. By the second inequality of \eqref{Xprime},
\begin{equation}\label{XpPh}
|X'_{ij}(\Phi(x,t))|\le C(1+\chi_1)^a\mathcal K(x,t).
\end{equation}

Let $I_5=\int_U \mathcal K |D^2 u|^2   |\nabla u|^{2s}  \zeta^2 dx$.

\medskip
\noindent\textit{Estimation of $I_1$.} We have
\begin{align*}
I_1
&= -\int_U X'_{im}(\Phi) ( \partial_m\partial _j u + \partial _j \mathcal Z_m u^2 + 2u \partial_j u \mathcal Z_m )    \partial_j \partial_i u  |\nabla u|^{2s}  \zeta^2 dx\\
&=I_{1,1}+I_{1,2}+I_{1,3},
\end{align*}
where
\begin{align*}
I_{1,1}&=-\int_U D(\partial_j u)  X'(\Phi) \nabla (\partial_j u )  |\nabla u|^{2s}  \zeta^2 dx,\\
I_{1,2}&=-\int_U X'_{im}(\Phi)  \partial _j \mathcal Z_m u^2   \partial_j \partial_i u  |\nabla u|^{2s}  \zeta^2 dx,\\
I_{1,3}&=-2 \int_U X'_{im}(\Phi) u \partial_j u \mathcal Z_m     \partial_j \partial_i u  |\nabla u|^{2s}  \zeta^2 dx.
\end{align*}

To estimate $I_{1,1}$, by applying \eqref{hXh} to $y=\Phi$, $\xi=\nabla(\partial_j u)$, we have
\begin{align*}
 I_{1,1}
&\le - c_8\chi_1^{-2} \int_U \mathcal K  \Big( \sum_{j=1}^n|\nabla(\partial_j u)|^2 \Big)  |\nabla u|^{2s}  \zeta^2 dx
=-c_8\chi_1^{-2} I_5.
 \end{align*}

To estimate $I_{1,2}$, using  inequality \eqref{XpPh} to estimate $|X'_{im}(\Phi)|$, identity \eqref{DZ} for $|\partial_j\mathcal Z_m|$, and then applying the Cauchy inequality to $u^2  |D^2 u |$, we obtain
\begin{align*}
|I_{1,2}|&\le C \int_U  (1+\chi_1)^a \mathcal K  \mu_{\mathcal Z} u^2  |D^2 u | |\nabla u|^{2s}  \zeta^2 dx\\
&\le \varepsilon I_5+ C\varepsilon^{-1}(1+\chi_1)^{2a} \mu_{\mathcal Z}^2 \int_U \mathcal K  u^4   |\nabla u|^{2s} \zeta^2 dx  .
\end{align*}

Similarly, we estimate $I_{1,3}$ by
\begin{align*}
|I_{1,3}|&\le C(1+\chi_1)^a \int_U  \mathcal K  M_{\mathcal Z} u  |D^2 u | |\nabla u|^{2s+1}  \zeta^2 dx\\
&\le \varepsilon I_5+ C\varepsilon^{-1}(1+\chi_1)^{2a} M_{\mathcal Z}^2\int_U \mathcal K  u^2   |\nabla u|^{2s+2} \zeta^2 dx  .
\end{align*}

Summing up, we obtain
\begin{equation}\label{estione}
\begin{aligned}
I_1&\le (2\varepsilon -c_8\chi_1^{-2} )I_5+ C\varepsilon^{-1}(1+\chi_1)^{2a} \mu_{\mathcal Z}^2 \int_U \mathcal K  u^4   |\nabla u|^{2s} \zeta^2 dx\\
&\quad + C\varepsilon^{-1}(1+\chi_1)^{2a} M_{\mathcal Z}^2\int_U \mathcal K  u^2   |\nabla u|^{2s+2} \zeta^2 dx.
\end{aligned}
\end{equation}

\medskip
\noindent\textit{Estimation of $I_2$.} We calculate
\begin{align*}
I_2
&=-2\int_U X'_{im}(\Phi)\Big ( \partial_m\partial _j u + \partial _j \mathcal Z_m u^2 + 2u \partial_j u \mathcal Z_m\Big )   \partial_j u \, |\nabla u|^{2s} \, \zeta \partial_i  \zeta dx\\
&=I_{2,1}+I_{2,2}+I_{2,3},
\end{align*}
where the integral is split along the sum $\partial_m\partial _j u + \partial _j \mathcal Z_m u^2 + 2u \partial_j u \mathcal Z_m$.

Using \eqref{XpPh} and Cauchy's inequality gives
\begin{align*}
|I_{2,1}| &\le C(1+\chi_1)^a\int_U  \mathcal K  |D^2 u||\nabla u|^{2s+1} \zeta |\nabla  \zeta| dx\\
&\le \varepsilon I_5+C\varepsilon^{-1}(1+\chi_1)^{2a} \int_U \mathcal K  |\nabla u|^{2s+2}  |\nabla  \zeta|^2 dx,
\end{align*}

Using \eqref{XpPh}, \eqref{DZ} and \eqref{MZ}, we obtain
\begin{align*}
|I_{2,2}| &\le C(1+\chi_1)^a\int_U  \mathcal K |D\mathcal Z| u^2|\nabla u|^{2s+1} \zeta |\nabla  \zeta| dx\\
&\le C(1+\chi_1)^a\mu_{\mathcal Z} \int_U  \mathcal K u^2|\nabla u|^{2s+1} \zeta |\nabla  \zeta| dx,
\end{align*}
and
\begin{align*}
|I_{2,3}| &\le C(1+\chi_1)^a\int_U \mathcal K |\mathcal Z| u|\nabla u|^{2s+2} \zeta |\nabla  \zeta| dx\\
&\le C(1+\chi_1)^aM_{\mathcal Z}\int_U \mathcal K  u|\nabla u|^{2s+2} \zeta |\nabla  \zeta| dx.
\end{align*}

Thus, we have
\begin{equation}\label{estitwo}
\begin{aligned}
&I_2\le \varepsilon I_5+C\varepsilon^{-1}(1+\chi_1)^{2a} \int_U \mathcal K  |\nabla u|^{2s+2}  |\nabla  \zeta|^2 dx\\
&\quad+C(1+\chi_1)^a\mu_{\mathcal Z} \int_U  \mathcal K u^2|\nabla u|^{2s+1} \zeta |\nabla  \zeta| dx+C(1+\chi_1)^aM_{\mathcal Z}\int_U \mathcal K  u|\nabla u|^{2s+2} \zeta |\nabla  \zeta| dx.
\end{aligned}
\end{equation}

\medskip
\noindent\textit{Estimation of $I_3$.} Using similar calculations to those for $I_2$, we have
\begin{align*}
I_3
&=-2s\int_U X'_{mi}(\Phi) \Big( \partial_m\partial _j u + \partial _j \mathcal Z_m u^2 + 2u \partial_j u \mathcal Z_m \Big)
\cdot \partial_j u   \, (|\nabla u|^{2s-2} \partial_i\partial_l u\partial_l u)\,     \zeta^2 dx\\
&=I_{3,1}+I_{3,2}+I_{3,3}.
\end{align*}
where the integral, again,  is split along the sum $\partial_m\partial _j u + \partial _j \mathcal Z_m u^2 + 2u \partial_j u \mathcal Z_m$.

Rewriting $I_{3,1}$ and applying \eqref{hXh} to $y=\Phi$ and $\xi=\frac{1}2 \nabla( |\nabla u|^2)$, we have
\begin{align*}
I_{3,1}
&=-2s\int_U X'_{im}(\Phi)\cdot  [ \partial_m\partial _j u \cdot \partial_j u ] \cdot[\partial_i\partial_l \cdot u\partial_l u] \cdot |\nabla u|^{2s-2}\,     \zeta^2 dx\\
&=-2s\int_U X'_{im}(\Phi)  \Big [\frac{1}2 \partial _m  |\nabla u|^2\Big ]  \Big[\frac{1}2 \partial_i |\nabla u|^2)\Big]\cdot  |\nabla u|^{2s-2}     \zeta^2 dx\\
&\le -2s\, c_8 \chi_1^{-2} \int_U  \mathcal K  \Big|\frac{1}2\nabla  (|\nabla u|^2)\Big|^2   |\nabla u|^{2s-2}      \zeta^2 dx \le 0.
\end{align*}

We estimate $I_{3,2}$ and $I_{3,3}$ similarly to $I_{2,1}$, $I_{2,2}$, $I_{2,3}$, and obtain
\begin{align*}
|I_{3,2}|&\le C(1+\chi_1)^a\int_U \mathcal K  |D\mathcal Z| u^2    \, |\nabla u|^{2s} \, |D^2 u|    \zeta^2 dx\\
&\le \varepsilon I_5 + C\varepsilon^{-1}(1+\chi_1)^{2a} \mu_{\mathcal Z}^2 \int_U \mathcal K  u^4    \, |\nabla u|^{2s}    \zeta^2 dx,
\end{align*}
and
\begin{align*}
|I_{3,3}|&\le C(1+\chi_1)^a M_{\mathcal Z}\int_U \mathcal K  u    \, |\nabla u|^{2s+1} \, |D^2 u|   \zeta^2 dx\\
&\le \varepsilon I_5 + C\varepsilon^{-1} (1+\chi_1)^{2a} M_{\mathcal Z}^2\int_U \mathcal K  u^2    \, |\nabla u|^{2s+2}    \zeta^2 dx.
\end{align*}

Therefore,
\begin{equation}\label{estithree}
\begin{aligned}
I_3&\le  2\varepsilon I_5+ C\varepsilon^{-1}(1+\chi_1)^{2a} \mu_{\mathcal Z}^2 \int_U \mathcal K  u^4    \, |\nabla u|^{2s}    \zeta^2 dx\\
&\quad + C\varepsilon^{-1} (1+\chi_1)^{2a} M_{\mathcal Z}^2\int_U \mathcal K  u^2    \, |\nabla u|^{2s+2}    \zeta^2 dx.
\end{aligned}
\end{equation}

Combining \eqref{esti} with the estimates \eqref{estione}, \eqref{estitwo} and \eqref{estithree}, we have
\begin{align*}
 &\frac{\phi}{2s+2} \frac{d}{dt} \int_U |\nabla u|^{2s+2} \zeta^2 dx +(c_8\chi_1^{-2}-5\varepsilon)I_5
 \le  C\varepsilon^{-1}(1+\chi_1)^{2a}\mu_{\mathcal Z}^2 \int_U \mathcal K  u^4   |\nabla u|^{2s} \zeta^2 dx  \\
 & +C\varepsilon^{-1} (1+\chi_1)^{2a}M_{\mathcal Z}^2\int_U \mathcal K  u^2   |\nabla u|^{2s+2} \zeta^2 dx
 +C\varepsilon^{-1}(1+\chi_1)^{2a} \int_U \mathcal K |\nabla u|^{2s+2}  |\nabla  \zeta|^2 dx\\
 & +C(1+\chi_1)^a\mu_{\mathcal Z}\int_U \mathcal K |u|^2|\nabla u|^{2s+1} \zeta |\nabla  \zeta| dx
 + C(1+\chi_1)^aM_{\mathcal Z} \int_U \mathcal K |u||\nabla u|^{2s+2} \zeta |\nabla  \zeta| dx
 +I_4.
\end{align*}

Using the Cauchy inequality, we have, for the fourth integral on the right-hand side,
\begin{equation*}
\begin{split}
(1+\chi_1)^a\mu_{\mathcal Z} |u|^2|\nabla u|^{2s+1} \zeta |\nabla  \zeta|
&\le \frac 1 2 \left [\varepsilon^{-1}(1+\chi_1)^{2a}\mu_{\mathcal Z}^2  u^4   |\nabla u|^{2s} \zeta^2+ \varepsilon |\nabla u|^{2s+2}  |\nabla  \zeta|^2  \right],
\end{split}
\end{equation*}
and, for the fifth integral on the right-hand side,
\begin{equation*}
\begin{split}
(1+\chi_1)^aM_{\mathcal Z}|u| \zeta |\nabla  \zeta|
&\le \frac 1 2\left[ \varepsilon^{-1} (1+\chi_1)^{2a}M_{\mathcal Z}^2 u^2\zeta^2+ \varepsilon  |\nabla  \zeta|^2 \right].
\end{split}
\end{equation*}

Therefore, we obtain
\begin{equation} \label{d10}
\begin{aligned}
& \frac{\phi}{2s+2} \frac{d}{dt} \int_U |\nabla u|^{2s+2} \zeta^2 dx +(c_8\chi_1^{-2}-5\varepsilon)I_5 \\
& \le  C\varepsilon^{-1}(1+\chi_1)^{2a}\mu_{\mathcal Z}^2 \int_U \mathcal K  |\nabla u|^{2s} u^4   \zeta^2 dx
   +C\varepsilon^{-1} (1+\chi_1)^{2a}M_{\mathcal Z}^2\int_U \mathcal K  u^2   |\nabla u|^{2s+2} \zeta^2 dx  \\
&\quad  +C(\varepsilon^{-1}(1+\chi_1)^{2a}+\varepsilon) \int_U \mathcal K |\nabla u|^{2s+2}  |\nabla  \zeta|^2 dx
  + C \int_U |\nabla u|^{2s+2} \zeta|\zeta_t| dx.
\end{aligned}
\end{equation}

Choosing $\varepsilon=c_8\chi_1^{-2}/10$, and integrating \eqref{d10} in time, we get
$$\phi \sup_{t\in[0,T]}\int_U |\nabla u(x,t)|^{2s+2} \zeta^2(x,t) dx \text{ and } (s+1)c_8\chi_1^{-2}\inttx{0}{T}{U} \mathcal K  |D^2 u|^2   |\nabla u|^{2s}  \zeta^2  dx dt$$
are bounded from above by
\begin{align*}
&\phi \int_U |\nabla u_0(x)|^{2s+2} \zeta^2(x,0) dx
+C\mu_{\mathcal Z}^2 (1+\chi_1)^{2(1+a)}\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s} u^4   \zeta^2 dx dt \\
&\quad + C ((1+\chi_1)^{2(1+a)}+\varepsilon)\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} (M_{\mathcal Z}^2 u^2\zeta^2 + |\nabla \zeta|^2)   dx dt \\
&\quad + C\inttx{0}{T}{U} |\nabla u|^{2s+2} \zeta|\zeta_t| dx dt.
\end{align*}
Then using the fact $\varepsilon\le c_8 \chi_0^{-2}/10\le C$ for the last $\varepsilon$, we obtain \eqref{iterate1}.

We now estimate $I_0$ further. We use \eqref{RMm}, \eqref{xichi} to estimate $M_{\mathcal Z}$, $\mu_{\mathcal Z}$, $\chi_1$, note, for $m=2,4$, that
$u^m \zeta^2\le \widetilde M^m \zeta^2$.
With these estimates, we have
\begin{equation} \label{JJ0}I_0\le C J_0.
\end{equation}
Hence, \eqref{irat0} directly follows the first estimate \eqref{iterate1}.

Similarly,  multiplying the second estimate in \eqref{iterate1} by $(s+1)^{-1}c_8^{-1}\chi_1^2$, then using  \eqref{xichi} and \eqref{JJ0}, we obtain \eqref{iterate4}.
\end{proof}

Next, we combine Lemma \ref{lem61} with the embedding in Theorem \ref{LUembed} to derive a bootstrapping estimate.

\begin{lemma}\label{lem62}
If $s\ge 0$ then
\begin{equation} \label{Kug3}
\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s+4} \zeta^2 dxdt+M_*^2 \inttx{0}{T}{U} \mathcal K  |D^2 u|^2   |\nabla u|^{2s}  \zeta^2  dx dt
\le  C(I_*+J_*),
\end{equation}
where
\begin{align*}
I_*&= \chi_*^2M_*^2 \phi \mathcal D_s + T\chi_*^{4(2s+3)}M_*^6 (M_*+1)^{8s+6} \\
&\quad +  \chi_*^{2(4+a)}M_*^2(M_*+1)^4\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} (\zeta^2 + |\nabla \zeta|^2)   dx dt\\
&\quad  + {\rm sgn}(s)M_*^2 \inttx{0}{T}{U} \mathcal K  |D^2 u|^2 \zeta^2 dxdt,
\end{align*}
and
\begin{align*}
J_*&= T\chi_*^{4(1+a)}M_*^4 (M_*+1)^{4a}\sup|\zeta_t|^2\\
&\quad + \chi_*^{4(1+a\cdot{\rm sgn}(s))}M_*^4 (M_*+1)^{4a\cdot{\rm sgn}(s)}\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} |\zeta_t|^2 dx dt.
\end{align*}
\end{lemma}
\begin{proof}
Denote
\begin{align*}
\alpha&=\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s+4} \zeta^2 dxdt,& \gamma&=\inttx{0}{T}{U} |\nabla u|^{2s+2} \zeta|\zeta_t| dx dt,\\
\beta&=\inttx{0}{T}{U} \mathcal K  |D^2 u|^2   |\nabla u|^{2s}  \zeta^2  dx dt,& \beta_0&=\inttx{0}{T}{U} \mathcal K  |D^2 u|^2  \zeta^2  dx dt.
\end{align*}

For $s\ge 0$, by applying \eqref{ibt} to $w(x)=u(x,t)$, $Q(x)=\mathcal Z(x,t)$, $s:=s+1$, and then integrating in $t$ from $0$ to $T$, we have
\begin{align*}
\alpha
&\le
 C M_*^2 \beta + CM_*^2(1+(M_*\mu_{\mathcal Z}+M_{\mathcal Z})^2M_*^2) \inttx{0}{T}{U} \mathcal K |\nabla u|^{2s+2}  (|\nabla \zeta|^2 + \zeta^2 ) dx dt \\
&\quad + C {\rm sgn}(s)M_*^2 \beta_0 + CT(M_*^2M_{\mathcal Z})^{2s+4}(1+M_*^2M_{\mathcal Z})^{2s+2}.
\end{align*}
Using \eqref{RMm} for upper bounds of $M_{\mathcal Z}$ and $\mu_{\mathcal Z}$, we have
\be
\begin{aligned}
\alpha
&\le C M_*^2 \beta+ C\chi_*^4M_*^2(M_*+1)^4 \inttx{0}{T}{U} \mathcal K |\nabla u|^{2s+2}  (|\nabla \zeta|^2 + \zeta^2 ) dx dt \\
&\quad + C {\rm sgn}(s)M_*^2 \beta_0 + CT\chi_*^{4(2s+3)}M_*^{4(s+2)}(M_*+1)^{4(s+1)}.
\end{aligned}
\end{equation}

We estimate $\beta$ by using \eqref{iterate4} and the fact $\widetilde M\le M_*$ to have
\begin{align*}
CM_*^2 \beta
&\le  C \chi_*^2M_*^2\phi \mathcal D_s
+C\chi_*^{2(4+a)}M_*^6\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s}    \zeta^2 dx dt \\
&\quad + C\chi_*^{2(4+a)}M_*^2(M_*+1)^2\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} ( \zeta^2 + |\nabla \zeta|^2)   dx dt
+ C\chi_*^2M_*^2 \gamma.
\end{align*}
Thus,
\begin{equation} \label{albe}
\begin{aligned}
2\alpha+M_*^2 \beta
&\le  C\chi_*^2M_*^2 \phi \mathcal D_s
+C\chi_*^{2(4+a)}M_*^6\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s} \zeta^2 dx dt \\
&\quad + C \chi_*^{2(4+a)}M_*^2(M_*+1)^4\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} (\zeta^2 + |\nabla \zeta|^2)   dx dt\\
&\quad  + C {\rm sgn}(s)M_*^2 \beta_0 + CT\chi_*^{4(2s+3)}M_*^{4(s+2)}(M_*+1)^{4(s+1)}+ C\chi_*^2M_*^2 \gamma.
\end{aligned}
\end{equation}

For the second term on the right-hand side of \eqref{albe}, the integral is bounded by
\begin{equation*}
\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s} \zeta^2 dx dt
\le \inttx{0}{T}{U} \mathcal K  (|\nabla u|^{2s+2}+1) \zeta^2 dx dt\le \inttx{0}{T}{U} \mathcal K  |\nabla u|^{2s+2}\zeta^2dx dt +CT.
\end{equation*}

Combining this with the third term on the right-hand side of \eqref{albe} gives
\begin{align*}
&2\alpha+M_*^2 \beta\le  C\chi_*^2M_*^2 \phi \mathcal D_s + C \chi_*^{2(4+a)}M_*^2(M_*+1)^4\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2s+2} (\zeta^2 + |\nabla \zeta|^2)   dx dt\\
&\quad + C {\rm sgn}(s)M_*^2 \beta_0 + CT\chi_*^{4(2s+3)}M_*^{4(s+2)}(M_*+1)^{4(s+1)}+CT\chi_*^{2(4+a)}M_*^6 + C\chi_*^2 M_*^2 \gamma.
\end{align*}

As far as the two $T$-terms in the last inequality are concerned, the first one has
 $$M_*^{4(s+2)}(M_*+1)^{4(s+1)}\le M_*^6(M_*+1)^{8s+6},$$
 while the second one has
 $$\chi_*^{2(4+a)}\le \chi_*^{10}\le \chi_*^{4(2s+3)}.$$

Therefore,
\begin{equation}
2\alpha+M_*^2 \beta\le CI_* + C\chi_*^2 M_*^2 \gamma.\label{ab0}
\end{equation}

We estimate the last term by using Cauchy's inequality to obtain
\begin{align*}
 C\chi_*^2 M_*^2 \gamma
 &\le  \inttx{0}{T}{U} \mathcal K|\nabla u|^{2s+4} \zeta^2 dx dt
 + C\chi_*^4M_*^4\inttx{0}{T}{U} \mathcal K^{-1}|\nabla u|^{2s}|\zeta_t|^2 dxdt\\
 &=\alpha
 + C\chi_*^4M_*^4\inttx{0}{T}{U} \mathcal K |\nabla u|^{2s} \mathcal K^{-2}|\zeta_t|^2 dxdt.
\end{align*}

In the last integral, we have
\begin{align*}
 |\nabla u|^{2s} \mathcal K^{-2}&\le |\nabla u|^{2s} (1+|\nabla u|+ M_*^2\chi_*^2)^{2a}\\
& \le C(|\nabla u|^{2s} +|\nabla u|^{2s+2a}+ M_*^{4a}\chi_*^{4a}|\nabla u|^{2s})\\
 &\le C[1+|\nabla u|^{2s+2}+ M_*^{4a}\chi_*^{4a}(1+{\rm sgn}(s)|\nabla u|^{2s+2})],
\end{align*}
which can be conveniently rewritten as
\begin{equation*}
 |\nabla u|^{2s} \mathcal K^{-2}\le C (M_*+1)^{4a}\chi_*^{4a}+C (M_*+1)^{4a\cdot{\rm sgn}(s)}\chi_*^{4a\cdot{\rm sgn}(s)}|\nabla u|^{2s+2}.
\end{equation*}

Thus,
\begin{equation}\label{ab}
\begin{aligned}
 C\chi_*^2M_*^2 \gamma
 &\le  \alpha
 + C\chi_*^{4(1+a\cdot{\rm sgn}(s))}M_*^4 (M_*+1)^{4a\cdot{\rm sgn}(s)}\inttx{0}{T}{U} \mathcal K |\nabla u|^{2s+2}|\zeta_t|^2 dxdt\\
&\quad + C\chi_*^{4(1+a)}M_*^4 (M_*+1)^{4a}\inttx{0}{T}{U} |\zeta_t|^2 dxdt\le  \alpha + CJ_*.
 \end{aligned}
\end{equation}

Combining \eqref{ab0} and \eqref{ab} yields
\begin{equation*}
2\alpha+M_*^2 \beta\le CI_*+\alpha+CJ_*,
\text{ which implies } \alpha+M_*^2 \beta\le C(I_*+J_*).
\end{equation*}
We have proved \eqref{Kug3}.
\end{proof}

As one can see from \eqref{Kug3} that the integral of higher power $2s+4$ of $|\nabla u|$, with the weight $\mathcal K(x,t)$, can be bounded by the corresponding integral of lower power $2s+2$. However, it still involves a second order term, which is the last integral of $I_*$. This term, as it turns out, can be estimated in \eqref{ab4} below.

\subsection{Estimates for the $L_{x,t}^{4-a}$-norm}\label{L4ma}

We start using inequality \eqref{Kug3} with the smallest possible value for $s$, i.e., $s=0$.
It will result in the $\mathcal K$-weighted $L_{x,t}^2$-estimate, and, consequently, the $L_{x,t}^{4-a}$ estimate for $|\nabla u|$.

\begin{proposition}\label{prop63}
One has
\begin{equation}\label{ab4}
\begin{aligned}
&\inttx{0}{T}{U} \mathcal K  |\nabla u|^{4} \zeta^2 dxdt+M_*^2\inttx{0}{T}{U} \mathcal K  |D^2 u|^2 \zeta^2 dxdt\\
&\le C(1+\sup|\nabla \zeta|^2+\sup |\zeta_t|^2)\Big\{ \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi( \|\bar u_0\|_{L^2}^2 +\mathcal D_0)\\
&\quad +  \chi_*^{4(4+a)}  T  M_*^4(M_*+1)^8  + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\Big\}.
\end{aligned}
\end{equation}
\end{proposition}
\begin{proof}
Denote by $I$ the sum on the left-hand side of \eqref{ab4}. It follows \eqref{Kug3} with $s=0$ that
\begin{align*}
I&\le  C\chi_*^2M_*^2 \phi \mathcal D_0  + CT\cdot \big [\chi_*^{12}M_*^6 (M_*+1)^{6}+\chi_*^{4(1+a)}M_*^4 (M_*+1)^{4a}\sup|\zeta_t|^2\big ] \\
&\quad + C \chi_*^{2(4+a)}M_*^2(M_*+1)^4\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2} (\zeta^2 + |\nabla \zeta|^2)   dx dt\\
&\quad + C \chi_*^{4}M_*^4\inttx{0}{T}{U} \mathcal K   |\nabla u|^{2} |\zeta_t|^2 dx dt.
\end{align*}
The second term on the right-hand side is bounded by
$$C(1+\sup|\zeta_t|^2)\chi_*^{12} T M_*^4(M_*+1)^8 ,$$
and the sum of the last two terms on the right-hand side is bounded by
\begin{equation*} C(1+\sup |\nabla\zeta|^2+\sup |\zeta_t|^2)  \chi_*^{2(4+a)} M_*^2(M_*+1)^4
\inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt.
\end{equation*}
Hence,
\begin{align*}
I&\le   C(1+\sup |\nabla\zeta|^2+\sup |\zeta_t|^2)\Big\{ \chi_*^2M_*^2 \phi \mathcal D_0+\chi_*^{12}  T  M_*^4(M_*+1)^8 \\
&\quad + \chi_*^{2(4+a)}M_*^2(M_*+1)^4  \inttx{0}{T}{U} \mathcal K|\nabla u|^2 dxdt\Big\}.
\end{align*}

Estimating the last integral by \eqref{gradu2} gives
\begin{align*}
I&\le   C(1+\sup |\nabla\zeta|^2+\sup |\zeta_t|^2)\Big\{ \chi_*^2M_*^2 \phi \mathcal D_0+\chi_*^{12}  T  M_*^4(M_*+1)^8 \\
&\quad + \chi_*^{2(4+a)}M_*^2(M_*+1)^4  \Big[ \chi_*^2\phi \|\bar u_0\|_{L^2}^2
 + \chi_*^{2(4+a)} T M_*^2(M_*+1)^2 + \chi_*^{2(2+a)} {\mathcal E}_*\Big]\Big\}.
\end{align*}

Grouping the like-terms on the right-hand side and using simple estimations yield inequality \eqref{ab4}.
\end{proof}

By selecting the cut-off function $\zeta$ in \eqref{ab4} appropriately, we derive the spatial, as well as the spatial-temporal, interior estimates for $|\nabla u|$.

\begin{xnotation}
For simplicity, we will write $V\Subset U$ to indicate that $V$ is an open, relatively compact subset of $U$.
\end{xnotation}

\begin{theorem}\label{thm64}
Let $U'\Subset U$.

{\rm (i)} One has
\begin{equation}\label{ab1}
\begin{aligned}
\inttx{0}{T}{U'} \mathcal K  |\nabla u|^{4} dxdt
&\le C\Big\{ \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi( \|\bar u_0\|_{L^2}^2 +\|\nabla u_0\|_{L^2}^2)\\
&\quad +  \chi_*^{4(4+a)}  T  M_*^4(M_*+1)^8  + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\Big\}.
\end{aligned}
\end{equation}

Consequently,
\begin{equation}\label{ab11}
\inttx{0}{T}{U'} \mathcal K  |\nabla u|^{4} dxdt
\le C\chi_*^{4(4+a)} M_*^2(M_*+1)^{10} {\mathcal N}_2.
\end{equation}

{\rm (ii)} If $T_0$ is any number in $(0,T)$, then
\begin{equation}\label{ab2}
\begin{aligned}
\int_{T_0}^{T}\int_{U'} \mathcal K  |\nabla u|^{4} dxdt
&\le C(1+T_0^{-1})^2\Big\{ \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi \|\bar u_0\|_{L^2}^2 \\
&\quad +  \chi_*^{4(4+a)}  T  M_*^4(M_*+1)^8  + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\Big\}.
\end{aligned}
\end{equation}

Consequently,
\begin{equation}\label{ab22}
\int_{T_0}^{T}\int_{U'} \mathcal K  |\nabla u|^{4} dxdt
\le C(1+T_0^{-1})^2\chi_*^{4(4+a)} M_*^2(M_*+1)^{10}{\mathcal N}_*.
\end{equation}

{\rm (iii)} If  $s\in[2,4]$, then
\begin{equation} \label{ab23}
\inttx{0}{T}{U'} \mathcal K  |\nabla u|^s dxdt \le C\chi_*^{(4+a)s}M_*^{s-2}(M_*+1)^{3s-2}{\mathcal N}_2,
\end{equation}
and, for $T_0\in(0,T)$,
\begin{equation} \label{ab24}
\int_{T_0}^T\int_{U'} \mathcal K  |\nabla u|^s dxdt \le C(1+T_0^{-1})^{s-2}\chi_*^{(4+a)s}M_*^{s-2}(M_*+1)^{3s-2}{\mathcal N}_*.
\end{equation}
\end{theorem}
\begin{proof}
(i) We fix a cut-off function $\zeta=\zeta(x)$ with $\zeta\equiv 1$ on $U'$.
We have $|\nabla\zeta|\le C$ and $\zeta_t\equiv 0$. Then, by using inequality \eqref{ab4}, we obtain
\begin{align*}
\inttx{0}{T}{U'} \mathcal K  |\nabla u|^{4}  dxdt
&\le \inttx{0}{T}{U} \mathcal K  |\nabla u|^{4}\zeta^2  dxdt\\
&\le C\Big\{ \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi( \|\bar u_0\|_{L^2}^2 +\mathcal D_0)\\
&\quad +  \chi_*^{4(4+a)}  T  M_*^4(M_*+1)^8  + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\Big\}.
\end{align*}
This proves \eqref{ab1}.
Now, note on the right-hand side of \eqref{ab1} that
\begin{equation} \label{oo}
\chi_*^{2(5+a)}M_*^2(M_*+1)^4,   \chi_*^{4(4+a)}  M_*^4(M_*+1)^8,  \chi_*^{4(3+a)}M_*^2(M_*+1)^4 \le \chi_*^{4(4+a)}M_*^2(M_*+1)^{10}.
\end{equation}
Utilizing these estimates, we obtain \eqref{ab11} from \eqref{ab1}.

(ii) We select a different cut-off function $\zeta=\zeta(x,t)$ such that $\zeta=0$ for $0\le t\le T_0/2$, and $\zeta=1$ on $U'\times [T_0,T]$, and its derivatives satisfy
\begin{equation} \label{zprop}
 |\nabla\zeta|\le C \text{ and }0\le \zeta_t \le C T_0^{-1},
\end{equation}
where $C>0$ is independent of $T_0,T$.

With this function $\zeta$, it is obvious from \eqref{Ds} that $\mathcal D_0=0$.
Then, by \eqref{ab4}, we have
\begin{align*}
\int_{T_0}^{T}\int_{U'} \mathcal K  |\nabla u|^{4}  dxdt
&\le \inttx{0}{T}{U}  \mathcal K  |\nabla u|^{4}\zeta^2   dxdt\\
&\le C(1+T_0^{-2})\Big\{ C\chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi \|\bar u_0\|_{L^2}^2\\
&\quad +  \chi_*^{4(4+a)}  T  M_*^4(M_*+1)^8  + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\Big\},
\end{align*}
which gives \eqref{ab2}.
Utilizing \eqref{oo} again for the right-hand side of \eqref{ab2}, we obtain \eqref{ab22}.

(iii) The inequalities \eqref{ab23} and \eqref{ab24} already hold for $s=2$ thanks to \eqref{gradu6a} and \eqref{NNNs}, and for $s=4$ thanks to \eqref{ab1} and \eqref{ab11}.

Consider $2<s<4$ now. By interpolation inequality \eqref{Lpinter}, we have
 \begin{align*}
 \inttx{0}{T}{U'} \mathcal K  |\nabla u|^s dxdt
 \le \Big( \inttx{0}{T}{U'} \mathcal K  |\nabla u|^2 dxdt\Big)^\frac{4-s}2\Big( \inttx{0}{T}{U'} \mathcal K  |\nabla u|^4 dxdt\Big)^\frac{s-2}2.
 \end{align*}

Applying  inequality \eqref{gradu6a}, respectively \eqref{ab11}, to estimate the first, respectively second, integral on the right-hand side, we obtain
\begin{align*}
 \inttx{0}{T}{U'} \mathcal K  |\nabla u|^s dxdt
 &\le C\Big(\chi_*^{2(4+a)}(M_*+1)^4 {\mathcal N}_*\Big)^\frac{4-s}2\Big( \chi_*^{4(4+a)} M_*^2(M_*+1)^{10} {\mathcal N}_2\Big)^\frac{s-2}2,
 \end{align*}
which yields \eqref{ab23}.

Similarly, by \eqref{Lpinter}, \eqref{gradu6a} and \eqref{ab22}, we have
\begin{align*}
 \int_{T_0}^T\int_{U'} \mathcal K  |\nabla u|^s dxdt
& \le \Big( \int_{T_0}^T\int_{U'} \mathcal K  |\nabla u|^2 dxdt\Big)^\frac{4-s}2\Big( \int_{T_0}^T\int_{U'} \mathcal K  |\nabla u|^4 dxdt\Big)^\frac{s-2}2  \\
&\le C \Big(\chi_*^{2(4+a)}(M_*+1)^4 {\mathcal N}_*\Big)^\frac{4-s}2\Big( (1+T_0^{-1})^2\chi_*^{4(4+a)} M_*^2(M_*+1)^{10} {\mathcal N}_*\Big)^\frac{s-2}2,
\end{align*}
which implies \eqref{ab24}.
\end{proof}

The estimates obtained in Theorem \ref{thm64} contain the weight $\mathcal K(x,t)$.
Below, we derive the estimates for the standard Lebesgue $L^{4-a}_{x,t}$-norm (without that weight).

\begin{corollary}\label{cor65}
Let $U'\Subset U$ and $T_0\in(0,T)$.

{\rm (i)} One has
\begin{equation}\label{ab31}
\begin{aligned}
\inttx{0}{T}{U'} |\nabla u|^{4-a} dxdt
&\le C\chi_*^{4(4+a)} (M_*+1)^{12} {\mathcal N}_2,
\end{aligned}
\end{equation}
and
\begin{equation}\label{ab32}
\begin{aligned}
\int_{T_0}^{T}\int_{U'} |\nabla u|^{4-a} dxdt
&\le C(1+T_0^{-1})^2\chi_*^{4(4+a)} (M_*+1)^{12}{\mathcal N}_*.
\end{aligned}
\end{equation}

{\rm (ii)} If $s$ is any number in $(2-a,4-a)$, then
\begin{equation} \label{ab33}
\inttx{0}{T}{U'}  |\nabla u|^s dxdt
\le C\chi_*^{(4+a)(s+a)} (M_*+1)^{4(s+a-1)}{\mathcal N}_2,
\end{equation}
and
\begin{equation} \label{ab34}
\int_{T_0}^T\int_{U'}  |\nabla u|^s dxdt
\le C(1+T_0^{-1})^{s+a-2}\chi_*^{(4+a)(s+a)} (M_*+1)^{4(s+a-1)}{\mathcal N}_*.
\end{equation}
 \end{corollary}
\begin{proof}
(i) Applying \eqref{kugs} to $s=4-a$ gives
 \begin{align*}
 |\nabla u|^{4-a}\le C \mathcal K |\nabla u|^4 +C (1+M_*\chi_*)^8.
 \end{align*}
Combining this with  \eqref{ab11}, respectively \eqref{ab22}, we obtain \eqref{ab31}, respectively \eqref{ab32}.

 (ii) Consider $2-a<s<4-a$. By interpolation inequality \eqref{Lpinter} and then using \eqref{gradu6b}, \eqref{NNNs} and \eqref{ab31}, we have
  \begin{align*}
 &\inttx{0}{T}{U'}  |\nabla u|^s dxdt
 \le \Big( \inttx{0}{T}{U'}  |\nabla u|^{2-a} dxdt\Big)^\frac{4-a-s}2\Big( \inttx{0}{T}{U'} |\nabla u|^{4-a} dxdt\Big)^\frac{s-(2-a)}2  \\
 &\le C\Big[\chi_*^{2(4+a)}(M_*+1)^4 {\mathcal N}_*\Big]^\frac{4-a-s}2  \Big[\chi_*^{4(4+a)} (M_*+1)^{12}{\mathcal N}_2\Big]^\frac{s-(2-a)}2 \\
 &\le C\chi_*^{(4+a)(s+a)} (M_*+1)^{4(s+a-1)}{\mathcal N}_2.
 \end{align*}
Thus, we obtain \eqref{ab33}.

Similarly, by \eqref{Lpinter}, \eqref{gradu6b} and \eqref{ab32} we have
 \begin{align*}
 &\int_{T_0}^T\int_{U'}  |\nabla u|^s dxdt
 \le \Big( \int_{T_0}^T\int_{U'}  |\nabla u|^{2-a} dxdt\Big)^\frac{4-a-s}2\Big( \int_{T_0}^T\int_{U'} |\nabla u|^{4-a} dxdt\Big)^\frac{s-(2-a)}2  \\
 &\le C\Big[\chi_*^{2(4+a)}(M_*+1)^4 {\mathcal N}_*\Big]^\frac{4-a-s}2  \Big[(1+T_0^{-1})^2\chi_*^{4(4+a)} (M_*+1)^{12}{\mathcal N}_*\Big]^\frac{s-(2-a)}2 \\
 &\le C(1+T_0^{-1})^{s+a-2}\chi_*^{(4+a)(s+a)} (M_*+1)^{4(s+a-1)}{\mathcal N}_*.
 \end{align*}
Thus, we obtain \eqref{ab34}.
 \end{proof}

 \begin{remark}
The estimate  \eqref{ab34} of the $L^s_{x,t}$-norm of  $\nabla u(x,t)$, for $t>0$, requires, as far as the initial data $u_0$ is concerned, at most the $L^\infty$-norm of $u_0$. Therefore, it shows the (formal) regularization effect of the PDE \eqref{ueq}.
This observation also applies to Corollary \ref{high1} and Theorem \ref{high5} below.
\end{remark}

\subsection{Estimates for higher $L_{x,t}^s$-norms}\label{Lhigher}

In this subsection, we have  estimates for the $L_{x,t}^s$-norms of $\nabla u$ with $s>4-a$.

\begin{lemma}\label{lem67}
Let $s>2$, and $V$ be an open subset of $U$.

{\rm (i)} If $\zeta=\zeta(x)$ with compact support in $V$, then
\begin{multline}\label{ab66}
\inttx{0}{T}{U} \mathcal K  |\nabla u|^{s+2}\zeta^2 dxdt
\le  C (1+\sup|\nabla \zeta|^2)\\
\cdot \Big\{ \chi_*^{4(s+2+a)}M_*^2 (M_*+1)^{4s+2} {\mathcal N}_s
+ \chi_*^{2(4+a)}M_*^2(M_*+1)^4 \inttx{0}{T}{V} \mathcal K |\nabla u|^s dx dt\Big\}.
\end{multline}

{\rm (ii)}  If $\zeta=\zeta(x,t)$ with $\zeta(x,0)\equiv 0$ and, for each $t\in[0,T]$, the mapping $\zeta(\cdot,t)$ has compact support in $V$,  then
\begin{multline}\label{ab77}
\inttx{0}{T}{U} \mathcal K  |\nabla u|^{s+2} \zeta^2 dxdt
\le  C(1+\sup|\nabla \zeta|^2+\sup|\zeta_t|^2)\\
\cdot \Big\{  \chi_*^{4(s+2+a)}M_*^2 (M_*+1)^{4s+2} {\mathcal N}_*
 + \chi_*^{2(4+a)}M_*^2(M_*+1)^6\inttx{0}{T}{V} \mathcal K   |\nabla u|^s  dx dt\Big\}.
\end{multline}
\end{lemma}
\begin{proof}
Denote $$I=\inttx{0}{T}{U} \mathcal K  |\nabla u|^{2(s+2)}\zeta^2 dxdt \text{ and }J=\inttx{0}{T}{V} \mathcal K |\nabla u|^{2(s+1)}dx dt.$$

(i) Consider $s>0$. We estimate $I$ by \eqref{Kug3}, neglecting the second term on the left-hand side. Note in this case that $\zeta_t=0$ and hence $J_*=0$.  We then use \eqref{ab4} to estimate the last term of $I_*$. The result is
\begin{align*}
I
&\le  C M_*^2\chi_*^2\phi \mathcal D_s + CT\chi_*^{4(2s+3)}M_*^6 (M_*+1)^{8s+6}  + C\chi_*^{2(4+a)}M_*^2(M_*+1)^4 (1+\sup|\nabla \zeta|^2)  J\\
&\quad +  C(1+\sup|\nabla \zeta|^2)\Big[  \chi_*^{2(5+a)}M_*^2(M_*+1)^4\phi (\|\bar u_0\|_{L^2}^2+\|\nabla u_0\|_{L^2}^2)
+ T \chi_*^{4(4+a)}M_*^4(M_*+1)^8 \\
&\quad +\chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_* \Big].
\end{align*}
For the terms containing the initial data, we estimate
\begin{align*}
M_*^2\chi_*^2\le  \chi_*^{2(5+a)}M_*^2(M_*+1)^4,
\end{align*}
and for the terms containing $T$, we use
\begin{equation}\label{Tcoef}
\chi_*^{4(2s+3)}M_*^6 (M_*+1)^{8s+6},\ \chi_*^{4(4+a)}M_*^4(M_*+1)^8\le \chi_*^{8(s+2)+4a}M_*^4 (M_*+1)^{8(s+1)}.
\end{equation}
Hence, we obtain
\begin{multline}\label{Kug6}
I\le  C (1+\sup|\nabla \zeta|^2) \Big\{ \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi(\|\bar u_0\|_{L^2}^2+ \|\nabla u_0\|_{L^2}^2+\|\nabla u_0\|_{L^{2s+2}}^{2s+2})\\
+ T\chi_*^{8(s+2)+4a}M_*^4 (M_*+1)^{8(s+1)}
+\chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*
+ \chi_*^{2(4+a)}M_*^2(M_*+1)^4 J \Big\}.
\end{multline}

Now, consider $s>2$. By replacing $2s+2$ in \eqref{Kug6} with $s$, noting that
$$I \text{ becomes }\inttx{0}{T}{U} \mathcal K  |\nabla u|^{s+2}\zeta^2 dxdt,\quad J
\text{ becomes }\inttx{0}{T}{V} \mathcal K |\nabla u|^s dx dt,
$$
the power $8(s+2)+4a$ becomes $4(s+2+a)$, and the power $8(s+1)$ becomes $4s$,
we obtain
\begin{multline}\label{ab6}
\inttx{0}{T}{U} \mathcal K  |\nabla u|^{s+2}\zeta^2 dxdt
\le  C (1+\sup|\nabla \zeta|^2)\\
\cdot \Big\{ \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi(\|\bar u_0\|_{L^2}^2+ \|\nabla u_0\|_{L^2}^2+\|\nabla u_0\|_{L^s}^s)
+ T\chi_*^{4(s+2+a)}M_*^4 (M_*+1)^{4s}\\
+\chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*
+ \chi_*^{2(4+a)}M_*^2(M_*+1)^4 \inttx{0}{T}{V} \mathcal K |\nabla u|^s dx dt\Big\}.
\end{multline}
On the right-hand side of \eqref{ab6}, in order to group the terms $\phi(\|\bar u_0\|_{L^2}^2+ \|\nabla u_0\|_{L^2}^2+\|\nabla u_0\|_{L^s}^s)$, $T$, ${\mathcal E}_*$ together, we estimate their coefficients by
$$\chi_*^{2(5+a)}M_*^2(M_*+1)^4,\chi_*^{4(s+2+a)}M_*^4 (M_*+1)^{4s},\chi_*^{4(3+a)}M_*^2(M_*+1)^4
\le \chi_*^{4(s+2+a)}M_*^2 (M_*+1)^{4s+2}.$$
Then inequality \eqref{ab66} follows \eqref{ab6}.

(ii) Consider $s>0$. Note that $\mathcal D_0=\mathcal D_s=0$.  We have from \eqref{Kug3} that
\begin{align*}
I&\le  C(1+\sup|\nabla \zeta|^2+\sup|\zeta_t|^2)
\Big\{  T\cdot\big[\chi_*^{4(2s+3)}M_*^6 (M_*+1)^{8s+6} + \chi_*^{4(1+a)}M_*^4 (M_*+1)^{4a}\big] \\
&\quad +\big[ \chi_*^{2(4+a)}M_*^2(M_*+1)^4 + \chi_*^{4(1+a)}M_*^4 (M_*+1)^{4a}\big]\cdot J\Big\}
  + M_*^2 \inttx{0}{T}{U} \mathcal K  |D^2 u|^2 \zeta^2 dxdt.
\end{align*}

We use \eqref{ab4} to estimate the last term $\displaystyle M_*^2 \inttx{0}{T}{U} \mathcal K  |D^2 u|^2 \zeta^2 dxdt$.
For the $T$-term, we use \eqref{Tcoef} again. For the $J$-term we use
\begin{equation*}
 \chi_*^{2(4+a)}M_*^2(M_*+1)^4 , \chi_*^{4(1+a)}M_*^4 (M_*+1)^{4a}
 \le  \chi_*^{2(4+a)}M_*^2(M_*+1)^6.
\end{equation*}
Combining these estimates gives
\begin{align*}
I&\le  C(1+\sup|\nabla \zeta|^2+\sup|\zeta_t|^2)
\Big\{  T\chi_*^{8(s+2)+4a}M_*^4 (M_*+1)^{8(s+1)}
+ \chi_*^{2(4+a)} M_*^2(M_*+1)^6 J\\
&\quad + \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi \|\bar u_0\|_{L^2}^2+ T \chi_*^{4(4+a)}  M_*^4(M_*+1)^8  + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\Big\}.
\end{align*}
Simplifying the right-hand side once more, we obtain
\begin{equation}\label{Kug7}
\begin{aligned}
I
&\le  C(1+\sup|\nabla \zeta|^2+\sup|\zeta_t|^2)
\Big\{  \chi_*^{2(5+a)}M_*^2(M_*+1)^4 \phi \|\bar u_0\|_{L^2}^2  \\
&\quad + T\chi_*^{8(s+2)+4a}M_*^4 (M_*+1)^{8(s+1)}
 + \chi_*^{4(3+a)}M_*^2(M_*+1)^4 {\mathcal E}_*\\
&\quad  + \chi_*^{2(4+a)}M_*^2(M_*+1)^6 J\Big\}.
\end{aligned}
\end{equation}

Same as in the proof of part (i), when $s>2$, replacing $2s+2$ in \eqref{Kug7} with $s$ yields \eqref{ab77}.
\end{proof}

\begin{theorem}\label{high0}
If $U'\Subset U$ and  $s\ge 4$, then
 \begin{equation}\label{ih0}
  \inttx{0}{T}{U'}\mathcal K |\nabla u|^{s}dxdt
  \le C \chi_*^{(4+a)(s+2)}M_*^2 (M_*+1)^{4s}{\mathcal N}_{s-2}.
 \end{equation}
\end{theorem}
\begin{proof}
(a) When $s=4$ the inequality \eqref{ih0} holds true thanks to the estimate \eqref{ab11}.
Hence we only focus on the case $s>4$.

(b) Consider the case $s=s_*+2m$ with $s_*>2$ and $m\in \mathbb N$.
Let $V$ be an open subset of $U$ such that $U'\Subset V\Subset U$.
We claim that
\begin{equation}\label{kug0}
\begin{aligned}
  \inttx{0}{T}{U'}\mathcal K |\nabla u|^{s}dxdt
& \le C\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{4m}  \inttx{0}{T}{V} \mathcal K |\nabla u|^{s_*}dxdt\\
&\quad  +C \chi_*^{(4+a)s}M_*^2 (M_*+1)^{4s-6}{\mathcal N}_{s-2}.
\end{aligned}
\end{equation}

\textit{Proof of \eqref{kug0}.}
Let $\{U_k\}_{k=0}^m$ be a family of smooth, open subsets of $U$ such that
 \begin{equation}\label{Usets}
 U'\subset U_m\Subset U_{m-1}\Subset U_{m-2}\Subset \ldots \Subset U_{1} \Subset U_0 \subset V\Subset U.
 \end{equation}

 Denote
$\displaystyle y_k=\inttx{0}{T}{U_k} \mathcal K |\nabla u|^{s_*+2k}dx dt$ for $0\le k\le m$.

Let $k\in\{0,1,2,\ldots, m-1\}$. Choose $\zeta=\zeta_k(x)$, a $C^2$ cut-off function
which is equal to $1$ on $U_{k+1}$ and has compact support in $U_k$.
Applying \eqref{ab66} to $s:=s_*+2k$, we have
\begin{equation}\label{yk2} y_{k+1}\le A y_k+B,
\end{equation}
where $A=C_k\chi_*^{2(4+a)} M_*^2(M_*+1)^4$
and $\displaystyle B= C_k\chi_*^{4(s_*+2k+2+a)}M_*^2(M_*+1)^{4(s_*+2k)+2}\widehat B$,
with
\begin{equation*}
\widehat B
=\phi(\|\bar u_0\|_{L^2}^2+ \|\nabla u_0\|_{L^2}^2+\|\nabla u_0\|_{L_{s_*+2k}}^{s_*+2k})+T+{\mathcal E}_*
\end{equation*}
for some $C_k>0$ independent of $T$.
Note that
\begin{align*}
\|\nabla u_0\|_{L_{s_*+2k}}^{s_*+2k}
=\int_U |\nabla u_0|^{s_*+2k}dx\le \int_U \big[|\nabla u_0|^2+|\nabla u_0|^{s_*+2(m-1)}\big]dx.
\end{align*}
Hence,
\begin{equation*}
\widehat B
\le \phi(\|\bar u_0\|_{L^2}^2+ 2\|\nabla u_0\|_{L^2}^2+\|\nabla u_0\|_{L^{s_*+2(m-1)}}^{s_*+2(m-1)})+T+{\mathcal E}_*
\le 2{\mathcal N}_{s-2}.
\end{equation*}

Let $C_*=2\max\{C_k:k=1,2,\ldots,m-1\}$.
Hence,
\begin{equation}\label{yk1} y_{k+1}\le A_* y_k+B_k,
\end{equation}
where
\begin{align*}
A_*=C_*\chi_*^{2(4+a)} M_*^2(M_*+1)^4,\quad
B_k= C_*\chi_*^{4(s_*+2k+2+a)}M_*^2(M_*+1)^{4(s_*+2k)+2}{\mathcal N}_{s-2}
= B_* S^k,
\end{align*}
with
$ S=\chi_*^8(M_*+1)^8$ and $B_*=A_*\chi_*^{4s_*+2a}(M_*+1)^{4s_*-2}{\mathcal N}_{s-2}$.

Iterating \eqref{yk1}, we obtain
\begin{align*}
 y_{k+1}
 &\le A_*( A_*y_{k-1}+B_{k-1})+B_{k}
 = A_*^2y_{k-1}+A_*B_{k-1}+B_{k}\\
 &\le A_*^3y_{k-2}+A_*^2B_{k-2}+A_*B_{k-1}+B_{k}\\
 &\le \cdots\le A_*^{k+1}y_{0}+\sum_{j=0}^{k-1} A_*^{k-j} B_j + B_{k}.
\end{align*}

Letting $k=m-1$, we then have
\begin{equation}\label{yk0}
y_m \le A_*^m y_0 + \sum_{j=0}^{m-2} A_*^{m-1-j}B_j+B_{m-1} .
\end{equation}

Dealing with the middle sum on the right-hand side of \eqref{yk0}, elementary calculations show, for $0\le j\le m-2$, that
\begin{align*}
A_*^{m-j-1}B_j
&=A_*^{m-j} \chi_*^{4s_*+2a+8j}(M_*+1)^{4s_*-2+8j}{\mathcal N}_{s-2}\\
&= C \chi_*^{4(s_*+2m)+2a(m-j+1)} M_*^{2(m-j)} (M_*+1)^{4(s_*+2m)-4(m-j)-2}{\mathcal N}_{s-2}\\
&\le C \chi_*^{4s+2a(m+1)}\cdot [ M_*^2 (M_*+1)^{2(m-j-1)} ]\cdot (M_*+1)^{4s-4(m-j)-2}{\mathcal N}_{s-2}\\
&= C \chi_*^{4s+2a(m+1)} M_*^2  (M_*+1)^{4s-2(m-j)-4}{\mathcal N}_{s-2}.
\end{align*}
Note that $m-j\ge 2$ and
\begin{equation}\label{sm}
s>2+2m.
\end{equation}
Then we have
\begin{equation}\label{abm}
A_*^{m-1-j}B_j
\le C \chi_*^{(4+a)s} M_*^2(M_*+1)^{4s-8}{\mathcal N}_{s-2}.
\end{equation}

For the last term in \eqref{yk0}, one has
\begin{align}\label{bm}
 B_{m-1}
 &=C\chi_*^{4(s_*+2m+a)}M_*^2(M_*+1)^{4(s_*+2m)-6}{\mathcal N}_{s-2}\notag\\
 &=C\chi_*^{4(s+a)}M_*^2(M_*+1)^{4s-6}{\mathcal N}_{s-2}.
\end{align}

Then combining \eqref{yk0} with \eqref{abm} and \eqref{bm} gives
\begin{align*}
y_m&\le C\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{4m}y_0+C (m-1)\chi_*^{(4+a)s} M_*^2(M_*+1)^{4s-8}{\mathcal N}_{s-2}\\
&\quad +C\chi_*^{4(s+a)}M_*^2(M_*+1)^{4s-6}{\mathcal N}_{s-2}\\
&\le C\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{4m}y_0+C \chi_*^{(4+a)s}M_*^2 (M_*+1)^{4s-6}{\mathcal N}_{s-2}.
\end{align*}
Therefore, estimate \eqref{kug0} follows.

(c) Consider the general case $s>4$ now. Then there exist  $s_*\in(2,4]$ and integer $m\ge 1$ such that $s=s_*+2m$.
We apply estimate \eqref{kug0} using the relation \eqref{sm}, and  have
\begin{align*}
  \inttx{0}{T}{U'}\mathcal K |\nabla u|^{s}dxdt
& \le C\chi_*^{(4+a)(s-2)}M_*^{2m}(M_*+1)^{2(s-2)}  \inttx{0}{T}{V} \mathcal K |\nabla u|^{s_*}dxdt\\
&\quad  +C \chi_*^{(4+a)s}M_*^2 (M_*+1)^{4s-6}{\mathcal N}_{s-2}.
\end{align*}
Using \eqref{sm} again,
$$M_*^{2m}\le M_*^2(M_*+1)^{2m-2}\le M_*^2(M_*+1)^{s-4}.$$
Then
\begin{align*}
  \inttx{0}{T}{U'}\mathcal K |\nabla u|^{s}dxdt
& \le C\chi_*^{(4+a)(s-2)}M_*^2(M_*+1)^{3s-8}  \inttx{0}{T}{V} \mathcal K |\nabla u|^{s_*}dxdt\\
&\quad  +C \chi_*^{(4+a)s}M_*^2 (M_*+1)^{4s-6}{\mathcal N}_{s-2}.
\end{align*}

Note, by Young's inequality and applying \eqref{ab11} to $U':=V$, that
\begin{align*}
&\inttx{0}{T}{V} \mathcal K |\nabla u|^{s_*}dxdt
\le C(T+\inttx{0}{T}{V} \mathcal K |\nabla u|^{4}dxdt) \\
&\le  C \chi_*^{4(4+a)}(M_*+1)^{12}\big[ \phi (\|\bar u_0\|_{L^2}^2+\|\nabla u_0\|_{L^2}^2)  +T+{\mathcal E}_*\big].
\end{align*}

Due to \eqref{NNNs} we can conclude that
\begin{equation*}
  \inttx{0}{T}{U'}\mathcal K |\nabla u|^{s}dxdt
\le C\chi_*^{(4+a)(s+2)}M_*^2(M_*+1)^{3s+4} {\mathcal N}_2  +C \chi_*^{(4+a)s}M_*^2 (M_*+1)^{4s-6}{\mathcal N}_{s-2},
\end{equation*}
and we obtain \eqref{ih0}.
\end{proof}

\begin{proposition}\label{high2}
If $U'\Subset V\Subset U$, and $s=s_*+2m$ with $s_*>2$ and  $m\in \mathbb N$, then
\begin{equation}\label{kug3}
\begin{aligned}
  \int_{T_0}^{T}\int_{U'}\mathcal K |\nabla u|^{s}dxdt
&  \le C(1+t_0^{-1})^{2m}\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{6m}  \int_{T_0-t_0}^{T}\int_{V}\mathcal K |\nabla u|^{s_*}dxdt\\
&\quad   +C(1+t_0^{-1})^{2m}\chi_*^{(4+a)s} M_*^2(M_*+1)^{4s-6}{\mathcal N}_*.
\end{aligned}
   \end{equation}
for any numbers $T_0$ and $t_0$ such that $0<t_0<T_0<T$.
\end{proposition}
\begin{proof}
Let $\{U_k\}_{k=0}^m$ be as in \eqref{Usets}.
Let $\tau_0=T_0-t_0<\tau_1<\tau_2<\ldots<\tau_m=T_0$ be evenly paced.
Define
$\displaystyle y_k=\int_{\tau_k}^{T}\int_{U_k} \mathcal K |\nabla u|^{s_*+2k} dxdt$ for $0\le k\le m$.

Given $k\in\{0,1,2,\ldots, m-1$\}. Let $\zeta_k(x,t)$ be a smooth cut-off function
 which is equal to one on $U_{k+1}\times [\tau_{k+1},T]$,  has compact support in $U_k\times[\tau_k,T]$, and satisfies
$$|\nabla \zeta_k|\le C'_k,\quad 0\le \zeta_{k,t}\le \frac2{\tau_{k+1}-\tau_k}=\frac2{mt_0},$$
where $C'_k>0$ is independent of $T,T_0,t_0$.
Then using $s:=s_*+2k$ and $\zeta=\zeta_k$ in \eqref{ab77}, we have  the same relation \eqref{yk2}, with the constants defined by
 \begin{align*}
 A&=C_k(1+t_0^{-2})\chi_*^{2(4+a)}M_*^2(M_*+1)^6,\\
 B&=C_k(1+t_0^{-2})\chi_*^{4(s_*+2k+2+a)}M_*^2(M_*+1)^{4(s_*+2k)+2}{\mathcal N}_*,
\end{align*}
for some positive constant $C_k$ independent of $T,T_0,t_0$.

Set $C_*=\max\{C_k:k=0,1,\ldots,m-1\}$, we obtain \eqref{yk1} where
\begin{align*}
A_*&=C_*(1+t_0^{-1})^2\chi_*^{2(4+a)}M_*^2(M_*+1)^6,\\
B_k&=C_*(1+t_0^{-1})^2\chi_*^{4(s_*+2k+2+a)}M_*^2(M_*+1)^{4(s_*+2k)+2}{\mathcal N}_*
= B_* S^k
\end{align*}
with the same $S=\chi_*^8(M_*+1)^8$, but
\begin{align*}
 B_*&=C_*(1+t_0^{-1})^2\chi_*^{4(s_*+2+a)}M_*^2(M_*+1)^{4s_*+2}{\mathcal N}_*=A_* \chi_*^{4s_*+2a}(M_*+1)^{4s_*-4}{\mathcal N}_*.
\end{align*}
Then we obtain \eqref{yk0} by iteration again.
For $0\le j\le m-2$,
\begin{align*}
A_*^{m-j-1}B_j
&=A_*^{m-j} \chi_*^{4s_*+2a+8j}(M_*+1)^{4s_*+8j-4}{\mathcal N}_*\\
&\le C(1+t_0^{-1})^{2(m-j)} \chi_*^{4(s_*+2m)+2a(m-j+1)} M_*^{2(m-j)} (M_*+1)^{4s_*+6m+2j-4}{\mathcal N}_*.
\end{align*}
Simply estimating $M_*^{2(m-j)}\le  M_*^2 (M_*+1)^{2(m-j-1)}$, we then have
\begin{align*}
A_*^{m-j-1}B_j
&\le  C(1+t_0^{-1})^{2(m-j)} \chi_*^{4(s_*+2m)+2a(m+1)} M_*^2(M_*+1)^{4s_*+8m-6}{\mathcal N}_*\\
&= C(1+t_0^{-1})^{2m} \chi_*^{(4+a)s} M_*^2(M_*+1)^{4s -6}{\mathcal N}_*.
\end{align*}
Also,
\begin{equation*}
B_{m-1}=C(1+t_0^{-1})^2\chi_*^{4(s+a)}M_*^2(M_*+1)^{4s-6}{\mathcal N}_*.
\end{equation*}
Thus, we have from \eqref{yk0} that
\begin{align*}
y_m&\le C(1+t_0^{-1})^{2m}\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{6m}y_0\\
&\quad +C (m-1)(1+t_0^{-1})^{2m} \chi_*^{(4+a)s} M_*^2(M_*+1)^{4s -6}{\mathcal N}_*\\
&\quad +C(1+t_0^{-1})^2\chi_*^{4(s+a)}M_*^2(M_*+1)^{4s-6}{\mathcal N}_*.
\end{align*}
Hence, we obtain \eqref{kug3}.
\end{proof}

\begin{theorem}\label{high3}
If $U'\Subset U$ and $s>4$, then one has, for any $T_0\in (0,T)$,  that
 \begin{equation} \label{kug4}
  \int_{T_0}^{T}\int_{U'}\mathcal K |\nabla u|^{s}dxdt
  \le C (1+T_0^{-1})^s \chi_*^{(4+a)(s+2)}M_*^2 (M_*+1)^{4s+2}{\mathcal N}_*.
 \end{equation}
\end{theorem}
\begin{proof}
There exist $2<s_*\le 4$ and integer $m\ge 1$ such that $s=s_*+2m$. Let $t_0:=T_0/2$, and $V$ be a set  with $U'\Subset V\Subset U$.
Applying \eqref{kug3}, we have
\begin{equation}\label{kug3b}
\begin{aligned}
  \int_{T_0}^{T}\int_{U'}\mathcal K |\nabla u|^{s}dxdt
&  \le C(1+T_0^{-1})^{2m}\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{6m}  y_*\\
&\quad   +C(1+T_0^{-1})^{2m}\chi_*^{(4+a)s} M_*^2(M_*+1)^{4s-6}{\mathcal N}_*.
\end{aligned}
   \end{equation}
where
$\displaystyle y_*=\int_{T_0/2}^{T}\int_V\mathcal K |\nabla u|^{s_*}dxdt$.
By Young's inequality and \eqref{ab22} applied to $U':=V$, we have
\begin{equation*} y_*\le C\Big(T+\int_{T_0/2}^{T}\int_V\mathcal K |\nabla u|^{4}dxdt\Big)
\le  C(1+T_0^{-1})^2 \chi_*^{4(4+a)}(M_*+1)^{12}{\mathcal N}_*.
\end{equation*}
Then
\begin{align*}
 C&(1+T_0^{-1})^{2m}\chi_*^{2(4+a)m}M_*^{2m}(M_*+1)^{6m}y_*
\le C(1+T_0^{-1})^{s-2}\chi_*^{(4+a)(s-2)}M_*^{2}(M_*+1)^{8m-2}y_*
\\
&\le C(1+T_0^{-1})^{s}\chi_*^{(4+a)(s+2)} M_*^2(M_*+1)^{4s+2}{\mathcal N}_*.
\end{align*}
Combining this with \eqref{kug3b} gives \eqref{kug4}.
\end{proof}

\begin{corollary}\label{high1} Let $U'\Subset U$ and $s> 4-a$.
Then
 \begin{equation}\label{ih1}
  \inttx{0}{T}{U'} |\nabla u|^{s}dxdt
    \le C \chi_*^{(4+a)(s+a+2)}(M_*+1)^{4(s+a+1/2)}{\mathcal N}_{s+a-2}.
 \end{equation}
Moreover, it holds, for any number $T_0\in(0,T)$, that
 \begin{equation} \label{ih2}
  \int_{T_0}^{T}\int_{U'}|\nabla u|^{s}dxdt
  \le C (1+T_0^{-1})^{s+a} \chi_*^{(4+a)(s+a+2)} (M_*+1)^{4(s+a+1)}{\mathcal N}_*.
 \end{equation}
 \end{corollary}
\begin{proof}
Using \eqref{kugs} and applying \eqref{ih0} with $s$ being substituted by $s+a$, we have
 \begin{align*}
   \inttx{0}{T}{U'}|\nabla u|^{s}dxdt
   &\le C\inttx{0}{T}{U'} (\mathcal K|\nabla u|^{s+a} + (\chi_*(M_*+1))^{2(s+a)})dxdt\\
  &\le C \chi_*^{(4+a)(s+a+2)}M_*^2 (M_*+1)^{4(s+a)}{\mathcal N}_{s+a-2} + CT\chi_*^{2(s+a)}(M_*+1)^{2(s+a)}.
 \end{align*}
 Note that $T\le  {\mathcal N}_{s+a-2}$. Then \eqref{ih1} follows.
 Similarly, using \eqref{kug4}, instead of \eqref{ih0}, we obtain \eqref{ih2}.
\end{proof}

\section{Gradient estimates (III)}\label{maxintime}
This section is focused on the estimates for the $L_t^\infty L_x^s$-norms of $\nabla u$.
For $s\ge 2$, replacing $s$ in \eqref{irat0} with $s/2-1$ gives
\begin{equation}\label{iterate2}
\begin{aligned}
I
&\stackrel{\rm def}{=} \phi \sup_{t\in[0,T]}\int_U |\nabla u(x,t)|^s \zeta^2(x,t) dx\\
&\le \phi \int_U |\nabla u_0(x)|^s \zeta^2(x,0) dx
 +C\chi_*^{2(3+a)}M_*^4\inttx{0}{T}{U} \mathcal K  |\nabla u|^{s-2}  \zeta^2 dx dt \\
&\quad + C \chi_*^{2(3+a)}(M_*+1)^2\inttx{0}{T}{U} \mathcal K   |\nabla u|^s (\zeta^2 + |\nabla \zeta|^2)   dx dt + C\inttx{0}{T}{U} |\nabla u|^s \zeta|\zeta_t| dx dt.
\end{aligned}
\end{equation}

\begin{theorem}\label{high4}
If $U'\Subset U$, then one has, for all $t\in[0,T]$, that
\begin{align}\label{pwtall}
 \phi \int_{U'} |\nabla u(x,t)|^s  dx
&\le \phi \int_U |\nabla u_0(x)|^s dx \notag \\
&\quad +C\begin{cases}
  \chi_*^{4(4+a)}(M_*+1)^{6}{\mathcal N}_0 &\text{ if }s=2,\\
  \chi_*^{(s+2)(4+a)}M_*^{s-2}(M_*+1)^{3s+2}{\mathcal N}_2 &\text{ if }2<s\le 4,\\
  \chi_*^{(s+4)(4+a)}M_*^2 (M_*+1)^{4(s+1)}{\mathcal N}_{s-2} &\text{ if }s>4.
         \end{cases}
\end{align}
\end{theorem}
\begin{proof}
Denote $\displaystyle J=\phi \sup_{t\in[0,T]}\int_{U'} |\nabla u(x,t)|^s  dx.$
Choose $\zeta$ to be the same function $\zeta(x)$ as in the proof of Theorem \ref{thm64}(i).
Then we have the relation
\begin{equation}\label{JI}
J\le I.
\end{equation}
We then bound $I$ by using inequality \eqref{iterate2}, noticing that the last integral of this inequality vanishes,
and the integrand of the second term on its right-hand side can be bounded by
\begin{align*}
\mathcal K  |\nabla u|^{s-2} \le \mathcal K  (1 + |\nabla u|^s)\le 1+\mathcal K |\nabla u|^s .
\end{align*}
After this, combining the two constants for the integrals involving $\mathcal K |\nabla u|^{s} $, we obtain
\begin{equation}\label{start7}
\begin{aligned}
J&\le \phi \int_U |\nabla u_0(x)|^s dx + C\chi_*^{2(3+a)}M_*^4 T\\
&\quad +C\chi_*^{2(3+a)}(M_*+1)^4 \inttx{0}{T}{U} \mathcal K |\nabla u|^s (\zeta^2 + |\nabla \zeta|^2)   dx dt.
\end{aligned}
\end{equation}

Consider $s=2$. Using \eqref{gradu4} to estimate the last integral in \eqref{start7}, we obtain
\begin{align*}
J&\le \phi \int_U |\nabla u_0(x)|^s dx + C\chi_*^{2(3+a)}M_*^4 T\\
&\quad +C\chi_*^{2(3+a)}(M_*+1)^4 \cdot \chi_*^{2(4+a)}(M_*+1)^2 {\mathcal N}_0.
\end{align*}
Making a generous bound $2(3+a)<2(4+a)$ for the first two exponents of $\chi_*$ above, we obtain the first estimate in \eqref{pwtall}.

Consider $2< s\le 4$. Using \eqref{ab23} to estimate the last integral in \eqref{start7}, we obtain
\begin{align*}
J&\le \phi \int_U |\nabla u_0(x)|^s dx + C\chi_*^{2(3+a)}M_*^4 T\\
&\quad +C\chi_*^{2(3+a)}(M_*+1)^4 \cdot \chi_*^{(4+a)s}M_*^{s-2}(M_*+1)^{3s-2}{\mathcal N}_2.
\end{align*}
Then the second  estimate in \eqref{pwtall} follows.

Consider $s> 4$. Using \eqref{ih0} to estimate the last integral in \eqref{start7}, we have
\begin{align*}
J&\le \phi \int_U |\nabla u_0(x)|^s dx + C\chi_*^{2(3+a)}M_*^4 T\\
&\quad +C\chi_*^{2(3+a)}(M_*+1)^4 \cdot \chi_*^{(4+a)( s+2)}M_*^2 (M_*+1)^{4s}
{\mathcal N}_{s-2}.
\end{align*}
With simple manipulations, we obtain from this the third estimate in \eqref{pwtall}.
\end{proof}

\begin{theorem}\label{high5}
Let $U'\Subset U$ and $T_0\in(0,T)$. Then it holds, for all $t\in[T_0,T]$, that
\begin{align}\label{pwtnew}
&\phi\int_{U'} |\nabla u(x,t)|^{s}dx
\le C \notag \\
& \cdot
\begin{cases}
\chi_*^{(4+a)^2}(1+T_0^{-1})^{1+a}(M_*+1)^{2(3+a)}\big\{ M_*^{a} (M_*+1)^{2+a} {\mathcal N}_*+{\mathcal N}_0\big\}
&\text{ if }s=2,\\
\chi_*^{(4+a)(s+a+2)} (1+T_0^{-1})^{s+a-1} M_*^{s-2} (M_*+1)^{3s+4a+2} {\mathcal N}_*
&\text{ if }2< s\le 4-a,\\
\chi_*^{(4+a)(s+a+4)} (1+T_0^{-1})^{s+a+1} M_*^{s-2} (M_*+1)^{3s+4a+10} {\mathcal N}_*
&\text{ if }4-a<s\le 4,\\
\chi_*^{(4+a)(s+a+4)}(1+T_0^{-1})^{s+a+1}M_*^2 (M_*+1)^{4s+4a+6} {\mathcal N}_*
&\text{ if }s>4.
\end{cases}
\end{align}

Consequently, one has, for all $s\ge 2$ and $t\in[T_0,T]$, that
 \begin{equation}\label{pwt6}
\phi\int_{U'} |\nabla u(x,t)|^{s}dx
\le  C  \chi_*^{(4+a)(s+a+4)}(1+T_0^{-1})^{s+a+1}(M_*+1)^{4(s+a+2)} {\mathcal N}_*.
\end{equation}
\end{theorem}
\begin{proof}
Choose $\zeta(x,t)$ to be the cut-off function in the proof of Theorem \ref{thm64}(ii) which satisfies additionally that
$\zeta$ has compact support in $V\times [T_0/2,T]$, where $U'\Subset V\Subset U$.

Let $J$ be the same as in Theorem \ref{high4}. Again, we have \eqref{JI}, and  use \eqref{iterate2} to estimate $I$.
Note, on the right-hand side of \eqref{iterate2}, that
\begin{equation*}
\mathcal K  |\nabla u|^{s-2}\le 1+|\nabla u|^s,\quad  \mathcal K  |\nabla u|^s\le  |\nabla u|^s.
\end{equation*}
Utilizing these properties as well as \eqref{zprop}, we have from \eqref{JI} and \eqref{iterate2} that
\begin{equation}\label{JJ}
J\le C\chi_*^{2(3+a)}M_*^4 T
 + C \chi_*^{2(3+a)}(M_*+1)^4(1+T_0^{-1}) \int_{T_0/2}^T\int_V  |\nabla u|^s  dx dt.
\end{equation}
Estimate the last integral in \eqref{JJ},
\begin{align*}
&\int_{T_0/2}^{T}\int_V |\nabla u|^s dx dt
 = \int_{T_0/2}^{T}\int_V \mathcal K|\nabla u|^s \mathcal K^{-1}dx dt\\
  &\le C\int_{T_0/2}^{T}\int_V \mathcal K(|\nabla u|^{s+a}+|\nabla u|^s\chi_*^{2a}(M_*+1)^{2a})dxdt\\
 &= C\int_{T_0/2}^{T}\int_V \mathcal K|\nabla u|^{s+a}dxdt+C\chi_*^{2a}(M_*+1)^{2a} \int_{T_0/2}^{T}\int_V \mathcal K|\nabla u|^s dxdt.
\end{align*}

Denote by $I_1$ and $I_2$ the last two double integrals. We estimate them, in calculations below, by using inequalities \eqref{ab24} and \eqref{kug4} with $T_0:=T_0/2$ and $U':=V$.

Case $s=2$. Applying \eqref{ab24} to $s:=2+a\in(2,4)$ to bound $I_1$, and applying \eqref{gradu4} to $s:=2$ to bound $I_2$ give
\begin{align*}
J
&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{2(3+a)}(M_*+1)^4(1+T_0^{-1})\\
  &\quad \cdot\Big\{  (1+T_0^{-1})^a \chi_*^{(4+a)(2+a)}M_*^a(M_*+1)^{4+3a}{\mathcal N}_*  + \chi_*^{2a}(M_*+1)^{2a} \cdot \chi_*^{2(4+a)}(M_*+1)^2 {\mathcal N}_0\Big\}\\
&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{(4+a)^2}(1+T_0^{-1})^{1+a}M_*^{a} (M_*+1)^{8+3a} {\mathcal N}_*\\
&\quad    + C \chi_*^{14+6a}(1+T_0^{-1})(M_*+1)^{6+2a} {\mathcal N}_0.
\end{align*}
We obtain the first estimate in \eqref{pwtnew}.

Case $2<s\le 4-a$. Estimating $I_1$ by \eqref{ab24} applied to $s:=s+a$, and estimating $I_2$ by \eqref{ab24}, we have
\begin{align*}
J&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{2(3+a)}(M_*+1)^4(1+T_0^{-1})\\
  &\quad \cdot\Big\{  (1+T_0^{-1})^{s+a-2}\chi_*^{(4+a)(s+a)}M_*^{s+a-2}(M_*+1)^{3(s+a)-2}{\mathcal N}_* \\
&\quad + \chi_*^{2a}(M_*+1)^{2a} \cdot (1+T_0^{-1})^{s-2}\chi_*^{(4+a)s}M_*^{s-2}(M_*+1)^{3s-2}{\mathcal N}_*\Big\}\\
&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{(4+a)(s+a+2)}(1+T_0^{-1})^{s+a-1} M_*^{s-2} (M_*+1)^{3s+4a+2}{\mathcal N}_*.
\end{align*}
We obtain the second estimate in \eqref{pwtnew}.

Case $4-a<s\le 4$. Estimating $I_1$ by \eqref{kug4} applied to $s:=s+a$, and estimating $I_2$ by  \eqref{ab24} yield
\begin{align*}
J&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{2(3+a)}(M_*+1)^4(1+T_0^{-1})\\
  &\quad \cdot\Big\{  (1+T_0^{-1})^{s+a} \chi_*^{(4+a)(s+a+2)}M_*^2 (M_*+1)^{4(s+a)+2}{\mathcal N}_* \\
&\quad + \chi_*^{2a}(M_*+1)^{2a} \cdot (1+T_0^{-1})^{s-2}\chi_*^{(4+a)s}M_*^{s-2}(M_*+1)^{3s-2}{\mathcal N}_*\Big\}\\
&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{(4+a)(s+a+4)}(1+T_0^{-1})^{s+a+1}M_*^{s-2} (M_*+1)^{3s+4a+10} {\mathcal N}_*.
\end{align*}
We obtain the third estimate in \eqref{pwtnew}.

Case $s>4$. Estimating $I_1$ by \eqref{kug4} for $s:=s+a$, and estimating $I_2$ by \eqref{kug4} result in
\begin{align*}
J&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{2(3+a)}(M_*+1)^4(1+T_0^{-1})\\
  &\quad \cdot\Big\{  (1+T_0^{-1})^{s+a} \chi_*^{(4+a)(s+a+2)}M_*^2 (M_*+1)^{4(s+a)+2}{\mathcal N}_* \\
&\quad + \chi_*^{2a}(M_*+1)^{2a} \cdot (1+T_0^{-1})^s \chi_*^{(4+a)(s+2)}M_*^2 (M_*+1)^{4s+2}{\mathcal N}_*\Big\}\\
&\le C\chi_*^{2(3+a)}M_*^4 T
  + C \chi_*^{(4+a)(s+a+4)}(1+T_0^{-1})^{s+a+1}M_*^2 (M_*+1)^{4s+4a+6} {\mathcal N}_*.
\end{align*}
We obtain the fourth estimate in \eqref{pwtnew}.

Finally, one can easily unify the estimates in \eqref{pwtnew} for all $s>2$ with \eqref{pwt6}. This can also be done for the case $s=2$ by comparing ${\mathcal N}_0$ with ${\mathcal N}_*$ using the last relation in \eqref{NNN0}.
\end{proof}

\begin{remark}\label{smlrmk2}
Similar to Remark \ref{smlrmk1}, when $u$, $u_0$, $\bar u_0$ are small in necessary norms, and ${\mathcal E}_*$ is small, then $M_*$ and ${\mathcal N}_0$ are small, which make the the right-hand sides of \eqref{pwtall} and \eqref{pwtnew} to be small.
\end{remark}

\bigskip
\noindent\textbf{\large Acknowledgments.} The authors would like to thank Dat Cao, Akif Ibragimov and Tuoc Phan for very helpful discussions.

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