# Existence and regularity results for anisotropic parabolic equations with
degenerate coercivity

Weilin Zou${}^{\mbox{a,}}$111 Corresponding author.
E-mail addresses<EMAIL_ADDRESS>Tel.: +8679183863755.  Yuanchun
Ren${}^{\mbox{a}}$ Wei Wang${}^{\mbox{b}}$
${}^{\mbox{a}}$ College of Mathematics and Information Science, Nanchang
Hangkong University, Nanchang, 330063, China
${}^{\mbox{b}}$ Party and Administration Office, Nanchang Hangkong University,
Nanchang, 330063, China

Abstract: This paper deals with a class of nonlinear anisotropic parabolic
equations with degenerate coercivity. Using the anisotropic Gagliardo-
Nirenberg-type inequality, we prove some existence and regularity results for
the solutions under the framework of anisotropic Sobolev spaces, which
generalize the previous results of [10, 18, 23].

Keywords: Existence and regularity; Anisotropic parabolic equations;
Degenerate coercivity; Anisotropic Gagliardo-Nirenberg inequality.

MSC:  35B65; 35K65; 35K55

##  1 Introduction and statement of the main results

This paper deals with the existence and regularity results of weak solutions
to the following anisotropic parabolic problem

$(P)\ \left\\{\begin{array}[]{ll}\frac{\partial u}{\partial
t}-\overset{N}{\underset{i=1}{\sum}}D_{i}(a_{i}(x,t,u,Du))=f&\mbox{in }Q,\\\
u=0&\mbox{on }\partial\Omega\times(0,T),\\\ u(x,0)=0&\mbox{in
}\Omega,\end{array}\right.$

where $f\in L^{m}(Q)$ with $m\geq 1$, $\Omega$ is an open bounded subset of
$\mathbb{R}^{N}(N\geq 2)$ , $T$ is a positive constant and
$Q=\Omega\times(0,T)$ with the lateral boundary $\partial\Omega\times(0,T)$.

Here $a_{i}:Q\times\mathbb{R}\times\mathbb{R}^{N}\rightarrow\mathbb{R}$
($i=1,2...,N$) are Carathéodory functions such that for almost all $(x,t)\in
Q$, for all $s\in\mathbb{R}$, and for all $\xi,\zeta\in\mathbb{R}^{N}$ with
$\xi\neq\zeta$,

$\displaystyle\sum_{i=1}^{N}a_{i}(x,t,s,\xi)\xi_{i}\geq\alpha\sum_{i=1}^{N}\frac{|\xi_{i}|^{p_{i}}}{(1+|s|)^{\theta(p_{i}-1)}},$
(1.1) $\displaystyle 0\leq\theta<1+\frac{\bar{p}}{N(\bar{p}-1)},$ (1.2)
$\displaystyle|a_{i}(x,t,s,\xi)|\leq\beta(1+|\xi_{i}|^{p_{i}-1}),\ \ \
i=1,2,\ldots,N,$ (1.3)

and

$\displaystyle\sum_{i=1}^{N}[a_{i}(x,t,s,\xi)-a_{i}(x,t,s,\zeta)][\xi_{i}-\zeta_{i}]>0,$
(1.4)

where $\alpha,\beta$ are two positive constants,
$p_{1},...,p_{N}\in(1,+\infty)$ and $\bar{p}$ is defined as in (2.1).

As $\theta=0$, the typical model of problem $(P)$ reduces to the so-called
anisotropic evolutionary $\vec{p}$-Laplacian equation. Indeed, assumption
(1.1) presents an interesting feature, namely an anisotropic diffusion with
orthotropic structure. These kinds of problems have attracted an increasing
amount of attention in recent years as they reflect the modeling of physical
and mechanical processes in anisotropic continuous medium. With no hope of
being thorough, we mention some papers regarding the study of these problems
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] (see also the references therein).

In the case that $\theta\neq 0$, the parabolic operator in problem $(P)$ may
degenerate as soon as the solution is unbounded. Indeed, the diffusion
coefficient may tend to zero as the solution $u$ becomes large, which means
that a slow diffusion effect may appear. There are already some results for
such kind of problems both in the stationary and evolution cases with
isotropic growth condition. For example, the isotropic elliptic case (with
$p_{i}\equiv p$ for $i=1,2,...,N$) associated to problem $(P)$ was firstly
studied in [12] and then developed in [17, 13, 14, 15, 16]; in the isotropic
parabolic case, the existence and regularity results of problem $(P)$ were
proved in [18](see also [19, 20, 21]), and the study of problem $(P)$ with
nonzero initial data can be found in [22].

Recently, the regularity results for weak solutions to the anisotropic
elliptic case of problem $(P)$ were investigated in [23]. The existence and
regularity of the entropy solutions for stationary problem $(P)$ with lower
order terms were considered in [24]. However, to the best of our knowledge,
there are no existence and regularity results concerning with the nonlinear
anisotropic parabolic equations with degenerate coercivity. Hence, the aim of
this paper is to extend the results of [18, 23] to the case of anisotropic
parabolic problems. To deal with this kind of problem, we shall look for
$u_{n}$ as a solution to the approximate problem with non-degenerate
coercivity, and establish some estimates on the weak solutions of these
problems under the framework of anisotropic Sobolev spaces. The proofs of
these estimates rely crucially on the anisotropic version of Gagliardo-
Nirenberg inequality and the interpolation inequality.

The remainder of this paper is organized as follows. In Section 2, we give
some preliminary results and state the main results. Section 3 is devoted to
proving some estimates of weak solutions to the approximate problem under
different assumptions on $f$. In Section 4, we briefly give the proofs of
Theorem 2.1-2.5.

##  2 Preliminaries and main results

Given $\vec{p}=(p_{1},p_{2},...,p_{N})$, we define

$\bar{p}=\left(\frac{1}{N}\sum\limits_{i=1}^{N}\frac{1}{p_{i}}\right)^{-1}.$
(2.1)

Assume $1<\bar{p}<N$, we denote

$\bar{p}^{*}=\frac{N\bar{p}}{N-\bar{p}}.$

Anisotropic Sobolev space. Let $\Omega$ be an open bounded subset of
$\mathbb{R}^{N}(N\geq 2)$, we define the following space

$W_{0}^{1,\vec{p}}(\Omega):=\left\\{g\in W_{0}^{1,1}(\Omega):D_{i}g\in
L^{p_{i}}(\Omega)\ \mbox{for\ every}\ i=1,2\ldots N\right\\},$

which is a Banach space under the norm

$\|g\|_{W_{0}^{1,\vec{p}}(\Omega)}:=\sum\limits_{i=1}^{N}\|D_{i}g\|_{L^{p_{i}}(\Omega)}.$

We also introduce the functional space
$L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))$ by

$L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega)):=\left\\{u\in
L^{1}\left(0,T;W_{0}^{1,1}(\Omega)\right):D_{i}u\in L^{p_{i}}(Q),\ \mbox{for\
every}\ i=1,2,\ldots,N\right\\}.$

The dual space of $L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))$ is denoted by
$L^{\vec{p^{\prime}}}(0,T;W^{-1,\vec{p^{\prime}}}(\Omega))$, where
$\vec{p^{\prime}}=(p_{1}^{\prime},p_{2}^{\prime},...,p_{N}^{\prime})$.

Next we recall the following anisotropic Sobolev embedding Theorem(see [25,
26]).

Lemma 2.1. _Let $u\in W_{0}^{1,\vec{p}}(\Omega)$. If $1<\bar{p}<N$, then there
exists a constant $C=C(N,\vec{p})>0$ such that_

$\|u\|_{L^{\bar{p}^{*}}(\Omega)}\leq
C\left(\prod_{i=1}^{N}\|D_{i}u\|_{L^{p_{i}}(\Omega)}\right)^{1/N}.$

Now we give the definition of weak solutions of problem $(P)$.

Definition 2.1. A measurable function $u\in L^{1}(0,T;W_{0}^{1,1}(\Omega))$ is
a weak solution of problem $(P)$ if $a_{i}(x,t,u,Du)\in L^{1}(Q)$ and

$-\iint_{Q}u\phi_{t}dxdt+\iint_{Q}\sum_{i=1}^{N}a_{i}(x,t,u,Du)D_{i}\phi
dxdt=\iint_{Q}f\phi dxdt,$

for every test function $\phi\in C^{\infty}(\bar{Q})$ such that $\phi=0$ in a
neighborhood of $\partial\Omega\times(0,T)\cup(\Omega\times\\{T\\})$.

The main results of this work are stated as follows.

Theorem 2.1. _Assume that ( 1.1)-(1.4) hold and $f\in L^{m}(Q)$ with
$m>\frac{N}{\bar{p}}+1$. Then there exists a weak solution $u\in
L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))\cap L^{\infty}(Q)$ to problem
$(P)$._

Remark 2.1. This result is independent of $\theta$ which corresponds to the
one that obtains in the isotropic case with $\theta=0$(see [27]). If
$p_{1}=p_{2}=...=p_{N}=2$, it is proved in [18] that problem $(P)$ admits at
least a bounded weak solution provided that $f$ belongs to $L^{m}(Q)$ with
$m>\frac{N}{2}+1$. Thus, Theorem 2.1 extends the results of [27] and [18].

Theorem 2.2. _Assume that ( 1.1)-(1.4) hold and $f\in L^{m}(Q)$ with
$m=\frac{N}{\bar{p}}+1$. Then there exists a constant $\lambda>0$ such that
problem $(P)$ admits at least a weak solution $u\in
L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))$ with
$e^{\lambda|u|^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\in
L^{1}(Q)$._

Remark 2.2. It is clear that
$e^{\lambda|u|^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\in
L^{1}(Q)$ implies $u\in L^{\tilde{r}}(Q)$ for any $\tilde{r}\in(1,+\infty)$.
This improves the result of Theorem 1.2 in [18].

Theorem 2.3. _Assume that ( 1.1)-(1.4) and $f\in L^{m}(Q)$ with $m$
satisfying_

$m_{1}\leq m<\frac{N}{\bar{p}}+1,$ (2.2)

_where_

$m_{1}:=\frac{N+2\bar{p}+N(\bar{p}-1)(1-\theta)+\theta\left[\left(\max\limits_{1\leq
i\leq
N}p_{i}-1\right)(N+\bar{p})\right]}{N(1-\theta)(\bar{p}-1)+2\bar{p}-\theta\bar{p}\left(\max\limits_{1\leq
i\leq N}p_{i}-1\right)}.$ (2.3)

_Then there exists a weak solution $u\in
L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))\cap L^{\tilde{r}}(Q)$ problem $(P)$
with_

$\tilde{r}=\frac{mN[(1-\theta)(\bar{p}-1)]+m\bar{p}}{N-m\bar{p}+\bar{p}}.$
(2.4)

Theorem 2.4. _Assume that ( 1.1)-(1.4) hold and $f\in L^{m}(Q)$ with $m$
satisfying_

$\begin{split}\frac{N+2+\theta(\bar{p}-1)}{\bar{p}(1+N)+(1-N)[1+\theta(\bar{p}-1)]}<m<m_{N}\end{split},$
(2.5)

_where_

$m_{N}:=\frac{N+2\bar{p}+N(\bar{p}-1)(1-\theta)+\theta\left[\left(\min\limits_{1\leq
i\leq
N}p_{i}-1\right)(N+\bar{p})\right]}{N(1-\theta)(\bar{p}-1)+2\bar{p}-\theta\bar{p}\left(\min\limits_{1\leq
i\leq N}p_{i}-1\right)}.$ (2.6)

_Then there exists a weak solution $u\in
L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))\cap L^{\tilde{r}}(Q)$ to problem
$(P)$ with $\vec{q}=(q_{1},q_{2},...,q_{N})$, where $\tilde{r}$ is defined as
in (2.4) and_

$q_{i}=\frac{mp_{i}[N(1-\theta)(\bar{p}-1)+\bar{p}]}{[N-\bar{p}(m-1)][\theta(p_{i}-1)+1]+N(\bar{p}-1)(1-\theta)+\bar{p}}.$
(2.7)

Remark 2.3. The case of $f\in L^{m}(\Omega)$ with $m_{N}\leq m<m_{1}$ is not
considered in Theorem 2.1- Theorem 2.4. This case is rather more complicate,
which will be discussed in Remark 4.1 of Section 4. If $1\leq
m<\frac{N+2+\theta(\bar{p}-1)}{\bar{p}(1+N)+(1-N)[1+\theta(\bar{p}-1)]}$, it
is possible to prove the existence of an entropy solution in the sense made
precisely below(see [29, 33, 28, 30, 31, 32]).

For this we need to introduce two usual functions defined as follows

$\begin{split}T_{k}(s)=\min\\{k,\max\\{-k,s\\}\\},\ \
S_{k}(s)=\int^{s}_{0}T_{k}(\tau)d\tau,\ \ \forall s\in\mathbb{R}.\end{split}$
(2.8)

Definition 2.2. A measurable function $u\in L^{\infty}(0,T;L^{1}(\Omega))$ is
called an entropy solution of problem $(P)$, if it satisfies the following two
conditions:

$T_{k}(u)\in L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega)),$ (2.9) $S_{k}(u)\in
C([0,T];L^{1}(\Omega)),$ (2.10)

and if

$\begin{split}&\int_{\Omega}S_{k}\left(u(T)-\phi(T)\right)dx-\int_{\Omega}S_{k}\left(-\phi(0)\right)dx+\int_{0}^{T}\left\langle\phi_{t},T_{k}(u-\phi)\right\rangle
dt\\\
&+\iint_{Q}\sum_{i=1}^{N}a_{i}(x,t,u,Du)D_{i}T_{k}(u-\phi)dxdt=\iint_{Q}fT_{k}(u-\phi)dxdt,\end{split}$
(2.11)

for every $\phi\in L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))\cap
L^{\infty}(Q)$ such that $\phi_{t}\in
L^{\vec{p^{\prime}}}(0,T;W^{-1,\vec{p^{\prime}}}(\Omega))\cap L^{1}(Q).$

Remark 2.4 It is well known that conditions (2.9) and (2.10) allow to define
the very weak gradient $Du$ almost everywhere in $Q$(see [29]). For any $k>0$,
we have

$DT_{k}(u)=v\chi_{\\{|u|<k\\}}\ \ \mbox{a.e. in }\ \ Q,$

where $\chi_{\\{|u|<k\\}}$ represents the characteristic function on the set
$\\{|u|<k\\}$.

Definition 2.3. (see [33, 34]) We say that a measurable function
$u:Q\rightarrow R$ belongs to the Marcinkiewicz space ${\cal M}^{q}(Q)$ with
$0<q<+\infty$, if there exists a constant $C$ such that

$[u]_{q}=\mathop{\mbox{sup}}\limits_{k>0}k\ \mbox{meas}\\{(x,t)\in
Q:|u(x,t)|>k\\}^{\frac{1}{q}}\leq C.$

For $r<q,$ one can deduce that ${\cal M}^{q}(Q)\subset{\cal M}^{r}(Q)$ and
$L^{q}(Q)\subset{\cal M}^{q}(Q)\subset L^{r}(Q)$(see [34, 35]).

Theorem 2.5. _Assume that ( 1.1)-(1.4) hold and $f\in L^{m}(Q)$ with_

$\begin{split}1\leq
m\leq\max\left\\{\frac{N+2+\theta(\bar{p}-1)}{\bar{p}(1+N)+(1-N)[1+\theta(\bar{p}-1)]},1\right\\}.\end{split}$
(2.12)

_Then there exists an entropy solution $u$ to problem $(P)$ with
$u\in\mathcal{M}^{\tilde{r}}(Q)$ and $|D_{i}u|\in\mathcal{M}^{q_{i}}(Q)$ for
$i=1,2\ldots N$, where $\tilde{r}$ and $q_{i}$ are defined as before. _

Remark 2.5. If $\theta=0$, Theorem 2.5 reduces to the corresponding result
obtained in [36] for parabolic equations with measure data, and [10] for
anisotropic parabolic equations with measure data.

##  3 The approximating problems and some useful estimates

For convenience, we shall use $|E|$ denoting the Lebesgue measure of
measurable set $E$. If not otherwise specified, we will denote by $C$ or
$C_{i}$(i=1,2…)several positive constants whose values may change from line to
line. These values will only depend on the data but they will never depend on
the indexes $n$ of the sequences.

To prove Theorem 2.1-2.5, first of all, we consider the following approximate
problem

$(P_{n})\ \left\\{\begin{array}[]{ll}\frac{\partial u_{n}}{\partial
t}-\sum\limits_{i=1}^{N}D_{i}\left(a_{i}(x,t,T_{n}(u_{n}),Du_{n}\right)=f_{n}&\mbox{in
}Q,\\\ u_{n}=0&\mbox{on }\partial\Omega\times(0,T),\\\ u_{n}(x,0)=0&\mbox{in
}\Omega,\end{array}\right.$

where $f_{n}\in C_{0}^{\infty}(Q)$ such that

$||f_{n}||_{L^{m}{(Q)}}\leq||f||_{L^{m}{(Q)}},\ \ \ \forall n,$ (3.1)
$f_{n}\rightarrow f\mbox{\ \ strongly \ \ in }L^{m}(Q).$ (3.2)

According to the classical result(see [37]), there exists at least a weak
solution $u_{n}\in C([0,T];L^{2}(\Omega))\cap
L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))$ to problem $(P_{n})$.

In order to get some estimates of $u_{n}$, we need the following three
results. The first one is the anisotropic version of Gagliardo-Nirenberg
inequality(see [7]) which is stated as follows.

Proposition 3.1. _Let $\vec{\sigma}=(\sigma_{1},\sigma_{2},...,\sigma_{N})$,
where $\sigma_{i}>1$ for $i=1,2,...,N$ with $\bar{\sigma}<N$. If $u\in
C\left(0,T;L^{2}(\Omega)\right)\cap
L^{\vec{\sigma}}\left(0,T;W_{0}^{1,\vec{\sigma}}(\Omega)\right)$. Then there
exists a constant $C>0$ that depends only on $N$ and $\sigma_{i}$ such that_

$\iint_{Q}|u|^{\frac{2\bar{\sigma}}{N}+\bar{\sigma}}dxdt\leq
C\left(\mathop{\mbox{ess\
sup}}_{t\in(0,T)}\int_{\Omega}|u|^{2}dx\right)^{\frac{\bar{\sigma}}{N}}\prod_{i=1}^{N}\left(\iint_{Q}|D_{i}u|^{\sigma_{i}}dxdt\right)^{\frac{\bar{\sigma}}{\sigma_{i}N}}.$

The second one is the following embedding result of Marcinkiewicz space.

Proposition 3.2. _Let $v$ be a measurable function belonging to ${\cal
M}^{\mu}(Q)$ for $\mu>0$, and suppose that there exist two constants
$\nu_{i}>\gamma\geq 0$ such that_

$\displaystyle\iint_{Q}|D_{i}T_{k}(v)|^{p_{i}}dxdt\leq
C(1+k)^{\gamma(p_{i}-1)}k^{\nu_{i}-\gamma(p_{i}-1)},\ \ \forall k>0,$ (3.3)

_where $C$ is a positive constant independent of $k$. Then_

$D_{i}v\in{\cal M}^{\delta_{i}}(Q),$

_where $\delta_{i}=\frac{p_{i}\mu}{\mu+\nu_{i}}$._

Proof. The proof is similar to the proof of Proposition 3.1 in [18](see also
[12]). For the sake of clarity and readability, we give the details below.

For every $k>0,$ we have

$\displaystyle|\\{|D_{i}v|>l\\}|\leq|\\{|v|>k\\}|+|\\{|D_{i}v|>l,|v|\leq
k\\}|,$ (3.4)

where $l$ is a fixed positive number.

In view of (3.3), it is easy to get

$|\\{|D_{i}v|>l,|v|\leq
k\\}|\leq\frac{1}{l^{p_{i}}}\iint_{Q}|D_{i}T_{k}(v)|^{p_{i}}dxdt\leq
C\frac{(1+k)^{\gamma(p_{i}-1)}k^{\nu_{i}-\gamma(p_{i}-1)}}{l^{p_{i}}}.$

This implies

$|\\{|D_{i}v|>l,|v|\leq k\\}|\leq
2^{\gamma(p_{i}-1)}C\frac{k^{\nu_{i}}}{l^{p_{i}}},\ \ \mbox{for}\ k\geq 1.$

Noticing that $v\in{\cal M}^{\mu}(Q)$, we have

$|\\{|v|>k\\}|\leq\frac{\tilde{C}}{k^{\mu}},$

where $\tilde{C}$ is a positive constant independent of $k$.

Combining the above two inequalities with (3.4), we obtain

$|\\{|D_{i}v|>l\\}|\leq\max\\{2^{\gamma(p_{i}-1)}C,\tilde{C}\\}\left(\frac{k^{\nu_{i}}}{l^{p_{i}}}+\frac{1}{k^{\mu}}\right),\
\ \mbox{for}\ k\geq 1.$

Minimizing with respect to $k$, namely taking
$k={(\frac{\mu}{\nu_{i}})}^{\frac{1}{\mu+\nu_{i}}}l^{\frac{p_{i}}{\mu+\nu_{i}}}$,
we get

$|\\{|D_{i}v|>l\\}|\leq\max\\{2^{\gamma(p_{i}-1)}C,\tilde{C}\\}\left[{(\frac{\mu}{\nu_{i}})}^{\frac{\nu_{i}}{\mu+\nu_{i}}}+1\right]\frac{1}{l^{\delta_{i}}},\
\ \mbox{for}\ l\geq{(\frac{\nu_{i}}{\mu})}^{\frac{1}{p_{i}}}.$

Since $Q$ is bounded, the above inequality obviously holds true for
$l<{(\frac{\nu_{i}}{\mu})}^{\frac{1}{p_{i}}}.$ Thus, the proof is finished.
$\Box$

Furthermore, by the proof of lemma 3.11 of [38], we easily get the following
result.

Proposition 3.3. _Let $r\geq 1$ and $g:Q\rightarrow\mathbb{R}$ be a measurable
function. If_

$\sum_{\tilde{k}=0}^{\infty}\tilde{k}^{r-1}|\\{|g|>\tilde{k}\\}|<+\infty,$

_then $g\in L^{r}(Q)$ and_

$\int_{0}^{T}\int_{\Omega}|g|^{r}dxdt\leq
2^{r}|Q|+2^{r-1}\left[|Q|+\sum_{\tilde{k}=0}^{\infty}\tilde{k}^{r-1}|\\{|g|>\tilde{k}\\}|\right].$

Lemma 3.1. _Suppose that ( 1.1)-(1.4) hold true and $f\in L^{m}(Q)$ with
$m>\frac{N}{\bar{p}}+1$. If $u_{n}$ is a weak solution of problem $(P_{n})$,
then_

$||u_{n}||_{L^{\infty}(Q)}\leq C_{1},$ (3.5)
$||u_{n}||_{L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))}\leq C_{1}.$ (3.6)

Proof. We only prove the case $\theta>0$, since the case $\theta=0$ is a well-
known result.

For $k>0$, denote $G_{k}(s)=s-T_{k}(s),\forall s\in\mathbb{R},k>0$. Taking
$G_{k}(u_{n}(x,t))\chi_{(0,\tau)}(t)$ as a test function in problem $(P_{n})$,
and using (1.2) and (3.1) together with Hölder’s inequality, we get for every
$\tau\in(0,T]$

$\begin{split}&\int_{A_{k}(\tau)}|G_{k}(u_{n}(\tau))|^{2}dx+\int_{0}^{\tau}\int_{A_{k}(t)}\sum_{i=1}^{N}\frac{\alpha|D_{i}u_{n}|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)}}dxdt\\\
&\leq
C\left(\int_{0}^{\tau}\int_{A_{k}(t)}|G_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{1}{m^{\prime}}},\end{split}$
(3.7)

where

$A_{k}(t)=\\{x\in\Omega:|u_{n}(x,t)|>k\\}.$ (3.8)

Setting

$\sigma_{i}=\frac{p_{i}\bar{p}(N+2-N\theta)+N\theta
p_{i}}{N\theta(p_{i}-\bar{p})+(N+2)\bar{p}},$

then it is easy to check that $\sigma_{i}\leq p_{i}.$ Furthermore, we have

$\frac{\theta(p_{i}-1)\sigma_{i}}{p_{i}-\sigma_{i}}=\frac{(N+2)\bar{\sigma}}{N},\
\ \ \ \ \frac{p_{i}-\sigma_{i}}{p_{i}}=\frac{\theta
N\sigma_{i}(p_{i}-1)}{(N+2)\bar{\sigma}p_{i}},$

and

$\bar{\sigma}=\frac{(N+2)\bar{p}-N\theta\bar{p}+N\theta}{N+2}.$ (3.9)

According to (LABEL:2.9) and Hölder’s inequality, we get

$\begin{split}&\int_{0}^{\tau}\int_{A_{k}(t)}|D_{i}G_{k}(u_{n})|^{\sigma_{i}}dxdt\\\
&\leq\left(\int_{0}^{\tau}\int_{A_{k}(t)}\frac{|D_{i}u_{n}|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)}}dxdt\right)^{\frac{\sigma_{i}}{p_{i}}}\left(\int_{0}^{\tau}\int_{A_{k}(t)}(1+|u_{n}|)^{\frac{\theta(p_{i}-1)\sigma_{i}}{p_{i}-\sigma_{i}}}dxdt\right)^{\frac{p_{i}-\sigma_{i}}{p_{i}}}\\\
&\leq
C\left(\int_{0}^{\tau}\int_{A_{k}(t)}|G_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{\sigma_{i}}{p_{i}m^{\prime}}}\left(\int_{0}^{\tau}\int_{A_{k}(t)}(k+|G_{k}(u_{n})|)^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\right)^{\frac{N\theta\sigma_{i}(p_{i}-1)}{(N+2)\bar{\sigma}p_{i}}}\\\
&\leq
C\left(\int_{0}^{\tau}\int_{A_{k}(t)}|G_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{\sigma_{i}}{p_{i}m^{\prime}}}\\\
&\ \ \ \
\times\left[k^{\frac{\theta\sigma_{i}(p_{i}-1)}{p_{i}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\frac{N\theta\sigma_{i}(p_{i}-1)}{(N+2)\bar{\sigma}p_{i}}}+\left(\int_{0}^{\tau}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\right)^{\frac{N\theta\sigma_{i}(p_{i}-1)}{(N+2)\bar{\sigma}p_{i}}}\right].\end{split}$

From proposition 3.1, (LABEL:2.9) and the above inequality, it follows that

$\begin{split}&\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\\\
&\leq C\left(\mathop{\mbox{ess\
sup}}_{t\in(0,T)}\int_{A_{k}(t)}|G_{k}(u_{n})|^{2}dx\right)^{\frac{\bar{\sigma}}{N}}\prod_{i=1}^{N}\left(\int_{0}^{T}\int_{A_{k}(t)}|D_{i}G_{k}(u_{n})|^{\sigma_{i}}dxdt\right)^{\frac{\bar{\sigma}}{\sigma_{i}N}}\\\
&\leq
C\left(\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{\bar{\sigma}(N+\bar{p})}{m^{\prime}N\bar{p}}}\\\
&\ \ \ \
\times\left[k^{\frac{\theta\bar{\sigma}(\bar{p}-1)}{\bar{p}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\frac{N\theta(\bar{p}-1)}{\bar{p}(N+2)}}+\left(\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\right)^{\frac{N\theta(\bar{p}-1)}{\bar{p}(N+2)}}\right].\end{split}$

Recalling that $m>\frac{N}{\bar{p}}+1$ and
$0\leq\theta<1+\frac{\bar{p}}{N(\bar{p}-1)},$ we get
$\frac{\bar{\sigma}(N+2)}{m^{\prime}N}>1.$ Applying Hölder’s inequality, we
have

$\begin{split}&\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\\\
&\leq
C\left(\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\right)^{\frac{N+\bar{p}}{\bar{p}(N+2)}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\frac{\bar{\sigma}(N+\bar{p})(m-1)}{\bar{p}Nm}-\frac{N+\bar{p}}{\bar{p}(N+2)}}\\\
&\ \ \ \
\times\left[k^{\frac{\theta\bar{\sigma}(\bar{p}-1)}{\bar{p}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\frac{N\theta(\bar{p}-1)}{\bar{p}(N+2)}}+\left(\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\right)^{\frac{N\theta(\bar{p}-1)}{\bar{p}(N+2)}}\right],\\\
\end{split}$

which implies that

$\begin{split}&\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\\\
&\leq
C\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\frac{\bar{\sigma}(N+\bar{p})(m-1)(N+2)}{Nm[N(\bar{p}-1)+\bar{p}]}-\frac{N+\bar{p}}{N(\bar{p}-1)+\bar{p}}}\\\
&\ \ \ \
\times\left[k^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}+\left(\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\right)^{\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}\right].\end{split}$

Therefore, by Young’s inequality and the fact that
$\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}<1,$ we obtain

$\displaystyle\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt$
$\displaystyle\leq
Ck^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\lambda}$
$\displaystyle\ \ \ \ \
+C\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\left[\frac{\bar{\sigma}(N+\bar{p})(m-1)(N+2)}{Nm[N(\bar{p}-1)+\bar{p}]}-\frac{N+\bar{p}}{N(\bar{p}-1)+\bar{p}}\right]\frac{N(\bar{p}-1)+\bar{p}}{N(\bar{p}-1)-N\theta(\bar{p}-1)+\bar{p}}},$
(3.10)

where

$\lambda=\frac{\bar{\sigma}(N+\bar{p})(m-1)(N+2)}{Nm[N(\bar{p}-1)+\bar{p}]}+\frac{N\theta(\bar{p}-1)-N-\bar{p}}{N(\bar{p}-1)+\bar{p}}.$

Since $m>\frac{N}{\bar{p}}+1$, it is easy to check that

$\displaystyle\left[\frac{\bar{\sigma}(N+\bar{p})(m-1)(N+2)}{Nm[N(\bar{p}-1)+\bar{p}]}-\frac{N+\bar{p}}{N(\bar{p}-1)+\bar{p}}\right]\frac{N(\bar{p}-1)+\bar{p}}{N(\bar{p}-1)-N\theta(\bar{p}-1)+\bar{p}}>\lambda>1,$

where we have used (3.9).

Hence, we get for $k\geq 1$

$\begin{split}&\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\\\
&\leq
Ck^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\lambda}\\\
&\ \ \ \
\mbox{+}C|Q|^{\left[\frac{\bar{\sigma}(N+\bar{p})(m-1)(N+2)}{Nm[N(p-1)+p]}-\frac{N+\bar{p}}{N(\bar{p}-1)+\bar{p}}\right]\frac{N(\bar{p}-1+\bar{p})}{N(\bar{p}-1)-N\theta(\bar{p}-1)+\bar{p}}-\lambda}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\lambda}\\\
&\leq
Ck^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\left(\int_{0}^{T}|A_{k}(t)|dt\right)^{\lambda},\\\
\end{split}$

which yields

$\varphi(h)\leq\frac{Ck^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\varphi(k)^{\lambda}}{(h-k)^{\frac{(N+2)\bar{\sigma}}{N}}},\
\ \ \forall h>k\geq 1,$ (3.11)

where

$\varphi(k)=\int_{0}^{T}|A_{k}(t)|dt.$

By Lemma A.2 in [12], one may conclude that there exists a constant $k^{*}>0$
independent of $n$ satisfying

$\varphi(k^{*})=0,$ (3.12)

from which, we get (3.5). Taking $u_{n}$ as a test function in problem
$(P_{n})$ and using (3.5), we easily get (3.6). $\Box$

Lemma 3.2. _Assume that ( 1.1)-(1.4) hold true and $f\in L^{m}(Q)$ with
$m=\frac{N}{\bar{p}}+1$. If $u_{n}$ is a weak solution of problem $(P_{n})$,
then_

$||u_{n}||_{L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))}\leq C_{2},$ (3.13)
$\|e^{\lambda|u_{n}|^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\|_{L^{1}(Q)}\leq
C_{2}.$ (3.14)

Proof. In the same way as in the proof of (3), we get

$\displaystyle\int_{0}^{T}\int_{A_{k}(t)}|G_{k}(u_{n})|^{\frac{(N+2)\bar{\sigma}}{N}}dxdt\leq
Ck^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\int_{0}^{T}|A_{k}(t)|dt+C\int_{0}^{T}|A_{k}(t)|dt,$

and as a result, one obtains

$\varphi(h)\leq\frac{Ck^{\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}}\varphi(k)}{(h-k)^{\frac{(N+2)\bar{\sigma}}{N}}},\
\ \ \forall h>k\geq 1.$

This, combined with lemma 3 of [39](here
$\tau_{1}=\frac{\theta(\bar{p}-1)\bar{\sigma}(N+2)}{N(\bar{p}-1)+\bar{p}}$ and
$\tau_{2}=\frac{(N+2)\bar{\sigma}}{N}$), gives

$\displaystyle\int_{0}^{T}|\\{|u_{n}|>k\\}|dt$ $\displaystyle\leq
e^{s_{0}+1}e^{-\left({\frac{k-1}{\lambda_{0}}}\right)^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\int_{0}^{T}|\\{|u_{n}|>1\\}|dt,\
\ \ \forall k\geq 1+\lambda_{0}s_{0}^{\frac{\tau_{2}}{\tau_{2}-\tau_{1}}},$

where
$\lambda_{0}=\mbox{max}\\{1,\left[(Ce)^{\frac{1}{\tau_{2}}}2^{\frac{\tau_{1}}{\tau_{2}}}\frac{\tau_{2}-\tau_{1}}{\tau_{2}}\right]^{\frac{\tau_{2}}{\tau_{2}-\tau_{1}}}\\}$,
and $s_{0}\in\mathbb{N}$ with
$s_{0}>\frac{2^{\frac{\tau_{2}-\tau_{1}}{\tau_{2}}}}{2^{\frac{\tau_{2}-\tau_{1}}{\tau_{2}}}-1}$.

Note that $k\geq 1+\lambda_{0}s_{0}^{\frac{\tau_{2}}{\tau_{2}-\tau_{1}}}\geq
2$ implies that $k-1\geq\frac{k}{2}.$ As a result, the above inequality shows
that

$\int_{0}^{T}|\\{|u_{n}|>k\\}|dt\leq
Ce^{-2\lambda{k}^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}},$

where
$2\lambda=(\frac{1}{2\lambda_{0}})^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}.$
Thus

$\displaystyle\int_{0}^{T}\left|\left\\{e^{\lambda|u_{n}|^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}>e^{\lambda
k^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\right\\}\right|dt$
$\displaystyle=\int_{0}^{T}|\\{|u_{n}|>k\\}|dt\leq
C\left(e^{\lambda{k}^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\right)^{-2},\
\ \forall k\geq 1+\lambda_{0}s_{0}^{\frac{\tau_{2}}{\tau_{2}-\tau_{1}}}.$
(3.15)

Now let us take
$k=\left(\frac{\mbox{ln}\tilde{k}}{\lambda}\right)^{\frac{1}{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}$
with $\tilde{k}\in\mathbb{N}$ satisfying

$\tilde{k}\geq\mbox{max}\\{e^{\lambda{2}^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}},e^{\lambda{(1+\lambda_{0}s_{0}^{\frac{\tau_{2}}{\tau_{2}-\tau_{1}}})}^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}\\}=\bar{k}.$

Then, by (3), it follows that

$\int_{0}^{T}\left|\left\\{e^{\lambda|u_{n}|^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}>\tilde{k}\right\\}\right|dt\leq\frac{|Q|e}{\tilde{k}^{2}},\
\ \forall\tilde{k}\geq\bar{k},$

which yields

$\sum_{\tilde{k}\in\mathbb{N},\tilde{k}\geq\bar{k}}\int_{0}^{T}\left|\left\\{e^{\lambda|u_{n}|^{1-\frac{N\theta(\bar{p}-1)}{N(\bar{p}-1)+\bar{p}}}}>\tilde{k}\right\\}\right|dt\leq|Q|e\sum_{\tilde{k}=1}^{\infty}\frac{1}{\tilde{k}^{2}}<+\infty.$

This, combined with proposition 3.3, gives (3.14).

Let us take
$\psi(u_{n})=[(1+|u_{n}|)^{l}-1]\mbox{sign}(u_{n})\chi_{(0,\tau)}(t)$ as a
test function of problem $(P_{n})$, where $l>1$ will be chosen later. It
follows from (1.1) and integrating by parts that

$\displaystyle\int_{\Omega}\Psi(u_{n}(x,\tau))dx+\alpha
l\sum_{i=1}^{N}\iint_{Q}\frac{|D_{i}u_{n}|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)+1-l}}dxdt$
$\displaystyle\leq\int_{0}^{\tau}\int_{\Omega}|f_{n}|(1+|u_{n}|^{l})dxdt,$
(3.16)

where $\Psi(s)=\int_{0}^{s}\psi(\xi)d\xi.$ Observing that

$\Psi(s)\geq\frac{1}{2l+1}|s|^{l+1}-C,\ \ \ \forall s\in\mathbb{R},$

and taking the supremum of both terms for $\tau\in(0,T)$ in (3), we obtain

$\begin{split}&\frac{1}{2l+1}\mathop{\mbox{ess\ sup}}_{0\leq t\leq
T}\int_{\Omega}|u_{n}(x,t)|^{l+1}dx+\alpha
l\sum_{i=1}^{N}\iint_{Q}\frac{|D_{i}u_{n}|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)+1-l}}dxdt\\\
&\leq\|f\|_{L^{m}(Q)}\left(\iint_{Q}(1+|u_{n}|)^{lm^{{}^{\prime}}}dxdt\right)^{\frac{1}{m^{{}^{\prime}}}}+C.\end{split}$
(3.17)

Taking $l>1+\theta(\mbox{max}\\{p_{1},p_{2},...,p_{N}\\}-1)$, using (3.14) and
(3.17), we have

$\begin{split}\sum_{i=1}^{N}\iint_{Q}|D_{i}u_{n}|^{p_{i}}dxdt&\leq\sum_{i=1}^{N}\iint_{Q}|D_{i}u_{n}|^{p_{i}}(1+|u_{n}|)^{l-1-\theta(p_{i}-1)}dxdt\\\
&\leq
C\left(\iint_{Q}(1+|u_{n}|)^{lm^{\prime}}dxdt\right)^{\frac{1}{m^{\prime}}}+C\\\
&\leq C,\end{split}$

from which, we get the desired result (3.13). $\Box$

Lemma 3.3. _Assume that ( 1.1)-(1.4) hold true and $f\in L^{m}(Q)$ with $m$
satisfying condition (2.2). If $u_{n}$ is a weak solution of problem
$(P_{n})$, then_

$||u_{n}||_{L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega))}\leq C_{3},$ (3.18)
$||u_{n}||_{L^{\tilde{r}}(Q)}\leq C_{3},$ (3.19)

_where $\tilde{r}$ is defined as in (2.4)._

Proof. We also use
$\psi=[(1+|u_{n}|)^{l}-1]\mbox{sign}(u_{n})\chi_{(0,\tau)}(t)$ as a test
function of problem $(P_{n})$, and then obtain (3.17), where $l>1$ will be
chosen later. Set

$q_{i}=\frac{[\bar{p}(N-N\theta+l+1)+N(\theta+l-1)]p_{i}}{N\theta
p_{i}+N(\lambda-1)p_{i}+\bar{p}(N-N\theta+l+1)},$

where $1<\lambda<l.$ A careful calculation shows that

$1<q_{i}<p_{i}\ \ \ \mbox{and}\ \ \
\frac{q_{i}[\theta(p_{i}-1)+1-l]}{p_{i}-q_{i}}+(\lambda-1)\frac{p_{i}q_{i}}{p_{i}-q_{i}}=\frac{N+\frac{l+1}{\lambda}}{N}\lambda\bar{q},$
(3.20)

where

$\bar{q}=\frac{\bar{p}(N-N\theta+l+1)+N(\theta+l-1)}{N\lambda+l+1}.$ (3.21)

Combining (3.17), (3.20) with Hölder’s inequality, we deduce

$\begin{split}&\iint_{Q}|D_{i}|u_{n}|^{\lambda}|^{q_{i}}dxdt=\lambda^{q_{i}}\iint_{Q}|u_{n}|^{(\lambda-1)q_{i}}|D_{i}u_{n}|^{q_{i}}dxdt\\\
&\leq\lambda^{q_{i}}\left(\iint_{Q}\frac{|D_{i}u_{n}|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)+1-l}}dxdt\right)^{\frac{q_{i}}{p_{i}}}\left(\iint_{Q}(1+|u_{n}|)^{[\frac{\theta(p_{i}-1)+1-l}{p_{i}}+(\lambda-1)]\frac{q_{i}p_{i}}{p_{i}-q_{i}}}dxdt\right)^{\frac{p_{i}-q_{i}}{p_{i}}}\\\
&\leq
C\left(\left(\iint_{Q}(1+|u_{n}|)^{lm^{\prime}}dxdt\right)^{\frac{q_{i}}{m^{\prime}p_{i}}}+1\right)\left(\iint_{Q}(1+|u_{n}|)^{\frac{N+\frac{l+1}{\lambda}}{N}\lambda\bar{q}}dxdt\right)^{\frac{p_{i}-q_{i}}{p_{i}}}.\end{split}$

Now let

$lm^{\prime}=\frac{N\lambda+l+1}{N}\bar{q},$

that is

$l=\frac{(m-1)[N(1-\theta)(\bar{p}-1)+\bar{p}]}{N+\bar{p}-\bar{p}m}.$ (3.22)

Hence, applying Young’s inequality in the above estimate gives

$\begin{split}&\iint_{Q}|D_{i}|u_{n}|^{\lambda}|^{q_{i}}dxdt\\\ &\leq
C\left[\left(\iint_{Q}(1+|u_{n}|)^{\frac{N\lambda+l+1}{N}\bar{q}}dxdt\right)^{\frac{q_{i}}{m^{\prime}p_{i}}+\frac{p_{i}-q_{i}}{p_{i}}}+1\right].\end{split}$
(3.23)

Denoting

$r=\frac{N\lambda+l+1}{N\lambda}\bar{q},$ (3.24)

we get the following interpolation inequality for $L^{r}$-norms of
$|u_{n}|^{\lambda}$

$\||u_{n}|^{\lambda}\|_{L^{r}(\Omega)}\leq\||u_{n}|^{\lambda}\|_{L^{\bar{q}^{*}}(\Omega)}^{\delta}\||u_{n}|^{\lambda}\|_{L^{\frac{l+1}{\lambda}}(\Omega)}^{1-\delta},$
(3.25)

with $\delta$ satisfying

$\frac{1}{r}=\frac{\delta}{\bar{q}^{*}}+\frac{\lambda(1-\delta)}{l+1}.$

This, combined with (3.24), gives

$\delta=\frac{N\lambda}{N\lambda+l+1},\ \ \ \ \delta r=\bar{q}.$ (3.26)

Exploiting (3.17), (3.25) and (3.26), and using Lemma 2.1, we get

$\begin{split}\int_{0}^{T}\left\||u_{n}|^{\lambda}\right\|_{L^{r}(\Omega)}^{r}dt&\leq\int_{0}^{T}\left\||u_{n}|^{\lambda}\right\|_{{L^{\bar{q}^{*}}}(\Omega)}^{\delta
r}\left\||u_{n}|^{\lambda}\right\|_{{L^{\frac{l+1}{\lambda}}}(\Omega)}^{(1-\delta)r}dt\\\
&\leq
C\left[\left(\iint_{Q}(1+|u_{n}|)^{lm^{{}^{\prime}}}dxdt\right)^{\frac{\lambda(1-\delta)r}{m^{{}^{\prime}}(l+1)}}+1\right]\int_{0}^{T}\prod_{i=1}^{N}\left\|D_{i}|u_{n}|^{\lambda}\right\|_{L^{q_{i}}(\Omega)}^{\frac{\bar{q}}{N}}dt.\end{split}$

Observing that $\sum\limits_{i=1}^{N}\frac{\bar{q}}{Nq_{i}}=1$, using general
Hölder’s inequality and Young’s inequality, then it follows from the above
estimate that

$\displaystyle\int_{0}^{T}\left\||u_{n}|^{\lambda}\right\|_{L^{r}(\Omega)}^{r}dt$
$\displaystyle\leq
C\left[\left(\iint_{Q}(1+|u_{n}|)^{lm^{{}^{\prime}}}dxdt\right)^{\frac{\lambda(1-\delta)r}{m^{{}^{\prime}}(l+1)}}+1\right]\prod_{i=1}^{N}\left(\iint_{Q}\left|D_{i}|u_{n}|^{\lambda}\right|^{q_{i}}dxdt\right)^{\frac{\bar{q}}{Nq_{i}}}$
$\displaystyle\leq
C\left[\left(\iint_{Q}(1+|u_{n}|)^{lm^{{}^{\prime}}}dxdt\right)^{\frac{\lambda(1-\delta)r}{m^{{}^{\prime}}(l+1)}}+1\right]\prod_{i=1}^{N}\left[\left(\iint_{Q}(1+|u_{n}|)^{lm^{\prime}}dxdt\right)^{\frac{q_{i}}{m^{\prime}p_{i}}+\frac{p_{i}-q_{i}}{p_{i}}}+1\right]^{\frac{\bar{q}}{Nq_{i}}}$
$\displaystyle\leq
C\left(\iint_{Q}(1+|u_{n}|)^{lm^{\prime}}dxdt\right)^{\frac{\lambda(1-\delta)r}{m^{\prime}(l+1)}+\sum\limits_{i=1}^{N}\left[{\frac{q_{i}}{m^{\prime}p_{i}}+\frac{p_{i}-q_{i}}{p_{i}}}\right]\frac{\bar{q}}{Nq_{i}}}+C.$
(3.27)

Let $\tilde{r}=r\lambda$, then it is apparent that
$\tilde{r}=lm^{\prime}=\frac{mN[(1-\theta)(\bar{p}-1)]+m\bar{p}}{N-m\bar{p}+\bar{p}}$.
Therefore, by (3), it follows that

$\iint_{Q}|u_{n}|^{\tilde{r}}dxdt\leq
C\left(\iint_{Q}|u_{n}|^{\tilde{r}}dxdt\right)^{\tilde{h}}+C,$

where

$\begin{split}\tilde{h}=\frac{l(1-\delta)}{l+1}+\frac{(m-1)\bar{q}}{m\bar{p}}+1-\frac{\bar{q}}{\bar{p}}.\end{split}$

By (3.21), (3.22), (3.26) and the condition $m<\frac{N}{\bar{p}}+1$ , one can
check that $\tilde{h}<1.$ Therefore, by Young’s inequality we get (3.19)
immediately.

Note that condition (2.2) gives $l\geq 1+\theta\left(\max\limits_{1\leq i\leq
N}p_{i}-1\right)$. This, combined with (3.17) and (3.19), gives (3.18). The
proof is completed. $\Box$

Lemma 3.4. _Assume that ( 1.1)-(1.4) hold true and $f\in L^{m}(Q)$ with $m$
satisfying condition (2.5). If $u_{n}$ is a weak solution of problem
$(P_{n})$, then_

$||u_{n}||_{L^{\widetilde{r}}(Q)}\leq C_{4},$ (3.28)
$||u_{n}||_{L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))}\leq C_{4},$ (3.29)

_with $\vec{q}=(q_{1},q_{2},...,q_{N})$, where $q_{i}$ and $\tilde{r}$ are
defined as in (2.4) and (2.7) respectively._

Proof. In the same way as in the proof of Lemma 3.3, taking
$\psi=[(1+|u_{n}|)^{l}-1]\mbox{sign}(u_{n})\chi_{(0,\tau)}(t)$ as a test
function of problem $(P_{n})$, we get (3.17), where $l$ will be fixed later.

Let us set

$q_{i}=\frac{p_{i}[\bar{p}(N+l+1)-N\theta\bar{p}+N(\theta+l-1)]}{N[\theta(p_{i}-1)-l+1]+\bar{p}(N+l+1)-N\theta\bar{p}+N(\theta+l-1)},$
(3.30)

then $q_{i}<p_{i}$ and

$\frac{[\theta(p_{i}-1)-l+1]q_{i}}{p_{i}-q_{i}}=\frac{N+l+1}{N}\bar{q},$
(3.31)

where

$\bar{q}=\frac{\bar{p}(N+l+1)-N\theta\bar{p}+N(\theta+l-1)}{N+l+1}.$ (3.32)

Thanks to (3.17) and (3.31), and using Hölder’s inequality, we obtain

$\begin{split}&\iint_{Q}|D_{i}u_{n}|^{q_{i}}dxdt\\\
&\leq\left(\iint_{Q}\frac{|D_{i}u_{n}|^{q_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)+1-l}}dxdt\right)^{\frac{q_{i}}{p_{i}}}\left(\iint_{Q}(1+|u_{n}|)^{\frac{[\theta(p_{i}-1)+1-l]q_{i}}{p_{i}-q_{i}}}dxdt\right)^{\frac{p_{i}-q_{i}}{p_{i}}}\\\
&\leq\left[C\left(\iint_{Q}|u_{n}|^{lm^{\prime}}dxdt\right)^{\frac{1}{m^{\prime}}}+C\right]^{\frac{q_{i}}{p_{i}}}\left(\iint_{Q}(1+|u_{n}|)^{{\frac{N+l+1}{N}}\bar{q}}dxdt\right)^{\frac{p_{i}-q_{i}}{p_{i}}}.\end{split}$

This, combined with Young’s inequality, shows

$\iint_{Q}|D_{i}u_{n}|^{q_{i}}dxdt\leq
C\left(\iint_{Q}|u_{n}|^{lm^{\prime}}dxdt\right)^{\frac{q_{i}}{m^{\prime}p_{i}}+\frac{p_{i}-q_{i}}{p_{i}}}+C,$
(3.33)

where we have set

$lm^{\prime}=\frac{(N+1+l)\bar{q}}{N}.$ (3.34)

From (3.30), (3.32) and (3.34), it follows that

$\begin{split}&l=\frac{(m-1)[N(1-\theta)(\bar{p}-1)+\bar{p}]}{N+\bar{p}-\bar{p}m},\\\
&\bar{q}=\frac{m[N(1-\theta)(\bar{p}-1)+\bar{p}]}{N+1-[1+\theta(\bar{p}-1)](m-1)},\end{split}$

and

$\begin{split}q_{i}=\frac{mp_{i}[N(1-\theta)(\bar{p}-1)+\bar{p}]}{[N-\bar{p}(m-1)][\theta(p_{i}-1)+1]+N(\bar{p}-1)(1-\theta)+\bar{p}}.\end{split}$

Now, setting

$\tilde{r}=\frac{N+l+1}{N}\bar{q}=\frac{\bar{p}(l+1)+N[l+(1-\theta)(\bar{p}-1)]}{N},$
(3.35)

we get the interpolation inequality for $L^{\tilde{r}}$-norms of $u_{n}$

$\|u_{n}\|_{L^{\widetilde{r}}(\Omega)}\leq\|u_{n}\|_{L^{\bar{q}^{*}}(\Omega)}^{\delta}\|u_{n}\|_{L^{l+1}(\Omega)}^{1-\delta},$
(3.36)

with $0<\delta<1$ such that

$\frac{1}{\tilde{r}}=\frac{\delta}{\bar{q}^{*}}+\frac{1-\delta}{l+1}.$

This, together with (3.35), indicates

$\delta=\frac{N^{2}(l+1)-N\bar{q}(N+l+1)}{[(N-\bar{q})(l+1)-N\bar{q}](N+l+1)}=\frac{N}{N+l+1}.$
(3.37)

In view of (3.35)-(3.37) and Lemma 2.1, and reasoning exactly as in the proof
of (3), we get

$\begin{split}\int_{0}^{T}\|u_{n}\|_{L^{\tilde{r}}(\Omega)}^{\tilde{r}}dt&\leq\int_{0}^{T}\|u_{n}\|_{L^{\bar{q}^{*}}(\Omega)}^{\delta\tilde{r}}\|u_{n}\|_{L^{l+1}(\Omega)}^{(1-\delta)\tilde{r}}dt\\\
&\leq
C\left[\left(\iint_{Q}|u_{n}|^{lm^{\prime}}dxdt\right)^{\frac{(1-\delta)\tilde{r}}{m^{\prime}(l+1)}}+1\right]\int_{0}^{T}\|u_{n}\|_{L^{\bar{q}^{*}}(\Omega)}^{\bar{q}}dt\\\
&\leq
C\left[\left(\iint_{Q}|u_{n}|^{lm^{\prime}}dxdt\right)^{\frac{(1-\delta)\tilde{r}}{m^{\prime}(l+1)}}+1\right]\int_{0}^{T}\prod_{i=1}^{N}\|D_{i}u_{n}\|_{L^{q_{i}}(\Omega)}^{\frac{\bar{q}}{N}}dt\\\
&\leq
C\left(\iint_{Q}|u_{n}|^{lm^{\prime}}dxdt\right)^{\frac{(1-\delta)\tilde{r}}{m^{\prime}(l+1)}+\sum\limits_{i=1}^{N}\left[\frac{q_{i}}{m^{\prime}p_{i}}+\frac{p_{i}-q_{i}}{p_{i}}\right]\frac{\bar{q}}{Nq_{i}}}+C.\end{split}$

Note that $lm^{\prime}=\tilde{r}$ and
$\frac{(1-\delta)\tilde{r}}{m^{\prime}(l+1)}+\sum\limits_{i=1}^{N}\left[\frac{q_{i}}{m^{\prime}p_{i}}+\frac{p_{i}-q_{i}}{p_{i}}\right]\frac{\bar{q}}{Nq_{i}}<1$.
Using Young’s inequality, we can conclude that (3.28) holds true.

By (3.28), (3.33) and using the fact $lm^{\prime}=\tilde{r}$, we get the
result (3.29). The proof is completed. $\Box$

Lemma 3.5. _Assume that ( 1.1)-(1.4) hold true and $f\in L^{m}(Q)$ with $m$
satisfying condition (2.12). If $u_{n}$ is a weak solution of problem
$(P_{n})$, then_

$||u_{n}||_{L^{\infty}(0,T;L^{1}(\Omega))}\leq C_{5},$ (3.38)
$\|D_{i}T_{k}(u_{n})\|_{L^{p_{i}}(\Omega)}\leq
C_{5}(1+k)^{\frac{\theta(p_{i}-1)+1}{p_{i}}},$ (3.39)
$|\left\\{|u_{n}|>k\right\\}|\leq\frac{C_{5}}{k^{\tilde{r}}},$ (3.40)
$|\left\\{|D_{i}u_{n}|>k\right\\}|\leq\frac{C_{5}}{k^{q_{i}}},$ (3.41)

_where $q_{i}$ and $\tilde{r}$ are defined as in (2.4) and (2.7)
respectively._

Proof. The proof is divided into two steps.

Step 1. In this step, we give the proof of the estimates of (3.40) and (3.41).

For all $\tau\in(0,T]$, choosing $T_{k}(u_{n}(x,t))\chi_{(0,\tau)}(t)$ as a
test function for problem $(P_{n})$, and using (1.2) and Hölder’s inequality,
we have

$\begin{split}&\mathop{\mbox{ess\ sup}}_{0\leq t\leq
T}\int_{\Omega}|T_{k}(u_{n}(x,\tau))|^{2}dx+\iint_{Q}\sum_{i=1}^{N}\frac{|D_{i}T_{k}(u_{n})|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)}}dxdt\\\
&\leq
C\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{1}{m^{\prime}}}.\end{split}$
(3.42)

Hence

$\begin{split}&\iint_{Q}|D_{i}T_{k}(u_{n})|^{p_{i}}dxdt\leq
C(1+k)^{\theta(p_{i}-1)}\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{1}{m^{\prime}}}.\end{split}$
(3.43)

The rest of the proof is divided into three cases.

case (i): $m>\frac{\bar{p}(N+2)}{\bar{p}(N+2)-N}.$

In this case, we have $m^{\prime}<\frac{\bar{p}(N+2)}{N}$. Hence we can choose
$\rho_{i}$ such that

$p_{i}\geq\rho_{i}\geq\frac{m^{\prime}p_{i}}{\theta(p_{i}-1)+m^{\prime}},\ \ \
\frac{\bar{\rho}(N+2)}{N}=m^{\prime},$

i.e.

$\frac{\theta(p_{i}-1)\rho_{i}}{p_{i}-\rho_{i}}\geq m^{\prime},\ \ \ \
\bar{\rho}=\frac{Nm}{(N+2)(m-1)}.$ (3.44)

Using proposition 3.1, Hölder’s inequality and (3.44), we obtain that for
$k\geq 1$

$\begin{split}&\iint_{Q}|T_{k}(u_{n})|^{\frac{(N+2)\bar{\rho}}{N}}dxdt\\\
&\leq C\left(\mathop{\mbox{ess\ sup}}_{0\leq t\leq
T}\int_{\Omega}|T_{k}(u_{n})|^{2}dx\right)^{\frac{\bar{\rho}}{N}}\prod_{i=1}^{N}\left[\iint_{Q}|D_{i}T_{k}(u_{n})|^{\rho_{i}}dxdt\right]^{\frac{\bar{\rho}}{\rho_{i}N}}\\\
&\leq
C\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{\bar{\rho}}{Nm^{\prime}}}\prod_{i=1}^{N}\left(\iint_{Q}\frac{|D_{i}T_{k}(u_{n})|^{p_{i}}}{(1+|T_{k}(u_{n})|)^{\theta(p_{i}-1)}}dxdt\right)^{\frac{\bar{\rho}}{p_{i}N}}\\\
&\ \ \ \
\times\left[\iint_{Q}(1+|T_{k}(u_{n})|)^{\frac{\theta(p_{i}-1)\rho_{i}}{p_{i}-\rho_{i}}-m^{\prime}}(1+|T_{k}(u_{n})|)^{m^{\prime}}dxdt\right]^{\frac{p_{i}-\rho_{i}}{p_{i}}\frac{\bar{\rho}}{\rho_{i}N}}\\\
&\leq
C\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{\bar{\rho}}{Nm^{\prime}}}\\\
&\ \ \ \
\times\prod_{i=1}^{N}(2k)^{\left[\frac{\theta(p_{i}-1)\rho_{i}}{p_{i}-\rho_{i}}-m^{\prime}\right]\frac{p_{i}-\rho_{i}}{p_{i}}\frac{\bar{\rho}}{\rho_{i}N}}\left[\iint_{Q}(1+|T_{k}(u_{n})|)^{m^{\prime}}dxdt\right]^{\frac{\bar{\rho}}{m^{\prime}p_{i}N}+\frac{p_{i}-\rho_{i}}{p_{i}}\frac{\bar{\rho}}{\rho_{i}N}}.\end{split}$

If $\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\geq 1$, then by the above
inequality we get

$\begin{split}\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt&\leq
C2^{\sum\limits_{i=1}^{N}\frac{\bar{\rho}}{m^{\prime}p_{i}N}+\frac{p_{i}-\rho_{i}}{p_{i}}\frac{\bar{\rho}}{\rho_{i}N}}k^{\sum\limits_{i=1}^{N}\left[\frac{\theta(p_{i}-1)\rho_{i}}{p_{i}-\rho_{i}}-m^{\prime}\right]\frac{p_{i}-\rho_{i}}{p_{i}}\frac{\bar{\rho}}{\rho_{i}N}}\\\
&\ \ \ \
\times\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{\bar{\rho}}{Nm^{\prime}}+\sum\limits_{i=1}^{N}\frac{\bar{\rho}}{m^{\prime}p_{i}N}+\frac{p_{i}-\rho_{i}}{p_{i}}\frac{\bar{\rho}}{\rho_{i}N}}\\\
&\leq
Ck^{\theta\bar{\rho}-\frac{\theta\bar{\rho}}{\bar{p}}-m^{\prime}+\frac{m^{\prime}\bar{\rho}}{\bar{p}}}\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{1+\frac{\bar{\rho}}{\bar{p}m^{\prime}}-\frac{\bar{\rho}}{\bar{p}}+\frac{\bar{\rho}}{Nm^{\prime}}},\\\
\end{split}$

where we have used the fact $m^{\prime}=\frac{(N+2)\bar{\rho}}{N}$.

It follows from the above inequality and (3.44) that for
$\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\geq 1$ and $k\geq 1$

$\begin{split}&\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\\\ &\leq
Ck^{\left[\theta\bar{\rho}-\frac{\theta\bar{\rho}}{\bar{p}}-m^{\prime}+\frac{m^{\prime}\bar{\rho}}{\bar{p}}\right]\frac{\bar{p}Nm^{\prime}}{(Nm^{\prime}-\bar{p}-N)\bar{\rho}}}\\\
&=Ck^{\frac{m[\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm]}{(N-\bar{p}m+\bar{p})(m-1)}}.\end{split}$
(3.45)

Recalling that $m>\frac{\bar{p}(N+2)}{\bar{p}(N+2)-N}$, by (1.2) and (2.12),
one can deduce that

$\frac{m[\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm]}{(N-\bar{p}m+\bar{p})(m-1)}>0.$

This, combined with (LABEL:3.70), gives that for any $k\geq 1$

$\begin{split}&\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\leq Ck^{\frac{m[\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm]}{(N-\bar{p}m+\bar{p})(m-1)}}.\end{split}$
(3.46)

If $k\leq 1,$ then using (2.12), we also have

$\begin{split}\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\leq|Q|k^{m^{\prime}}\leq|Q|k^{\frac{m[\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm]}{(N-\bar{p}m+\bar{p})(m-1)}}.\end{split}$

Combing the above two inequalities, it follows that for any $k>0$

$\begin{split}\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\leq Ck^{\frac{m[\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm]}{(N-\bar{p}m+\bar{p})(m-1)}}.\end{split}$
(3.47)

As a result, we obtain

$\begin{split}|\left\\{(x,t)\in Q:|u_{n}(x,t)|>k\right\\}|&\leq
Ck^{\frac{m[\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm]}{(N-\bar{p}m+\bar{p})(m-1)}-m^{\prime}}\\\
&=Ck^{-\frac{m[(N+2)\bar{p}-\theta
N(\bar{p}-1)-N-\bar{p}]}{N-\bar{p}m+\bar{p}}}.\end{split}$ (3.48)

Thus, we get (3.40).

By (LABEL:3.65) and (3.47), it is easy to see that

$\begin{split}\iint_{Q}|D_{i}T_{k}(u_{n})|^{p_{i}}dxdt\leq
C(1+k)^{\theta(p_{i}-1)}k^{\frac{\theta
N(\bar{p}-1)(m-1)-(m-1)(N+2)\bar{p}+Nm}{N-\bar{p}m+\bar{p}}}.\end{split}$
(3.49)

From (3.49) and proposition 3.2, we get the desired result (3.41). This
completes the proof of case (i).

Case (ii): $1<m\leq\frac{\bar{p}(N+2)}{\bar{p}(N+2)-N}.$

In this case, we have $m^{\prime}\geq\frac{\bar{p}(N+2)}{N},$ and as a result,
one obtains

$\begin{split}\left(\iint_{Q}|T_{k}(u_{n})|^{m^{\prime}}dxdt\right)^{\frac{1}{m^{\prime}}}\leq
k^{1-\frac{\bar{p}(N+2)}{Nm^{\prime}}}\left(\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\right)^{\frac{1}{m^{\prime}}}.\end{split}$
(3.50)

Using (LABEL:3.64), (LABEL:3.65), (3.50) and proposition 3.1, we get

$\begin{split}&\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\\\ &\leq
C\left(\mathop{\mbox{ess\ sup}}_{0\leq t\leq
T}\int_{\Omega}|T_{k}(u_{n})|^{2}dx\right)^{\frac{\bar{p}}{N}}\prod_{i=1}^{N}\left(\iint_{Q}|D_{i}u_{n}|^{p_{i}}dxdt\right)^{\frac{\bar{p}}{p_{i}N}}\\\
&\leq
C\left[k^{1-\frac{\bar{p}(N+2)}{Nm^{\prime}}}\left(\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\right)^{\frac{1}{m^{\prime}}}\right]^{\frac{\bar{p}}{N}}\\\
&\ \ \ \
\times\prod_{i=1}^{N}\left[(1+k)^{\theta(p_{i}-1)}k^{1-\frac{\bar{p}(N+2)}{Nm^{\prime}}}\left(\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\right)^{\frac{1}{m^{\prime}}}\right]^{\frac{\bar{p}}{p_{i}N}}\\\
&=C(1+k)^{{\theta}(\bar{p}-1)}k^{\left[1-\frac{\bar{p}(N+2)}{Nm^{\prime}}\right](1+\frac{\bar{p}}{N})}\left(\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\right)^{\frac{\bar{p}+N}{m^{\prime}N}}.\end{split}$
(3.51)

Since $m\leq\frac{\bar{p}(N+2)}{\bar{p}(N+2)-N},$ one can easily check that
$\frac{\bar{p}+N}{m^{\prime}N}\leq\frac{\bar{p}+N}{\bar{p}(N+2)}<1$ and
$\frac{\bar{p}(N+2)}{Nm^{\prime}}\leq 1.$ Therefore, we have

$\begin{split}&\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\\\
&\leq\left[C(1+k)^{\theta(\bar{p}-1)}k^{\left[1-\frac{\bar{p}(N+2)}{Nm^{\prime}}\right](1+\frac{\bar{p}}{N})}\right]^{\frac{1}{1-\frac{\bar{p}+N}{m^{\prime}N}}}\\\
&\leq C(1+k)^{\frac{\theta
mN(\bar{p}-1)}{N-\bar{p}m+\bar{p}}}k^{\frac{(N+\bar{p})[Nm-\bar{p}(m-1)(N+2)]}{N(N-\bar{p}m+\bar{p})}},\end{split}$
(3.52)

which yields

$\begin{split}&\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\\\ &\leq
C2^{\frac{\theta
mN(\bar{p}-1)}{N-\bar{p}m+\bar{p}}}k^{{\frac{(N+\bar{p})[Nm-\bar{p}(m-1)(N+2)]+\theta
mN^{2}(\bar{p}-1)}{N(N-\bar{p}m+\bar{p})}}},\ \ \mbox{for}\ k\geq
1.\end{split}$ (3.53)

By (1.2) and (LABEL:3.81), we arrive at

$\begin{split}\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt&\leq|Q|k^{\frac{\bar{p}(N+2)}{N}}\\\
&\leq|Q|k^{\frac{(N+\bar{p})[Nm-\bar{p}(m-1)(N+2)]+\theta
mN^{2}(\bar{p}-1)}{N(N-\bar{p}m+\bar{p})}},\ \ \mbox{for}\ k<1.\end{split}$

Combining the above inequality with (LABEL:3.82), one can deduce that

$\begin{split}\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\leq
Ck^{\frac{(N+\bar{p})[Nm-\bar{p}(m-1)(N+2)]+\theta
mN^{2}(\bar{p}-1)}{N(N-\bar{p}m+\bar{p})}}.\end{split}$ (3.54)

This implies that (3.40) holds true. Finally, by (LABEL:3.64), (LABEL:3.65),
(3.50), (3.54) and proposition 3.2, we obtain the desired result (3.41).

Case (iii): $m=1.$

In this case, choosing $T_{k}(u_{n}(x,t))\chi_{(0,\tau)}(t)$ as a test
function for problem $(P_{n})$, we have

$\begin{split}&\mathop{\mbox{ess\ sup}}_{0\leq t\leq
T}\int_{\Omega}|T_{k}(u_{n}(x,\tau))|^{2}dx+\iint_{Q}\sum_{i=1}^{N}\frac{|D_{i}T_{k}(u_{n})|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)}}dxdt\\\
&\leq k\|f\|_{L^{1}(Q)}\end{split}$ (3.55)

and

$\begin{split}&\iint_{Q}|D_{i}T_{k}(u_{n})|^{p_{i}}dxdt\leq
C(1+k)^{\theta(p_{i}-1)}k.\end{split}$ (3.56)

Arguing as in the proof of (3.54), by (LABEL:p352), (LABEL:p353) and
proposition 3.1, we obtain

$\begin{split}\iint_{Q}|T_{k}(u_{n})|^{\frac{\bar{p}(N+2)}{N}}dxdt\leq
Ck^{\frac{N+\bar{p}+\theta N(\bar{p}-1)}{N}},\end{split}$ (3.57)

which shows that (3.40) holds true. The result (3.41) follows from
(LABEL:p352)-(3.57) and proposition 3.2 immediately.

Step 2. In this step, we give the proof of the estimates of (3.38) and (3.39).

Taking $T_{1}(u_{n}){\chi(0,\tau)}(t)$ as a test function for problem
$(P_{n})$, from (1.1) and Hölder’s inequality, we obtain

$\begin{split}\int_{\Omega}\int_{0}^{u_{n}(x,\tau)}T_{1}(s)dsdx+\alpha\int_{0}^{\tau}\int_{\Omega}\sum\limits_{i=1}^{N}\frac{|D_{i}T_{1}(u_{n})|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)}}dxdt\leq\|f_{n}\|_{L^{1}(Q)}.\end{split}$

This, together with the fact
$|s|-\frac{1}{2}\leq\int_{0}^{s}T_{1}(\eta)d\eta\leq|s|,$ gives

$\mathop{\mbox{ess\ sup}}_{0\leq t\leq T}\int_{\Omega}|u_{n}(x,t)|dx\leq C,$
(3.58)

which is the desired result (3.38).

From (LABEL:p352) and (LABEL:p353), it follows that (3.39) holds true. Thus,
the proof is completed. $\Box$

##  4 Proofs of Theorem 2.1-2.5

In this section, we prove the main results. Since the proofs of Theorem 2.2,
Theorem 2.3 and Theorem 2.4 are very similar to that of Theorem 2.1, we only
give the proofs of Theorem 2.1 and Theorem 2.5.

Proof of Theorem 2.1. By Lemma 3.1 and the equation of problem $(P_{n})$, we
conclude that

$\frac{\partial u_{n}}{\partial t}\ \mbox{is\ bounded\ in}\
L^{\vec{p^{\prime}}}(0,T;W^{-1,\vec{p^{\prime}}}(\Omega))+L^{m}(Q).$

Therefore, using a compactness result (see [40]), we deduce that there exist a
subsequence of $\\{u_{n}\\}$ (still denoted by $\\{u_{n}\\}$) and a measurable
function $u$ such that

$u_{n}\longrightarrow u\mbox{ a.e\ \ in }Q,$ (4.1) $u_{n}\rightharpoonup
u\mbox{ weakly\ \ in }L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega)),$ (4.2)

and

$u_{n}\rightharpoonup u\ \mbox{ weakly}*\ \mbox{\ in }L^{\infty}(Q).$ (4.3)

Arguing as in the proof of Theorem 3.3 of [36], we have

$Du_{n}\longrightarrow Du\ \mbox{ a.e.\ \ in }Q.$ (4.4)

By (1.3), (1.4), (4.1), (4.4) and using Vitali’s Theorem, it follows that

$a_{i}(x,t,T_{n}(u_{n}),Du_{n})\longrightarrow a_{i}(x,t,u,Du)\ \mbox{
strongly\ \ in }L^{p^{\prime}_{i}}(Q).$ (4.5)

Passing to the limit as $n\rightarrow\infty$ in the weak formulation of
$(P_{n})$, using (LABEL:2.2), (3.1), (3.2), (4.2) and (4.5), we infer that $u$
is a weak solution of problem $(P)$. $\Box$

Proof of Theorem 2.5. The proof is similar to [33], so we only sketch it here.
Taking $T_{k}(u_{n}){\chi_{(0,\tau)}}(t)$ as a test function for problem
$(P_{n})$, we get

$\begin{split}&\int_{\Omega}\int_{0}^{u_{n}(x,\tau)}T_{k}(s)dsdx+\alpha\int_{0}^{\tau}\int_{\Omega}\sum\limits_{i=1}^{N}\frac{|D_{i}T_{k}(u_{n})|^{p_{i}}}{(1+|u_{n}|)^{\theta(p_{i}-1)}}dxdt\leq
k\|f\|_{L^{1}(Q)}.\end{split}$ (4.6)

It can be deduced from (LABEL:4.6) and the equation of $(P_{n})$ that for any
$S\in W^{2,\infty}(\mathbb{R})$ such that $S^{\prime}$ is compact

$\frac{\partial S(u_{n})}{\partial t}\ \mbox{is\ bounded\ in}\
L^{\vec{p^{\prime}}}(0,T;W^{-1,\vec{p^{\prime}}}(\Omega))+L^{1}(Q).$

Arguing again as in [41], the above two estimates imply that there exist a
subsequence of $\\{u_{n}\\}$ (still denoted by $\\{u_{n}\\}$) and a measurable
function $u$ such that

$u_{n}\longrightarrow u\mbox{ a.e\ \ in }Q,$ (4.7)

and

$T_{k}(u_{n})\rightharpoonup T_{k}(u)\ \mbox{ weakly in
}L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega)).$ (4.8)

Then, proceeding as in [33, 42], we get

$Du_{n}\longrightarrow Du\ \mbox{ a.e.\ \ in }Q.$ (4.9)

Arguing as in Theorem 3.2 of [33], we obtain that $u\in
L^{\infty}(0,T;L^{1}(\Omega))$ is a renormalized solution of $(P)$, i.e. $u$
satisfies

$T_{k}(u)\in L^{\vec{p}}(0,T;W_{0}^{1,\vec{p}}(\Omega)),\ \ S_{k}(u)\in
C([0,T];L^{1}(\Omega)),$ (4.10)
$\begin{split}&\underset{n\rightarrow\infty}{\mbox{lim}}\iint_{\\{n\leq|u|\leq
n+1\\}}\sum\limits_{i=1}^{N}a_{i}(x,t,u,Du)D_{i}udxdt=0\end{split}$ (4.11)

and

$\begin{split}&(S(u))_{t}-\mbox{div}\left(S^{\prime}(u)a(x,t,u,Du)\right)\\\
&+\sum\limits_{i=1}^{N}a_{i}(x,t,u,Du)D_{i}uS^{\prime\prime}(u)=fS^{\prime}(u),\end{split}$
(4.12)

where $a(s,t,u,Du)=(a_{1}(x,t,u,Du),a_{2}(x,t,u,Du),\ldots,a_{N}(x,t,u,Du))$.

Let us choose $S(s)=H_{n}(s)=\int_{0}^{s}1-|T_{1}(\tau-T_{n}(\tau))|d\tau$,
and take $T_{k}(H_{n}(u)-\phi)$ as test function in (LABEL:4.10). Passing to
the limit as $n\rightarrow\infty$, one can deduce that $u$ is a entropy
solution of problem $(P)$.

Furthermore, by (3.40), (3.41) and Fatou’s lemma, we get $u\in{\cal
M}^{\tilde{r}}(Q)$ and $|D_{i}u|\in{\cal M}^{\beta_{i}}(Q).$ Hence the proof
of Theorem 2.5 is completed. $\Box$

Remark 4.1 Without loss of generality, we may assume that $p_{1}\geq p_{2}\geq
p_{3}\geq...\geq p_{N}$. If $f\in L^{m}(\Omega)$ with
$m_{2}:=\frac{\bar{p}[(2\theta-N-2)p_{1}-\theta
p_{2}(p_{1}+1)]}{\bar{p}[(2\theta-N-2)p_{1}-\theta
p_{2}(p_{1}+1)]+Np_{1}[1+\theta(p_{2}-1)]}\leq m<m_{1},$ we have
$1+\theta(p_{2}-1)\leq l<1+\theta(p_{1}-1).$ Arguing as in the proof of Lemma
3.4, we conclude that problem $(P_{n})$ admits at least a weak solution
$u_{n}\in L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))\cap L^{\tilde{r}}(Q)$
with $\vec{q}=(q_{1},q_{2},...,q_{N})$ and
$\tilde{r}=\frac{N+l+1}{N}\bar{q}=lm^{\prime}$ satisfying

$\|u_{n}\|_{L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))\cap
L^{\tilde{r}}(Q)}\leq C,$

where

$q_{i}=\left\\{\begin{array}[]{ll}\frac{(N+l+1)\bar{p}p_{1}-\bar{p}p_{1}\theta-\bar{p}(1-l-\theta)}{\bar{p}p_{1}(N+l+1)-\bar{p}(p_{1}\theta+1-l-\theta)+Np_{1}[\theta(p_{1}-1)+1-l]}&i=1,\\\
p_{i}&i\geq 2,\\\ \end{array}\right.$

and

$l=\frac{\bar{p}\left[\theta+(1-\theta)\frac{1}{p_{1}}-N-1\right]}{\bar{p}(1+\frac{1}{p_{1}})-m^{\prime}N}.$

Similar to the proof of Theorem 2.1, we can infer that there exists at least a
solution $u\in L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))\cap
L^{\tilde{r}}(Q)$ to problem $(P)$, where $\vec{q}$ and $\tilde{r}$ are
defined as above.

If $f\in L^{m}(\Omega)$ with
$m_{3}:=\frac{\bar{p}[p_{1}p_{2}(2\theta-N-2)-\theta(p_{1}+p_{2})p_{3}]}{\bar{p}[p_{1}p_{2}(2\theta-N-2)-\theta(p_{1}+p_{2})p_{3}]+Np_{1}p_{2}[1+\theta(p_{3}-1)]}\leq
m<m_{2},$ we have $1+\theta(p_{3}-1)\leq l<1+\theta(p_{2}-1).$ Arguing as
before, we infer that problem $(P)$ admits at least a weak solution $u\in
L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))\cap L^{\tilde{r}}(Q)$ with
$\vec{q}=(q_{1},q_{2},...,q_{N})$ and
$\tilde{r}=\frac{N+l+1}{N}\bar{q}=lm^{\prime}$, where

$q_{i}=\left\\{\begin{array}[]{ll}\frac{\bar{p}p_{i}p_{1}p_{2}(N+l+1)-\bar{p}p_{i}[2\theta
p_{1}p_{2}+(1-l-\theta)(p_{1}+p_{2})]}{Np_{1}p_{2}[\theta(p_{i}-1)+1-l]+p_{1}p_{2}\bar{p}(N+l+1)-\bar{p}[2\theta
p_{1}p_{2}+(1-l-\theta)(p_{1}+p_{2})]}&i=1,2,\\\ p_{i}&i\geq 3,\\\
\end{array}\right.$

and
$l=\frac{\bar{p}\left[2\theta+(1-\theta)(\frac{1}{p_{1}}+\frac{1}{p_{2}})-N-1\right]}{\bar{p}(1+\frac{1}{p_{1}})+\frac{1}{p_{2}}-m^{\prime}N}.$
Thus, using exactly the same procedure presented as above, we shall divide
this interval $[m_{N},m_{1})$ into $N-1$ subintervals and obtain the
corresponding results for each subintervals. For the last interval, we have
the following result.

If $f\in L^{m}(\Omega)$ with $m_{N}\leq m<m_{N-1}$, we have
$1+\theta\left(\min\limits_{1\leq i\leq N}p_{i}-1\right)\leq
l<1+\theta(p_{N-1}-1),$ where
$m_{N-1}=\frac{\bar{p}[(N\theta-N-2+\theta)\bar{p}p_{N}-\theta
p_{N-1}(Np_{N}-\bar{p}+\bar{p}p_{N})]}{\bar{p}[(N\theta-N-2+\theta)\bar{p}p_{N}-\theta
p_{N-1}(Np_{N}-\bar{p}+\bar{p}p_{N})]+N\bar{p}p_{N}[1+\theta(p_{N-1}-1)]}.$ In
this case, we may conclude that problem $(P)$ admits at least a weak solution
$u\in L^{\vec{q}}(0,T;W_{0}^{1,\vec{q}}(\Omega))\cap L^{\tilde{r}}(Q)$ with
$\vec{q}=(q_{1},q_{2},...,q_{N})$ and
$\tilde{r}=\frac{N+l+1}{N}\bar{q}=lm^{\prime}$, where

$q_{i}=\left\\{\begin{array}[]{ll}\frac{(N+l+1)\bar{p}p_{i}-\bar{p}p_{i}\left[(N-1)\theta+(1-l-\theta)(\frac{N}{\bar{p}}-\frac{1}{p_{N}})\right]}{N[\theta(p_{i}-1)+1-l]+(N+l+1)\bar{p}-\bar{p}\left[(N-1)\theta+(1-l-\theta)\left(\frac{N}{\bar{p}}-\frac{1}{p_{N}}\right)\right]}&i=1,2,...,N-1,\\\
p_{i}&i=N,\\\ \end{array}\right.$

and
$l=\frac{\bar{p}\left[N\theta+(1-\theta)\left(\frac{N}{\bar{p}}-\frac{1}{p_{N}}\right)-N-1\right]}{\bar{p}(1+\frac{N}{\bar{p}}-\frac{1}{p_{N}})-m^{\prime}N}.$

Acknowledgment

This research is supported by National Natural Science Foundation of
China(No.11801259, No.11461048), Natural Science Foundation of Jiangxi
Province (No.20202BABL201009, No. 20202ACBL211002), Academic and Technical
Leaders Training Plan of Jiangxi Province (No. 20212BCJ23027) and the
foundation of Education Department of Jiangxi Province(no.GJJ190514).

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