# Interior Estimates for Generalized Forchheimer Flows of Slightly
Compressible Fluids

Luan T. Hoanga,∗ and Thinh T. Kieub

###### Abstract

The generalized Forchheimer flows are studied for slightly compressible fluids
in porous media with time-dependent Dirichlet boundary data for the pressure.
No restrictions on the degree of the Forchheimer polynomial are imposed. We
derive, for all time, the interior $L^{\infty}$-estimates for the pressure and
its partial derivatives, and the interior $L^{2}$-estimates for its Hessian.
The De Giorgi and Ladyzhenskaya-Uraltseva iteration techniques are used taking
into account the special structures of the equations for both pressure and its
gradient. These are combined with the uniform Gronwall-type bounds in
establishing the asymptotic estimates when time tends to infinity.

aDepartment of Mathematics and Statistics, Texas Tech University, Box 41042,
Lubbock, TX 79409–1042, U. S. A.

bDepartment of Mathematics, University of North Georgia, Gainesville Campus,
3820 Mundy Mill Rd., Oakwood, GA 30566, U. S. A.
Email addresses<EMAIL_ADDRESS><EMAIL_ADDRESS>
∗Corresponding author

##  1 Introduction

In this paper, we study the generalized Forchheimer flows for slightly
compressible fluids in porous media. Forchheimer equations are used to
describe the fluids’ dynamics when the ubiquitous Darcy’s law is not
applicable. They are nonlinear relations between the fluid’s velocity and
gradient of pressure which were realized by Darcy [3] and Dupuit [7] in early
works, formulated by Forchheimer [9, 10], and were studied more extensively
afterward in physics and engineering, see [17, 24, 2, 18, 21] and references
therein. Mathematics of Darcy flows has been studied thoroughly with a vast
literature for a long time dated back to the 1960s, see the treaty [23] for
references. In contrast, Forchheimer flows were investigated mathematically
much later in the 1990s. Even less are the mathematical works on compressible
Forchheimer flows. The reader is referred to [1, 11, 12] for more information
about this topic.

Generalized Forchheimer equations were proposed and studied in our previous
works [1, 11, 12, 13, 15] in order to cover a large class of fluid flows
formulated from experiments. For compressible fluids, they form a new class of
degenerate parabolic equations with their own characteristics compared to
other models of porous medium equations. Among a small number of papers on
these flows, recent work [15] is focused on studying the pressure and its time
derivative in space $L^{\infty}$, the pressure gradient in $L^{s}$ for
$s\in[1,\infty)$ and the pressure’s Hessian in $L^{2-\delta}$ for
$\delta\in(0,1)$. However, it requires the so-called Strict Degree Condition
(SDC), that is, the degree of the Forchheimer polynomial is less than
$4/(n-2)$, where $n$ is the spatial dimension. Another related paper [13], in
contrast, does not require (SDC), but the analysis is mainly for pressure in
space $L^{s}$, pressure gradient in space $L^{2-a}$ for a specific number
$a\in(0,1)$, and pressure’s time derivative in $L^{2}$. Our current work is to
unite the two approaches and develop them further without any restrictions on
the Forchheimer polynomials. Specifically, we consider the initial boundary
value problem (IBVP) for the pressure in a bounded domain with time-dependent
Dirichlet boundary data. The interior $W^{1,\infty}$ and $W^{2,2}$ norms of
the solutions are estimated, particularly for large time. Such estimates for
degenerate parabolic equations usually require much work, see, for e.g., [22,
19, 20, 6]. Improving on [15], we adapt and refine techniques by De Giorgi
[4], Ladyzhenskaya-Uraltseva [16], and also DiBenedetto [5] for parabolic
equations, and combine them with those used for Navier-Stokes equations [8].
These techniques are utilized successfully here thanks to the special
structure of our equation.

The paper is organized as follows. We recall the model, equations and basic
facts in section 2. In section 3, we derive the uniform Gronwall-type
estimates that sharpen the previous results in [11, 13] especially for large
time. In section 4, the interior $L^{\infty}$-estimates for pressure are
established by using $L^{\alpha}$-based De Giorgi iteration with sufficiently
large $\alpha$. This is different from [15] which is based on the
$L^{2}$-estimate and, hence, requires (SDC). As a result, no restrictions on
the Forchheimer polynomials are needed for our estimates. In section 5, we
first use Ladyzhenskaya-Uraltseva-type imbedding and iteration to estimate the
pressure gradient in $L^{s}$ for any $s\geq 1$. Our estimates only require the
initial data to be in the spaces $W^{1,2-a}$ and $L^{\alpha}$ for a fixed
$\alpha$. When $s$ is large, this requirement is much less than the $W^{1,s}$
condition in [15]. In subsection 5.2, we derive the $W^{1,\infty}$ estimates
for the pressure, which were not previously studied in [13, 15]. We make use
of the equation (5.21) for the pressure gradient, which it is highly nonlinear
but has a special structure. Thanks to this and previous $W^{1,s}$
($s<\infty$) bounds, we utilize the De Giorgi technique to obtain the
$L^{\infty}$-estimates for the pressure gradient. Section 6 contains the
$L^{\infty}$-estimates for the time derivative of the pressure, while section
7 contains new $L^{2}$-bounds for the pressure’s Hessian. It is noteworthy
that, thanks to the uniform Gronwall-type inequalities in section 3, the
asymptotic bounds obtained as time goes to infinity depend only on the
asymptotic behavior of the boundary data. This paper is focused primarily on
the interior estimates. The issue of the solutions’ estimates on the entire
domain is addressed in another work of ours [14].

##  2 Background

Consider a fluid in a porous medium occupying a bounded domain $U$ with
boundary $\Gamma=\partial U$ in space $\mathbb{R}^{n}$. For physics problem
$n=3$, but here we consider any $n\geq 2$. Let $x\in\mathbb{R}^{n}$ and
$t\in\mathbb{R}$ be the spatial and time variables. The fluid flow has
velocity $v(x,t)\in\mathbb{R}^{n}$, pressure $p(x,t)\in\mathbb{R}$ and density
$\rho(x,t)\in[0,\infty)$.

The generalized Forchheimer equations studied in [1, 11, 12, 13, 15] are of
the the form:

$g(|v|)v=-\nabla p,$ (2.1)

where $g(s)\geq 0$ is a function defined on $[0,\infty)$. When
$g(s)=\alpha,\alpha+\beta s,\alpha+\beta s+\gamma
s^{2},\alpha+\gamma_{m}s^{m-1}$, where $\alpha,\beta,\gamma,m,\gamma_{m}$ are
empirical constants, we have Darcy’s law, Forchheimer’s two-term, three-term
and power laws, respectively.

In this paper, we study the case when the function $g$ in (2.1) is a
generalized polynomial with non-negative coefficients, that is,

$g(s)=a_{0}s^{\alpha_{0}}+a_{1}s^{\alpha_{1}}+\ldots+a_{N}s^{\alpha_{N}}\quad\text{for}\quad
s\geq 0,$ (2.2)

where $N\geq 1$, the powers $\alpha_{0}=0<\alpha_{1}<\ldots<\alpha_{N}$ are
fixed real numbers (not necessarily integers), the coefficients
$a_{0},a_{1},\ldots,a_{N}$ are non-negative with $a_{0}>0$ and $a_{N}>0$. This
function $g(s)$ is referred to as the Forchheimer polynomial in equation
(2.1), and $\alpha_{N}$ is the degree of $g$.

From (2.1) one can solve $v$ implicitly in terms of $\nabla p$ and derives a
nonlinear version of Darcy’s equation:

$v=-K(|\nabla p|)\nabla p,$ (2.3)

where the function $K:[0,\infty)\to[0,\infty)$ is defined by

$K(\xi)=\frac{1}{g(s(\xi))},\text{ with }s=s(\xi)\geq 0\text{ satisfying
}sg(s)=\xi,\ \text{ for }\xi\geq 0.$ (2.4)

In addition to (2.1), we have the continuity equation

$\phi\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho v)=0,$ (2.5)

where number $\phi\in(0,1)$ is the constant porosity. Also, for slightly
compressible fluids, the equation of state is

$\frac{d\rho}{dp}=\frac{\rho}{\kappa},\quad\text{with }\kappa=const.>0.$ (2.6)

From (2.3), (2.5) and (2.6) one derives a scalar equation for the pressure:

$\phi\frac{\partial p}{\partial t}=\kappa\nabla\cdot(K(|\nabla p|)\nabla
p)+K(|\nabla p|)|\nabla p|^{2}.$ (2.7)

On the right-hand side of (2.7), the constant $\kappa$ is very large for most
slightly compressible fluids in porous media [17], hence we neglect its second
term and by scaling the time variable, we study the reduced equation

$\frac{\partial p}{\partial t}=\nabla\cdot(K(|\nabla p|)\nabla p).$ (2.8)

Note that this reduction is commonly used in engineering.

Our aim is to study the IBVP for equation (2.8) in a bounded domain. Here
afterward $U$ is a bounded open connected subset of $\mathbb{R}^{n}$,
$n=2,3,\ldots$ with $C^{2}$ boundary $\Gamma=\partial U$. Throughout, the
Forchheimer polynomial $g(s)$ is fixed. The following number is frequently
used in our calculations:

$a=\frac{\alpha_{N}}{1+\alpha_{N}}\in(0,1).$ (2.9)

The function $K(\xi)$ in (2.4) has the following properties (c.f. [1, 11]): it
is decreasing in $\xi$ mapping $\xi\in[0,\infty)$ onto $(0,1/a_{0}]$, and

$\frac{C_{1}}{(1+\xi)^{a}}\leq K(\xi)\leq\frac{C_{2}}{(1+\xi)^{a}},$ (2.10)
$C_{3}(\xi^{2-a}-1)\leq K(\xi)\xi^{2}\leq C_{2}\xi^{2-a},$ (2.11)
$-aK(\xi)\leq K^{\prime}(\xi)\xi\leq 0,$ (2.12)

where $C_{1}$, $C_{2}$, $C_{3}$ are positive constants depending on $g$. As in
previous works, we use the function $H(\xi)$ defined by

$H(\xi)=\int_{0}^{\xi^{2}}K(\sqrt{s})ds\quad\hbox{for }\xi\geq 0.$

It satisfies $K(\xi)\xi^{2}\leq H(\xi)\leq 2K(\xi)\xi^{2}$, thus, by (2.11),

$C_{3}(\xi^{2-a}-1)\leq H(\xi)\leq 2C_{2}\xi^{2-a}.$ (2.13)

The following parabolic Poincaré-Sobolev inequalities are needed for our
study. For each $T>0$, denote $Q_{T}=U\times(0,T)$. We define a threshold
exponent

$\alpha_{*}=\frac{an}{2-a}.$ (2.14)

######  Lemma 2.1.

Assume

$\alpha\geq 2\quad\text{and}\quad\alpha>\alpha_{*}.$ (2.15)

Let

$p=\alpha\Big{(}1+\frac{2-a}{n}\Big{)}-a.$ (2.16)

Then

$\|u\|_{L^{p}(Q_{T})}\leq C(1+\delta T)^{1/p}[[u]],$ (2.17)

where $\delta=1$ in general, $\delta=0$ in case $u$ vanishes on the boundary
$\partial U$, and

$[[u]]=\mathop{\mathrm{ess\,sup}}_{[0,T]}\|u(t)\|_{L^{\alpha}(U)}+\Big{(}\int_{0}^{T}\int_{U}|u|^{\alpha-2}|\nabla
u|^{2-a}dxdt\Big{)}^{\frac{1}{\alpha-a}}.$ (2.18)

In case $U=B_{R}$ – a ball of radius $R$ – the inequality (2.17) holds with

$[[u]]=R^{\frac{n}{p}-\frac{n}{\alpha}}\mathop{\mathrm{ess\,sup}}_{[0,T]}\|u(t)\|_{L^{\alpha}(B_{R})}+R^{\frac{n}{p}-\frac{n-(2-a)}{\alpha-a}}\Big{(}\int_{0}^{T}\int_{B_{R}}|u|^{\alpha-2}|\nabla
u|^{2-a}dxdt\Big{)}^{\frac{1}{\alpha-a}}$ (2.19)

and the constant $C$ independent of $R$.

The proof of Lemma 2.1 is given in Appendix A. The next is a particular
embedding with spatial weights from Lemma 2.4 of [15] (see inequality (2.28)
with $m=2$ there).

######  Lemma 2.2 (cf. [15], Lemma 2.4).

Given $W(x,t)>0$ on $Q_{T}$. Let $r$ be a number that satisfies

$\frac{2n}{n+2}<r<2.$ (2.20)

Set

$\varrho=\varrho(r)\mathbin{\buildrel\rm
def\over{\mathbin{=\kern-2.0pt=}}}4(1-1/r^{*}).$ (2.21)

Then

$\|u\|_{L^{\varrho}(Q_{T})}\leq C[[u]]_{2,W;T}\Big{\\{}\delta
T^{\frac{1}{\varrho}}+\mathop{\mathrm{ess\,sup}}_{t\in[0,T]}\Big{(}\int_{U}W(x,t)^{-\frac{r}{2-r}}\chi_{{\rm
supp}\,u}(x,t)dx\Big{)}^{\frac{2-r}{\varrho r}}\Big{\\}},$ (2.22)

where $\delta=1$ in general, $\delta=0$ in case $u$ vanishes on the boundary
$\partial U$, and

$[[u]]_{2,W;T}=\mathop{\mathrm{ess\,sup}}_{[0,T]}\|u(t)\|_{L^{2}(U)}+\Big{(}\int_{0}^{T}\int_{U}W(x,t)|\nabla
u|^{2}dxdt\Big{)}^{\frac{1}{2}}.$ (2.23)

The following is a generalization of the convergence of fast decay geometry
sequences in Lemma 5.6, Chapter II of [16]. It will be used in the De Giorgi
iterations.

######  Lemma 2.3 (cf. [15], Lemma A.2).

Let $\\{Y_{i}\\}_{i=0}^{\infty}$ be a sequence of non-negative numbers
satisfying

$Y_{i+1}\leq\sum_{k=1}^{m}A_{k}B_{k}^{i}Y_{i}^{1+\mu_{k}},\quad
i=0,1,2,\ldots,$

where $A_{k}>0$, $B_{k}>1$ and $\mu_{k}>0$ for $k=1,2,\ldots,m$. Let
$B=\max\\{B_{k}:1\leq k\leq m\\}$ and $\mu=\min\\{\mu_{k}:1\leq k\leq m\\}$.
If $Y_{0}\leq\min\\{(m^{-1}A_{k}^{-1}B^{-\frac{1}{\mu}})^{1/\mu_{k}}:1\leq
k\leq m\\}$ then $\lim_{i\to\infty}Y_{i}=0$.

##  3 Uniform Gronwall-type estimates

We study the following IBVP for $p(x,t)$:

$\begin{cases}\displaystyle\frac{\partial p}{\partial t}=\nabla\cdot(K(|\nabla
p|)\nabla p)&\text{in }U\times(0,\infty),\\\ p(x,0)=p_{0}(x)&\text{in }U,\\\
p(x,t)=\psi(x,t)&\text{on }\Gamma\times(0,\infty).\end{cases}$ (3.1)

In order to deal with the non-homogeneous boundary condition, the data
$\psi(x,t)$ with $x\in\Gamma$ and $t>0$ is extended to a function $\Psi(x,t)$
with $x\in\bar{U}$ and $t\geq 0$. Throughout, our results are stated in terms
of $\Psi$ instead of $\psi$. Nonetheless, corresponding results in terms of
$\psi$ can be retrieved as performed in [11]. The function $\Psi$ is always
assumed to have adequate regularities for all calculations in this paper.

It is proved in Section 3 of [13] that (3.1) possesses a weak solution
$p(x,t)$ for all $t>0$. It, in fact, has more regularity in spatial and time
variables, see [5]. For the current study, we assume that solution $p(x,t)$
has sufficient regularities both in $x$ and $t$ variables such that our
calculations hereafterward can be performed legitimately.

In this section, we obtain improved estimates for solutions of (3.1). We
emphasize the asymptotic estimates as $t\to\infty$ in terms of the asymptotic
behavior of $\Psi(x,t)$.

Notation. (a) Hereafter, symbol $C$ is used to denote a positive number
independent of the initial and boundary data, and the time variables $t$,
$T_{0}$, $T$; it may depend on many parameters, namely, exponents and
coefficients of polynomial $g$, the spatial dimension $n$ and domain $U$,
other involved exponents $\alpha$, $s$, etc. in calculations. The value of $C$
may vary from place to place, even on the same line. (b) For partial
derivative notation, we will alternatively use $\partial p/\partial
t=\partial_{t}p=p_{t}$, and $\partial p/\partial
x_{m}=\partial_{m}p=p_{x_{m}}$. (c) The Lebesgue norm $\|\cdot\|_{L^{s}}$
means $\|\cdot\|_{L^{s}(U)}$. (d) For a function $f(x,t)$, we denote by $f(t)$
the function $x\to f(x,t)$.

For $\alpha\geq 1$, we define

$A(\alpha,t)=\Big{[}\int_{U}|\nabla\Psi(x,t)|^{\frac{\alpha(2-a)}{2}}dx\Big{]}^{\frac{2(\alpha-a)}{\alpha(2-a)}}+\Big{[}\int_{U}|\Psi_{t}(x,t)|^{\alpha}dx\Big{]}^{\frac{\alpha-a}{\alpha(1-a)}}$
(3.2)

for $t\geq 0$, and

$A(\alpha)=\limsup_{t\to\infty}A(\alpha,t)\quad\text{and}\quad\beta(\alpha)=\limsup_{t\to\infty}[A^{\prime}(\alpha,t)]^{-}.$
(3.3)

Also, define for $\alpha>0$ the number

$\widehat{\alpha}=\max\big{\\{}\alpha,2,\alpha_{*}\big{\\}}.$ (3.4)

Whenever $\beta(\alpha)$ is in use, it is understood that the function $t\to
A(\alpha,t)$ belongs to $C^{1}((0,\infty))$.

For a function $f:[0,\infty)\to\mathbb{R}$, we denote by $Envf$ a continuous
and increasing function $F:[0,\infty)\to\mathbb{R}$ such that $F(t)\geq f(t)$
for all $t\geq 0$.

Let $p(x,t)$ be a solution to IBVP (3.1). Denote $\overline{p}=p-\Psi$ and
$\bar{p}_{0}=p_{0}-\Psi(\cdot,0)$. We recall relevant results from [13] below.

######  Theorem 3.1 (cf. [13], Theorem 4.3).

Let $\alpha>0$.

(i) For all $t\geq 0$,

$\int_{U}|\bar{p}(x,t)|^{\alpha}dx\leq
C\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\widehat{\alpha}}dx+[EnvA(\widehat{\alpha},t)]^{\frac{\widehat{\alpha}}{\widehat{\alpha}-a}}\Big{)}.$
(3.5)

(ii) If $A(\widehat{\alpha})<\infty$ then

$\limsup_{t\rightarrow\infty}\int_{U}|\bar{p}(x,t)|^{\alpha}dx\leq
C\big{(}1+A(\widehat{\alpha})^{\frac{\widehat{\alpha}}{\widehat{\alpha}-a}}\big{)}.$
(3.6)

(iii) If $\beta(\widehat{\alpha})<\infty$ then there is $T>0$ such that

$\int_{U}|\bar{p}(x,t)|^{\alpha}dx\leq
C\big{(}1+\beta(\widehat{\alpha})^{\frac{\widehat{\alpha}}{\widehat{\alpha}-2a}}+A(\widehat{\alpha},t)^{\frac{\widehat{\alpha}}{\widehat{\alpha}-a}}\big{)}\quad\text{for
all }t\geq T.$ (3.7)

For gradient and time derivative estimates, we denote

$\displaystyle G_{1}(t)$
$\displaystyle=\int_{U}|\nabla\Psi(x,t)|^{2}dx+\Big{[}\int_{U}|\Psi_{t}(x,t)|^{r_{0}}dx\Big{]}^{\frac{2-a}{r_{0}(1-a)}}+\Big{[}\int_{U}|\Psi_{t}(x,t)|^{r_{0}}dx\Big{]}^{\frac{1}{r_{0}}},$
$\displaystyle G_{2}(t)$
$\displaystyle=\int_{U}|\nabla\Psi_{t}(x,t)|^{2}dx+\int_{U}|\Psi_{t}(x,t)|^{2}dx,$
$\displaystyle G_{3}(t)$ $\displaystyle=G_{1}(t)+G_{2}(t),\quad
G_{4}(t)=G_{3}(t)+\int_{U}|\Psi_{tt}|^{2}dx.$

with $r_{0}=\frac{n(2-a)}{(2-a)(n+1)-n}$. For $t\geq 0$, recall from (4.20) in
[13] and from (3.25) in [11] that

$\displaystyle\int_{0}^{t}\int_{U}H(|\nabla p(x,\tau)|)dxd\tau$
$\displaystyle\leq
C\int_{U}\bar{p}^{2}_{0}(x)dx+C\int_{0}^{t}G_{1}(\tau)d\tau,$ (3.8)
$\displaystyle\int_{U}H(|\nabla
p(x,t)|)dx+\int_{0}^{t}\int_{U}|\bar{p}_{t}(x,\tau)|^{2}dxd\tau$
$\displaystyle\leq\int_{U}\big{[}H(|\nabla
p_{0}(x)|)+\bar{p}^{2}_{0}(x)\big{]}dx+C\int_{0}^{t}G_{3}(\tau)d\tau.$ (3.9)

Let $\alpha\geq\widehat{2}$. For $t>0$, applying Theorem 4.5 in [13] with
$t_{0}=t$ and following Remark 4.8 there to replace $\widehat{2}$ by $\alpha$,
we have

$\int_{U}|\bar{p}_{t}(x,t)|^{2}dx\leq
C\Big{\\{}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+t^{-1}\Big{(}\int_{U}|\bar{p}_{0}(x)|^{2}+H(|\nabla
p_{0}(x)|)dx+\int_{0}^{t}G_{3}(\tau)d\tau\Big{)}\Big{\\}}\\\
+C\int_{0}^{t}e^{-d_{0}(t-\tau)}\big{(}[EnvA(\alpha,\tau)]^{\frac{\alpha}{\alpha-a}}+G_{4}(\tau)\big{)}d\tau,$
(3.10)

where $d_{0}>0$ is independent of $\alpha$. Since $EnvA(\alpha,t)$ is
increasing in $t$, and $G_{3}\leq G_{4}$ we simplify (3.10) as

$\int_{U}|\bar{p}_{t}(x,t)|^{2}dx\leq
C(1+t^{-1})\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+\int_{U}H(|\nabla
p_{0}(x)|)dx\\\
+[EnvA(\alpha,t)]^{\frac{\alpha}{\alpha-a}}+\int_{0}^{t}G_{4}(\tau)d\tau\Big{)}.$
(3.11)

Below are improved estimates for time $t$ large by using uniform Gronwall-type
inequalities.

######  Lemma 3.2.

For $t\geq 1$,

$\int_{t-1}^{t}\int_{U}H(|\nabla p(x,\tau)|)dxd\tau\leq
C\int_{U}\bar{p}^{2}(x,t-1)dx+C\int_{t-1}^{t}G_{1}(\tau)d\tau,$ (3.12)
$\int_{U}H(|\nabla
p(x,t)|)dx+\frac{1}{2}\int_{t-1/2}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau\leq
C\int_{U}\bar{p}^{2}(x,t-1)dx+C\int_{t-1}^{t}G_{3}(\tau)d\tau,$ (3.13)
$\int_{U}\bar{p}_{t}^{2}(x,t)dx\leq
C\int_{U}\bar{p}^{2}(x,t-1)dx+C\int_{t-1}^{t}G_{4}(\tau)d\tau.$ (3.14)

###### Proof.

The proof follows Theorems 4.4 and 4.5 in [12].

Proof of (3.12). Using inequality (3.4) of Lemma 3.1 in [11], one has

$\frac{d}{dt}\int_{U}\bar{p}^{2}(x,t)dx\leq-C\int_{U}H(|\nabla
p|)dx+CG_{1}(t).$ (3.15)

Dropping the negative term on the right-hand side of (3.15), and integrating
it from $t-1$ to $t$, we obtain (3.12).

Proof of (3.13). Using (3.17) of Lemma 3.3 in [11] with $\varepsilon=1$, one
has

$\frac{d}{dt}\int_{U}H(|\nabla
p|)dx\leq-\int_{U}\bar{p}_{t}^{2}(x,t)dx+\int_{U}H(|\nabla p|)dx+CG_{2}(t).$
(3.16)

Let $s\in[t-1,t]$. Integrating (3.16) from $s$ to $t$ we have

$\int_{U}H(|\nabla
p(x,t)|)dx+\int_{s}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau\\\
\leq\int_{U}H(|\nabla p|)(x,s)dx+\int_{s}^{t}\int_{U}H(|\nabla
p|)dxd\tau+C\int_{s}^{t}G_{2}(\tau)d\tau.$

Thus,

$\int_{U}H(|\nabla
p(x,t)|)dx+\int_{s}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau\\\
\leq\int_{U}H(|\nabla p|)(x,s)dx+\int_{t-1}^{t}\int_{U}H(|\nabla
p|)dxd\tau+C\int_{t-1}^{t}G_{2}(\tau)d\tau.$ (3.17)

Integrating (3.17) in $s$ from $t-1$ to $t$, and using (3.12), we obtain

$\int_{U}H(|\nabla
p(x,t)|)dx+\int_{t-1}^{t}\int_{s}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau
ds\leq 2\int_{t-1}^{t}\int_{U}H(|\nabla
p|)dxd\tau+C\int_{t-1}^{t}G_{2}(\tau)d\tau\\\ \leq
C\int_{U}\bar{p}^{2}(x,t-1)dx+C\int_{t-1}^{t}G_{1}(\tau)d\tau+C\int_{t-1}^{t}G_{2}(\tau)d\tau.$
(3.18)

Observe that

$\displaystyle\int_{t-1}^{t}\int_{s}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau
ds=\int_{t-1}^{t}\int_{t-1}^{\tau}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxdsd\tau$
$\displaystyle\geq\int_{t-1/2}^{t}\int_{t-1}^{\tau}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxdsd\tau\geq\frac{1}{2}\int_{t-1/2}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau.$

Therefore (3.13) follows from (3.18).

Proof of (3.14). From (3.37) of Proposition 3.11 in [11] we have

$\frac{d}{dt}\int_{U}\bar{p}_{t}^{2}dx\leq-C\int_{U}K(|\nabla p|)|\nabla
p_{t}|^{2}dx+C\int_{U}|\nabla\Psi_{t}|^{2}dx-\int_{U}\bar{p}_{t}\Psi_{tt}dx.$

Dropping the first term on the right-hand side and using Cauchy’s inequality
give

$\frac{d}{dt}\int_{U}\bar{p}_{t}^{2}dx\leq
C\int_{U}|\nabla\Psi_{t}|^{2}dx+\frac{1}{2}\int_{U}|\bar{p}_{t}|^{2}dx+\frac{1}{2}\int_{U}|\Psi_{tt}|^{2}dx.$
(3.19)

For $s\in[t-1/2,t]$, integrating (3.19) from $s$ to $t$ gives

$\displaystyle\int_{U}\bar{p}_{t}^{2}(x,t)dx$
$\displaystyle\leq\int_{U}\bar{p}_{t}^{2}(x,s)dx+C\int_{s}^{t}\int_{U}|\nabla\Psi_{t}|^{2}+|\Psi_{tt}|^{2}dxd\tau+\frac{1}{2}\int_{s}^{t}\int_{U}|\bar{p}_{t}(x,\tau)|^{2}dxd\tau,$
$\displaystyle\int_{U}\bar{p}_{t}^{2}(x,t)dx$
$\displaystyle\leq\int_{U}\bar{p}_{t}^{2}(x,s)dx+C\int_{t-1}^{t}\int_{U}|\nabla\Psi_{t}|^{2}+|\Psi_{tt}|^{2}dxd\tau+\frac{1}{2}\int_{t-1/2}^{t}\int_{U}|\bar{p}_{t}(x,\tau)|^{2}dxd\tau.$

Integrating the last inequality in $s$ from $t-1/2$ to $t$ yields

$\frac{1}{2}\int_{U}\bar{p}_{t}^{2}(x,t)dx\leq\frac{5}{4}\int_{t-1/2}^{t}\int_{U}\bar{p}_{t}^{2}(x,s)dxds+C\int_{t-1}^{t}\int_{U}|\nabla\Psi_{t})|^{2}+|\Psi_{tt}|^{2}dxd\tau.$

Using (3.13) for the first integral on the right-hand side, we have

$\int_{U}\bar{p}_{t}^{2}(x,t)dx\leq
C\int_{U}\bar{p}^{2}(x,t-1)dx+C\int_{t-1}^{t}G_{3}(\tau)d\tau+C\int_{t-1}^{t}\int_{U}|\nabla\Psi_{t})|^{2}+|\Psi_{tt}|^{2}dxd\tau.$
(3.20)

Using the fact $\int_{U}|\nabla\Psi_{t}|^{2}dx\leq CG_{2}(t)\leq CG_{3}(t)$,
we obtain (3.14) from (3.20). The proof is complete. ∎

Combining the above, we have the following specific estimates which will be
used conveniently in subsequent sections.

######  Corollary 3.3.

Let $\alpha\geq\widehat{2}$.

(i) For $t\geq 1$,

$\int_{t-1}^{t}\int_{U}H(|\nabla p(x,\tau)|)dxd\tau\leq
C\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+[EnvA(\alpha,t)]^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)},$
(3.21) $\int_{t-1/2}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau\leq
C\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+[EnvA(\alpha,t)]^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}.$
(3.22)

(ii) If $A(\alpha)<\infty$ then

$\limsup_{t\rightarrow\infty}\int_{U}H(|\nabla p(x,t)|)dx\leq
C\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)},$
(3.23) $\limsup_{t\rightarrow\infty}\int_{U}\bar{p}_{t}^{2}(x,t)dx\leq
C\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{t-1}^{t}G_{4}(\tau)d\tau\Big{)},$
(3.24) $\limsup_{t\rightarrow\infty}\int_{t-1}^{t}\int_{U}H(|\nabla
p(x,\tau)|)dxd\tau\leq
C\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)},$
(3.25)
$\limsup_{t\rightarrow\infty}\int_{t-1/2}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau\leq
C\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}.$
(3.26)

(iii) If $\beta(\alpha)<\infty$ then there is $T>0$ such that one has for all
$t\geq T$ that

$\int_{U}H(|\nabla p(x,t)|)dx\leq
C\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)},$
(3.27) $\int_{U}\bar{p}_{t}^{2}(x,t)dx\leq
C\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{4}(\tau)d\tau\Big{)},$
(3.28) $\int_{t-1}^{t}\int_{U}H(|\nabla p(x,\tau)|)dxd\tau\leq
C\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)},$
(3.29) $\int_{t-1/2}^{t}\int_{U}\bar{p}_{t}^{2}(x,\tau)dxd\tau\leq
C\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}.$
(3.30)

###### Proof.

Note that $\widehat{\alpha}=\alpha$.

(i) For (3.21), resp. (3.22), we combine (3.12), resp. (3.13), with inequality

$\int_{U}\bar{p}^{2}(x,t-1)dx\leq\int_{U}(1+|\bar{p}(x,t-1)|^{\alpha})dx,$
(3.31)

and the estimate (3.5) for $t-1$.

(ii) We combine estimates in Lemma 3.2 with (3.31) and the limit (3.6).

(iii) This part is similar to part (ii) with the use of (3.7) in place of
(3.6). ∎

##  4 Interior $L^{\infty}$-estimates for pressure

In this section we estimate the $L^{\infty}$-norm of the pressure. Hereafter,
$\alpha$ is a number that satisfies (2.15). We use $\kappa_{j}$ to denote a
number of exponents that depend on $\alpha$. Let

$\kappa_{0}=\alpha(1+(2-a)/n)-a,\quad\kappa_{1}=1/\delta_{1}\quad\text{and}\quad\kappa_{2}=1/\delta_{2},$
(4.1)

where $\delta_{1}=1-\alpha/\kappa_{0}$ and
$\delta_{2}=(1-a/\alpha)(1-\alpha_{*}/\alpha)$. Note that
$\delta_{1},\delta_{2}\in(0,1)$, hence, $\kappa_{1},\kappa_{2}>1$.

We start with estimating the $L^{\infty}$-norm in terms of the
$L^{\alpha}$-norm.

######  Theorem 4.1.

Let $U^{\prime}\Subset U$. If $T_{0}\geq 0$, $T>0$ and $\theta\in(0,1)$ then

$\sup_{[T_{0}+\theta T,T_{0}+T]}\|{p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C(1+T)^{\frac{\kappa_{1}}{\kappa_{0}}}\Big{(}1+(\theta
T)^{-1}\Big{)}^{\frac{\kappa_{1}}{\alpha-a}}\Big{(}1+\|p\|_{L^{\alpha}(U\times(T_{0},T_{0}+T))}\Big{)}^{\kappa_{2}}.$
(4.2)

###### Proof.

Without loss of generality, we assume $T_{0}=0$. For $k\geq 0$, define
$p^{(k)}=\max\\{p-k,0\\}$ and denote by $\chi_{k}(x,t)$ the characteristic
function of the set $\operatorname{supp}p^{(k)}$. Let $\phi_{1}(x)$ and
$\phi_{2}(t)$ be cut-off functions with $\phi_{1}=1$ on $U^{\prime}$,
$\phi_{1}=0$ on a neighborhood of $\partial U$, and $\phi_{2}(0)=0$. Let
$\zeta(x,t)=\phi_{1}(x)\phi_{2}(t)$. Multiplying the first equation in (3.1)
by $|p^{(k)}|^{\alpha-1}\zeta^{2}$ and integrating over $U$, we have

$\displaystyle\frac{1}{\alpha}\frac{d}{dt}\int_{U}|p^{(k)}|^{\alpha}\zeta^{2}dx+(\alpha-1)\int_{U}K(|\nabla
p^{(k)}|)|\nabla p^{(k)}|^{2}|p^{(k)}|^{\alpha-2}\zeta^{2}dx$
$\displaystyle=\frac{2}{\alpha}\int_{U}|p^{(k)}|^{\alpha}\zeta\zeta_{t}dx+2\int_{U}K(|\nabla
p^{(k)}|)(\nabla p^{(k)}\cdot\nabla\zeta)|p^{(k)}|^{\alpha-1}\zeta dx.$

By Cauchy’s inequality, we have for the last integral that

$\displaystyle 2|K(|\nabla p^{(k)}|)(\nabla
p^{(k)}\cdot\nabla\zeta)|p^{(k)}|^{\alpha-1}\zeta|\leq\frac{\alpha-1}{2}K(|\nabla
p^{(k)}|)|\nabla p^{(k)}|^{2}|p^{(k)}|^{\alpha-2}\zeta^{2}+CK(|\nabla
p^{(k)}|)|p^{(k)}|^{\alpha}|\nabla\zeta|^{2}.$

Combining the above, we obtain

$\displaystyle\frac{1}{\alpha}\frac{d}{dt}\int_{U}|p^{(k)}|^{\alpha}\zeta^{2}dx+\frac{\alpha-1}{2}\int_{U}K(|\nabla
p^{(k)}|)|\nabla p^{(k)}|^{2}|p^{(k)}|^{\alpha-2}\zeta^{2}dx$
$\displaystyle\quad\leq
C\int_{U}|p^{(k)}|^{\alpha}\zeta|\zeta_{t}|dx+C\int_{U}K(|\nabla
p^{(k)}|)|p^{(k)}|^{\alpha}|\nabla\zeta|^{2}dx.$

Using (2.11), we then have

$\displaystyle\frac{d}{dt}\int_{U}|p^{(k)}|^{\alpha}\zeta^{2}dx+\int_{U}|\nabla
p^{(k)}|^{2-a}|p^{(k)}|^{\alpha-2}\zeta^{2}dx$ $\displaystyle\quad\leq
C\int_{U}|p^{(k)}|^{\alpha}\zeta|\zeta_{t}|dx+C\int_{U}K(|\nabla
p^{(k)}|)|p^{(k)}|^{\alpha}|\nabla\zeta|^{2}dx+C\int_{U}|p^{(k)}|^{\alpha-2}\zeta^{2}dx.$

Using the boundedness of $K(\cdot)$ in the second to last integral, and
applying Young’s inequality to the last terms yield

$\displaystyle\frac{d}{dt}\int_{U}|p^{(k)}|^{\alpha}\zeta^{2}dx+\int_{U}|\nabla
p^{(k)}|^{2-a}|p^{(k)}|^{\alpha-2}\zeta^{2}dx$ $\displaystyle\quad\leq
C\int_{U}|p^{(k)}|^{\alpha}(\zeta^{2}+\zeta|\zeta_{t}|+|\nabla\zeta|^{2})dx+C\int_{U}\chi_{k}\zeta^{2}dx.$

For $t\in[0,T]$, integrating from $0$ to $T$, and taking supremum in $t$, we
obtain

$\displaystyle\sup_{[0,T]}\int_{U}|p^{(k)}|^{\alpha}\zeta^{2}dx+\int_{0}^{T}\int_{U}|\nabla
p^{(k)}|^{2-a}|p^{(k)}|^{\alpha-2}\zeta^{2}dxdt$ (4.3) $\displaystyle\leq
C\int_{0}^{T}\int_{U}|p^{(k)}|^{\alpha}(\zeta^{2}+\zeta|\zeta_{t}|+|\nabla\zeta|^{2})dx+C\int_{0}^{T}\int_{U}\chi_{k}\zeta^{2}dx.$

Let $x_{0}$ be any given point in $\bar{U}^{\prime}$. Denote $\rho={\rm
dist}(\bar{U}^{\prime},\partial U)>0$. Let $M_{0}>0$ be fixed which will be
determined later. For $i\geq 0$, define

$k_{i}=M_{0}(1-2^{-i}),\quad t_{i}=\theta
T(1-2^{-i}),\quad\rho_{i}=\frac{1}{4}\rho(1+2^{-i}).$ (4.4)

Then $t_{0}=0<t_{1}<\ldots<\theta T$ and
$\rho_{0}=\rho/2>\rho_{1}>\ldots>\rho/4>0.$ Note that

$\lim_{i\to\infty}t_{i}=\theta
T\quad\text{and}\quad\lim_{i\to\infty}\rho_{i}=\rho/4.$ (4.5)

Let $U_{i}=\\{x:\|{x-x_{0}}\|<\rho_{i}\\}$ then $U_{i+1}\Subset U_{i}$ for
$i=0,1,2,\ldots$. For $i,j\geq 0$, we denote

$\mathcal{Q}_{i}=U_{i}\times(t_{i},T)\quad\text{ and }\quad
A_{i,j}=\\{(x,t)\in\mathcal{Q}_{j}:p(x,t)>k_{i}\\}.$ (4.6)

For each $\mathcal{Q}_{i}$, we use a cut-off function $\zeta_{i}(x,t)$ with
$\zeta_{i}\equiv 1$ in $\mathcal{Q}_{i+1}$ and $\zeta_{i}\equiv 0$ on
$Q_{T}\setminus\mathcal{Q}_{i}$, and

$|(\zeta_{i})_{t}|\leq\frac{C}{t_{i+1}-t_{i}}=\frac{C2^{i+1}}{\theta
T},\quad\text{ and
}\quad|\nabla\zeta_{i}|\leq\frac{C}{\rho_{i}-\rho_{i+1}}=\frac{C2^{i+1}}{4\rho}.$
(4.7)

for some $C>0$. Applying (4.3) with $k=k_{i+1}$ and $\zeta=\zeta_{i}$ gives

$\displaystyle\sup_{[0,T]}\int_{U}|p^{(k_{i+1})}|^{\alpha}\zeta_{i}^{2}dx+\int_{0}^{T}\int_{U}|\nabla
p^{(k_{i+1})}|^{2-a}|p^{(k_{i+1})}|^{\alpha-2}\zeta_{i}^{2}dxdt$ (4.8)
$\displaystyle\quad\quad\leq\int_{0}^{T}\int_{U}|p^{(k_{i+1})}|^{\alpha}(\zeta_{i}^{2}+\zeta_{i}|\zeta_{it}|+|\nabla\zeta_{i}|^{2})dxdt+C\mathcal{\int}_{0}^{T}\int_{U}\chi_{k_{i+1}}\zeta_{i}^{2}dxdt.$

Define

$F_{i}\mathbin{\buildrel\rm
def\over{\mathbin{=\kern-2.0pt=}}}\sup_{[t_{i+1},T]}\int_{U_{i+1}}|p^{(k_{i+1})}|^{\alpha}dx+\int_{t_{i+1}}^{T}\int_{U_{i+1}}|\nabla
p^{(k_{i+1})}|^{2-a}|p^{(k_{i+1})}|^{\alpha-2}dxdt.$

Then (4.8) yields

$\displaystyle F_{i}$
$\displaystyle\leq\int_{t_{i}}^{T}\int_{U}|p^{(k_{i+1})}|^{\alpha}(\zeta_{i}^{2}+\zeta_{i}|\zeta_{it}|+|\nabla\zeta_{i}|^{2})dxdt+C\Big{(}\int_{t_{i}}^{T}\int_{U_{i}}\chi_{k_{i+1}}dxdt\Big{)}$
(4.9) $\displaystyle\leq C4^{i}((\theta
T)^{-1}+1)\|p^{(k_{i+1})}\|_{L^{\alpha}(A_{i+1,i})}^{\alpha}+C|A_{i+1,i}|.$

Since
$\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}\geq\|p^{(k_{i})}\|_{L^{\alpha}(A_{i+1,i})}\geq(k_{i+1}-k_{i})|A_{i+1,i}|^{1/\alpha}$,
we have

$|A_{i+1,i}|\leq(k_{i+1}-k_{i})^{-\alpha}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}^{\alpha}\leq
C2^{\alpha i}M_{0}^{-\alpha}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}^{\alpha}.$
(4.10)

This and (4.9) imply

$\displaystyle F_{i}$ $\displaystyle\leq C4^{i}(1+(\theta
T)^{-1})\|p^{(k_{i+1})}\|_{L^{\alpha}(A_{i+1,i})}^{\alpha}+C2^{\alpha
i}M_{0}^{-\alpha}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}^{\alpha}$ (4.11)
$\displaystyle\leq C2^{\alpha i}\Big{(}1+(\theta
T)^{-1}+M_{0}^{-\alpha}\Big{)}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}^{\alpha}.$

Note that $\kappa_{0}$ is the exponent defined in (2.16). Applying Lemma 2.1
to domain $U_{i}$, interval $[t_{i+1},T]$ and function $p^{(k_{i+1})}$ with
the use of (2.19), we have

$\|p^{(k_{i+1})}\|_{L^{\kappa_{0}}(A_{i+1,i+1})}=\|p^{(k_{i+1})}\|_{L^{\kappa_{0}}(\mathcal{Q}_{i+1})}\leq
C(1+T)^{\frac{1}{\kappa_{0}}}(F_{i}^{1/\alpha}+F_{i}^{1/(\alpha-a)}).$ (4.12)

Note that the constant $C$ depends on $\rho$. Holder’s inequality gives

$\|p^{(k_{i+1})}\|_{L^{\alpha}(A_{i+1,i+1})}\leq\|p^{(k_{i+1})}\|_{L^{\kappa_{0}}(A_{i+1,i+1})}|A_{i+1,i+1}|^{1/\alpha-1/\kappa_{0}}\leq\|p^{(k_{i+1})}\|_{L^{\kappa_{0}}(A_{i+1,i+1})}|A_{i+1,i}|^{1/\alpha-1/\kappa_{0}}.$

It follows from this, (4.12), (4.11), and (4.10) that

$\displaystyle\|p^{(k_{i+1})}\|_{L^{\alpha}(A_{i+1,i+1})}$ $\displaystyle\leq
C(1+T)^{\frac{1}{\kappa_{0}}}(F_{i}^{1/\alpha}+F_{i}^{1/(\alpha-a)})|A_{i+1,i}|^{1/\alpha-1/\kappa_{0}}$
$\displaystyle\leq C(1+T)^{\frac{1}{\kappa_{0}}}B^{i}\Big{\\{}\Big{(}1+(\theta
T)^{-1}+M_{0}^{-\alpha}\Big{)}^{1/\alpha}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}$
$\displaystyle\quad+\Big{(}1+(\theta
T)^{-1}+M_{0}^{-\alpha}\Big{)}^{1/(\alpha-a)}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i})}^{\alpha/(\alpha-a)}\Big{\\}}M_{0}^{-1+\alpha/\kappa_{0}}\|p^{(k_{i})}\|_{L^{2}(A_{i,i})}^{1-\alpha/\kappa_{0}},$

where $B=2^{\alpha}$. Now selecting $M_{0}\geq 1$ we have

$\displaystyle\|p^{(k_{i})}\|_{L^{\alpha}(A_{i+1,i+1})}$ $\displaystyle\leq
C(1+T)^{\frac{1}{\kappa_{0}}}B^{i}\Big{\\{}\Big{(}1+(\theta
T)^{-1}\Big{)}^{1/\alpha}M_{0}^{-1+\alpha/\kappa_{0}}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}^{2-\alpha/\kappa_{0}}$
$\displaystyle\quad+\Big{(}1+(\theta
T)^{-1}\Big{)}^{1/(\alpha-a)}M_{0}^{-1+\alpha/\kappa_{0}}\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}^{\alpha/(\alpha-a)+1-\alpha/\kappa_{0}}\Big{\\}}.$

Denote $Y_{i}=\|p^{(k_{i})}\|_{L^{\alpha}(A_{i,i})}.$ Then

$Y_{i+1}\leq
CB^{i}\Big{(}D_{1}Y_{i}^{1+\delta_{1}}+D_{2}Y_{i}^{1+\nu_{1}}\Big{)},$

where $\delta_{1}>0$ is defined in (4.1),
$\nu_{1}=\alpha/(\alpha-a)-\alpha/\kappa_{0}>0$,

$D_{1}=(1+T)^{\frac{1}{\kappa_{0}}}\big{(}1+(\theta
T)^{-1}\big{)}^{1/\alpha}M_{0}^{-\delta_{1}},\quad\text{and}\quad
D_{2}=(1+T)^{\frac{1}{\kappa_{0}}}\big{(}1+(\theta
T)^{-1}\big{)}^{1/(\alpha-a)}M_{0}^{-\delta_{1}}.$

Take $M_{0}$ sufficiently large such that

$Y_{0}\leq C\min\\{D_{1}^{-1/\delta_{1}},D_{2}^{-1/\nu_{1}}\\}.$ (4.13)

Then by Lemma 2.3, $\lim_{i\to\infty}Y_{i}=0$, consequently, $\int_{\theta
T}^{T}\int_{B_{\rho/4}(x_{0})}|p^{(M_{0})}|^{\alpha}dxdt=0,$ that is

$p(x,t)\leq M_{0}\quad\text{a.e. in }B_{\rho/4}(x_{0})\times(\theta T,T).$

Since $Y_{0}\leq\|p\|_{L^{\alpha}(Q_{T})}$, condition (4.13) is met if
$\|p\|_{L^{\alpha}(Q_{T})}\leq
C\min\\{D_{1}^{-1/\delta_{1}},D_{2}^{-1/\nu_{1}}\\}.$ This, in turn, is
satisfied if

$M_{0}\geq C(1+T)^{\frac{1}{\kappa_{0}\delta_{1}}}\Big{\\{}(1+(\theta
T)^{-1})^{1/(\alpha\delta_{1})}\|p\|_{L^{\alpha}(Q_{T})}+(1+(\theta
T)^{-1})^{1/((\alpha-a)\delta_{1})}\|p\|_{L^{\alpha}(Q_{T})}^{\nu_{1}/\delta_{1}}\Big{\\}}.$

Note that $\nu_{1}/\delta_{1}=1/\delta_{2}=\kappa_{2}>1$. Combining this and
condition $M_{0}\geq 1$ we choose

$M_{0}=C(1+T)^{\frac{\kappa_{1}}{\kappa_{0}}}(1+\theta
T)^{-1})^{\frac{\kappa_{1}}{\alpha-a}}(1+\|p\|_{L^{\alpha}(Q_{T})})^{\kappa_{2}},$

with an appropriate positive constant $C$. By using a finite covering of
$\bar{U}^{\prime}$, we obtain

$p(x,t)\leq M_{0}\quad\text{a.e. in }U^{\prime}\times(\theta T,T).$

Repeating the argument for $-p$ instead of $p$, we obtain $|p(x,t)|\leq M_{0}$
a.e. in $U^{\prime}\times[\theta T,T]$, and hence (4.2) follows. The proof is
complete. ∎

Combining Theorem 4.1 with estimates in Section 3, we obtain the following
specific estimates for the $L^{\infty}$-norm.

######  Theorem 4.2.

Let $U^{\prime}\Subset U$.

(i) If $t\in(0,1)$ then

$\displaystyle\|{p(t)}\|_{L^{\infty}(U^{\prime})}$ $\displaystyle\leq
Ct^{-\frac{\kappa_{1}}{\alpha-a}}\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\kappa_{2}}.$
(4.14)

If $t\geq 1$ then

$\|{p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\kappa_{2}}.$
(4.15)

(ii) If $A(\alpha)<\infty$ then

$\limsup_{t\rightarrow\infty}\|{p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\big{)}^{\kappa_{2}}.$
(4.16)

(iii) If $\beta(\alpha)<\infty$ then there is $T>0$ such that

$\|{p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\|A(\alpha,\cdot)\|_{L^{\frac{\alpha}{\alpha-a}}(t-1,t)}^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\big{)}^{\kappa_{2}}$
(4.17)

for all $t\geq T$.

###### Proof.

(i) Note that $\alpha=\widehat{\alpha}$. Let $t\in(0,1)$. Applying (4.2) for
$T=t$ and $\theta=1/2$, we have

$\displaystyle\|p(t)\|_{L^{\infty}(U^{\prime})}$ $\displaystyle\leq
C(1+t^{-\frac{\kappa_{1}}{\alpha-a}})(1+\|p\|_{L^{\alpha}(U\times(0,t))}^{\kappa_{2}})$
$\displaystyle\leq
Ct^{-\frac{\kappa_{1}}{\alpha-a}}(1+\|\bar{p}\|_{L^{\alpha}(U\times(0,t))}+\|\Psi\|_{L^{\alpha}(U\times(0,t))})^{\kappa_{2}}.$

Using (3.5) to estimate $\|\bar{p}\|_{L^{\alpha}(U\times(0,t))}$ and the fact
that $EnvA(\alpha,t)$ is increasing in $t$, we obtain (4.14).

For $t\geq 1$, applying (4.2) with $T_{0}=t-1$, $T=1$ and $\theta=1/2$ we
obtain

$\|{p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+\|p\|_{L^{\alpha}(U\times(t-1,t))}^{\kappa_{2}}\Big{)}\leq
C\Big{(}1+\|\bar{p}\|_{L^{\alpha}(U\times(t-1,t))}+\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\kappa_{2}}.$
(4.18)

Again, using inequality (3.5) to estimate
$\|\bar{p}\|_{L^{\alpha}(U\times(t-1,t))}$, we obtain (4.15).

(ii) From (4.18) we have

$\limsup_{t\to\infty}\|{p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+\limsup_{t\to\infty}\|\bar{p}\|_{L^{\alpha}(U\times(t-1,t))}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\kappa_{2}}.$

By (3.6),

$\limsup_{t\to\infty}\|\bar{p}\|_{L^{\alpha}(U\times(t-1,t))}^{\alpha}\leq\limsup_{t\to\infty}\int_{U}|\bar{p}(x,t)|^{\alpha}dx\leq
C(1+A(\alpha))^{\alpha/(\alpha-a)}.$

Thus (4.16) follows.

(iii) By (3.7), we have for large $t$ that

$\int_{t-1}^{t}\int_{U}|\bar{p}(x,\tau)|^{\alpha}dxd\tau\leq
C\big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+\int_{t-1}^{t}A(\alpha,\tau)^{\frac{\alpha}{\alpha-a}}d\tau).$
(4.19)

From (4.18) and (4.19), we obtain (4.17). ∎

It is worthy to mention that the quantities on the right-hand sides of
estimates (4.16) and (4.17) only depend on the boundary data’s large time
behavior. This feature will be re-established in many of our estimates below
for the gradient, time derivative and Hessian of the pressure.

##  5 Interior estimates for pressure gradient

We will first estimate the pressure gradient in $L^{s}$-norm (for $s<\infty$)
in subsection 5.1, and then in $L^{\infty}$-norm in subsection 5.2.

###  5.1  $L^{s}$-estimates

The following imbedding lemma from [15] is a suitable extension of Lemma 5.4
on page 93 in [16].

######  Lemma 5.1 (cf. [15], Lemma 3.2).

For each $s\geq 1$, there exists a constant $C>0$ depending on $s$ such that
for each smooth cut-off function $\zeta(x)\in C_{c}^{\infty}(U)$, the
following inequality holds

$\displaystyle\int_{U}K(|\nabla p|)|\nabla p|^{2s+2}\zeta^{2}dx$
$\displaystyle\leq C\max_{\text{\rm supp}\zeta}|p|^{2}\Big{[}\int_{U}K(|\nabla
p|)|\nabla p|^{2s-2}|\nabla^{2}p|^{2}\zeta^{2}dx$
$\displaystyle\quad\quad\quad+\int_{U}K(|\nabla p|)|\nabla
p|^{2s}|\nabla\zeta|^{2}dx\Big{]},$

for every sufficiently regular function $p(x)$ such that the right hand side
is well-defined.

We establish the basic step for the Ladyzhenskaya-Uraltseva iteration.

######  Lemma 5.2.

For each $s\geq 0$, if $T_{0}\geq 0$, $T>0$, and $\zeta(x,t)$ is a smooth
function with $\zeta(\cdot,T_{0})=0$ and
$\operatorname{supp}\zeta(\cdot,t)\subset U$ for all $t\in[T_{0},T_{0}+T]$,
then

$\sup_{[T_{0},T_{0}+T]}\int_{U}|\nabla
p(x,t)|^{2s+2}\zeta^{2}dx+\int_{T_{0}}^{T_{0}+T}\int_{U}K(|\nabla
p|)|\nabla^{2}p|^{2}|\nabla p|^{2s}\zeta^{2}dxdt\\\ \leq
C\int_{T_{0}}^{T_{0}+T}\int_{U}K(|\nabla p|)|\nabla
p|^{2s+2}|\nabla\zeta|^{2}dxdt+C\int_{T_{0}}^{T_{0}+T}\int_{U}|\nabla
p|^{2s+2}\zeta|\zeta_{t}|dx.$ (5.1)

###### Proof.

Without loss of generality, assume $T_{0}=0$. Multiplying the equation (3.1)
by $-\nabla\cdot(|\nabla p|^{2s}\zeta^{2}\nabla p)$, integrating the resultant
over $U$ and using integration by parts, we obtain

$\displaystyle\frac{1}{2s+2}\frac{d}{dt}\int_{U}|\nabla
p|^{2s+2}\zeta^{2}dx=-\int_{U}\partial_{j}(K(|\nabla
p|)\partial_{i}p)\,\partial_{i}(|\nabla
p|^{2s}\partial_{j}p\zeta^{2})dx+\frac{1}{s+1}\int_{U}|\nabla
p|^{2s+2}\zeta\zeta_{t}dx.$

Calculated as in Lemma 3.1 of [15] above equation is rewritten as

$\displaystyle\frac{1}{2s+2}\frac{d}{dt}\int_{U}|\nabla
p|^{2s+2}\zeta^{2}dx=-\int_{U}\Big{[}\partial_{y_{l}}(K(|y|)y_{i})\Big{|}_{y=\nabla
p}\partial_{j}\partial_{l}p\Big{]}\partial_{j}\partial_{i}p|\nabla
p|^{2s}\zeta^{2}dx$
$\displaystyle\quad-2\int_{U}\Big{[}\partial_{y_{l}}(K(|y|)y_{i})\Big{|}_{y=\nabla
p}\partial_{j}\partial_{l}p\Big{]}\partial_{j}p\,|\nabla
p|^{2s}\,\zeta\partial_{i}\zeta dx$
$\displaystyle\quad-2s\int_{U}\Big{[}\partial_{y_{l}}(K(|y|)y_{i})\Big{|}_{y=\nabla
p}\partial_{j}\partial_{l}p\Big{]}\partial_{j}p\,(|\nabla
p|^{2s-2}\partial_{i}\partial_{m}p\partial_{m}p)\,\zeta^{2}dx+\frac{1}{s+1}\int_{U}|\nabla
p|^{2s+2}\zeta\zeta_{t}dx.$

We denote the four terms on the right-hand side by $I_{1}$, $I_{2},I_{3}$ and
$I_{4}$. It follows from the calculations in Lemma 3.1 of [15] that

$\displaystyle I_{1}$ $\displaystyle\leq-(1-a)\sum_{j}\int_{U}K(|\nabla
p|)|\nabla(\partial_{j}p)|^{2}|\nabla p|^{2s}\zeta^{2}dx,$
$\displaystyle|I_{2}|$ $\displaystyle\leq 2(1+a)\int_{U}K(|\nabla
p|)|\nabla^{2}p||\nabla p|^{2s+1}\zeta|\nabla\zeta|dx,$

and

$\displaystyle I_{3}\leq-2(1-a)s\int_{U}K(|\nabla
p|)\Big{|}\nabla\Big{(}\frac{1}{2}|\nabla p|^{2}\Big{)}\Big{|}^{2}|\nabla
p|^{2s-2}\zeta^{2}dx\leq 0.$

Combining these estimates with Young’s inequality, we find that

$\displaystyle\frac{1}{2s+2}\frac{d}{dt}\int_{U}|\nabla
p|^{2s+2}\zeta^{2}dx+\frac{1-a}{2}\int_{U}K(|\nabla
p|)|\nabla^{2}p|^{2}|\nabla p|^{2s}\zeta^{2}dx$ (5.2) $\displaystyle\quad\leq
C\int_{U}K(|\nabla p|)|\nabla
p|^{2s+2}|\nabla\zeta|^{2}dx+\frac{1}{s+1}\int_{U}|\nabla
p|^{2s+2}\zeta|\zeta_{t}|dx.$

Inequality (5.1) follows directly by integrating (5.2) from $0$ to $T$. ∎

Next, we reduce estimates for $W^{1,s}$-norm, with large $s$, down to
$W^{1,2-a}$ and $L^{\infty}$ norms.

######  Proposition 5.3.

Let $U^{\prime}\Subset V\Subset U$, $T_{0}\geq 0$, $T>0$ and $\theta\in(0,1)$.
If $s\geq 2$ then

$\int_{T_{0}+\theta T}^{T_{0}+T}\int_{U^{\prime}}K(|\nabla p|)|\nabla
p|^{s}dxdt\leq C\Big{(}1+(\theta
T)^{-1}\Big{)}^{s-2}\big{(}1+\sup_{[T_{0}+\theta
T/2,T_{0}+T]}\|p\|^{2}_{L^{\infty}(V)}\big{)}^{s-2}\\\
\cdot\int_{T_{0}}^{T_{0}+T}\int_{U}(1+K(|\nabla p|)|\nabla p|^{2})dxdt,$ (5.3)

and

$\sup_{t\in[T_{0}+\theta T,T_{0}+T]}\int_{U^{\prime}}|\nabla
p(x,t)|^{s}dxdt\leq C\Big{(}1+(\theta
T)^{-1}\Big{)}^{s+a-1}\big{(}1+\sup_{[T_{0}+\theta
T/2,T_{0}+T]}\|p\|^{2}_{L^{\infty}(V)}\big{)}^{s-2+a}\\\
\cdot\int_{T_{0}}^{T_{0}+T}\int_{U}(1+K(|\nabla p|)|\nabla p|^{2})dxdt,$ (5.4)

where constant $C>0$ is independent of $T_{0}$, $T$, and $\theta$.

###### Proof.

Without loss of generality, assume $T_{0}=0$. First, we prove a more general
version of (5.3).

Claim. For $0<\theta^{\prime}<\theta<1$ we have

$\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla p|^{s}dxdt\leq
C\Big{(}1+[(\theta-\theta^{\prime})T]^{-1}\Big{)}^{s-2}\big{(}1+\sup_{[\theta^{\prime}T,T]}\|p\|^{2}_{L^{\infty}(V)}\big{)}^{s-2}\\\
\cdot\int_{0}^{T}\int_{U}(1+K(|\nabla p|)|\nabla p|^{2})dxdt,$ (5.5)

where constant $C>0$ is independent of $T$, $\theta$, and $\theta^{\prime}$.

Note that (5.5) holds trivially when $s=2$. Hence, we will only prove (5.5)
for $s>2$ in a number of steps.

Step 1. Let $\zeta(x,t)$ be the cut-off function with $\zeta(x,t)=0$ for
$(x,t)\not\in V\times(\theta^{\prime}T,T]$ and $\zeta(x,t)=1$ for $(x,t)\in
U^{\prime}\times(\theta T,T]$. By applying Lemma 5.1 with $s+1$ in place of
$s$, we have for $s\geq 0$ that

$\displaystyle\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{2s+4}\zeta^{2}dxdt\leq C\max_{\rm{supp}\zeta}|p|^{2}$
$\displaystyle\Big{[}\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{2s}|\nabla^{2}p|^{2}\zeta^{2}dxdt$
$\displaystyle+\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{2s+2}|\nabla\zeta|^{2}dxdt\Big{]}.$

Let $N_{0}=\sup_{V\times[\theta^{\prime}T,T]}|p|^{2}$. Using (5.1) to estimate
the first integral on the right-hand side, we find that

$\displaystyle\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla p|^{2s+4}\zeta^{2}dxdt$
(5.6) $\displaystyle\quad\leq CN_{0}\Big{[}\int_{0}^{T}\int_{U}K(|\nabla
p|)|\nabla p|^{2s+2}|\nabla\zeta|^{2}dxdt+\int_{0}^{T}\int_{U}|\nabla
p|^{2s+2}\zeta|\zeta_{t}|dx\Big{]}.$

By the boundedness and property (2.10) of function $K(\xi)$,

$K(|\nabla p|)|\nabla p|^{2s+2}\leq C|\nabla p|^{2s+2}\leq CK(|\nabla
p|)(1+|\nabla p|^{a})|\nabla p|^{2s+2}\\\ \leq CK(|\nabla p|)(1+|\nabla
p|^{2s+2+a})\leq C(1+K(|\nabla p|)|\nabla p|^{2s+2+a}).$ (5.7)

Hence, inequality (5.6) leads to

$\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla p|^{2s+4}\zeta^{2}dxdt\leq
CN_{0}\int_{0}^{T}\int_{U}(1+K(|\nabla p|)|\nabla
p|^{2s+2+a})(|\nabla\zeta|^{2}+\zeta|\zeta_{t}|)dxdt.$ (5.8)

Step 2. Let $s=0$ in (5.6). For $\varepsilon>0$, applying Cauchy’s inequality
to the last integral of (5.6) gives

$\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla p|^{4}\zeta^{2}dx\leq\varepsilon
N_{0}\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{4}\zeta^{2}dx+C\varepsilon^{-1}N_{0}\int_{0}^{T}\int_{U}K(|\nabla
p|)^{-1}\zeta_{t}^{2}dxdt\\\ +CN_{0}\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{2}|\nabla\zeta|^{2}dxdt\Big{]}.$ (5.9)

Here, the cut-off function $\zeta$ satisfies $|\nabla\zeta|\leq C$ and
$0\leq\zeta_{t}\leq C[(\theta-\theta^{\prime})T]^{-1}$. Taking
$\varepsilon=(2(N_{0}+1))^{-1}$ in (5.9) and using Young’s inequality yield

$\displaystyle\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla
p|^{4}dxdt$ $\displaystyle\leq
C(N_{0}+1)N_{0}((\theta-\theta^{\prime})T)^{-2}\int_{\theta^{\prime}T}^{T}\int_{U}(1+|\nabla
p|)^{a}dxdt$
$\displaystyle\quad+CN_{0}\int_{\theta^{\prime}T}^{T}\int_{U}K(|\nabla
p|)|\nabla p|^{2}dxdt$ $\displaystyle\leq
C(1+N_{0})^{2}\Big{(}1+\frac{1}{(\theta-\theta^{\prime})T}\Big{)}^{2}\int_{\theta^{\prime}T}^{T}\int_{U}(1+K(|\nabla
p|)|\nabla p|^{2})dxdt.$

This implies (5.5) when $s=4$.

Step 3. When $s\in(2,4)$, let $\beta$ be a number in $(0,1)$ such that
$\frac{1}{s}=\frac{1-\beta}{2}+\frac{\beta}{4}$. Then, by interpolation
inequality:

$\displaystyle\Big{(}\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla
p|^{s}dxdt\Big{)}^{\frac{1}{s}}$ $\displaystyle\leq\Big{(}\int_{\theta
T}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{2}dxdt\Big{)}^{\frac{1-\beta}{2}}\Big{(}\int_{\theta
T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla
p|^{4}dxdt\Big{)}^{\frac{\beta}{4}}.$

Note that $\beta s/4=s/2-1$. Using (5.9) to estimate the last double integral,
we obtain

$\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla p|^{s}dxdt\leq
C\Big{(}1+\frac{1}{(\theta-\theta^{\prime})T}\Big{)}^{s-2}(1+N_{0})^{s-2}\int_{\theta^{\prime}T}^{T}\int_{U}(1+K(|\nabla
p|)|\nabla p|^{2})dx.$ (5.10)

This implies (5.5) for $s\in(2,4)$. Therefore, we have proved (5.5) for
$s\in(2,4]$.

Step 4. When $s>4$, let $m$ be the positive integer that satisfies
$\frac{s-4}{2-a}\leq m<\frac{s-4}{2-a}+1$, then $s-m(2-a)\in(2+a,4]$. Then,
let $\\{U_{k}\\}_{k=0}^{m}$ and $\\{V_{k}\\}_{k=0}^{m}$ be two families of
open subsets of $U$ such that

$U^{\prime}=U_{0}\Subset V_{0}\Subset U_{1}\Subset V_{1}\Subset U_{2}\Subset
V_{2}\Subset\ldots\Subset U_{m-1}\Subset V_{m-1}\Subset U_{m}\Subset
V_{m}=V\Subset U.$

For each $k=0,1,\ldots,m+1$, let
$\theta_{k}=\theta-(\theta-\theta^{\prime})k/(m+1)$. Note that
$\theta=\theta_{0}>\theta_{1}>\theta_{2}>\ldots>\theta_{m}>\theta_{m+1}=\theta^{\prime}$
and $\theta_{k}-\theta_{k+1}=(\theta-\theta^{\prime})/(m+1)$. For $1\leq k\leq
m$, let $\zeta_{k}(x,t)$ be a smooth cut-off function which is equal to one on
$U_{k}\times(t_{k},T]$ and zero on $Q_{T}\setminus V_{k}\times(t_{k-1},T]$,
and satisfies $|\nabla\zeta_{k}|\leq C_{m}$ and $0\leq\zeta_{k,t}\leq
C_{m}[(\theta-\theta^{\prime})T]^{-1}$. Here and in the following calculations
in this proof, $C_{m}$ denotes a generic positive constant depending on all
$U_{k}$ and $V_{k}$, $k=1,2,\ldots,m$.

We denote $N_{0}=\sup_{V\times[\theta^{\prime}T,T]}|p|^{2}$ still. Applying
(5.8) to $V_{0}$, $U_{1}$, $\theta_{1}$ in place of $V$, $U$, $\theta$, we
have

$\displaystyle\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla
p|^{s}dxdt$ $\displaystyle\leq
C_{m}(1+[(\theta-\theta^{\prime})T]^{-1})N_{0}\int_{\theta_{1}T}^{T}\int_{U_{1}}(1+K(|\nabla
p|)|\nabla p|^{s-(2-a)})dxdt$
$\displaystyle=C_{m}(1+[(\theta-\theta^{\prime})T]^{-1})N_{0}\Big{[}T+\int_{\theta_{1}T}^{T}\int_{U_{1}}K(|\nabla
p|)|\nabla p|^{s-(2-a)}dxdt\Big{]}.$

Above, we used the fact that $\sup_{V_{k}\times[\theta_{k-1}T,T]}|p|^{2}\leq
N_{0}$ for $1\leq k\leq m$. Applying (5.8) recursively to $U_{k}\Subset
V_{k}\Subset U_{k+1}$ and $\theta_{k}>\theta_{k+1}$, we obtain

$\displaystyle\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla
p|^{s}dxdt$ $\displaystyle\leq
C_{m}T\sum_{i=1}^{m}d^{i}+Cd^{m}\int_{\theta_{m}T}^{T}\int_{U_{m}}K(|\nabla
p|)|\nabla p|^{s-m(2-a)}dxdt$ $\displaystyle\leq
C_{m}d^{m}\Big{[}T+\int_{\theta_{m}T}^{T}\int_{U_{m}}K(|\nabla p|)|\nabla
p|^{s-m(2-a)}dxdt\Big{]},$

where $d=(1+[(\theta-\theta^{\prime})T]^{-1})(1+N_{0})$. Since
$s-m(2-a)\in(2,4]$, estimating the last integral by (5.10) gives

$\int_{\theta T}^{T}\int_{U^{\prime}}K(|\nabla p|)|\nabla p|^{s}dxdt\leq
Cd^{m}T+Cd^{s-m(2-a)-2+m}\int_{\theta^{\prime}T}^{T}\int_{U}K(|\nabla
p|)|\nabla p|^{2}dx.$ (5.11)

Since $d\geq 1$ and $m,s-m(2-a)-2+m\leq s-2$, the desired estimate (5.5)
follows (5.11). This completes the proof of (5.5) for all $s\geq 2$.

The inequality (5.3) then is an obvious consequence of (5.5) with
$\theta^{\prime}=\theta/2$.

Now, we turn to the proof of (5.4). For $s>2-a$, it follows from (5.1) that

$\displaystyle\sup_{[0,T]}\int_{U}|\nabla p|^{s}\zeta^{2}dx$
$\displaystyle\leq C\int_{0}^{T}\int_{U}K(|\nabla p|)|\nabla
p|^{s}|\nabla\zeta|^{2}dxdt+C\int_{0}^{T}\int_{U}|\nabla
p|^{s}\zeta|\zeta_{t}|dx$ $\displaystyle\leq C\int_{0}^{T}\int_{U}|\nabla
p|^{s}(|\nabla\zeta|^{2}+\zeta|\zeta_{t}|)dxdt.$

Same as (5.7), we have $|\nabla p|^{s}\leq C(1+K(|\nabla p|)|\nabla
p|^{s+a})$, and hence

$\sup_{[0,T]}\int_{U}|\nabla p|^{s}\zeta^{2}dx\leq
C\int_{0}^{T}\int_{U}(1+K(|\nabla p|)|\nabla
p|^{s+a})(|\nabla\zeta|^{2}+\zeta|\zeta_{t}|)dx.$ (5.12)

Let $V^{\prime}$ be an open set such that $U^{\prime}\Subset V^{\prime}\Subset
V$. Choose $\zeta(x,t)$ such that $\zeta=0$ for $t\leq 3\theta T/4$ or
$x\not\in V^{\prime}$, and $\zeta=1$ for $(x,t)\in U^{\prime}\times[\theta
T,T]$. Then we have from (5.12) that

$\sup_{[\theta T,T]}\int_{U}|\nabla p|^{s}dx\leq C(1+(\theta
T)^{-1})\int_{3\theta T/4}^{T}\int_{V^{\prime}}(1+K(|\nabla p|)|\nabla
p|^{s+a})dxdt.$ (5.13)

To estimate the last integral, we apply (5.5) with parameters $s$,
$\theta^{\prime}$, $\theta$, $U^{\prime}$ being replaced by $s+a$, $\theta/2$,
$3\theta/4$, $V^{\prime}$. Therefore, we obtain

$\displaystyle\sup_{[\theta T,T]}\int_{U^{\prime}}|\nabla p|^{s}dx$
$\displaystyle\leq C\Big{\\{}((1+(\theta
T)^{-1})^{s+a-1}(1+N_{0})^{s+a-2}\int_{0}^{T}\int_{U}1+K(|\nabla p|)|\nabla
p|^{2}dxdt\Big{\\}},$

hence proving (5.4). The proof is complete. ∎

Now, we combine Proposition 5.3 with estimates in section 3 to express the
bounds in terms of the initial and boundary data.

######  Theorem 5.4.

Let $U^{\prime}\Subset U$ and $s\geq 2$. If $t\in(0,2)$ then

$\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx\leq
Ct^{-\mu_{1}}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\mu_{2}+2}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\mu_{2}}\\\
\quad\cdot\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)},$ (5.14)

where

$\mu_{1}=\mu_{1}(\alpha,s)\mathbin{\buildrel\rm
def\over{\mathbin{=\kern-2.0pt=}}}\big{[}1+\frac{2\kappa_{1}}{\alpha-a}\big{]}(s+a-2)+1\quad\text{and}\quad\mu_{2}=\mu_{2}(\alpha,s)\mathbin{\buildrel\rm
def\over{\mathbin{=\kern-2.0pt=}}}2\kappa_{2}(s-2+a).$ (5.15)

If $t\geq 2$ then

$\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\mu_{2}+\alpha}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\mu_{2}+\alpha}\\\
\cdot\Big{(}1+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}.$ (5.16)

###### Proof.

Let $t\in(0,2)$. Applying (5.4) to $T_{0}=0$, $T=t$ and $\theta=1/2$, and
using (4.14), (3.8) and relation (2.13), we obtain

$\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx\leq
Ct^{-\mu_{1}}\cdot\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{2\kappa_{2}(s-2+a)}\\\
\quad\cdot\Big{(}1+\|\bar{p}_{0}\|_{L^{2}}^{2}+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}.$

Then (5.14) follows. Let $t\geq 2$. Applying (5.4) with $T_{0}=t-1$, $T=1$,
$\theta=1/2$, then combining it with (4.15) and (3.21), we obtain

$\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx\leq
C\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\mu_{2}}\\\
\cdot\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}^{\alpha}+EnvA(\alpha,t)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}.$

Then (5.16) follows. ∎

For large time estimates, we have:

######  Theorem 5.5.

Let $U^{\prime}\Subset U$, $s\geq 2$, and let $\mu_{2}$ be defined as in
Theorem 5.4.

(i) If $A(\alpha)<\infty$ then

$\limsup_{t\to\infty}\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\mu_{2}+\alpha}\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}.$
(5.17)

(ii) If $\beta(\alpha)<\infty$ then there is $T>0$ such that for all $t>T$,

$\displaystyle\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx$ $\displaystyle\leq
C\Big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-2,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\mu_{2}+\alpha}\Big{(}1+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}.$
(5.18)

###### Proof.

For $t\geq 1$ Applying (5.4) with $T_{0}=t-1$, $T=1$, $\theta=1/2$,

$\displaystyle\int_{U^{\prime}}|\nabla p(x,t)|^{s}dxdt\leq
C\big{(}1+\sup_{[t-1,t]}\|p\|^{2}_{L^{\infty}(V)}\big{)}^{s-2+a}\cdot\int_{t-1}^{t}\int_{U}(1+K(|\nabla
p|)|\nabla p|^{2})dxdt.$ (5.19)

(i) Taking limit superior as $t\to\infty$ and using (4.16), (3.25) give

$\displaystyle\limsup_{t\to\infty}\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx$
$\displaystyle\leq
C\big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\big{)}^{\mu_{2}}$
$\displaystyle\quad\cdot\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}.$

Then estimate (5.17) follows.

(ii) By combining (5.19) with (4.17) and (3.29), we have

$\displaystyle\int_{U^{\prime}}|\nabla p(x,t)|^{s}dx$ $\displaystyle\leq
C\Big{\\{}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-2,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{\\}}^{\mu_{2}}$
$\displaystyle\quad\cdot\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}.$

Then (5.18) follows. ∎

###  5.2  $L^{\infty}$-estimates

In this subsection, we obtain interior $L^{\infty}$-estimates for the gradient
of pressure. For each $m=1,2,\ldots,n,$ denote $u_{m}=p_{x_{m}}$ and
$u=(u_{1},u_{2},\ldots,u_{n})=\nabla p$. We have

$\frac{\partial u_{m}}{\partial
t}=\partial_{m}(\nabla\cdot(K(|u|)u))=\nabla\cdot(K(|u|)\partial_{m}u)+\nabla\cdot\Big{[}K^{\prime}(|u|)\frac{\sum_{i}u_{i}\partial_{m}u_{i}}{|u|}u\Big{]}.$
(5.20)

Since $\partial_{i}u_{m}=\partial_{m}u_{i}$, we have
$\partial_{m}u=(\partial_{m}u_{1},\ldots,\partial_{m}u_{n})=(\partial_{1}u_{m},\ldots,\partial_{n}u_{m})=\nabla
u_{m}$, and
$\sum_{i}u_{i}\partial_{m}u_{i}=\sum_{i}u_{i}\partial_{i}u_{m}=u\cdot\nabla
u_{m}$. Therefore, we rewrite (5.20) as

$\frac{\partial u_{m}}{\partial t}=\nabla\cdot(K(|u|)\nabla
u_{m})+\nabla\cdot\Big{[}K^{\prime}(|u|)\frac{u\cdot\nabla
u_{m}}{|u|}u\Big{]}.$ (5.21)

We will apply De Giorgi’s technique to equation (5.21). In the following, we
fix a number $s_{0}$ such that $r=s_{0}$ satisfies (2.20). Note that
$s_{0}^{*}>2$. We will also use $s_{j}$ for $j\geq 1$ to denote some exponents
that depend on $s_{0}$ but are independent of $\alpha$. Let

$s_{1}=(1-2/s_{0}^{*})^{-1}>1.$

######  Theorem 5.6.

Let $U^{\prime}\Subset V\Subset U$. For any $T_{0}\geq 0$, $T>0$, and
$\theta\in(0,1)$, if $t\in[T_{0}+\theta T,T_{0}+T]$ then

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq C(1+(\theta
T)^{-1})^{\frac{s_{1}+1}{2}}\lambda^{\frac{s_{1}}{2}}\|{\nabla
p}\|_{L^{2}(V\times(T_{0}+\theta T/2,T_{0}+T))},$ (5.22)

where

$\lambda=\lambda(T_{0},T,\theta;V)=\Big{(}\int_{T_{0}+\theta
T/2}^{T_{0}+T}\int_{V}(1+|\nabla
p|)^{\frac{as_{0}}{2-s_{0}}}dxdt\Big{)}^{\frac{2-s_{0}}{s_{0}}}.$ (5.23)

and constant $C>0$ is independent of $T_{0}$, $T$, and $\theta$.

###### Proof.

Without loss of generality, assume $T_{0}=0$. Fix $m\in\\{1,2,\ldots,n\\}$. We
will show for $t\in[\theta T,T]$ that

$\|{p_{x_{m}}(t)}\|_{L^{\infty}(U^{\prime})}\leq C(1+(\theta
T)^{-1})^{\frac{s_{1}+1}{2}}\lambda^{\frac{s_{1}}{2}}\|p_{x_{m}}\|_{L^{2}(V\times(T_{0}+\theta
T/2,T_{0}+T))}.$ (5.24)

Let $\zeta(x,t)=\phi(x)\varphi(t)$ be a cut-off function with $\varphi(t)=0$
for $t\leq\theta T/2$, and supp $\phi\subset U$. We define for $k\geq 0$

$\displaystyle u_{m}^{(k)}=\max\\{u_{m}-k,0\\},\quad S_{k}(t)=\\{x\in
U:u_{m}^{(k)}(x,t)\geq 0\\},$

and denote by $\chi_{k}(x,t)$ the characteristic function of $S_{k}(t)$.

Multiplying (5.21) by $u_{m}^{(k)}\zeta^{2}$ and integrating over $U$, we have

$\frac{1}{2}\frac{d}{dt}\int_{U}|u_{m}^{(k)}|^{2}\zeta^{2}dx=\int_{U}|u_{m}^{(k)}|^{2}\zeta|\zeta_{t}|dx-\int_{U}K(|u|)|\nabla
u_{m}^{(k)}|^{2}\zeta^{2}dx-2\int_{U}K(|u|)(\nabla
u_{m}^{(k)}\cdot\nabla\zeta)u_{m}^{(k)}\zeta dx\\\
-\int_{U}\frac{1}{|u|}K^{\prime}(|u|)(u\cdot\nabla
u_{m}^{(k)})(u\cdot\nabla(u_{m}^{(k)}\zeta^{2}))dx.$ (5.25)

By the product rule,

$-\frac{1}{|u|}K^{\prime}(|u|)(u\cdot\nabla
u_{m}^{(k)})(u\cdot\nabla(u_{m}^{(k)}\zeta^{2}))=-\frac{1}{|u|}K^{\prime}(|u|)(u\cdot\nabla
u_{m}^{(k)})\Big{(}u\cdot\nabla u_{m}^{(k)}\zeta^{2}+2u_{m}^{(k)}\zeta
u\cdot\nabla\zeta\Big{)}.$

It follows property (2.12) that

$-\frac{1}{|u|}K^{\prime}(|u|)(u\cdot\nabla(u_{m}^{(k)})^{2}\zeta^{2}\leq\frac{aK(|u|)}{|u|^{2}}|u|^{2}|\nabla
u_{m}^{(k)}|^{2}\zeta^{2}=aK(|u|)|\nabla u_{m}^{(k)}|^{2}\zeta^{2}$

and

$-\frac{2}{|u|}K^{\prime}(|u|)u^{2}u_{m}^{(k)}\zeta\nabla
u_{m}^{(k)}\cdot\nabla\zeta\leq\frac{2}{|u|}|K^{\prime}(|u|)|u|^{2}|\nabla
u_{m}^{(k)}||u_{m}^{(k)}|\zeta|\nabla\zeta|\leq CK(|u|)|\nabla
u_{m}^{(k)}||u_{m}^{(k)}|\zeta|\nabla\zeta|.$

Then we obtain from (5.25) that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{U}|u_{m}^{(k)}|^{2}\zeta^{2}$
$\displaystyle\leq\int_{U}|u_{m}^{(k)}|^{2}\zeta|\zeta_{t}|dx-(1-a)\int_{U}K(|u|)|\nabla
u_{m}^{(k)}|^{2}\zeta^{2}dx$ $\displaystyle\quad+C\int_{U}K(|u|)|\nabla
u_{m}^{(k)}||u_{m}^{(k)}|\zeta|\nabla\zeta|dx.$

Applying Cauchy’s inequality to the last term in previous inequality yields

$\frac{1}{2}\frac{d}{dt}\int_{U}|u_{m}^{(k)}|^{2}\zeta^{2}\leq\int_{U}|u_{m}^{(k)}|^{2}\zeta|\zeta_{t}|dx-\frac{1-a}{2}\int_{U}K(|u|)|\nabla
u_{m}^{(k)}|^{2}\zeta^{2}dx+C\int_{U}K(|u|)|u_{m}^{(k)}|^{2}|\nabla\zeta|^{2}dx.$

Since $|\zeta\nabla
u_{m}^{(k)}|^{2}=|\nabla(u_{m}^{(k)}\zeta)-u_{m}^{(k)}\nabla\zeta|^{2}\geq\frac{1}{2}|\nabla(u_{m}^{(k)}\zeta)|^{2}-|u_{m}^{(k)}\nabla\zeta|^{2}$,
we obtain

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{U}|u_{m}^{(k)}\zeta|^{2}dx$
$\displaystyle+\frac{1-a}{4}\int_{U}K(|u|)|\nabla(u_{m}^{(k)}\zeta)|^{2}dx$
(5.26)
$\displaystyle\leq\int_{U}|u_{m}^{(k)}|^{2}\zeta|\zeta_{t}|dx+C\int_{U}K(|u|)|u_{m}^{(k)}|^{2}|\nabla\zeta|^{2}dx.$

Integrating (5.26) from $0$ to $t$ for $t\in[0,T]$, and then taking supremum
in $t$ give

$\displaystyle\max_{[0,T]}\int_{U}|u_{m}^{(k)}\zeta|^{2}dx$
$\displaystyle+C\int_{0}^{T}\int_{U}K(|u|)|\nabla(u_{m}^{(k)}\zeta)|^{2}dxdt$
(5.27)
$\displaystyle\leq\int_{0}^{T}\int_{U}|u_{m}^{(k)}|^{2}\zeta|\zeta_{t}|dx+C\int_{0}^{T}\int_{U}K(|u|)|u_{m}^{(k)}|^{2}|\nabla\zeta|^{2}dxdt.$

Let

$\nu_{2}=4(1-1/s_{0}^{*})>2.$ (5.28)

Applying Lemma 2.2 to function $u_{m}^{(k)}\zeta$ which vanishes on the
boundary, weight $W=K(|u|)$ and exponents $r=s_{0}$,
$\varrho=\varrho(s_{0})=\nu_{2}$, we have

$\displaystyle\|{u_{m}^{(k)}\zeta}\|_{L^{\nu_{2}}(Q_{T})}$ $\displaystyle\leq
C\Big{[}\mathop{\mathrm{ess\,sup}}_{t\in[0,T]}\|{u_{m}^{(k)}\zeta}\|_{L^{2}(U)}+\Big{(}\int_{0}^{T}\int_{U}K(|u|)|\nabla(u_{m}^{(k)}\zeta)|^{2}dxdt\Big{)}^{\frac{1}{2}}\Big{]}$
$\displaystyle\quad\cdot\Big{[}\int_{0}^{T}\int_{{\rm
supp}\zeta}K(|u|)^{-\frac{s_{0}}{2-s_{0}}}dxdt\Big{]}^{\frac{2-s_{0}}{\nu_{2}s_{0}}}.$

Using (2.11), we have $K(|u|)^{-\frac{s_{0}}{2-s_{0}}}\leq
C(1+|u|)^{\frac{as_{0}}{2-s_{0}}}$, hence

$\|{u_{m}^{(k)}\zeta}\|_{L^{\nu_{2}}(Q_{T})}\leq
C\lambda^{1/\nu_{2}}\Big{[}\max_{t\in[0,T]}\int_{U}|u_{m}^{(k)}\zeta|^{2}dx+\int_{0}^{T}\int_{U}K(|u|)|\nabla(u_{m}^{(k)}\zeta)|^{2}dxdt\Big{]}^{\frac{1}{2}}.$
(5.29)

By (5.29), (5.27) and the boundedness of function $K(\cdot)$ we find that

$\displaystyle\|{u_{m}^{(k)}\zeta}\|_{L^{\nu_{2}}(Q_{T})}$ $\displaystyle\leq
C\lambda^{1/\nu_{2}}\Big{(}\max_{[0,T]}\int_{U}|u_{m}^{(k)}\zeta|^{2}dx+C\int_{0}^{T}\int_{U}K(|u|)|\nabla(u_{m}^{(k)}\zeta)|^{2}dxdt\Big{)}^{\frac{1}{2}}$
(5.30) $\displaystyle\leq
C\lambda^{1/\nu_{2}}\Big{(}\int_{0}^{T}\int_{U}|u_{m}^{(k)}|^{2}\zeta|\zeta_{t}|dx+\int_{0}^{T}\int_{U}|u_{m}^{(k)}|^{2}|\nabla\zeta|^{2}dxdt\Big{)}^{\frac{1}{2}}.$

Here, we use the same notation $x_{0}$, $\rho$, $M_{0}$, $k_{i}$, $\rho_{i}$,
$U_{i}$, $\mathcal{Q}_{i}$ and $\zeta_{i}$ as introduced in the proof of
Theorem 4.1 from (4.4) to (4.7). Also, the sets $A_{i,j}$ are defined by (4.6)
with $p$ being replaced by $u_{m}$.

Define $F_{i}=\|{u_{m}^{(k_{i+1})}\zeta_{i}}\|_{L^{\nu_{2}}(A_{i+1,i})}$.
Applying (5.30) with $k=k_{i+1}$ and $\zeta=\zeta_{i}$ gives

$F_{i}\leq
C\lambda^{1/\nu_{2}}\Big{\\{}\int_{0}^{T}\int_{U}|u_{m}^{(k_{i+1})}|^{2}\zeta_{i}|(\zeta_{i})_{t}|dx+\int_{0}^{T}\int_{U}|u_{m}^{(k_{i+1})}|^{2}|\nabla\zeta_{i}|^{2}\Big{)}dxdt\Big{\\}}^{1/2}.$
(5.31)

Using (4.7), we obtain

$F_{i}\leq C2^{i}\lambda^{1/\nu_{2}}(1+\frac{1}{\theta
T})^{1/2}\|u_{m}^{(k_{i+1})}\|_{L^{2}(A_{i+1,i})}\leq C2^{i}(1+\frac{1}{\theta
T})^{1/2}\lambda\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i,i})}.$

Since $\nu_{2}>2$, it follows from Hölder’s inequality that

$\displaystyle\|u_{m}^{(k_{i+1})}\zeta_{i}\|_{L^{2}(A_{i+1,i+1})}$
$\displaystyle\leq\|u_{m}^{(k_{i+1})}\zeta_{i}\|_{L^{\nu_{2}}(A_{i+1,i+1})}|A_{i+1,i+1}|^{1/2-1/{\nu_{2}}}$
(5.32)
$\displaystyle\leq\|u_{m}^{(k_{i+1})}\zeta_{i}\|_{L^{\nu_{2}}(A_{i+1,i+1})}|A_{i+1,i}|^{1/2-1/{\nu_{2}}}\leq
CF_{i}|A_{i+1,i}|^{1/2-1/{\nu_{2}}}.$

Note that
$\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i,i})}\geq\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i+1,i})}\geq(k_{i+1}-k_{i})|A_{i+1,i}|^{1/2}$.
Thus,

$|A_{i+1,i}|\leq(k_{i+1}-k_{i})^{-2}\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i})}^{2}\leq
C4^{i}M_{0}^{-2}\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i,i})}^{2}.$ (5.33)

Then it follows (5.32), (5.31) and (5.33) that

$\displaystyle\|u_{m}^{(k_{i+1})}\|_{L^{2}(A_{i+1,i+1})}$ $\displaystyle\leq
C2^{i}(1+\frac{1}{\theta
T})^{1/2}\lambda^{1/\nu_{2}}\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i})}2^{i-\frac{2i}{\nu_{2}}}M_{0}^{-1+2/{\nu_{2}}}\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i,i})}^{1-2/{\nu_{2}}}$
$\displaystyle\leq C4^{i}(1+\frac{1}{\theta
T})^{1/2}\lambda^{1/\nu_{2}}M_{0}^{-1+2/{\nu_{2}}}\|{u_{m}^{(k_{i})}}\|_{L^{2}(A_{i,i})}^{2-2/{\nu_{2}}}.$

Denote $\nu_{3}=1-2/\nu_{2}$. Let
$Y_{i}=\|u_{m}^{(k_{i})}\|_{L^{2}(A_{i,i})}$, $B=4$ and $D=C(1+\frac{1}{\theta
T})^{1/2}\lambda^{1/\nu_{2}}M_{0}^{-\nu_{3}}.$ We obtain

$Y_{i+1}\leq DB^{i}Y_{i}^{1+\nu_{3}}\quad\text{for all }i\geq 0.$

We now determine $M_{0}$ so that $Y_{0}\leq
D^{-1/\nu_{3}}B^{-1/{\nu_{3}^{2}}}$. This condition is met if

$M_{0}\geq C\Big{[}\lambda^{1/\nu_{2}}(1+\frac{1}{\theta
T})^{1/2}\Big{]}^{1/\nu_{3}}Y_{0}=C\lambda^{s_{1}/2}(1+\frac{1}{\theta
T})^{\frac{1+s_{1}}{2}}Y_{0}.$

Since
$Y_{0}=\|{u_{m}^{(k_{0})}}\|_{L^{2}{(A_{0,0})}}\leq\|u_{m}\|_{L^{2}(V\times(\theta
T/2,T))},$ it suffices to choose $M_{0}$ as

$M_{0}=C\lambda^{s_{1}/2}(1+\frac{1}{\theta
T})^{\frac{s_{1}+1}{2}}\|u_{m}\|_{L^{2}(V\times(\theta T/2,T))}.$ (5.34)

Then Lemma 2.3 gives $\displaystyle{\lim_{i\to\infty}}Y_{i}=0$. Hence,

$\int_{\theta T}^{T}\int_{B(x_{0},\rho/4)}|u_{m}^{(M_{0})}|^{2}dxdt=0.$

Thus, $u_{m}(x,t)\leq M_{0}$ a.e. in $B(x_{0},\rho/4)\times(0,T)$. Replace
$u_{m},u$ by $-u_{m},-u$ and use the same argument we obtain $|u_{m}(x,t)|\leq
M_{0}$ a.e. in $B(x_{0},\rho/4)\times(0,T)$. Now by covering $U^{\prime}$ by
finitely many such balls $B(x_{0},\rho/4)$, we come to conclusion

$|u_{m}(x,t)|\leq M_{0}\quad\text{a.e. in }U^{\prime}\times(\theta T,T).$
(5.35)

By the choice of $M_{0}$ we obtain from (5.35) that

$|u_{m}(x,t)|\leq C(1+\frac{1}{\theta
T})^{\frac{s_{1}+1}{2}}\lambda^{s_{1}/2}\|u_{m}\|_{L^{2}(V\times(\theta
T/2,T))}$

for all $m=1,\ldots,n.$ Then (5.24) follows. ∎

We will combine Theorem 5.6 with the high integrability of $\nabla p$ in
subsection 5.1 to obtain the $L^{\infty}$-estimates. Let

$\displaystyle s_{2}$
$\displaystyle=\max\Big{\\{}2,\frac{as_{0}}{2-s_{0}}\Big{\\}},\quad
s_{3}=s_{1}(2-s_{0})/s_{0}+1,$ $\displaystyle\kappa_{3}$
$\displaystyle=(s_{2}+a-2)\big{(}1+\frac{2\kappa_{1}}{\alpha-a}\big{)},\quad\kappa_{4}=1+s_{1}+\kappa_{3}s_{3},\quad\kappa_{5}=\kappa_{2}(s_{2}-2+a).$

######  Theorem 5.7.

If $t\in(0,2)$ then

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
Ct^{-\kappa_{4}/2}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{s_{3}(\kappa_{5}+1)}\Big{(}1+EnvA(\alpha,t)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{s_{3}\kappa_{5}}\\\
\cdot\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}^{s_{3}/2}.$ (5.36)

If $t\geq 2$ then

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{s_{3}(\kappa_{5}+\alpha/2)}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{s_{3}(\kappa_{5}+\alpha/2)}\\\
\cdot\Big{(}1+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{s_{3}/2}.$ (5.37)

###### Proof.

First, we have from (5.22) that

$\sup_{[T_{0}+\theta T,T_{0}+T]}\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C(1+(\theta T)^{-1})^{\frac{1+s_{1}}{2}}\Big{(}\int_{T_{0}+\theta
T/2}^{T_{0}+T}\int_{V}(1+|\nabla p|)^{s_{2}}dxdt\Big{)}^{s_{3}/2}.$ (5.38)

Let $t\in(0,2)$, applying (5.38) with $T_{0}=0$, $T=t$, and $\theta=1/2$, we
obtain

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
Ct^{-\frac{1+s_{1}}{2}}\Big{(}\int_{t/4}^{t}\int_{V}(1+|\nabla
p|)^{s_{2}}dxdt\Big{)}^{s_{3}/2}.$ (5.39)

We apply (5.14) with $s=s_{2}$ and $U^{\prime}=V$. Note from formulas in
(5.15) that

$\mu_{1}(\alpha,s_{2})=\kappa_{3}+1\quad\text{and}\quad\mu_{2}(\alpha,s_{2})=2\kappa_{5}.$
(5.40)

We obtain

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
Ct^{-\frac{1+s_{1}}{2}}\Big{\\{}t\cdot
t^{-(\kappa_{3}+1)}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{2(\kappa_{5}+1)}\\\
\cdot\Big{(}1+EnvA(\alpha,t)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{2\kappa_{5}}\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}\Big{\\}}^{s_{3}/2}\\\
\leq
Ct^{-\frac{1+s_{1}+\kappa_{3}s_{3}}{2}}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{s_{3}(\kappa_{5}+1)}\\\
\cdot\Big{(}1+EnvA(\alpha,t)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{s_{3}\kappa_{5}}\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}^{s_{3}/2}.$

Then (5.36) follows. Let $t\geq 2$. Applying (5.38) with $T_{0}=t-3/4$,
$T=3/4$, $\theta=2/3$, then

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+\int_{t-1/2}^{t}\int_{V}|\nabla p|^{s_{2}}dxdt\Big{)}^{s_{3}/2}.$
(5.41)

Thanks to (5.16) with $s=s_{2}$, we obtain (5.37). ∎

Combining (5.41) with Theorem 5.5, we have the following asymptotic estimates.

######  Theorem 5.8.

(i) If $A(\alpha)<\infty$ then

$\limsup_{t\to\infty}\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{{s_{3}}(\kappa_{5}+\alpha/2)}\\\
\cdot\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{s_{3}/2}.$
(5.42)

(ii) If $\beta(\alpha)<\infty$ then there is $T>0$ such that when $t>T$ we
have

$\|{\nabla p(t)}\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-3,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-3,t))}\Big{)}^{{s_{3}}(\kappa_{5}+\alpha/2)}\\\
\cdot\Big{(}1+\int_{t-2}^{t}G_{1}(\tau)d\tau\Big{)}^{s_{3}/2}.$ (5.43)

##  6 Interior estimates for time derivative of pressure

In this section, we estimate the $L^{\infty}$-norm of $p_{t}(x,t)$ for $t>0$.
Let $q=p_{t}$. Then

$\frac{\partial q}{\partial t}=\nabla\cdot\big{(}K(|\nabla p|)\nabla
p\big{)}_{t}.$ (6.1)

Using (6.1), we first derive a local-in-time estimate for $L^{\infty}$-norm of
$p_{t}$.

######  Proposition 6.1.

Let $U^{\prime}\Subset V\Subset U$. If $T_{0}\geq 0$, $T>0$ and
$\theta\in(0,1)$, then

$\sup_{[T_{0}+\theta T,T_{0}+T]}\|p_{t}\|_{L^{\infty}(U^{\prime})}\leq
C\lambda^{\frac{s_{1}}{2}}(1+(\theta
T)^{-1})^{\frac{s_{1}+1}{2}}\|p_{t}\|_{L^{2}(U\times(T_{0},T_{0}+T))},$ (6.2)

where $s_{1}$ and $\lambda=\lambda(T_{0},T,\theta;V)$ are defined in Theorem
5.6, and constant $C>0$ is independent of $T_{0}$, $T$, and $\theta$.

###### Proof.

Without loss of generality, assume $T_{0}=0$. For $k\geq 0$, let
$q^{(k)}=\max\\{q-k,0\\}$ and $S_{k}(t)=\\{x\in U:q(x,t)>k\\}$, and
$\chi_{k}(x,t)$ be the characteristic function of set $\\{(x,t)\in
U\times(0,T):q(x,t)>k\\}$. On $S_{k}(t)$, we have $(\nabla p)_{t}=\nabla
q=\nabla q^{(k)}$.

Let $\zeta=\zeta(x,t)$ be the cut-off function on $U\times[0,T]$ satisfying
$\zeta(\cdot,0)=0$ and $\zeta(\cdot,t)$ having compact support in $U$. We will
use test function $q^{(k)}\zeta^{2}$, noting that
$\nabla(q^{(k)}\zeta^{2})=\zeta[\nabla(q^{(k)}\zeta)+q^{(k)}\nabla\zeta]$.
Multiplying (6.1) by $q^{(k)}\zeta^{2}$ and integrating the resultant on $U$,
we get

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{U}|q^{(k)}\zeta|^{2}dx$
$\displaystyle=\int_{U}|q^{(k)}|^{2}\zeta\zeta_{t}dx-\int_{U}(K(|\nabla
p|))_{t}\nabla p\cdot[\nabla(q^{(k)}\zeta)+q^{(k)}\nabla\zeta]\zeta dx$ (6.3)
$\displaystyle\quad-\int_{U}K(|\nabla p|)(\nabla
p)_{t}\cdot[\nabla(q^{(k)}\zeta)+q^{(k)}\nabla\zeta]\zeta dx.$

For the last integral of (6.3), put
$z=\zeta[\nabla(q^{(k)}\zeta)+q^{(k)}\nabla\zeta]$. We have

$(\nabla p)_{t}\cdot z=\zeta\nabla
q^{(k)}\cdot[\nabla(q^{(k)}\zeta)+q^{(k)}\nabla\zeta]=|\nabla(q^{(k)}\zeta)|^{2}-|q^{(k)}\nabla\zeta|^{2}.$

For the second integral on the right-hand side of (6.3), taking into account
(2.12),

$\displaystyle|(K(|\nabla p|))_{t}\nabla p\cdot z|\ $
$\displaystyle=|K^{\prime}(|\nabla p|)|\frac{|\nabla p\cdot\nabla
p_{t}|}{|\nabla p|}|\nabla p\cdot z|\leq aK(|\nabla p|)|\nabla q||z|.$

Note that

$\displaystyle|\nabla q||z|$
$\displaystyle=|\zeta\nabla\bar{q}^{(k)}||\nabla(q^{(k)}\zeta)+q^{(k)}\nabla\zeta|\leq\big{\\{}|\nabla(q^{(k)}\zeta)|+|q^{(k)}||\nabla\zeta|\big{\\}}^{2}$
$\displaystyle=|\nabla(q^{(k)}\zeta)|^{2}+2|q^{(k)}||\nabla\zeta||\nabla(q^{(k)}\zeta)|+|q^{(k)}\nabla\zeta|^{2},$

and by applying Cauchy’s inequality to the second to last term

$\displaystyle a|\nabla q||z|\leq
a|\nabla(q^{(k)}\zeta)|^{2}+\frac{1-a}{2}|\nabla(q^{(k)}\zeta)|^{2}+\frac{2a^{2}}{1-a}|\bar{q}^{(k)}\nabla\zeta|^{2}+a|q^{(k)}\nabla\zeta|^{2}.$

It follows the above calculations that

$\displaystyle\frac{d}{dt}\int_{U}|q^{(k)}\zeta|^{2}dx$
$\displaystyle+(1-a)\int_{U}K(|\nabla p|)|\nabla(q^{(k)}\zeta)|^{2}dx$
$\displaystyle\quad\leq
2\int_{U}|q^{(k)}|^{2}\zeta|\zeta_{t}|dx+C\int_{U}K(|\nabla
p|)|q^{(k)}\nabla\zeta|^{2}dx.$

Integrating this inequality from $0$ to $T$, we obtain

$\displaystyle\max_{[0,T]}\int_{U}|q^{(k)}\zeta|^{2}dx$
$\displaystyle+\int_{0}^{T}\int_{U}K(|\nabla
p|)|\nabla(q^{(k)}\zeta)|^{2}dxdt$ $\displaystyle\leq
C\Big{[}\int_{0}^{T}\int_{U}|q^{(k)}|^{2}\zeta|\zeta_{t}|dxdt+\int_{0}^{T}\int_{U}K(|\nabla
p|)|q^{(k)}\nabla\zeta|^{2}dxdt\Big{]}$ $\displaystyle\leq
C\int_{0}^{T}\int_{U}|q^{(k)}|^{2}(\zeta|\zeta_{t}|+|\nabla\zeta|^{2})dxdt.$

The last inequality uses the fact that function $K(\cdot)$ is bounded above.
Applying Lemma 2.2 to $q^{(k)}\zeta$ with $W=K(|\nabla p|)$ and
$\varrho=\nu_{2}$ defined by (5.28), we have

$\displaystyle\|{q^{(k)}\zeta}\|_{L^{\nu_{2}}(Q_{T})}\leq
C\lambda^{1/\nu_{2}}\cdot\left\\{\max_{[0,T]}\int_{U}|q^{(k)}\zeta|^{2}dx+\int_{0}^{T}\int_{U}K(|\nabla
p|)|\nabla(q^{(k)}\zeta)|^{2}dxdt\right\\}^{1/2}.$

and hence,

$\begin{split}\|{q^{(k)}\zeta}\|_{L^{\nu_{2}}(Q_{T})}&\leq
C\lambda^{1/\nu_{2}}\Big{\\{}\int_{0}^{T}\int_{U}|q^{(k)}|^{2}(|\zeta_{t}|\zeta+|\nabla\zeta|^{2})dxdt\Big{\\}}^{1/2}.\end{split}$
(6.4)

This is similar to inequality (5.30). Then by following arguments of Theorem
5.6 applied for $q^{(k)}$ instead of $u_{m}^{(k)}$, we obtain (6.2). The proof
is complete. ∎

The next theorem contains the estimates in terms of the initial and boundary
data for all $t>0$. Let

$\displaystyle\kappa_{6}$
$\displaystyle=1+s_{1}+\kappa_{3}(s_{3}-1),\quad\kappa_{7}=(s_{3}-1)(\kappa_{5}+1)+1,$
$\displaystyle\kappa_{8}$
$\displaystyle=(s_{3}-1)\kappa_{5},\quad\kappa_{9}=(s_{3}-1)(\kappa_{5}+\alpha/2)+\alpha/2.$

######  Theorem 6.2.

Let $U^{\prime}\Subset U$. For $t\in(0,2)$,

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}$ $\displaystyle\leq
Ct^{-\kappa_{6}/2}\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}\Big{)}^{\kappa_{7}}\Big{(}1+\int_{U}H(|\nabla
p_{0}(x)|)dx\Big{)}^{1/2}.$ (6.5)
$\displaystyle\quad\cdot\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\kappa_{8}}\Big{(}1+\int_{0}^{t}G_{3}(\tau)d\tau)\Big{)}^{s_{3}/2}.$

For $t\geq 2$,

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}$ $\displaystyle\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\kappa_{9}}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\kappa_{9}}$
(6.6)
$\displaystyle\quad\cdot\Big{(}1+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{s_{3}/2}.$

###### Proof.

Let $t\in(0,2)$, apply (6.2) with $T_{0}=0$, $T=t$ and $\theta=1/2$. By (5.23)
and (5.3), we have

$\displaystyle\lambda^{s_{1}/2}$
$\displaystyle=\Big{(}\int_{t/2}^{t}\int_{V}(1+|\nabla
p|)^{\frac{as_{0}}{2-s_{0}}}dxd\tau\Big{)}^{\frac{(2-s_{0})s_{1}}{2s_{0}}}\leq\Big{(}\int_{t/2}^{t}\int_{V}(1+|\nabla
p|)^{s_{2}}dxd\tau\Big{)}^{\frac{\nu_{4}}{2}},$

where

$\nu_{4}=(2-s_{0})s_{1}/s_{0}=s_{3}-1.$ (6.7)

By (5.14) and relation (5.40),

$\displaystyle\lambda^{s_{1}/2}$ $\displaystyle\leq C\Big{\\{}t\cdot
t^{-(\kappa_{3}+1)}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{2(\kappa_{5}+1)}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{2\kappa_{5}}$
$\displaystyle\quad\quad\cdot\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}\Big{\\}}^{\nu_{4}/2}\leq
Ct^{\frac{-\kappa_{3}\nu_{4}}{2}}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{(\kappa_{5}+1)\nu_{4}}$
$\displaystyle\quad\quad\cdot\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\kappa_{5}\nu_{4}}\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}^{\nu_{4}/2}.$

Combining this with (6.2) and (3.9) gives

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}\leq
C\lambda^{\frac{s_{1}}{2}}t^{-\frac{1+s_{1}}{2}}\|p_{t}\|_{L^{2}(U\times(0,t))}$
$\displaystyle\leq
Ct^{\frac{-\kappa_{3}\nu_{4}-(1+s_{1})}{2}}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{(\kappa_{5}+1)\nu_{4}}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\kappa_{5}\nu_{4}}$
$\displaystyle\quad\cdot\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}^{\nu_{4}/2}$
$\displaystyle\quad\cdot\Big{(}\int_{U}[H(|\nabla
p_{0}(x)|)+\bar{p}^{2}_{0}(x)]dx+\int_{0}^{t}G_{3}(\tau)d\tau+\int_{0}^{t}\int_{U}|\Psi_{t}(x,\tau)|^{2}dxd\tau\Big{)}^{1/2}.$

Therefore,

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}$ $\displaystyle\leq
Ct^{-\kappa_{6}/2}\Big{(}1+\|\bar{p}_{0}\|_{L^{\alpha}}\Big{)}^{(\kappa_{5}+1)\nu_{4}+1}\Big{(}1+\int_{U}H(|\nabla
p_{0}(x)|)dx\Big{)}^{1/2}.$
$\displaystyle\quad\cdot\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\kappa_{5}\nu_{4}}\Big{(}1+\int_{0}^{t}G_{3}(\tau)d\tau)\Big{)}^{(\nu_{4}+1)/2}.$

We obtain (6.5).

Now consider $t\geq 2$. We apply (6.2) with $T_{0}=t-1$, $T=1$ and
$\theta=1/2$. Then by (5.23) and (5.16),

$\displaystyle\lambda^{s_{1}/2}$ $\displaystyle\leq
C\Big{(}\int_{t-1}^{t}\int_{V}(1+|\nabla
p|)^{s_{2}}dxd\tau\Big{)}^{\nu_{4}/2}$ $\displaystyle\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\nu_{4}(\kappa_{5}+\alpha/2)}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\nu_{4}(\kappa_{5}+\alpha/2)}$
$\displaystyle\quad\cdot\Big{(}1+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{\nu_{4}/2}.$

Combining this with (6.2) and (3.22) gives

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}\leq
C\lambda^{\frac{s_{1}}{2}}\|p_{t}\|_{L^{2}(U\times(t-1/2,t))}$
$\displaystyle\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\nu_{4}(\kappa_{5}+\alpha/2)}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\nu_{4}(\kappa_{5}+\alpha/2)}$
$\displaystyle\quad\cdot\Big{(}1+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{\nu_{4}/2}\cdot\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+[EnvA(\alpha,t)]^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{1/2}.$

Thus,

$\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\nu_{4}(\kappa_{5}+\alpha/2)+\alpha/2}\\\
\cdot\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\nu_{4}(\kappa_{5}+\alpha/2)+\alpha/2}\Big{(}1+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{\nu_{4}/2+1/2},$

and we obtain (6.6). The proof is complete. ∎

For large time or asymptotic estimates, we have:

######  Theorem 6.3.

Let $U^{\prime}\Subset U$.

(i) If $A(\alpha)<\infty$ then

$\displaystyle\limsup_{t\to\infty}\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}$
$\displaystyle\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\kappa_{9}}$
(6.8)
$\displaystyle\quad\cdot\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{s_{3}/2}.$

(ii) If $\beta(\alpha)<\infty$ then there is $T>0$ such that for all $t>T$,

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}$ $\displaystyle\leq
C\Big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-3,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-3,t))}\Big{)}^{\kappa_{9}}$
(6.9)
$\displaystyle\quad\cdot\Big{(}1+\int_{t-2}^{t}G_{3}(\tau)d\tau\Big{)}^{s_{3}/2}.$

###### Proof.

(i) For large $t$, we apply (6.2) with $T_{0}=t-1$, $T=1$ and $\theta=1/2$. We
have

$\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}\leq
C\lambda^{s_{1}/2}\|p_{t}\|_{L^{2}(U\times(t-1/2,t))}\leq
C\Big{(}\int_{t-1}^{t}\int_{V}(1+|\nabla
p|)^{s_{2}}dxd\tau\Big{)}^{\nu_{4}/2}\|p_{t}\|_{L^{2}(U\times(t-1/2,t))},$
(6.10)

where $\nu_{4}$ is defined by (6.7). Take the limit superior and using (5.17)
with $s=s_{2}$, and (3.26), we obtain

$\displaystyle\limsup_{t\to\infty}\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}$
$\displaystyle\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\nu_{4}(\kappa_{5}+\alpha/2)}\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{\nu_{4}/2}$
$\displaystyle\quad\cdot\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{t-1}^{t}\int_{U}|\Psi_{t}(x,\tau)|^{2}dxd\tau+\limsup_{t\to\infty}\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{1/2}$
$\displaystyle\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\nu_{4}(\kappa_{5}+\alpha/2)+\alpha/2}\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{\nu_{4}/2+1/2}.$

We obtain (6.8).

(ii) Using (6.10), (5.18), and (3.30), we obtain

$\displaystyle\|p_{t}(t)\|_{L^{\infty}(U^{\prime})}\leq
C\Big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-3,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-3,t))}\Big{)}^{\nu_{4}(\kappa_{5}+\alpha/2)}\Big{(}1+\int_{t-2}^{t}G_{1}(\tau)d\tau\Big{)}^{\nu_{4}/2}$
$\displaystyle\quad\cdot\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}\int_{U}|\Psi_{t}(x,\tau)|^{2}dxd\tau+\int_{t-1}^{t}G_{3}(\tau)d\tau\Big{)}^{1/2}.$

Therefore (6.9) follows. ∎

##  7 Interior estimates for pressure’s Hessian

In this section, we estimate the $L^{2}$-norm of the Hessian
$\nabla^{2}p=(p_{x_{i}x_{j}})_{i,j=1,2,\ldots,n}$.

######  Lemma 7.1.

Let $U^{\prime}\Subset V\Subset U$. For $t>0$,

$\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq C\Big{(}1+\|{\nabla
p(t)}\|_{L^{\infty}(V)}\Big{)}^{a}\Big{(}\int_{U}[|\nabla
p|^{2-a}+|p_{t}|^{2}]dx\Big{)}^{1/2}.$ (7.1)

###### Proof.

From (5.2) of Lemma 5.2 with $\zeta=\zeta(x)$ being a cut-off function in
space, we have

$\displaystyle\frac{1}{2s+2}\frac{d}{dt}\int_{U}|\nabla
p|^{2s+2}\zeta^{2}dx+\frac{1-a}{2}\int_{U}K(|\nabla
p|)|\nabla^{2}p|^{2}|\nabla p|^{2s}\zeta^{2}dx$ (7.2) $\displaystyle\leq
C\int_{U}K(|\nabla p|)|\nabla p|^{2s+2}|\nabla\zeta|^{2}dx.$

Clearly,

$\frac{1}{2s+2}\frac{d}{dt}\int_{U}|\nabla
p|^{2s+2}\zeta^{2}dx=-\int_{U}p_{t}\nabla\cdot(|\nabla p|^{2s}\nabla
p\zeta^{2})dx.$ (7.3)

Combining (7.2) and (7.3) with $s=0$, we have

$\displaystyle\int_{U}K(|\nabla p|)|\nabla^{2}p|^{2}\zeta^{2}dx\leq
C\int_{U}K(|\nabla p|)|\nabla
p|^{2}|\nabla\zeta|^{2}dx+C\int_{U}|p_{t}\nabla\cdot(\nabla p\zeta^{2})|dx.$

Since

$\displaystyle|p_{t}\nabla\cdot(\nabla p\zeta^{2})|$
$\displaystyle\leq|p_{t}||\nabla^{2}p|\zeta^{2}+2|p_{t}||\nabla
p|\zeta|\nabla\zeta|$ $\displaystyle\leq 1/2K(|\nabla
p|)|\nabla^{2}p|^{2}\zeta^{2}+CK(|\nabla p|)|\nabla
p|^{2}|\nabla\zeta|^{2}+C|p_{t}|^{2}K^{-1}(|\nabla p|)\zeta^{2},$

and, by (2.11), $K^{-1}(\xi)\leq C(1+\xi)^{a}$ we find that

$\displaystyle\int_{U}K(|\nabla p|)|\nabla^{2}p|^{2}\zeta^{2}dx\leq
C\int_{U}K(|\nabla p|)|\nabla
p|^{2}|\nabla\zeta|^{2}dx+C\int_{U}|p_{t}|^{2}(1+|\nabla p|)^{a}\zeta^{2}dx.$
(7.4)

Constructing appropriate $\zeta$ in (7.4) with $\zeta\equiv 1$ on $U^{\prime}$
and supp $\zeta\subset V$ we obtain

$\int_{U^{\prime}}K(|\nabla p|)|\nabla^{2}p|^{2}dx\leq C(1+\|\nabla
p\|_{L^{\infty}(V)})^{a}\int_{V}[|\nabla p|^{2-a}+|p_{t}|^{2}]dx.$ (7.5)

For $x\in U^{\prime}$,

$K(|\nabla p|)\geq C(1+|\nabla p|)^{-a}\geq C((1+\|{\nabla
p}\|_{L^{\infty}(V)})^{-a}$ (7.6)

From (7.5), (7.6) and Young’s inequality we obtain (7.1). ∎

Now, we combine Lemma 7.1 with estimates in section 3 and subsection 5.2 to
obtain particular bounds for the Hessian. Let

$\displaystyle s_{4}$ $\displaystyle=as_{3}+1,$ $\displaystyle\kappa_{10}$
$\displaystyle=a\kappa_{4}+1,\quad\kappa_{11}=as_{3}(\kappa_{5}+\alpha/2)+\alpha/2,\quad\kappa_{12}=as_{3}\kappa_{5}+\alpha/2.$

######  Theorem 7.2.

Let $U^{\prime}\Subset U$.

(i) If $t\in(0,2)$ then

$\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq
Ct^{-\kappa_{10}/2}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\kappa_{11}}\Big{(}1+\int_{U}H(|\nabla
p_{0}(x)|)dx\Big{)}^{1/2}\\\
\cdot\Big{(}1+EnvA(\alpha,t)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{\kappa_{12}}\Big{(}1+\int_{0}^{t}G_{4}(\tau)d\tau\Big{)}^{s_{4}/2}.$
(7.7)

(ii) If $t\geq 2$ then

$\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{\kappa_{11}}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{\kappa_{11}}\\\
\cdot\Big{(}1+\int_{t-1}^{t}G_{4}(\tau)d\tau\Big{)}^{s_{4}/2}.$ (7.8)

(iii) If $A(\alpha)<\infty$ then

$\limsup_{t\to\infty}\|{\nabla^{2}p}\|_{L^{2}(U^{\prime})}\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{\kappa_{11}}\\\
\cdot\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{4}(\tau)d\tau\Big{)}^{s_{4}/2}.$
(7.9)

(iv) If $\beta(\alpha)<\infty$ then there is $T>0$ such that for $t>T$,

$\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq
C\Big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-3,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-3,t))}\Big{)}^{\kappa_{11}}\\\
\cdot\Big{(}1+\int_{t-2}^{t}G_{4}(\tau)d\tau\Big{)}^{s_{4}/2}.$ (7.10)

###### Proof.

(i) Using (7.1), (5.36), (3.9) and (3.11):

$\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq
Ct^{-a\kappa_{4}/2}(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{as_{3}(\kappa_{5}+1)}\Big{(}1+EnvA(\alpha,t)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(0,t))}\Big{)}^{as_{3}\kappa_{5}}\\\
\cdot\Big{(}1+\int_{0}^{t}G_{1}(\tau)d\tau\Big{)}^{as_{3}/2}\Big{\\{}t^{-1}\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+\int_{U}H(|\nabla
p_{0}(x)|)dx\\\
+[EnvA(\alpha,t)]^{\frac{\alpha}{\alpha-a}}+\int_{U}|\Psi_{t}(x,t)|^{2}dx+\int_{0}^{t}G_{4}(\tau)d\tau\Big{)}\Big{\\}}^{1/2}.$

Note that $\alpha\geq 2$, and

$\int_{U}|\Psi_{t}(x,t)|^{2}dx\leq
C\Big{(}\int_{U}|\Psi_{t}(x,t)|^{\alpha}dx\Big{)}^{\frac{2}{\alpha}}\leq
CA(\alpha,t)^{\frac{2(1-a)}{\alpha-a}}\leq
C(1+A(\alpha,t)^{\frac{\alpha}{\alpha-a}}).$

Then (7.7) follows.

(ii) Using (7.1), (5.37), (3.13), (3.14) and (3.5):

$\displaystyle\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq
C(1+\|\bar{p}_{0}\|_{L^{\alpha}})^{as_{3}(\kappa_{5}+\alpha/2)}\Big{(}1+[EnvA(\alpha,t)]^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-2,t))}\Big{)}^{as_{3}(\kappa_{5}+\alpha/2)}$
$\displaystyle\quad\cdot\Big{(}1+\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{as_{3}/2}$
$\displaystyle\quad\cdot\Big{(}1+\int_{U}|\bar{p}_{0}(x)|^{\alpha}dx+[EnvA(\alpha,t-1)]^{\frac{\alpha}{\alpha-a}}+\int_{U}|\Psi_{t}(x,t-1)|^{2}dx+\int_{t-1}^{t}G_{4}(\tau)d\tau\Big{)}^{1/2}.$

Then (7.8) follows.

(iii) Using (7.1), (5.42), (3.23) and (3.24):

$\limsup_{t\to\infty}\|{\nabla^{2}p}\|_{L^{2}(U^{\prime})}\leq
C\Big{(}1+A(\alpha)^{\frac{1}{\alpha-a}}+\limsup_{t\to\infty}\|\Psi\|_{L^{\alpha}(U\times(t-1,t))}\Big{)}^{as_{3}(\kappa_{5}+\alpha/2)}\\\
\cdot\Big{(}1+\limsup_{t\to\infty}\int_{t-1}^{t}G_{1}(\tau)d\tau\Big{)}^{as_{3}/2}\Big{(}1+A(\alpha)^{\frac{\alpha}{\alpha-a}}+\limsup_{t\to\infty}\int_{U}|\Psi_{t}(x,t)|^{2}dx+\limsup_{t\to\infty}\int_{t-1}^{t}G_{4}(\tau)d\tau\Big{)}^{1/2}.$

Then (7.9) follows.

(iv) Using (7.1), (5.43), (3.27) and (3.28):

$\|{\nabla^{2}p(t)}\|_{L^{2}(U^{\prime})}\leq
C\Big{(}1+\beta(\alpha)^{\frac{1}{\alpha-2a}}+\sup_{[t-3,t]}A(\alpha,\cdot)^{\frac{1}{\alpha-a}}+\|\Psi\|_{L^{\alpha}(U\times(t-3,t))}\Big{)}^{as_{3}(\kappa_{5}+\alpha/2)}\\\
\cdot\Big{(}1+\int_{t-2}^{t}G_{1}(\tau)d\tau\Big{)}^{as_{3}/2}\Big{(}1+\beta(\alpha)^{\frac{\alpha}{\alpha-2a}}+A(\alpha,t-1)^{\frac{\alpha}{\alpha-a}}+\int_{t-1}^{t}G_{4}(\tau)d\tau+\int_{U}|\Psi_{t}(x,t)|^{2}dx\Big{)}^{1/2}.$

Then (7.10) follows. ∎

##  Appendix A Appendix

###### Proof of Lemma 2.1.

In this proof, we use notation $\|\cdot\|_{L^{p}}$ to denote
$\|\cdot\|_{L^{p}(U)}$. First, we have the following Poincaré-Sobolev
inequality

$\|\phi\|_{L^{r^{*}}}\leq C\|\nabla\phi\|_{L^{r}}+C\delta\|\phi\|_{L^{1}}.$
(A.1)

Let $r=2-a$ and $\phi=|u|^{m}$, where $m=(\alpha-a)/(2-a)\geq 1$. Applying
(A.1) to $\phi$, we obtain

$\|u\|_{L^{mr^{*}}}\leq C\Big{(}\int_{U}|u|^{\alpha-2}|\nabla
u|^{2-a}dx\Big{)}^{\frac{1}{\alpha-a}}+C\delta\|u\|_{L^{m}}\leq
C\Big{(}\int_{U}|u|^{\alpha-2}|\nabla
u|^{2-a}dx\Big{)}^{\frac{1}{\alpha-a}}+C\delta\|u\|_{L^{\alpha}}.$ (A.2)

The last inequality comes from Hölder’s inequality and the fact $\alpha>m$.
Consider a number $p\in(\alpha,mr^{*})$. Let $\theta\in(0,1)$ such that

$\frac{1}{p}=\frac{\theta}{\alpha}+\frac{1-\theta}{mr^{*}}.$ (A.3)

Combining interpolation inequality and (A.2) yields

$\displaystyle\|u\|_{L^{p}}^{p}$ $\displaystyle\leq\|u\|_{L^{\alpha}}^{\theta
p}\|u\|_{L^{mr^{*}}}^{(1-\theta)p}\leq C\|u\|_{L^{\alpha}}^{\theta
p}\Big{\\{}\Big{(}\int_{U}|u|^{\alpha-2}|\nabla
u|^{2-a}dx\Big{)}^{\frac{1}{\alpha-a}}+C\delta\|u\|_{L^{\alpha}}\Big{\\}}^{(1-\theta)p}$
$\displaystyle\leq C\delta\|u\|_{L^{\alpha}}^{p}+\|u\|_{L^{\alpha}}^{\theta
p}\Big{(}\int_{U}|u|^{\alpha-2}|\nabla
u|^{2-a}dx\Big{)}^{\frac{(1-\theta)p}{\alpha-a}}.$

In the preceding inequality, selecting $(1-\theta)p=\alpha-a$, letting
$u=u(x,t)$, and integrating in $t$ from $0$ to $T$ give

$\displaystyle\int_{0}^{T}\|u(t)\|_{L^{p}}^{p}dt$ $\displaystyle\leq C\delta
T\mathop{\mathrm{ess\,sup}}_{[0,T]}\|u\|_{L^{\alpha}}^{p}+C\mathop{\mathrm{ess\,sup}}_{[0,T]}\|u\|_{L^{\alpha}}^{\theta
p}\int_{0}^{T}\int_{U}|u|^{\alpha-2}|\nabla u|^{2-a}dxdt$ $\displaystyle\leq
C\delta T[[u]]^{p}+C[[u]]^{\theta p}[[u]]^{(1-\theta)p}=C(1+\delta
T)[[u]]^{p}.$

Therefore (2.17) follows. It remains to calculate $p$ and check sufficient
conditions. Note that $\theta p=p-\alpha+a$, and from (A.3), we derive

$1=\frac{\theta
p}{\alpha}+\frac{(1-\theta)p}{mr^{*}}=\frac{p-\alpha+a}{\alpha}+\frac{\alpha-a}{mr^{*}}.$

Solving for $p$ from this gives
$p=\alpha+(\alpha-a)(1-\frac{\alpha}{mr^{*}})$. Simple calculations yield $p$
in the form of (2.16). It is elementary to verify that the second condition in
(2.15) guarantees $\alpha<p<mr^{*}$.

In case $U$ is a ball $B_{R}(x_{0})$, the inequality with (2.19) follows from
the scaling $y=(x-x_{0})/R$ for $x\in B_{R}(x_{0})$ and (2.17) for $B_{1}(0)$.
∎

Acknowledgement. The authors would like to thank Tuoc Phan and Akif Ibragimov
for their suggestions and helpful discussions. L. H. acknowledges the support
by NSF grant DMS-1412796.

## References

  * [1] E. Aulisa, L. Bloshanskaya, L. Hoang, and A. Ibragimov.  Analysis of generalized Forchheimer flows of compressible fluids in porous media.  J. Math. Phys., 50(10):103102, 44, 2009.
  * [2] J. Bear.  Dynamics of Fluids in Porous Media.  American Elsevier Pub. Co., New York, 1972.
  * [3] H. Darcy.  Les Fontaines Publiques de la Ville de Dijon.  Dalmont, Paris, 1856.
  * [4] E. De Giorgi.  Sulla differenziabilità e l’analiticità delle estremali degli integrali multipli regolari.  Mem. Accad. Sci. Torino. Cl. Sci. Fis. Mat. Nat. (3), 3:25–43, 1957\.
  * [5] E. DiBenedetto.  Degenerate parabolic equations.  Universitext. Springer-Verlag, New York, 1993.
  * [6] L. Dung.  Ultimately uniform boundedness of solutions and gradients for degenerate parabolic systems.  Nonlinear Anal., 39(2, Ser. A: Theory Methods):157–171, 2000.
  * [7] J. Dupuit.  Mouvement de l’eau a travers le terrains permeables.  C. R. Hebd. Seances Acad. Sci., 45:92–96, 1857.
  * [8] C. Foias, O. Manley, R. Rosa, and R. Temam.  Navier-Stokes equations and turbulence, volume 83 of Encyclopedia of Mathematics and its Applications.  Cambridge University Press, Cambridge, 2001.
  * [9] P. Forchheimer.  Wasserbewegung durch Boden.  Zeit. Ver. Deut. Ing., 45:1781–1788, 1901.
  * [10] P. Forchheimer.  Hydraulik.  Number Leipzig, Berlin, B. G. Teubner. 1930.  3rd edition.
  * [11] L. Hoang and A. Ibragimov.  Structural stability of generalized Forchheimer equations for compressible fluids in porous media.  Nonlinearity, 24(1):1–41, 2011.
  * [12] L. Hoang and A. Ibragimov.  Qualitative Study of Generalized Forchheimer Flows with the Flux Boundary Condition.  Adv. Diff. Eq., 17(5–6):511–556, 2012.
  * [13] L. Hoang, A. Ibragimov, T. Kieu, and Z. Sobol.  Stability of solutions to generalized Forchheimer equations of any degree.  J. Math. Sci., 2015.  accepted.
  * [14] L. Hoang and T. Kieu.  Global estimates for generalized Forchheimer flows of slightly compressible fluids.  Journal d’Analyse Mathematique, 2015.  accepted.
  * [15] L. T. Hoang, T. T. Kieu, and T. V. Phan.  Properties of generalized Forchheimer flows in porous media.  J. Math. Sci., 202(2):259–332, 2014.
  * [16] O. A. Ladyženskaja, V. A. Solonnikov, and N. N. Ural′ceva.  Linear and quasilinear equations of parabolic type.  Translated from the Russian by S. Smith. Translations of Mathematical Monographs, Vol. 23. American Mathematical Society, Providence, R.I., 1968.
  * [17] M. Muskat.  The flow of homogeneous fluids through porous media.  McGraw-Hill Book Company, inc., 1937.
  * [18] D. A. Nield and A. Bejan.  Convection in porous media.  Springer-Verlag, New York, fourth edition, 2013.
  * [19] F. Ragnedda, S. Vernier Piro, and V. Vespri.  Asymptotic time behaviour for non-autonomous degenerate parabolic problems with forcing term.  Nonlinear Anal., 71(12):e2316–e2321, 2009.
  * [20] F. Ragnedda, S. Vernier Piro, and V. Vespri.  Large time behaviour of solutions to a class of non-autonomous, degenerate parabolic equations.  Math. Ann., 348(4):779–795, 2010.
  * [21] B. Straughan.  Stability and wave motion in porous media, volume 165 of Applied Mathematical Sciences.  Springer, New York, 2008.
  * [22] J. L. Vázquez.  Smoothing and decay estimates for nonlinear diffusion equations, volume 33 of Oxford Lecture Series in Mathematics and its Applications.  Oxford University Press, Oxford, 2006.  Equations of porous medium type.
  * [23] J. L. Vázquez.  The porous medium equation.  Oxford Mathematical Monographs. The Clarendon Press Oxford University Press, Oxford, 2007.  Mathematical theory.
  * [24] J. C. Ward.  Turbulent flow in porous media.  Journal of the Hydraulics Division, Proc. Am. Soc. Civ. Eng., 90(HY5):1–12, 1964.